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Chapter 4 Introduction to the Surreal Number Field No
117 CHAPTER 4
I N T R O D U C T I O N T O THE SURREAL NUMBER F I E L D No
4.00
SURREAL NUMBERS
In J . H . Conway's book, On Numbers and Games C241, t h e b a s i c c o n s t r u c t i o n o f numbers i s t h e f o l l o w i n g : (0)
I f L a n d R a r e two s e t s of n u m b e r s , a n d i f no member of L is t any
member of R , t h e n ( L I R } i s a number.
A l l numbers are c o n s t r u c t e d i n
t h i s way [ 2 4 , p . 41.
How t h e n d o e s o n e g e t s t a r t e d c o n s t r u c t i n g n u m b e r s u s i n g C o n w a y ' s
construction?
The empty s e t is a s e t of numbers which we know e ists.
L and R be empty. (01,
Note t h a t no member of L is 2 any member of R
[LIR] i s a number.
Let u s c a l l t h i s number 0 .
Let
t h u s , by
Conway C24, p . 41
adopted t h e following n o t a t i o n a l convention:
If x
=
( L I R } w e w r i t e xL f o r a t y p i c a l member of L , a n d
t y p i c a l member of R ; t h u s x e, f , e, f ,
... ) , ... 1 .
option _---
of x.
we mean t h a t x
L R {x Ix 1 .
= =
If we write x = ( a , b , c ,
... I d ,
and R
=
Id,
x L i s c a l l e d a l e f t o p t i o n of x , a n d x R i s c a l l e d a r i g h t If L ( r e s p . R ) i s empty, we may i n d i c a t e t h i s by l e a v i n g t h e
p l a c e where L ( r e s p . R ) would a p p e a r b l a n k . =
... }
( L I R ) , where L = [ a , b , c ,
xR for a
[I).
Hence ( ( 0 1 l a }
=
([O}
I],
and 0
I n K n u t h ' s m a t h e m a t i c a l n o v e l l a o n s u r r e a l n u m b e r s [52] he u s e s s l i g h t l y d i f f e r e n t n o t a t i o n i n t h e body of t h e t e x t . writes x
=
(X
X 1. L' R
For example, Knuth
We have c h o s e n t o a d o p t most of Conway's n o t a t i o n .
is n o t o n l y v e r y compact a n d e a s y t o u s e , b u t i t s u g g e s t s
feels
-
t h e r i g h t way t o t h i n k a b o u t t h e s u b j e c t .
-
It
t h e author
118
Norman L . A l l i n g
4.00
Conway t h e n d e f i n e s o r d e r between numbers a s f o l l o w s :
(1)
( 1 ) x 6 y i f and O n l y i f ( i i ) no y R 2 x and x S no x
L
.
Note t h a t ( 1 , i ) is a s t a t e m e n t about n u m b e r s , a n d t h a t ( 1 , i i ) i s a s t a t e m e n t a b o u t s e t s of n u m b e r s .
Conway d e s c r i b e s 0 as t h e t l s i m p l e s t t t
number t h a t was ttborn on day 0" [24, p.
111.
T h i s seems f i t t i n g i n d e e d ,
{*I.]. The numbers 1 = Conway s a y s of them t h a t
s i n c e i t i s b u i l t up fran t h e empty s e t u s i n g o n l y
{Ol) and -1
-
(10) a r e a l i t t l e more c o m p l e x .
t h e y were e a c h " b o r n o n d a y 1 " [ 2 4 , verify t h a t ( 1 , i i ) holds. and t h a t ( b ) 0 < j l ) ,
p . 111.
To s e e t h a t 0 2 1, w e m u s t
To do t h a t i t s u f f i c e s t o show t h a t ( a ) lo)
<
0,
S i n c e ( a ) and ( b ) a r e both t r u e , we see t h a t 0 2 1 .
Conway goes on t o make t h e f o l l o w i n g d e f i n i t i o n s : (2)
(i)
y 2 x i f and o n l y i f x 6 y ,
x = y i f and o n l y i f x 6 y and y S x , x < y i f and o n l y i f x 6 y and i f x + y , and ( i v ) y > x i f and o n l y i f x < y.
(ii)
(iii)
Perhaps t h e o n l y s u r p r i s e i s t h a t ( 2 , i i ) is a definition.
Conway
y, -x,
and xy
ends h i s s h o r t l i s t of remarkable s t a t e m e n t s by d e f i n i n g x
+
i n d u c t i v e l y f o r all numbers x and y as f o l l o w s . L R R I x L + y , x + y Ix + y , x + y 1 .
(3)
x + Y
(4)
-x = (-x
(5)
x y - ( x y + x y
=
R
L
1-x I .
L
L x Y
+
L L R R R R - x y , x y + x y - x y J R L R R L X Y - x y , x y + xy - x Ry L ) *
L
A t f i r s t g l a n c e t h e s e d e f i n i t i o n s may l o o k c i r c u l a r .
Note, f o r
example, i n ( 4 ) i f we know how t o form t h e n e g a t i v e of a l l t h e o p t i o n s of x used t o d e f i n e x , t h e n (4) i s n o n - c i r c u l a r .
S i m i l a r l y , i n ( 3 ) i f we c a n
p r e f o r m a l l t h e i n d i c a t e d a d d i t i o n s among o p t i o n s of x and y and y and x
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.00
t h e n w e c a n compute t h e s e t s on t h e l e f t i n ( 3 ) .
119
The same may b e s a i d of
(5). Conway a l s o showed C24, p p . 16-17] t h a t , i f x
(6)
L
<
[XI
4.01
<
=
[LIR] then
R.
CONWAY'S CONSTRUCTION
C o n w a y ' s c o n s t r u c t i o n a n d most of t h e p r o o f s h e g i v e s , a r e b y induction.
One of Conway's v e r y u s e f u l i d e a s i s t h a t of t h e b i r t h o r d e r of
s u r r e a l numbers.
As we w i l l s e e t h i s i s o n e of t h e most i m p o r t a n t Given aEOn, Conway d e f i n e s
p r o p e r t i e s of surreal numbers. (0)
0 as t h e s e t of a l l numbers b o r n before day a , a M
as t h e s e t of a l l numbers b o r n o n o r b e f o r e day a t a n d
N
as t h e s e t of a l l numbers b o r n o n d a y a C24, p . 291.
<
S i n c e t h e r e are no o r d i n a l s a
0, 0 ,
=
The s e t s L a n d R which
0.
a r e a v a i l a b l e t o make n u m b e r s o n day 0 a r e v e r y few: o n l y t h e empty s e t . Thus t h e o n l y Conway c u t ( 1 . 2 0 ) i n 0 is ( 0 , 0 ) .
We c a n t h i n k of a number as
a n e q u i v a l e n c e c l a s s o f Conway c u t s ( L , R ) i n NO, u n d e r t h e e q u i v a l e n c e relation (4.00:2,ii). (4.00:O).
Thus we s e e t h a t M ,
=
No
=
(01,
0 being
Ill
Now t h a t t h e n u m b e r s o n day 0 h a v e been c r e a t e d , t h e c a l e n d a r
a d v a n c e s , as i t were, a day t o day 1 .
On d a y 1 t h e r e a r e two s e t s o f n u m b e r s :
t h e e m p t y s e t 0 and [ O } .
Thus t h e r e a r e two Conway c u t s i n 0 , , ({01,0) a n d ( 0 , { 0 1 ) .
we w i l l d e f i n e t o be [Ol], and - 1 , t h e elements i n N , .
C l e a r l y 0,
=
Thus 1 , which
which we w i l l d e f i n e t o b e
M,
= [0,+11; thus
(lo],
are a l l
we a r e r e a d y t o b e g i n
t o c o n s i d e r t h e numbers c r e a t e d o n day 2. Conway d e f i n e s t h e c l a s s of a l l n u m b e r s c r e a t e d i n t h i s way a s No [24,
(I)
p . 41.
He shows [ 2 4 , p . 301 t h a t
g i v e n any xcNo t h e r e e x i s t s a u n i q u e aEOn s u c h t h a t X E N ~ .
Norman L. A l l i n g
120
4.01
Let a be c a l l e d t h e b i r t h d a y of x , and l e t i t be denoted by b ( x ) . w i l l c a l l b the b i r t h order function.
y if b(x)
<
Conway writes t h a t x i s s i m p l e r t h a n
Since On i s well-ordered t h e p h r a s e t h a t sane element i s
b(y).
" t h e s i m p l e s t element such t h a t
..
.I1
makes s e n s e .
Conway g i v e s t h e follow-
i n g very i l l u m i n a t i n g d e s c r i p t i o n of t h e c r e a t i o n p r o c e s s . numbers w i t h L
<
No, I L I R I
R in
=
G i v e n s e t s of
x
i s t h e s i m p l e s t element of No s u c h t h a t L
(2)
We
< {XI <
R.
Conway r e f e r s t o t h i s as "The S i m p l i c i t y Theorem" C24, Theorem 1 1 , p . 231.
Henceforth we w i l l r e f e r t o (2) as ttConwayls S i m p l i c i t y Theorem".
is a v i t a l i n g r e d i e n t i n many of
we w i l l s e e , Conway's S i m p l i c i t y Theorem
our c o n s i d e r a t i o n s .
Note a l s o t h a t P =
As
( Na
aeon
i s a p a r t i t i o n of No.
T h i s p a r t i t i o n can a l s o be g i v e n by g i v e n a map b , which maps each element
t o t h e index a .
in N
Given b , t h e n N
=
b
-1
( a ) . b c a n b e t h o u g h t of a s
a s s i g n i n g t h e b i r t h o r d e r t o t h e e l e m e n t s of No. more d e t a i l s . ) day 1 .
-2,
Thus 0 i s born f i r s t , on day 0.
-1/2,
( S e e [5, 384-3851 f o r
1 and -1 are born n e x t , on
1/2, and 2 are born n e x t , on d a y 2, e t c .
One of t h e t h i n g s t h a t Conway had t o d e a l w i t h was t h e f o l l o w i n g :
"A
m o s t i m p o r t a n t comment whose l o g i c a l e f f e c t s w i l l be d i s c u s s e d l a t e r i s that
the n o t a t i o n of
equality
is a
defined relation.
Thus a p p a r e n t l y
d i f f e r e n t d e f i n i t i o n s w i l l produce t h e same number, and we m u s t d i s t i n g u i s h
form
{LIR] of a number a n d t h e number i t s e l f . I t C24, p . 51 U s i n g ( 2 ) we c a n g i v e a d r a m a t i c i l l u s t r a t i o n of t h i s , n a m e l y t h e
between t h e following:
(3)
Let L and R be s u b s e t s of No s u c h t h a t L
< {O] <
R ; then
0 = ILIRf.
S i n c e t h e class of a l l o r d i n a l numbers On i s , i n a very n a t u r a l way, a s u b c l a s s of No C24, pp. 27-281, we see t h a t (4)
No i s a proper class.
One of t h e n a t u r a l t h i n g s t o d o , in t r y i n g t o c o n s t r u c t No i n a more c l a s s i c a l manner w i t h i n a conventional s e t t h e o r y , would be t o c o n s i d e r t h e
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.01
121
v a r i o u s Conway c u t s f r m which x c a n be d e f i n e d , s a y a s o r d e r e d p a i r s , a n d then pass t o equivalence c l a s s e s .
The d i f f i c u l t y of d o i n g t h i s can be s e e n
i n ( 3 1 , s i n c e 0 h a s a p r o p e r c l a s s of Conway c u t s ( L , R ) s u c h t h a t 0
F u r t h e r , ( 3 ) i s n o t a n i s o l a t e d o c c u r r e n c e , as we w i l l now show.
{LIR).
Let x = I L ( R 1 , where (L,R) i s
>
t h a t f o r a l l f3cOn w i t h 8 ( r e s p . R ) union [ 8 } .
(5)
=
x
=
[L (R
$
6
a Conway c u t i n No.
a, (-6)
<
L and R
<
There exists acOn s u c h
(8).
Let L
8
(resp R ) be L
B
Then
1, f o r each B > a .
Hence. we see ( 5 ) t h a t e a c h XENO has a p r o p e r c l a s s o f Conway c u t s (L',R')
such t h a t {L'IR') 4.02
=
x.
THE CUESTA DUTARI CONSTRUCTION OF No
Let T be an o r d e r e d s e t .
R e c a l l (1.20) t h a t a C u e s t a Dutari c u t i n T
is a p a i r of s u b s e t s (L,R) of T , s u c h t h a t ( i ) L L and R i s T .
<
R and ( i i ) t h e u n i o n o f
Let CD(T) = { C u e s t a D u t a r i c u t s i n T I .
S i n c e ( 0 , T ) and
(T, 0) a r e Cuesta Dutari cuts, (0)
C D ( T ) i s never empty.
Assume t h a t M is an o r d e r e d set which c o n t a i n s T , s u c h t h a t t h e o r d e r
on M i n d u c e s t h e o r i g i n a l o r d e r on T: i . e . , (M,6) (1.10).
Let (L,R)ECD(T). X E M w i l l be s a i d t o
i s a n e x t e n s i o n of (T,S)
rill (L,R)
in M if L
< [XI <
R.
Let x ( T ) , t h e C u e s t a D u t a r i c o m p l e t i o n o f T , b e t h e u n i o n of T a n d CD(T), o r d e r e d a s f o l l o w s : (1)
(i)
i f x and y a r e i n T , l e t them be o r d e r e d as t h e y were i n T ;
( i i ) i f XET and y (iii) i f x
=
=
(L,R), y
(L,R)€CD(T). =
s u b s e t of L ' . (2)
x ( T ) i s an o r d e r e d s e t .
x
<
y i f X E L , and y
(L',R')&CD(T), t h e n x
<
< x
i f XER;
y i f L is a p r o p e r
Norman L. A l l i n g
122
Let x , y , and z be i n x ( T ) , w i t h x < y a n d y
PROOF.
that x
(3)
< z
4.02
< z.
c o n s i d e r t h e e i g h t e a s i l y proven c a s e s s e p a r a t e l y .
< t, in < c, i n
For all t ,
(i)
( i i ) For a l l c,
To show
o
T , t h e r e e x i s t s CECD(T) with t o < c C D ( T ) , t h e r e exists
tET
< t,. < t < C,.
with c,
( i i i ) ( 0 , T ) i s t h e l e a s t and ( T , 0 ) i s a g r e a t e s t element of x ( T ) .
PROOF.
Let t o
< c <
then t o
Let t c L ,
-
be elements i n T.
< c,.
w i t h c,
(L,,R,)&CD(T),
L,.
< tl,
t,, establishing L o ; then c ,
(i).
Let c
Let c,
=
= ((-m,to],(to,+-));
and c,
(L,,Ro)
=
Then, by d e f i n i t i o n , Lo is a proper s u b s e t of
< t <
c,, establishing (ii).
I f T i s empty
t h e n x ( T ) h a s o n l y o n e p o i n t i n i t , namely ( 0 , 0 ) ; e s t a b l i s h i n g ( i i i ) i n case T
=
0.
Assume now t h a t T is non-empty.
For a n y ~ E T ,( 0 , T )
< t <
(T,0). (4)
c
=
f i l l s t h e Cuesta D u t a r i c u t ( L , R ) i n x ( T ) .
(L,R)ECD(T),
PROOF. NOTE.
L
R
By d e f i n i t i o n , f o r a l l x EL and a l l x ER, xL
<
<
c
x
R
.
o
Even though each Dedekind c u t i n T i s a C u e s t a D u t a r i c u t i n
T , t h e C u e s t a Dutari completion x ( T ) of T p l a y s a very d i f f e r e n t r o l e t h a n
Dedekind used gaps i n t h e r a t i o n a l
does t h e Dedekind completion 6(T) of T .
numbers Q t o d e f i n e i r r a t i o n a l n u m b e r s , a n d t h u s d e f i n e R.
Since R is
Dedekind-complete i t has no gaps; t h u s t h e Dedekind completion of R, i s R. S i n c e C D ( T ) i s n e v e r empty (01, t h e C u e s t a D u t a r i completion x ( T ) of T always c o n t a i n s T as a proper s u b s e t .
I n p a r t i c u l a r , R is a p r o p e r s u b s e t
of x ( R ) . Let T o be t h e empty s e t .
d e f i n e d T , t o be x(T,,),
Cuesta D u t a r i [251 a n d H a r z h e i m C431 t h e n
and noted t h a t T , = [ ( 0 , 0 ) ) .
Assume t h a t f o r sane
BEOn t h a t a f a m i l y (Ta)a
then define T
d i n a l , than d e f i n e T
set.
t o be x ( T , ) ; 8 t o be t h e union of
6 F u r t h e r , Harzheim proved t h a t
and i f 8 is a non-zero l i m i t o r -
Note t h a t T
WO
is a n n o -
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.02
(5)
T
is a n
IRI
+
<
-set 143, p.1191.
5
Let L a n d R be s u b s e t s of T
PROOF. ILI
q
WE'
=
(avcT
6:
a r e s u b s e t s of T
5
C'E
and ( i i )
s u c h t h a t L and R are s u b s e t s of T Clearly L'
Let R' be d e f i n e d t o be T
6'
such t h a t L Then
(L',Rf)eCD(T6).
(6)
< w5
< R,
is r e g u l a r and s i n c e
5
t h e r e e x i s t s acL s u c h t h a t a t 5 a } .
s a t u r a t e d s u b s e t of T
i n Tw 5
f o r which ( i ) L
S i n c e , by a s s u m p t i o n ( 1 . O O : O ) , w
( i i ) holds, there e x i s t s 6
L'
123
<
R,
T6+,, and L
6
-
L'.
is a lower
S i n c e L and R
R i s a s u b s e t of R I .
< {cl} <
Let
6'
Let c '
=
By c o n s t r u c t i o n , c' is
R.
. By i n d u c t i o n
is defined.
F o l l o w i n g Conway [24,
p . 291, l e t Oa ( t h e s e t o f " o l d n u m b e r s " ) b e
d e f i n e d t o be Ta, l e t N
a
( t h e set of Ifnew numbers") b e T
a+l
-
Ta, a n d l e t
Mu ( t h e s e t of "made numbersTf)b e Ta + l ' N o t e t h a t Ma i s t h e u n i o n o f 0 a a n d Nu C24, p . 291. F i n a l l y , l e t No be d e f i n e d t o be t h e u n i o n of (Oa)aEOn [24, p . 41.
See a l s o C60, p . 491 o n s u c h u n i o n s .
N o t e t h a t No i s a l s o t h e u n i o n o f (M M
a
and t h a t i f a
a adn'
<
6, t h e n
i s a p r o p e r s u b s e t of M6; t.hus g i v e n X E N O , t h e r e e x i s t s a l e a s t BEOn
s u c h t h a t XEM
6
(= T6+l).
t h e b i r t h d a y of x.
Let b ( x ) d e f i n e d t o be f3, a n d l e t i t be c a l l e d
b w i l l be c a l l e d t h e b i r t h - o r d e r
t h a t b ( x ) is t h e u n i q u e e l e m e n t i n On s u c h t h a t X E N , , ( ~ ) ( - Tg+l C l e a r l y f o r each acOn, b - l ( [ O , a ) ) Note t h a t (NaIaeOn
=
Oa, b - l ( [ O , a ] )
i s a p a r t i t i o n o f No.
=
Note
f u n c t i o n o n No.
Ma, a n d b - l ( { a } )
-
T6). =
N
a'
I n [5, p p . 384-3851 we c a l l e d
s u c h a p a r t i t i o n a Conway p a r t i t i o n , a n d c o n s i d e r e d s o m e o f t h e i r p r o p e r ties.
The f o l l o w i n g i s of g r e a t i m p o r t a n c e .
Norman L . A l l i n g
124
CONWAY'S SIMPLICITY THEOREM. L
<
Let I
R.
=
Let L and R be s u b s e t s of No f o r which
< {y) <
(YENO: L
4.02
R).
Then ( i ) I is non-empty, a n d ( i i )
t h e r e e x i s t s a u n i q u e x c I s u c h t h a t b ( x ) < b ( y ) , f o r all Y E I[24,
is an
11
5
i s r e g u l a r , a n d f o r which U is a s u b s e t of 0
-set (51, I i n t e r s e c t Ow
5
-
least element.
b ( J ) i s a n o n - e m p t y s u b c l a s s of On; l e t 5 be i t s
Let x a n d x' be i n J s u c h t h a t b ( x ) = B
f o r a moment, t h a t x
t <
XI.
T h e r e e x i s t s a u n i q u e XEJs u c h
{XI implies b(x) < b(y).
PROOF ( o f ( 7 ) .
T5.
Since
is a non-empty.
Let J be a n o n - m p t y i n t e r v a l i n No,
-
.
5
that y d
T6+,
Since U i s a s e t t h e r e e x i s t s
Let U be t h e u n i o n of L and R .
a CEOn s u c h t h a t w
(7)
(Cf.
Theorem 1 1 , p . 231.) PROOF.
0
{XI.
<
=
b(t)
6'
<
Assume
S i n c e b ( x ) = 6 = b ( x ' ) , x a n d x ' are i n N
X I .
By ( 3 , i i ) we know t h a t t h e r e i s a n e l e m e n t t E T
Since tcT
b(x').
6'
6
=
such t h a t x
<
S i n c e J is a n i n t e r v a l , t i s i n J ; which i s
6.
absurd, proving ( 7 ) . Apply ( 7 ) t o I , a n d t h e Conway S i m p l i c i t y Theoran i s p r o v e d .
o
Let ( L , R ) b e a Conway c u t i n No ( 1 . 2 0 ) ; i . e . , l e t L and R b e s u b s e t s of No f o r w h i c h L
<
R.
By C o n w a y ' s S i m p l i c i t y T h e o r e m , t h e r e e x i s t s a
u n i q u e X E N O of minimal b i r t h d a y s u c h t h a t L
<
{x)
<
R.
Since x is uniquely
d e t e r m i n e d by L a n d R , l e t us f o l l o w Conway [24, p . 41 a n d use t h e symbol (LIR) t o d e n o t e x .
We w i l l a l s o s a y t h a t t h e Conway c u t ( L , R ) r e p r e s e n t s
x , a n d t h a t ( L , R ) i s a Conway cut r e p r e s e n t a t i o n of x.
(8)
Given any Conway c u t ( L , R ) a n d ( L ' , R ' )
< R)
=
PROOF,
(yeNo: L'
<
(y)
<
R'),
-
of No f o r which (YENO: L
t h e n (LIR)
Apply Conway's S i m p l i c i t y Theorem.
-
x
< (y}
(L'IR').
o
We w i l l s a y t h a t L a n d L ' ( r e s p . R and R ' ) a r e m u t u a l l y c o f i n a l
4.02
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
125
( r e s p . m u t u a l l y c o i n i t i a l ) i f f o r a l l aEL t h e r e e x i s t s a'EL' s u c h t h a t a 5 a ' , a n d f o r a l l a f E L * t h e r e e x i s t s acL s u c h t h a t a' 5 a ( r e s p . i f f o r a l l CER t h e r e exists c ' E R '
such t h a t c 5 c').
such t h a t c' 2 c , and f o r a l l c'ER'
t h e r e e x i s t s CER
We w i l l s a y t h a t t h e Conway c u t s ( L , R ) a n d ( L f , R ' ) a r e
equivalent i f L and L ' are m u t u a l l y c o f i n a l and R and R ' Clearly (8) implies t h e following:
coinitial. (9)
If ( L , R ) a n d ( L ' , R ' )
EXAMPLE 0 . and R '
=
are mutually
Let L
a r e equivalent, then {L'IR'] =
and R
{-I}
11/21; t h e n 0 = { L f l R f ] .
cofinal.
=
[ l ] ; then 0
=
=
[LIR}.
(LIR].
C l e a r l y L and L '
Let L '
[-21
are not mutually
are not mutually c o i n i t i a l .
F u r t h e r , R and R '
=
Thus [LIR]
=
[ L f l R t } , and yet ( L , R ) and ( L f , R ' ) are not equivalent. A Conway c u t r e p r e s e n t a t i o n ( L , R ) of x w i l l be c a l l e d t i m e l y i f L and
R are s u b s e t s of 0
b ( x ) ( = Tb(x))*
Note t h a t t h e Conway c u t r e p r e s e n t a t i o n s
of 0 i n E x a m p l e 0 a r e n o t t i m e l y .
o
s e n t a t i o n of (10)
Also n o t e t h a t t h e o n l y t i m e l y r e p r e -
i s 10101.
Each XENO has a t i m e l y r e p r e s e n t a t i o n . PROOF.
Let b ( x ) = 8 ; t h e n XEN
a n element ( L , R ) i n CD(T ) . 0
8'
i.e.,
x is
in T
B+1
- TO.
Hence x i s
(L,R) i s t h e n a t i m e l y r e p r e s e n t a t i o n of x.
Another way t o d e s c r i b e s u c h a Conway c u t ( L , R ) i s t o n o t e t h a t L [ ~ E O ~ ( y~ <) x] : and that R ( ( Y E O ~ ( ~ ) y:
s e n t a t i o-n _------
=
[y€Ob(,):
y
>
XI [ 2 4 , p . 291.
x [24, p . 291.
=
Let us d e f i n e
< X ) , [ Y E O ~ ( ~y ) >: X I ) t o b e t h e C u e s t a D u t a r i
of
o
cut
repre-
Note also t h a t t h e Cuesta Dutari cut
r e p r e s e n t a t i o n of x is a C u e s t a D u t a r i c u t i n 0
S i n c e we h a v e b u i l t b(x)' up No, i n t h i s s e c t i o n , u s i n g C u e s t a Dutari c u t s have t h e f o l l o w i n g r e s u l t .
(12)
Let ( L , R ) a n d ( L ' , R f 1 b e t i m e l y Conway c u t r e p r e s e n t a t i o n s i n No,
Norman L. A l l i n g
126
4.02
such t h a t [LIR} = [ L t l R f } ; t h e n ( L , R ) and ( L ' , R ' )
a r e equivalent.
Assume f i r s t t h a t ( L , R ) i s t h e C u e s t a D u t a r i c u t r e p r e -
PROOF. s e n t a t i o n of x .
Let b ( x )
> x}.
6; then XECD(O 1, L
=
B
<
[x)
<
{ Y E O ~ ( ~y )<:
X I and
R
=
and s i n c e ( L ' , R ' )
i s t i m e l y , L ' is
a s u b s e t of L and R ' is a s u b s e t of R ; t h u s f o r a l l a ' c L '
t h e r e e x i s t s acL
( Y E O ~ ( ~ )y:
Since L '
=
R',
s u c h t h a t a ' 5 a , and f o r a l l c ' E R ' Since x L and R '
=
[L'IR'),
t h e r e e x i s t s C E R such t h a t c S c ' .
< [ y } < R') is empty;
{ Y E O ~ ( ~ )L: '
i s c o i n i t i a l i n R : i , e . , f o r a l l acL t h e r e e x i s t s a ' E L '
such t h a t
such t h a t c' 5 c .
a 5 a ' , and f o r a l l CER t h e r e e x i s t s c ' E R ' these
t h u s L ' is c o f i n a l i n
Combining
f i n d i n g s , we see t h a t L and L ' are m u t u a l l y c o f i n a l and R and R' are Now d r o p t h e a s s u m p t i o n t h a t [ L I R ] i s t h e C u e s t a
mutually c o i n i t i a l .
D u t a r i c u t r e p r e s e n t a t i o n of x .
Fran t h e above argument we see t h a t L and
L ' are both m u t u a l l y c o f i n a l w i t h {yEOb(x): y
m u t u a l l y c o f i n a l with each other.
< XI,
and hence t h e y a r e
S i m i l a r l y , one can show t h a t R and R '
a r e mutually coinitial.
(13)
(i) x S y iff ( i i ) x
<
R
f o r a l l y R , and xL
y
Assume f i r s t t h a t ( i ) h o l d s .
PROOF.
<
y for all x
Then x 5 y
<
y
R
,
L
.
and xL
< x
S
L
y , f o r a l l yR a n d x ; e s t a b l i s h i n g ( i i ) . C o n v e r s e l y , assume t h a t ( i i )
holds.
Assume, f o r t h e moment, t h a t ( i ) i s f a l s e : i . e . , t h a t y
we s e e t h a t t h e f o l l o w i n g hold: ( a ) y
.
x R , f o r a l l x L , x R , y L , and y R ( a ) we s e e t h a t b ( y ) we see t h a t b ( x )
<
<
L
<
y
< x <
y R , and ( b ) x
<
x.
L
<
Thus y
<
x
<
Using t h e Conway S i m p l i c i t y Theorem a n d Using t h e Conway S i m p l i c i t y Theorem and ( b )
b(r).
b ( y ) ; which i s a b s u r d .
The c o n t r a p o s i t i v e of (13) i s t h e f o l l o w i n g :
<
L
y iff (ii)x 5 y
, for
sane y L ,
or xR 6 y , f o r sane x
R
.
(14)
(i)x
(15)
U n l e s s s t a t e d t o t h e c o n t r a r y , w e w i l l assume h e n c e f o r t h t h a t a l l r e p r e s e n t a t i ons under c o n s i d e r a t i on are t i me1y
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.02
127
Ccmbining ( 9 ) and ( 1 2 ) we have t h e f o l l o w i n g : (16)
L e t ( L , R ) a n d (L',R')
b e Conway c u t s i n No.
e q u i v a l e n t i f and o n l y i f { L ' I R ' )
=
( L , R ) and ( L ' , R ' ) a r e
(LIRJ.
Note t h a t we c a n d e s c r i b e t h e C u e s t a Dutari c o n s t r u c t i o n No of t h e
s u r r e a l numbers as f o l l o w s . (17)
XENO i f and o n l y i f x i s t h e f o l l o w i n g o r d e r e d p a i r of sets: ((yENo(<,b(x)): y
< XI, (yENo(<,b(x)):
B I B L I O G R A P H I C NOTE.
y
>
XI).
Much of what was g i v e n i n f i r s t p a r t of t h i s
s e c t i o n i s at most a s l i g h t r e w o r k i n g of C u e s t a D u t a r i ' s p a p e r [ 2 5 1 of 1954.
( 5 ) c a n b e f o u n d i n Harzheim's e l e g a n t paper 143, pp. 116-1201, of
1964.
These i d e a s were t h e n a p p l i e d t o i d e a s developed by Conway C24, p p .
4-29],
i n 1976.
Cuesta-Dutari
N o t e a l s o how many s i m i l a r i t i e s t h e r e a r e between t h e
c o n s t r u c t i o n and von Neumann's c o n s t r u c t i o n of t h e c l a s s On
of a l l o r d i n a l n u m b e r s , ( 1 . 0 2 ) .
The new r e s u l t s of t h i s s e c t i o n a r e a
r e s u l t of j o i n t work w i t h P h i l i p E h r l i c h , much of w h i c h a p p e a r e d , w i t h o u t many of t h e p r o o f s , i n [6].
4.03
AN ABSTRACT CHARACTERIZATION OF A FULL CLASS OF SURREAL NUMBERS
Thus f a r i n t h i s monograph we have d i s c u s s e d two s l i g h t l y d i f f e r e n t c o n s t r u c t i o n s of c l a s s e s of s u r r e a l numbers: Conway's [241 ( S e c t i o n s (4.00) and ( 4 . 0 1 ) ) , and t h e one g i v e n i n (4.02) u s i n g Cuesta Dutari c u t s .
In
Chapter 5 we w i l l g i v e a n o t h e r o n e , u s i n g Conway's n o t i o n of t h e s i g n e x p a n s i o n of a surreal number as i t s d e f i n i t i o n .
We have yet t o show t h a t
Conway's surreal numbers a r e e s s e n t i a l l y t h e same as t h o s e c o n s t r u c t e d i n (4.02).
This s t a t e of a f f a i r s resembles, t o some d e g r e e , t h e s t a t u s of t h e
v a r i o u s c o n s t r u c t i o n s of t h e real numbers; as Dedekind cuts i n t h e r a t i o n a l n u m b e r s , a s l e a s t upper bounds of bounded s u b s e t s of t h e r a t i o n a l numbers, a s e q u i v a l e n c e c l a s s e s of Cauchy sequences of r a t i o n a l numbers, as i n f i n i t e decimals, e t c .
Each of t h e s e c o n s t r u c t i o n s has i t s u s e s .
Once i t h a s been
shown t h a t t h e f i e l d of a l l real numbers is, u p t o i s o m o r p h i s m , t h e o n l y c o m p l e t e o r d e r e d f i e l d , t h e n o n e c a n p a s s f r e e l y back and f o r t h between t h e s e v a r i o u s c o n s t r u c t i o n s , as need o r i n c l i n a t i o n s u g g e s t s .
Norman L. A l l i n g
1 28
4.03
I n t h i s s e c t i o n we w i l l d e v e l o p a n a b s t r a c t d e s c r i p t i o n o f w h a t we w i l l c a l l a f u l l c l a s s F of s u r r e a l n u m b e r s , a n d show t h a t any two s u c h
classes have an o r d e r - p r e s e r v i n g i s a n o r p h i s m between them t h a t p r e s e r v e s T h i s shows t h a t t h e two c o n s t r u c t i o n s g i v e n s o f a r of
birth-order.
No a r e
indeed isomorphic. I t p r o v e s u s e f u l t o g e n e r a l i z e t h i s d i s c u s s i o n a l i t t l e f u r t h e r by
c h o o s i n g a n i n i t i a l segment S of On. S = [O,B)
If S i s a p r o p e r s u b c l a s s of On, t h e n
f o r a u n i q u e 8 ~ O n . If n o t t h e n l e t
L e t ( a ) F be a n ordered class.
On; t h e n a g a i n S
=
[O,B).
Assume ( b ) t h a t a f u n c t i o n b h a s been
d e f i n e d which maps F o n t o t h e s u b c l a s s [O,B) birth-order function. -___
=
of On, t h a t w i l l be c a l l e d a
Assume ( c ) t h a t , g i v e n any Conway c u t ( L , R ) i n F f o r
< a < 6, t h e n t h e r e e x i s t s a u n i q u e XEF f o r which b ( x ) i s m i n i m a l , s u c h t h a t L < ( X I < R . Such a t r i p l e { F , < , b . f i ] w i l l b e c a l l e d a which b ( L ) , b(R)
class ___-
of s-u r r e a l -____ n u m b e r s -of ---h e i g h t 8.
I
If B = O n , l e t u s a l s o d e n o t e
I F , < , b , B ] by ( F , < , b } , a n d l e t u s c a l l i t a c l a s s of. s u r r e a l n u m b e r s . C o n d i t i o n ( c ) w i l l a l s o b e referred t o as Conway's S i m p l i c i t y Theorem. Let ( F , < , b , B } b e a class of surreal numbers of h e i g h t B.
t h a t s i n c e b maps F o n t o [ O , B ) , t h e n F is a proper class.
Next, n o t e t h a t s i n c e Conway's S i m p l i c i t y
Theorem h o l d s , x i s u n i q u e l y d e t e r m i n e d by L and R . t h e symbol ( L I R I be used
Note f i r s t
t h a t i f BEOn, t h e n 161 2 I F I ; and i f B = On
t o denote x.
F o l l o w i n g Conway, l e t
C o n t i n u i n g t o u s e Conway's n o t a t i o n L
a n d c o n v e n t i o n s , ( 4 . 0 0 ) o r C24, p . 41, we w i l l write x as { x Ix L (x
(0)
=
R
L and ( x 1
R.
Let y be i n No, w i t h y
( i ) x 2 y i f f (ii) x
PROOF. Similarly x ( i i ) holds.
x.
=
R
1 , where
L R ( y ( y 1.
=
< yR f o r a l l yR, a n d xL < y for a l l xL .
Assume t h a t ( i ) h o l d s .
Then x 2 y
<
yR, and hence x
L
< yR
.
< x S y , a n d h e n c e xL < y ; t h u s ( i i ) h o l d s . Assume now t h a t Also assume, f o r a moment, t h a t ( i ) i s f a l s e : i . e . , t h a t y <
Thus we s e e t h a t t h e f o l l o w i n g h o l d : ( f ) yL
< y < x < y R , a n d (**I
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.03
xL < y
< x <
R
xR, for a l l xL, x
L R y , and y
,
<
Theorem a n d ( * ) we s e e t h a t b ( y )
<
Theorem and ( * * ) we see t h a t b ( x )
.
b(x).
129
Using t h e Conway S i m p l i c i t y U s i n g t h e Conway S i m p l i c i t y
b ( y ) ; which i s a b s u r d .
n
The c o n t r a p o s i t i v e o f ( 0 ) i s t h e f o l l o w i n g :
(i) x
(1)
<
y i f f ( i i ) x 6 y L , f o r sane y L , o r x
F o r e a c h a ~ [ O , 8 ) ,l e t and l e t F ( = , a ) (2)
=
F(<,a)
=
b-'([O,a)),
R
2 y , f o r sane x
let F(6,a)
=
b
-1
R
.
([O,a]),
b-'([a)).
{F,<,b,B] is
full
i f f o r a l l Conway c u t s (L,R) i n F ( < , a ) , [LIR) i s i n
F ( < ,a).
i s a f u l l class of s u r r e a l numbers of
{ N o , < , b } , as d e f i n e d i n (4.021, h e i g h t On.
i s a l s o a f u l l c l a s s of s u r r e a l numbers of
C o n w a y ' s c l a s s No [ 2 4 ] h e i g h t On. p.
T h i s may be s e e n by c o n s u l t i n g Conway's c o n s t r u c t i o n o f No C24,
41, h i s d e f i n i t i o n o f o r d e r a n d e q u a l i t y o n No 1 2 4 , p . 41 ( w h i c h i n -
spired (0) and (4.02:13), c o n s u l t i n g [24,
pp.
a n d i s i n c o m p l e t e a g r e e m e n t w i t h e a c h ) , by
15-17],
by n o t i n g t h a t Theorem 1 1 1 2 4 , p . 231 is what
we h a v e c a l l e d Conway's S i m p l i c i t y Theorem, a n d by r e a d i n g [24, pp. 29-30]. Let [ S , < , b ] d e n o t e t h e o r d e r e d c l a s s of a l l s i g n - e x p a n s i o n s , as g i v e n by Conway [24, pp. 30-311.
See a l s o ( 5 . 3 0 ) .
t h a t t h e s i g n - e x p a n s i o n map X E N O map t h a t p r e s e r v e s b i r t h - o r d e r ; n u m b e r s of h e i g h t O n .
+
Conway s h o w e d C24, p . 301
(x)ES is a s u r j e c t i v e order-preserving
t h u s ( S , < , b ] i s a f u l l c l a s s of s u r r e a l
T h i s c a n a l s o b e s h o w n d i r e c t l y by w o r k i n g o n
is,<, b } . Let { F , < , b , B ) be a full c l a s s of s u r r e a l numbers.
= a
<
8.
then L(x)
Let L ( x )
< {XI <
=
ItcF(<,a): t
R(x).
d e f i n e d t o be [ L ( x ) ( R ( x ) ] .
< XI
and l e t R(x)
=
Let X E F , w i t h b ( x ) itsF(<,a):
Assume t h a t L ( x ) a n d R ( x ) a r e s e t s . By ( 2 1 , z i s i n F ( S , a ) .
~ ( x )a n d R ( x ) i s F ( < , a ) , b ( z )
= a.
t
>
x);
Let z be
S i n c e t h e u n i o n of
S i n c e C o n w a y ' s S i m p l i c i t y Theorem
Norman L. A l l i n g
130
holds, z
=
4.03
f o r No, t h e
Let ( L ( x ) , R ( x ) ) b e c a l l e d , as i t w a s i n ( 4 . 0 2 )
x.
Cuesta D-u t a r i --cut -~
r e p r e s e n t a t i o n of x i n I F , < , b , f 3 } .
1 _ 1
T h u s we h a v e p r o v e d
the following: Each XEF f o r which L ( x ) a n d R ( x ) a r e s e t s , x h a s a unique Cuesta
(3)
D u t a r i c u t r e p r e s e n t a t i o n i n { F , <, b , B } . Let [ F , < , b , B ) , { F ' , < , b * , B I , a n d [ F t t t , < , b l l , B l b e f u l l c l a s s e s of
L e t f be a map form F i n t o F' and l e t h be a
s u r r e a l numbers of h e i g h t B. m a p p i n g of F' i n t o F". YEF i m p l i e s f ( x )
f w i l l be c a l l e d a s u r r e a l monomorphism i f ( i ) x
< f ( y ) h o l d s , and
<
f o r a l l XEF.
(ii) i f b(x) = b ' ( f ( x ) ) ,
C l e a r l y we have t h e f o l l o w i n g . If f and h a r e s u r r e a l monomorphisms, t h e n h - f is a surreal monomor-
(4)
phi s m
.
( i ) There e x i s t s a s u r r e a l monomorphism of F i n t o F'.
LEMMA.
Let f a n d g b e a s u r r e a l monomorphism of F i n t o F ' , t h e n f = g . s u r r e a l monomorphism f of F i n t o F ' , m a p s F o n t o F ' .
Let
aE[O,e)
( i i i ) Any
( i v ) F o r a l l aEOn
is a s e t .
s u c h t h a t a 5 8, b-'([O,a)) PROOF.
(ii)
a n d assume t h a t t h e r e e x i s t s a s u r r e a l monomor-
p h i s m f of F ( < a ) o n t o F ' ( < & ) , a n d t h a t F ( < a ) a n d F ' ( < a ) a r e s e t s . t h a t 0 has t h i s p r o p e r t y . )
Let x , ~ F ( = a ) , a n d l e t ( L , , R , )
Dutari cut representation.
Clearly (f(L,),f(R,))
F'.
Let x,'
n u m b e r s of h e i g h t 6 ( 2 ) , b ' ( x , ' ) x,;
be its Cuesta
is a Cuesta Dutari cut i n
S i n c e [ F t , < , b l , B } i s a f u l l c l a s s of s u r r e a l
= {f(Lo)lf(Ro)].
i s equal t o F ' ( < a ) , b ' ( x , ' )
I a.
S i n c e t h e u n i o n of f ( L , ) a n d f ( R , )
2 a ; thus b'(x,')
=
Let x , E F ( = a ) , w i t h x o
a.
a n d l e t ( L , , R , ) b e t h e Cuesta D u t a r i c u t r e p r e s e n t a t i o n of x , .
( f ( L , ) , f ( R , ) ) i s a C u e s t a Dutar i c u t i n F 1 . S i n c e xo
< x,,
R
f(xl)
>
< xl,
<
Clearly
Let x l t = [ f ( L , ) l f ( R , ) l .
s i n c e x, and x , t F ( = a ) , a n d hence a l l x o R and a l l x l L are i n
F ( < a ) , we may i n v o k e ( 1 ) t o c o n c l u d e t h a t ( a ) x p ( b ) x,
(Note
f o r some x,
L f ( x l )cf(R,).
R
.
< xlL, for
some x l L , o r L
Assume t h a t ( a ) h o l d s ; t h e n x i ER,,
Since f ( R , ) R
and thus
> ( f ( x , ) ] , we s e e t h a t f ( x , ) > f ( x , ) .
Assume t h a t ( b ) h o l d s ; t h e n x, E L , , a n d t h u s f ( x , )
<
R f ( x , )cP(L,).
Since
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.03
<
f(L,)
{ f ( x , ) } , we s e e t h a t f ( x o )
<
131
Since each x , ~ F ( = c r ) is
f(x,).
u n i q u e l y d e t e r m i n e d by a C u e s t a D u t a r i c u t i n t h e s e t F ( < a ) , F(6a) i s a g a i n
a set.
Let z ' E F ' ( = a ) , w i t h z '
cut i n F'(
Let z
F(
=
=
{L'IR'),
where ( L ' , R ' )
i s a Cuesta D u t a r i
i s a Cuesta D u t a r i cut (L.R) i n
Clearly (f-'(L*),f-'(R'))
{LIR}; t h e n z EF ( =a ) a n d f ( z ) =
Since each z'cF'(=a) i s
2'.
u n i q u e l y d e t e r m i n e d by a C u e s t a D u t a r i c u t i n t h e s e t F ' ( < a ) , F'(Sa) i s again a s e t .
As t o ( i i ) , a s s u m e t h a t t h e r e s t r i c t i o n of f t o F ( < a ) ) ,
f l F ( < a ) , is equal t o glF(
=
( g ( L o ) , g ( R , ) ) i s a Cuesta Dutari c u t i n F'(
{ f ( L o ) ( f ( R o ) )= ( g ( L o ) ( g ( R o ) ) = g ( x o ) . THEOREM.
Thus f ( X o )
0
Assume t h a t { F , < , b , B ) a nd { F ' , < , b ' , B )
s u r r e a l numbers of h e i g h t B.
Clearly
a r e f u l l c l a s s e s of
The r e exists a u n i q u e surreal monomorphism f
of F o n t o F ' . COROLLARY.
T h e r e e x i s t s a u n i q u e s u r r e a l monomorphism, which maps
t h e c l a s s of Conway's s u r r e a l numbers C241 onto t h e c l a s s of s u r r e a l numb e r s c o n s t r u c t e d i n (4.02) BIBLIOGRAPHIC NOTE.
( a n d i n C71).
I n [ 7, ( 1 1 1 t h e a u t h o r a n d P h i l i p E h r l i c h m a d e ,
what h e r e i s a theorem ( 4 . 0 3 : 0 ) ,
into a condition.
I t w a s E h r l i c h who
f i r s t s u s p e c t e d t h a t r a t h e r t h a n b e i n g a c o n d i t i o n i t was, i n f a c t , a
theorem.
He w e n t o n t o show t h a t ( 0 , i ) i m p l i e d ( O , i i ) , w h i l e t h e a u t h o r
showed t h a t ( 0 , i i ) i m p l i e d ( 0 . i ) . [6],
b o t h authors - separately
-
s u r r e a l numbers of h e i g h t 6 a g a i n .
Having m i s s e d t h i s f a c t when we w r o t e c o n s i d e r e d t h e axioms f o r a f u l l class of The r e s u l t s of t h e a u t h o r s r e f l e c t i o n s
may be found i n S e c t i o n 4.60. 4.04
SUBTRACTION I N No
Assume t h a t No i s t h e c o n s t r u c t i o n of t h e s u r r e a l numbers by means of C u e s t a D u t a r i c u t s , t h a t we c o n s i d e r e d i n S e c t i o n 4.02. f u l l c l a s s of s u r r e a l n u m b e r s ( o f h e i g h t O n ) .
Then (N o , < , b } i s a
By T h e o re m 4 . 0 3 we can
t r a n s f e r a l l the or e m s t h a t a r e s t a t e d p u r e l y i n terms of t h e l i n e a r o r d e r and t h e b i r t h - o r d e r f u n c t i o n , t o any o t h e r f u l l c l a s s of s u r r e a l numbers.
132
Norman L . A l l i n g
4.04
Following Conway [ 2 4 , p . 51, we w i l l c o n s i d e r e x p r e s s i o n s { L I R ) , f o r which L and R a r e s u b s e t s of No, and c a l l s u c h an e x p r e s s i o n s games.
a game { L I R ) , t h e n i t i s a number i f and o n l y i f L Let x
( L I R } b e i n No.
=
<
Then, by assumption (4.02:15),
t i m e l y Conway c u t r e p r e s e n t a t i o n of x.
Given
R.
(L,R) is a
Following Conway, d e f i n e s u b t r a c -
t i o n as f o l l o w s : l e t -x be d e f i n e d t o be { - R l - L ] . (0)
(i)
-x i s an element i n No, which i s independent of r e p r e s e n t a t i o n ,
(ii)
-(-XI
=
x,
( i i i ) b(-x)
=
b ( x ) , and
x <
(iv)
y i f and o n l y i f -y
< -x. i s a game.
I n i t i a l l y a l l we know i s t h a t ( - R l - L )
PROOF.
We w i l l
p r o v e ( 0 ) by i n d u c t i o n , b y a s s u m i n g t h a t ( 0 ) h o l d s on Oa ( = b - ' ( [ O , a ) ) . Note t h a t i f a = 0 , t h e n Oa i s e m p t y , t h u s ( 0 ) h o l d s o n O a ; h e n c e o u r i n d u c t i o n i s launched.
-
a.
By ( O , i v ) , - x R
<
thus b(x)
(= b-'(a));
< xR.
To see t h a t (0) holds on Ma ( = b - l ( [ O , a l ) , F i r s t n o t e t h a t each x L
-x ; t h u s ( - R 1 - L )
L
let X E N ~
and xR i s i n Oa, and x
i s a number.
I n i t i a l l y we may
assume t h a t ( L , R ) i s t h e C u e s t a D u t a r i c u t r e p r e s e n t a t i o n o f x. ( L o , R o ) b e any Conway c u t i n 0
CL
t h a t r e p r e s e n t s x.
L o and L are m u t u a l l y c o f i n a l a n d R ,
(4.02:16),
coinitial.
Since ( 0 ) holds on Oa,
-(-x)
=
=
I-RI-L);
-{-RI-L)
=
. a
Thus by (4.02:16)
Since ( 0 , i i ) h o l d s on Oa,
{ L I R ] = x; showing t h a t ( 0 , i i ) h o l d s on Ma,
t h e element of No of minimal b i r t h d a y s u c h t h a t L and conclude t h a t t h e r e i s no element ~ b ( - x ) 2: b ( x ) .
and -R are m u t u a l l y
we know t h a t -R,
e s t a b l i s h i n g ( 0 , i ) on M
Let
Then, as we saw i n
and R are m u t u a l l y
c o f i n a l , and t h a t -Lo and -L are m u t u a l l y c o i n i t i a l . (-Rol-Lo}
L
<
x
<
€ such 0 ~t h a t -R
Since x is
R , we may a p p l y ( 0 )
<
(z}
<
-L.
Thus
S i n c e (No,<,b} i s a full class of surreal numbers (4.03:2),
b ( - x ) 5 a , showing t h a t b(-x) = b ( x ) , and hence showing t h a t ( 0 , i i i ) h o l d s o n Ma. on M
Ci
Now l e t ycM
, we
a'
w i t h y = ( L t l R t ] , and w i t h x
<
y.
Since ( 0 , i ) holds
may assume, without l o s s of g e n e r a l i t y , t h a t ( L t , R t )i s t h e
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.04
C u e s t a D u t a r i c u t r e p r e s e n t a t i o n of y. =
(L,R), y
=
F i r s t assume t h a t b ( y )
< y.
Since x
(L',R')ECD(T~).
is a p r o p e r s u b s e t of - R ;
y is i n R.
t h u s -y
<
a ; then x As
U s i n g ( 0 ) o n O a , we know t h a t Assume now t h a t b ( y )
-x.
and we s e e t h a t - y
Hence - y c ( - R ) ,
=
L i s a p r o p e r s u b s e t of L'.
a c o n s e q u e n c e , R ' is a p r o p e r s u b s e t of R . -R'
133
<
-x.
<
a ; then
T h u s we h a v e s h o w n
t h a t (0) h o l d s o n Ma.
A D D I T I O N I N No
4.05
Conway d e f i n e s a d d i t i o n i n No as f o l l o w s C24, p . 51:
x + y = Ix
(0)
L
L
R
R
+ y , x + y Ix
+ y , x t y }.
L e t u s s e e sane of t h e m o t i v a t i o n b e h i n d t h i s d e f i n i t i o n .
< x <
xL
x
R
and y
L
<
y
<
y
.
R
S i n c e we want No t o be a n o r d e r e d g r o u p u n d e r
<
a d d i t i o n , we m u s t h a v e ( i ) xL + y
<
<
xR + y , a n d ( i v ) x + y
x + y
R
.
b i r t h d a y s u c h t h a t i n e q u a l i t i e s (1)
[OL
-
(ii) 0
+
1 and 1 + 0 a r e b o t h t h e number 1 .
PROOF.
Since 0 = L
R
0 10
+
0, and there are no 1
L
1)
=
I0
+
0, 0
R
.
+
there are no 0 0
R
1
=
x + y, (iii) x + y
( i v ) hold.
0 i s t h e number 0, a n d
[I},
<
Thus x + y i s t h e e l e m e n t of smallest
+
+ 0, 0 +
I0 + 1
x + y , ( i i ) x + yL
0
(i)
(1)
Note t h a t
(1)
= 0.
L
R and no 0 ; t h u s 0
Recall t h a t 1
=
+
0 =
[Ol); t h u s l L
=
R
Thus 0 + 1 = [OL + 1 , 0 + lLIOR + 1 , 0 + 1 1 =
0 1 1 = I011 =1.
0
On o c c a s i o n w e w i l l p r o c e e d b y i n d u c t i o n o n t h e n a t u r a l ,
or
H e s s e n b e r g , sum of t h e o r d i n a l n u m b e r s b ( x ) a n d b ( y ) , b ( x ) + b ( y ) , e t c . Recall t h a t g i v e n a n o r d i n a l number a
>
0 i t has t h e following unique
e x p a n s i o n , which i s c a l l e d C a n t o r ' s normal form C18, p . 2371:
134
Norman L. A l l i n g
w
(2)
“1
B,
+
...
w
+
“n
Bn, w i t h
... > q n ,
>
q,
4.05
>
and w
... ,
B1,
Bn > 0.
R e g a r d i n g s u c h s u m s as f o r m a l power s e r i e s a l l o w s u s t o a d d a n d
Let u s a l s o r e g a r d t h e o r d i n a l number 0 as t h e empty expan-
m u l t i p l y them.
s i o n ( 0 1 , a n d l e t i t b e a d d e d a n d m u l t i p l i e d a s a formal power series. Such sums and p r o d u c t s g i v e r i s e t o t h e n a t u r a l numbers.
a n d p r o d u c t of o r d i n a l
C55, pp. 246-2611.)
(See e.g.,
Note t h a t we have y e t t o p r o v e t h a t x we know t h a t i t is a game ( 4 . 0 4 ) .
t i o n of a d d i t i o n between games.
+
y i s a l w a y s a number; however
F u r t h e r , we may r e g a r d (0) as a d e f i n i -
As s u c h we c a n e s t a b l i s h p r o p e r t i e s a b o u t
F i r s t note that
it.
(3)
0
+
x
=
PROOF.
x , f o r a l l XENO. Assume t h a t 0
+
u = u is t r u e f o r a l l UEO
U’
and l e t b ( x ) = a.
(Note that i f a
=
0, t h e i n d u c t i o n h y p o t h e s i s i s empty, x
noted i n ( 2 ) , 0
+
x
+
xR I .
Since 0
x.)
By d e f i n i t i o n , 0
+
x
(0
=
+
xLIO
L R h y p o t h e s i s ) i s {x Ix 1, which i s x .
x
(4)
+
y
=
PROOF.
+
x = {OL
+
x, 0
+
L R x 10
+
x, 0
( 1 1 , t h e r e a r e n o 0L , o r 0 R : i . e . , 0 h a s n o o p t i o n s
=
Thus 0
(4.00).
=
0 , a n d , as
=
+
R x 1 , which ( u s i n g o u r i n d u c t i o n
o
y + x , f o r a l l x , YENO. Let x a n d y be chosen s o t h a t f o r a l l U , V E N O ,
n a t u r a l sum, b ( u )
+
(Note t h a t i f
b ( v ) , i s less t h a n b ( x )
f o r which t h e
b(y) = a; then u
v
v
=
(Y
=
0, t h e n t h i s i s t h e empty i n d u c t i o n h y p o t h e s i s ,
x
y , and hence x
+
y
+
+
+
u.
= 0 =
x.)
S i n c e any o p t i o n z of x ( r e s p . y ) i s s i m p l e r
than x (resp. y): i.e., b ( z )
< b ( x ) ( r e s p . b ( z ) < b ( y ) ) , we c a n u s e t h e
=
y
+
i n d u c t i o n h y p o t h e s i s t o show t h a t x [y
t
L
x , y
L + x I y + x R , yR
+ XI
=
y
+
y
+
x.
=
IxL 0
+
y, x
+
yLlxR
+
y,
x
+
yR)
=
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.05
135
Let u s now c o n s i d e r two s t a t e m e n t s which we w i l l p r o v e f o r a l l u, v ,
a n d z i n No.
w,
The f i r s t o f t h e s e , we w i l l c a l l P ( u , v : w , z )
is the
following: ( i ) u 2 v a n d ( i i ) w 5 z, implies ( i i i ) u
(5)
w 2 v
+
(iv) Strict
z.
+
i n e q u a l i t y i n ( i ) or ( i i ) i m p l i e s s t r i c t i n e q u a l i t y i n ( i i i ) . O u r i n d u c t i o n w i l l be w i t h r e s p e c t t o rnax.(b(u) + b ( v ) , b(w) + b ( z ) ) , Let N ( x , y ) b e t h e s t a t e m e n t t h a t
which we w i l l d e f i n e t o be b ( P ( u , v : w , z ) ) .
x
(6)
+
L
y i s a number: i . e . , t h a t {x
+
Let b ( N ( x , y ) ) be d e f i n e d t o be b ( x )
y, x +
+
L
y 1
< {xR
t
R
y , x + y }.
b ( y ) , and consider t h e f o l l o w -
ing statement:. ( 5 ) a n d ( 6 ) h o l d , f o r a l l u, v , w , z , x , a n d y i n No.
(7)
PROOF.
To e s t a b l i s h ( 7 ) , we w i l l p r o c e e d b y i n d u c t i o n o n
max.(b(P(u,v:w,z)),
b ( N ( x , y ) ) ) , w h i c h we w i l l c a l l b ( Q ( u , v : w , z : x , y ) ) .
(Note t h a t i f b ( Q ( u , v : w . z : x , y ) ) (5) and ( 6 ) a r e t r u e . )
Let a
>
= 0,
then u
=
v = w
=
z =
x
=
y
=
0, and
0 , a n d assume t h a t f o r a l l u, v , w , z , x ,
<
a n d y i n No, f o r which b ( Q ( u , v : w , z : x , y ) )
a, t h e n (5) and ( 6 ) holds.
Now
l e t b ( Q ( u , v : w , z : x , y ) ) = a. L
<
<
x
u 5 v iff u
<
Note t h a t ( a ) u R
z < z , ( e ) xL (4.02:13),
x iff w
<
R
x
<
x
a n d wL
u R
< u R , ( b ) v L < v < v R , ( c ) w L < w < w R , ( d ) zL <
, and
v
R
(f) y
x
<
y
<
y
R
.
Note a l s o t h a t , b y
< v , f o r a l l v R a n d a l l uL ; and
a n d uL
< x, for a l l
L
R
and w
L
.
that w 2
S i n c e ( e ) a n d ( f ) h o l d , and
s i n c e t h e r e l e v a n t i n d u c t i o n h y p o t h e s i s h o l d s , we see by (51, t h a t x
xR (8)
t
y, x x
+
t
yL< x
t
y R , xL + y
<
x
t
y R , and x + y L
<
y is a number, as a l s o are u + w a n d v + x .
xR + y: t h u s
L
+
y
<
Norman L . A l l i n g
136
t e l l s us that u
Now n o t e t h a t (4.02:13)
<
u + w
( v + zIR and ( u
wIL < v
+
4.05
w 5 v
+
(v
+ z, for a l l
z if and o n l y i f
+
zIR, a n d a l l ( u
+
L
w) ,
+
But by ( 8 ) an d ( 0 1 , we h a v e
(9)
u
+
w 5 v + z iff u + w < v
u
+
wL < v
+ z, f o r a l l v
R
R
+
z, u + w
, zR , u L ,
<
v
and w
z R , uL
+
+
w < v
+
z, a n d
.
L
Then, u s i n g ( a ) - ( d ) , we know t h a t
Assume t h a t ( 5 , i ) a n d ( 5 , i i ) h o l d . t h e following holds:
u
(i)
(10)
L
(iii) u
< u
5 v
R
< v ,
< vR ,
(iv) w
and ( i i ) w
< zR ,
L
( v ) uL
< w
<
2 z
R
<
z ; which i m p l y t h a t
v, and ( v i ) w
L
< z.
Thus, we may i n v o k e t h e i n d u c t i o n h y p o t h e s i s , a p p l y i t t o ( l O , ( i i i ) ( v i ) ) u s i n g (51, t o e s t a b l i s h t h e r i g h t hand s i d e o f ( 9 ) .
With t h e a i d of
( 9 ) we h av e t h u s p r o v e d ( 5 , i i i ) .
Now assume t h a t s t r i c t i n e q u a l i t y h o l d s i n ( 5 , i ) o r ( 5 , i i ) .
that u
<
t h a t ( i ) uR S v , f o r s a n e u
see t h a t u
+
(4.02:14)
R
.
Assume f i r s t t h a t ( h ) h o l d s .
L
L
.
w 2 vL + z ; t h u s u + w 5 ( v + z I L , f o r some ( v + z ) w e see t h a t u + v < v + z . Assume, l a s t l y , t h a t ( i ) h o l d s .
w 5 v
t h u s ( u + w ) S~ v + z ,
+ z;
we s e e t h a t u
+
w
<
v
+ z,
,
or
U s i n g ( 5 ) we
+
(4.02:14),
(5), u R
we know t h a t ( h ) u 5 v L , f o r some v
Applying (4.02:14)
v.
Since
we may a s s u m e , w i t h o u t l o s s of g e n e r a l i t y ,
a d d i t i o n i s commutative ( 4 ) ,
f o r some ( u + w)
s h o wi n g t h a t ( 5 , i v ) h o l d s .
R
.
By By By
By i n d u c -
t i o n we h av e e s t a b l i s h e d ( 7 ) .
(x
(11)
+ y) + z =
PROOF.
“x
+
Y)
+ 2,
( y + z ) , f o r a l l x, y , an d z i n No.
+
Let u s p r o c e e d by i n d u c t i o n o n b ( x ) + b ( y ) + b ( z )
as we d i d ab o v e. L
x
(x
Thus (x + y ) + z = +
Y)
+
2
L )(x
+
y)
R
+
z, (x
+
y)
+
z
R
1
=
=
a , much
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.05
I(x L (xR
y) +
+
+
y)
+
L
137
LI
z, (x
+
y
z, (x
+
y)
+
z
z, (x
+
y ) + z, (x
+
y)
+
z R ] . We may now a p p l y t h e i n d u c t i o n
+
R
h y p o t h e s i s an d c o n c l u d e t h a t t h e l a s t number i s
(13)
x
+
(-x)
PROOF.
0, f o r a l l xcNo.
=
x
+
(-XI
(xL IxR
+
(-x R I-xL )
=
[xL
+
(-XI, x
+
(-XI L I
(-xR )1xR + (-XI, x + (-xL 1). Let u s p r o c e e d by i n d u c t i o n by a s s u m i n g t h a t u + (-u) = 0 , f o r a l l u s u c h t h a t R b ( u ) < a , w h e r e b ( x ) = a . We know t h a t ( i ) xL < x < x We h a v e s e e n R L L ( 4 . 0 4 : 0 , i v ) t h a t ( i ) i m p l i e s t h e f o l l o w i n g : ( i i ) -x < -x < -x , Adding x
xR
+
(-XI, x
+
(-XI
RI
-
=
(xL
+
(-XI,
x
+
.
t o t h e r i g h t - h a n d s i d e of ( i i ) , u s i n g t h e i n d u c t i o n h y p o t h e s i s , a n d ( 5 ) g i v e s us ( i i i ) xL
+
(-x)
<
0. Adding -xR t o t h e r i g h t - h a n d s i d e of
u s i n g t h e induction hypothesis, and (5) g i v e s us ( i v ) x
+
(i),
(-x R ) < 0.
Adding xR t o t h e l e f t - h a n d s i d e of ( i i ) , u s i n g t h e i n d u c t i o n h y p o t h e s i s , and (5) gives us (v) 0
< xR
+
(-XI.
F i n a l l y , a d d i n g -xL t o t h e l e f t hand
s i d e of ( i ) , u s i n g t h e i n d u c t i o n h y p o t h e s i s , a n d ( 5 ) g i v e s u s ( v i ) 0
L
(-x 1.
< x
+
Taken t o g e t h e r , ( i i i ) - ( v i ) show t h a t t h e f o l l o w i n g h o l d s :
S i n c e 0 i s t h e s i m p l e s t element i n No, ( 1 3 ) i s p r o v e d . Canbining ( 3 ) , proved t h e following:
( 4 1 , ( 7 1 , ( 8 ) , ( 1 1 1 , a n d ( 1 3 1 , we s e e t h a t we h a v e
4.05
Norman L. A l l i n g
138
Under a d d i t i o n No i s an o r d e r e d A b e l i a n g r o u p .
THEOREM.
The i d e a of u s i n g t h e n a t u r a l or H e s s e n b e r g sum
B I B L I O G R A P H I C NOTE.
i n t h e i n d u c t i o n , as we d i d a b o v e , c a n b e f o u n d i n G o n s h o r ' s b o o k o n t h e s u b j e c t [38].
Of c o u r s e t h e i d e a of n a t u r a l sums a n d p r o d u c t s p e r m e a t e s A f t e r a l l , Conway's normal form f o r a s u r r e a l number
Conway's work [24].
w h i c h we w i l l c o n s i d e r i n C h a p t e r 6 , i s a g e n e r a l i z a t i o n
[ 2 4 , pp. 32-33],
Normal f o r m s were a l s o u s e d by S i k o w s k i 1661, i n
of C a n t o r ' s normal form.
1 9 4 9 , t o c o n s t r u c t a n i n t e r e s t i n g f i e l d , which c a n be r e g a r d e d i n a v e r y n a t u r a l way as a s u b f i e l d of No. MULTIPLICATION I N No
4.06
Conway's d e f i n i t i o n of m u l t i p l i c a t i o n ([24, p . 51
Or
However, Conway g i v e s C24, p . 61
at first glance, a l i t t l e surprising.
sane more m o t i v a t i o n f o r i t , which we w i l l now expand o n .
<
i n No; t h e n xL
(0)
x - x
L
,
x
-
y
< xR
yL, xR
-
L
<
y
x , and yR
-
and y
<
y
R
( 4 . 0 0 : 5 ) ) seems,
.
Let x a n d y b e
Hence
y a r e a l l greater t h a n z e r o .
We w a n t t o d e f i n e a m u l t i p l i c a t i o n o n No u n d e r which i t w i l l be an o r d e r e d f i e l d ; t h u s w e must have t h e f o l l o w i n g :
(1)
L
L
L
(i)
O < ( x - x ) ( y - y ) = x y - x y - x y
(ii)
o <
(iii)0
(iv)
(xR
-
x)(y
R
-
R
-
-
R
R R
=
xy
< ( x - x L ) ( yR - y )
=
-xy
+
L x y
=
-xy
+
R x y +xyL
(xR - x ) ( y
-
yL
xy
L L x y ,
+
y)
o <
x y
L
+ x y
xyR
-
,
xLyR, a n d
- x Ry L ,
A s a consequence we s e e t h a t
(2)
(i)
L x y
( i i i ) xy
+
<
-
xyL
L
x y
+
L L x y
<
R xy, ( i i ) x y L R
xyR - x y
,
+
and ( i v ) xy
xy
<
R
R
R R x Y
x y + xy
< L
XY.
-
R L x y
.
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.06
139
Using ( 1 ) a n d ( 2 ) we s e e t h a t f o r No t o b e a n o r d e r e d f i e l d t h e f 011owing i nequal i t y must hol d: L L L L R R R R I x y + x y - x y , x y t x y - x y ) < x y < L R L R R Ix Y XY - x y , x y i. xyL - x R y L ) .
(3)
+
Hence by d e f i n i t i o n of x y , [24, p . 51, i s L
(4)
L L R R R R - x y , x y t x y - x y l R L R R L R L Y - x y , x y + x y - x y } .
I X Y + X Y
L
X Y + X
L
Thus xy is t h e s i m p l e s t element i n No s u c h t h a t (3) h o l d s . S i n c e ( 1 ) ( o r e q u i v a l e n t l y (2) o r ( 3 ) ) h a s y e t t o b e p r o v e d , we do n o t know t h a t xy i s a number, w e know o n l y t h a t i t i s a game.
As such t h e
f o l l o w i n g h o l d s f o r a l l x , YENO.
PROOF.
( xL0 + x o L
(1)
=
Recall t h a t 0
- x L0 L , x R 0
+
=
xOR
( 1 1;
-
t h u s t h e r e a r e n o O L o r a n y 0R
x R 0R
I
x L0
xOR
i.
-
0 , e s t a b l i s h i n g ( i ) . To prove ( i i ) , r e c a l l t h a t 1
R 0 , a n d t h e r e i s no 1
.
L {x 1
+
xlL
+
L (x 1
+
x0
-
-
R XLlL, x 1
L R x OIx 1
i n d u c t i o n on b ( x )
+
+
x0
-
xLOR, x R o + xoL
- {Ol 1 ;
.
x0 =
XROLJ
=
t h u s 1L
=
Proceeding by i n d u c t i o n on b ( x ) , we see t h a t x l = xlR
-
xRIRI xL1
- x R 01
=
(x
L
Ix
R
+
xlR
}
x.
-
-
x L I R , xR1 + x l L
-
x R1L 1 =
To prove ( i i i ) , proceed by
By d e f i n i t i o n ( 4 ) we know t h a t xy i s equal t o
b(y).
-
xRyR
I
x y
+
xy
R
-
-
R R y x
I
L y x
+
yxR
-
LR R R L x y , x y + xyL - x y I = L L L L R R R R R R L R L L R (yx + y x - y x , y x + y x - y x I y x L + y x - y x , y x + y x - y x } = L
+ xyL
-
xLyL, xRy
t y L x + yxL
-
yLxL, y R x + y x
{x y
yx.
+
xyR
R
y L x R , y R x + yxL
-
y Rx L 1
=
L a s t l y , t o e s t a b l i s h ( i v ) , l e t us proceed by i n d u c t i o n on b ( x ) + b ( y ) .
Norman L. A l l i n g
140
4.06 R
By t h e d e f i n i t i o n of s u b t r a c t i o n (4.041, I(-xR)y
-
(-x)yL
+
(-xR)yL, (-xL)y
+
L
L
( - x ) y = I-x I-x I I y Iy ( - x ) y R , - ( - x L ) yR I (-x ) y +
-
( - x R ) y R , ( - x L )Y
I-x R y - xy L -(xy).
which i s
(-x)y
R
+
=
(6)
(x
(-y)x
x Ry L , -x L y - xy R =
L
1(x (x
(x
y)z L
+
y z , f o r a l l x , y , and z i n No.
+
(x
+
-
y)2L
+
y)z
(xL
+
L
(x + y)zL
-
(x + y )z
y)z
+
(x
y)zL
-
(xR
R y )z
+
(x + y)zL
-
(x
+
+
y,
+
Rlz =
Y)ZL,
+
+
+ L Ix R
(xL + y ,
=
y )z
+ +
=
BY d e f i n i t i o n , ( x
t
+
R
L R R R R R x y I-x y - xy + x y , -xLy - xyL t xLyL} - ( y x ) ; t h u s u s i n g what we have j u s t p r o v e d , +
x(-y); establishing ( i v ) .
y)z = xz
+
PROOF.
(x
+
By ( i i i ) , - ( x y )
-(xy)
(7)
I
( - x ) y L - ( - x L ) yL 1. Using t h e i n d u c t i o n h y p o t h e s i s , and we s e e t h a t t h i s last game i s e q u a l t o t h e f o l l o w i n g game:
(4.05).
=
R
L
,
Y)ZL,
R
y )z
L
I.
We w i l l p r o v e ( 6 ) by i n d u c t i o n on b ( x )
+
b(y)
+ b(z).
Then, making
t h e usual i n d u c t i o n h y p o t h e s i s , ( 7 ) i s e q u a l t o t h e f o l l o w i n g :
(8)
I XL Z + Y Z + X Z ~ L- XL ,Z x z + yL z + y zL - y Lz L, y zR - y Rz R 1
xR z
+
yz
+ XZR
-
XRZR,
xz
+
y Rz
xL z
+
yz
+ XZR
-
XLZR,
xz
+
yL2 + y zR
-
yL z R ,
xR2
+
yz
+
xzL - xRzL, x z
+
y Rz
-
y R z L 1.
Now expanding x z
+
t
+
yzL
yz, we f i n d t h a t i t
is e q u a l t o t h e f o l l o w i n g :
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.06
L {x z
+
-
xzL
xLzLt xRz
x L z + x z R - xL zR , x R z + x z L - xR zL 1 +
R R xzR - x z
.+
L L L R R R R tyLz+yz - y z , y z + y z - y z
(9)
L Ix z
+
xz
yLz
+
L x z
xz
.+
.+
xzL .+
xzR
L y z
-
.+
-
L L x z
yzL
.+
R yz, x z
L L y z , xz
-
L R x z
+
.+
+
yLz + y z R
xzR
R y z
R y z , x z + xzL
yzR - yLzR, x z
.+
141
R y z
+
.+
-
y L zR , y R z
.+
y zL - y R zL l =
R R x z + yz,
yz
R
-
xRzL yzL
-
R R y z +
I
yz,
R L y z I.
S i n c e t h e games i n ( 8 ) a n d ( 9 ) a r e e q u a l , ( 5 ) i s p r o v e d .
4.07
ORDER A N D MULTIPLICATION I N No
L e t x l , x,,
y l , a n d y, be e l e m e n t s i n No.
Conway s t a t e s a n d p r o v e s
t h e f o l l o w i n g t h r e e r e s u l t s [ 2 4 , Theorem 8, p p . 19-20], THEOREM. xy is a number. (i)
x , S x, a n d ( i i ) y 1 5 y 2 i m p l y ( i i i ) x 1 y 2
.+
x,y,
( i v ) I f ( i ) and ( i i ) are strict t h e n ( i i i ) is strict I f xl
=
x, t h e n x l y
We w i l l p r o v e ( 0 )
-
=
5 xlyl
.
.+
x2y2.
x,y.
(2) t o g e t h e r , by i n d u c t i o n .
Conway r e f e r s t o t h e i n e q u a l i t y o f ( 1 , i i i ) a s P ( x l , x 2 : y 1 , y 2 ) C24, p 191.
I f ( 1 , i i i ) h o l d s t h e n we w i l l write t h a t t t P ( x , , x , : y l , y , )
holds".
( 1 , i i i ) i s a s t r i c t i n e q u a l i t y we w i l l w r i t e t h a t " P ( x , , x 2 : y 1 , y , )
strictly".
(3)
If
holds
Now assume t h a t x , 5 x, 6 x , , a n d t h a t y 1 5 y2 6 y B .
I f ( i ) P(x1,x2:y1,y2) and (ii) P(x2,x,:yl,y,) P(x,,x,:yl,y2)
holds.
hold, then ( i i i )
F u r t h e r , ( i v ) i f a t l e a s t o n e of ( i ) or ( i i )
holds s t r i c t l y , then ( i i i ) holds s t r i c t l y .
Norman L. A l l i n g
142
4.07
Assume t h a t ( 3 , i ) a n d ( 3 , 1 1 1 h o l d ; t h e n we h a v e t h e
PROOF of ( 3 ) .
following:
(4)
( i ) xlyZ
+
x2Y1 5 X , Y ,
x 2 y 2 , and ( i i ) x2y2
+
X,Y,
+
5 x2y1
x,y,.
+
Adding ( 4 . i ) a n d ( 4 , i i ) g i v e s u s
C a n c e l i n g t h e sum of t h e c e n t e r two terms o n e a c h s i d e of ( 5 ) shows that P(xI,x3:y1,y2) holds;
proving ( 3 , i i i ) .
I f ( 3 . i ) or ( 3 , i i ) is s t r i c t ,
t h e n s o is (5); thus ( 3 , i v ) h o l d s , hence (3) is proved.
(6)
If
( i ) P ( x , , x , : Y , , y 2 1 a n d ( i i ) P ( x , , x z : Y z , ~ ,) h o l d , t h e n ( i i i )
P(x1,x2:y1,y3) holds.
F u r t h e r , ( i v ) i f a t l e a s t o n e of ( i ) or ( i i )
holds s t r i c t l y , then ( i i i ) holds s t r i c t l y . PROOF of
(6).
Assume t h a t ( 6 , i ) a n d (6,111 h o l d ; t h e n we h a v e t h e
fo l l o w i n g :
(7)
(i)
xIy2
+
xZyl 5 xly,
+
x 2 y Z r and ( i i ) xlyB
t
x,y,
2 x1y2
t
x,y,.
Adding ( 7 , i ) a n d ( 7 , i i ) g i v e s us
C a n c e l i n g t h e two i d e n t i c a l terms i n ( 8 ) g i v e s
US
P(X,
,X2
:yl ,Y3),
e s t a b l i s h i n g (71, a n d h e n c e c o m p l e t i n g t h e p r o o f of (6).
PROOF of (O), ( 1 1 , a n d ( 2 ) . n a l 2 ( b ( x ) + b ( y ) ) t o prove ( 0 ) ;
Let u s p r o c e e d by i n d u c t i o n on t h e o r d i on b ( x l ) + b ( x , ) + b ( y , ) b(y,) t o prove
( 1 ) ; and on 2 ( b ( x , ) + b ( y ) ) ( w h i c h e q u a l s 2 ( b ( x , )
+
+
prove ( 2 ) .
To p r o v e ( O ) ,
we m u s t e s t a b l i s h t h e f o l l o w i n g :
b ( y ) ) i f X1
-
X,),
to
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.07
143
T h a t i s , t h e f o l l o w i n g i n e q u a l i t i e s m u s t be e s t a b l i s h e d :
Assume f i r s t t h a t x L ’ 5 x L 3 .
Then, u s i n g t h e i n d u c t i o n h y p o t h e s i s ,
we may a p p l y ( 1 ) a n d c o n c l u d e t h a t P ( x L 1 , x L 3 : y L 1 , y ) h o l d s a n d t h a t
P(xL3,x:yL1 , y R 3 ) holds s t r i c t l y .
Adding t h e two a s s o c i a t e d i n e q u a l i t i e s
together gives us t h e following:
On c a n c e l i n g o u t x L 3 y L 1 a n d s u b t r a c t i n g xL1yL1
+
x L 3 y R 3 f r a n ( 1 11, we
obtain (10,i).
Assume, o n t h e o t h e r h a n d , t h a t xL3 5 xL h y p o t h e s i s , we may a p p l y ( 1 )
’.
Then, u s i n g t h e i n d u c t i o n
and conclude t h a t P ( x L 1 , x : y L 1 , y R 3 ) holds
R s t r o n g l y and t h a t P ( x L 3 , x L 1 : y , y 3 , h o l d s .
A d d i n g t h e two a s s o c i a t e d i n -
e q u a l i t i e s together gives us t h e following:
(12)
xLlyR3 + xyLl
+
xL3yR3
+
xLl Y
<
XL’yL1 + xyR3 +
On c a n c e l i n g o u t xL1yR3 a n d s u b t r a c t i n g xL1yL1 obtain ( 1 0 , i ) again.
+
xL3y
+
,LlyR,.
x L 3 y R 3 fran (121, we
144
Norman L. A l l i n g
4.07
U s i n g t h e i n d u c t i o n h y p o t h e s i s , we m a y a p p l y ( 1 ) a n d c o n c l u d e t h a t P(xL1,x:yL1,y) and P(x,xR4:yL4,y) hold strongly.
Adding the two a s s o c i a t e d
i n e q u a l i t i e s t o g e t h e r gives us t h e f o l l o w i n g :
L (13) x 'y
+
xyL1 + x y
t
xR4yL4
<
xLiyLi
+
xy
+
xyL4
On c a n c e l i n g o u t xy a n d s u b t r a c t i n g x L 1 y L 1
+
+
xR4y.
x R 4 y L 4 from ( 1 3 ) ,
we
obtain (10,ii). Using t h e i n d u c t i o n h y p o t h e s i s , w e may a p p l y ( 1 ) a n d c o n c l u d e t h a t P(x,xR2:y,yR2) and P(xL3,x:y,yR3) hold strongly.
Adding t h e two a s s o c i a t e d
i n e q u a l i t i e s t o g e t h e r g i v e s us t h e f o l l o w i n g :
(14)
xyR2
+
x R 2 y + xL3yR3 + xy
< xy
+
,RzyR2
+
L
3y
+
xyR,*
On c a n c e l i n g o u t xy a n d s u b t r a c t i n g x L 3 y R 3 + x R z y R 2 from ( 1 4 ) ,
we
obtain (10,iii).
Assume t h a t x R z 4 x R 4 . apply
T h e n , u s i n g t h e i n d u c t i o n h y p o t h e s i s , we may
and conclude t h a t
(1)
P(x,xR2:yL4 ,yRz) h o l d s s t r o n g l y .
P(xR2,xRr:yL4,y) holds and t h a t
Adding t h e two a s s o c i a t e d i n e q u a l i t i e s
t o g e t h e r gives u s t h e f o l l o w i n g :
On c a n c e l i n g o u t x R z y L 4 a n d s u b t r a c t i n g x R 2 y R 2 + x R 4 y L 4 fran ( 1 5 1 , we obtain (10,iv).
Assume, on t h e o t h e r h a n d , t h a t x R 4 5 x R 2 .
the induction hypothesis,
Then, u s i n g
we m a y a p p l y ( 1 ) a n d c o n c l u d e t h a t
P ( x , x R 4 : y L 4 , y R z ) h o l d s s t r o n g l y a n d t h a t P ( x R 4, x R 2 : y , y R 2 ) h o l d s . t h e s e two i n e q u a l i t i e s t o g e t h e r gives us t h e f o l l o w i n g :
(16)
xyR2
+
xR4yL4 + xR4yR2
t xR2y
xyL4
+
,hyR2
R $y
+
xR2yR2.
Adding
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.07
On c a n c e l i n g o u t x R 4 y R 2a n d s u b t r a c t i n g x R z y R z + xR'yL' obtain (10,iv) again.
145
f r m ( 1 6 1 , we
T h i s c o m p l e t e s t h e i n d u c t i v e proof of ( 0 1 , t o t h e
n e x t h i g h e r l e v e l of t h e i n d u c t i o n . L e t u s now t u r n o u r a t t e n t i o n t o ( 1 ) .
(17)
Assume t h a t ( 1 . i ) a n d ( 1 , i i ) a r e b o t h s t r i c t : i . e . , ( 1 , i ) x,
t h a t we h a v e
< x, and ( 1 , i i ) y 1 < y,.
By ( 4 . 0 2 : 1 4 ) a n d ( 1 , i ) t h e r e i s a n x Z L w i t h x, 5 x z L o r t h e r e i s a n
x l R w i t h x l R 5 x,:
(18)
i.e.,
( i ) t h e r e i s a n x p L w i t h x i 5 x Z L < x,,
R
( i f )t h e r e i s a n x1
. with x I
Assume t h a t ( 1 8 , i ) h o l d s . f a c t t h a t y1
(19)
<
x,
R
or
6 x,.
We may u s e t h e i n d u c t i o n h y p o t h e s i s ; t h e
< y z ; a n d (3) t o c o n c l u d e t h e f o l l o w i n g :
P(x, ,x,:y, ,y,)
holds s t r o n g l y , s i n c e P(x, ,xZL:y, , y 2 ) h o l d s and
L
P(x, , x z : y l , y z ) h o l d s s t r i c t l y . Assume t h a t ( 1 8 , i f ) h o l d s . f a c t t h a t y1
(20)
We may u s e t h e i n d u c t i o n h y p o t h e s i s ;
the
< y,; and (3) t o c o n c l u d e t h a t R
P ( x I , x p : y l , y a ) h o l d s s t r o n g l y , since P ( x , , x l : y l , y z ) h o l d s s t r i c t l y
R and P(xl , x z : y 1 , y 2 ) holds. Now assume t h a t
(21)
( 1 , i ) i s s t r i c t and t h a t ( 1 , i i ) h o l d s , b u t i s n o t s t r i c t : i . e . , t h a t ( 1 , i ) x,
<
x, and ( 1 , l i ) y1 = y 2 .
146
Norman L. A l l i n g
By ( 4 . 0 2 : 1 4 )
a n d ( 1 , i ) there is a n x p L w i t h x , 5 x Z L or t h e r e i s a n
x I R w i t h x l R 5 x,:
(22)
i.e.,
t h e r e i s a n x Z L w i t h x , 4 x,
(i)
<
( i ' ) there i s a n x l R w i t h x,
Assume t h a t ( 2 2 , i ) h o l d s . f a c t t h a t y , 6 y,;
(23)
4.07
x1
L R
<
x,,
or
5 x,.
We may u s e t h e i n d u c t i o n h y p o t h e s i s t h e
and (3) t o conclude t h e f o l l o w i n g :
P(x, ,x,:yl , y 2 ) h o l d s , s i n c e P ( x , , x Z L : y , ,Y,) a n d P(x,Llx,:Yl
,Y,)
h o i d. Assume t h a t ( 2 2 , i ' ) h o l d s .
We may u s e t h e i n d u c t i o n h y p o t h e s i s ; t h e
f a c t t h a t y , 5 y 2 ; and (3) t o c o n c l u d e t h a t
Now a s s u m e t h a t (25)
( 1 , i ) a n d ( 1 , i i ) h o l d , b u t are n o t s t r i c t : i.e.,
and ( 1 , i i ) y l
=
that ( 1 , i ) x,
x,y,
+
x,y,;
x,
y,.
Note t h a t ( i , i i i ) i s , i n t h i s c o n t e x t , t h e s t a t e m e n t t h a t x 1 y 2 =
=
which follows i m m e d i a t e l y f r a n ( 2 ) .
+
x2yI
Thus l e t u s p r o c e e d t o
prove (2).
We know t h a t x 1
=
R
{ x I L I x , 1 , x,
F u r t h e r , assume t h a t x,
=
x,.
=
{xZLI x,
1, a n d y
that
( i ) { x i L ) a n d [ x , ~ ) are m u t u a l l y c o f i n a l , and t h a t
(ii)
R
R
=
{y
L
I
R
y 1.
By ( 4 . 0 2 : 1 6 ) we know t h a t t h e Conway c u t s
L ( x l L , x l R ) and (x, , x Z R ) are e q u i v a l e n t : i.e.,
(26)
R
{ x , 1 a n d {x, 1 a r e m u t u a l l y c o i n i t i a l .
4.07
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
-
Using t h e i n d u c t i o n h y p o t h e s i s we may a p p l y ( 0 )
147
(2) t o e s t a b l i s h
the following.
(27)
If
(i)
L
X,
( i i ) If x1 ( i i i ) If x,
(iv)
If
XI
R
L R
2 x P L then
R
5 x, 2 x,
L
L
x, y
then
X,
then
X,
R
L
R
y + x2y
R
y + x2y
R
5 x Z R t h e n x, y + x l y
Assume t h a t x1 L 5 x, L
PROOF.
L
+ xly
.
L
L L
-
x1
-
x2 Y
-
x2 Y
-
R R
L R
R L
x1 y
S i n c e yL
2
x2
L +
R
5 x, Y
5
x1
L
Y
+
+
R
x2 Y
+
X2Y X1Y
XlY X2Y
L R R
L
-
.
L L
x, y
R R
- x 1 y . L R
-x1 Y
*
R L
- x , y .
< y , and s i n c e t h e i n d u c t i o n
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( x , L , x Z L : y L , y ) h o l d s : i.e.,
(28)
L x, y
+
L L L L x, y 5 x1 Y
+
L x2 y.
Adding x,yL = x 2 y L t o b o t h sides of ( 2 8 ) we a r r i v e a t t h e f o l l o w i n g . L L L L (29) x , y + x , y + x 2
x,
L L +
x2y
L
L
+
x, y.
R e a r r a n g i n g terms i n ( 2 9 ) g i v e s ( 2 7 , i ) . A s s u m e t h a t x , R 5 x, R
.
Since y
<
yR, and s i n c e t h e i n d u c t i o n
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( x , R , x , R :y,yR) holds: i.e., R R R R (30) x , y + x 2 y 5 x,
+
R R
,
Adding x2yR = x , y R t o b o t h s i d e s of (30) we a r r i v e at t h e f o l l o w i n g . R R
(31) x , y
+
x2y
R +
R R R R R x2 Y 5 x1 y + x l y + x i ? Y .
Rearranging terms in (31) g i v e s ( 2 7 , i i ) .
148
Norman L . A l l i n g Assume t h a t x 1 L S x, L
.
Since y
4.07
yR , and s i n c e t h e i n d u c t i o n
<
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( x , L , ~ , ~ : y R, )y holds: i.e.,
(32)
L R x1 Y
+
L L x 2 y 6 x1 y
+ x2
L R
,
Adding x 2 y R = x l y R t o b o t h s i d e s of ( 3 2 ) we a r r i v e a t t h e f o l l o w i n g .
(33)
L R R L L x1 y + x2y + x 2 y S x 1 y
X,Y
+
L R R + x , y .
R e a r r a n g i n g terms i n ( 3 3 ) g i v e s ( 2 7 , i i i ) . Assume t h a t x 1R 6 x, R
.
S i n c e yL
<
y, and s i n c e t h e induction
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( ~ , ~ , x , ~ : y ~ , y ) holds: i.e.,
(34)
R R L x1 y + x, y 5 Adding x l y L
R
(35) x, y + xly
L
=
R L +
R
x2 y .
x 2 y L t o b o t h s i d e s o f ( 3 4 ) we a r r i v e a t t h e f o l l o w i n g .
R L
+ X , Y
R L < x , y
L +
X2Y
+
x,
R
y.
R e a r r a n g i n g terms i n (35) g i v e s ( 2 7 , i v ) ; p r o v i n g a l l
Of
(27).
0
Combining ( 2 6 ) a n d ( 2 7 ) we see t h a t we h a v e p r o v e d t h e f o l l o w i n g .
(36)
(i)
L (xl Y
+
L
(x, Y (ii)
L
txl y L
tx, y
+
+
xly X2Y
L L
xlyR
x,yR coinitial. +
-
L L
x1 y
L L
,
x1
R
t
R
-
x2 Y I x 2 Y L R R x1 Y I x , Y
-
x2
L R R Y I x2 Y
xly
R
+
x2yR L x1y
+
X,Y
+
L
-
xlRyR} and
-
x, y I a r e m u t u a l l y c o f i n a l .
-
R L x1 y 1 a n d
R R
xZRyLa ~re m u t u a l l y
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.07
By ( 4 . 0 2 : 1 6 )
149
a n d ( 3 6 ) we s e e t h a t ( 2 ) h o l d s a t t h e n e x t s t a g e i n t h e
induction.
We may now u s e ( 2 ) a t t h e l e v e l ( w h i c h h a s j u s t b e e n p r o v e d ) t o p r o v e (1).
T h u s ( 1 ) i s e s t a b l i s h e d a t t h e n e x t h i g h e r l e v e l of t h e i n d u c t i o n .
-
Thus, by i n d u c t i o n , ( 0 ) (37)
If 0
< x
=
xy.
<
y , i n No, t h e n 0
<
xy.
By ( 1 1 , P ( O , x : O , y ) h o l d s s t r i c t l y ; t h u s 0
PROOF. xy
and 0
( 2 ) have been proved.
=
Oy + x 0
<
0.0
+
0
Conway g a v e t h e f o l l o w i n g i l l u m i n a t i n g v e r s i o n of t h e d e f i n i t i o n of xy C24, p . 191:
(39)
xy
=
(xy - (x xy
4.08
+
- x
L L
(x - x )(Y
L y 1,
-
)(Y R
-
XY
Y ) , xy
+
(xR
-
(xR -
X)(Y X)(Y
R
-
Y)J y
L
11.
THE ASSOCIATIVE LAW FOR MULTIPLICATION I N No
LEMMA.
L e t x , y , a n d z b e i n No; t h e n ( x y ) z
=
x ( y z ) , which w i l l be
d e f i n e d t o b e x y z , as u s u a l . PROOF.
We w i l l p r o c e e d by i n d u c t i o n , on b ( x )
d e f i n i t i o n we s e e t h a t we h a v e t h e f o l l o w i n g .
+
b(y)
+
b z).
BY
Norman L . A l l i n g
150
4.08
Expanding t h e e x p r e s s i o n s f o u n d i n ( 0 ) ( i n ( 1 ) t h r o u g h ( 8 ) ) a n d ( 0 ' ) ( i n ( 9 ) t h r o u g h (1611, a n d i n d u c t i o n g i v e s r i s e t o t h e f o l l o w i n g .
(1)
x Ly z
+
xy Lz
x y zL
-
x Ly L z
-
xLyzL
-
xyLzL + x Ly L 2 L ,
(2)
x Ry z
+
xy R z + x y z L
-
x Ry R z
-
xRyzL
-
xyRzL + x Ry R z L ,
(3)
x Ly z
+
xy R z
+
xyzR
-
xL y R z
-
xLyzR - x y R z R + x L y R z R ,
(4)
x Ry z
+
xy L z
+
xyzR
-
x Ry L z
-
x R y z R - xyLzR
(5)
x Ly z
+
xy L z + x y z R
-
x Ly L z
-
xLyzR
-
(6)
x Ry z
+
xy R z
+
xyzR
-
x Ry R z
-
xRyzR
- xyRzR + x Ry R zR ,
(7)
x Ly z
+
xy R 2
+
xyzL
-
xLy R z
-
xLyzL
-
xyRzL
(8)
x Ry z
+
x y Lz
+
x y zL
-
x Ry L z
-
xRyzL
-
xyLzL + x Ry L z L ,
(9)
x Ly z
+
xy L z + x y z L
- xLy L z -
xLyzL
- xyLzL
(10)
L x yz
+
xy
R2
+
xyzR
- xLyR z -
xLyzR
-
xyRzR + ,LyRzR,
(11)
x Ry z
+
xy L z
+
xyzR
-
xRyLz
-
xRyzR
-
xyLzR + x Ry L zR ,
(12)
x Ry z
+
xy R z
+
xyzL
-
x Ry R z
-
xRyzL
-
xyRzL + x Ry R z L ,
(131 x L YZ
+
xy L z
+
x y zR
- xLyL z - xLyzR - xyLzR
(14)
x LYZ
+
xy R z
+
x y zL
-
x LyR z
- xLyzL -
(15)
x Ry z
+
XY L Z
+
x y zL
-
x Ry L z
-
+
xRyzL
-
+
x Ry L z R ,
xyLzR + x L y L zR ,
+
xLyRzL,
+ x Ly LzL ,
+ x LyL zR ,
xyRzL + x Ly R z L , xyLzL
t
x Ry LzL ,
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.08
(16)
R
x yz
+
R
xy z
+
xyzR
-
R R
x y z
-
R R R
xRyzR - xyRzR + x y z
151
.
The r e a d e r s h o u l d n o t e t h a t each o f t h e s e l a s t 1 6 expressions i s
determined by t h e c h o i c e of a n " L "
o r a n "Rfl o v e r e a c h o f x , y , a n d z.
( T h i s may most e a s i l y be s e e n by l o o k i n g a t t h e s u p e r s c r i p t s on t h e extreme
r i g h t hand term of each e x p r e s s i o n . ) (i,j,k),
(L,L,L),
These e x p r e s s i o n s w i l l t h e n be c a l l e d
where i , j , and k a r e e i t h e r L o r R. (2) is the expression ( R , R , L ) ,
etc.
Thus ( 1 ) i s t h e e x p r e s s i o n From t h i s a n a l y s i s we s e e
t h a t t h e following holds:
On t h e o t h e r hand,
S i n c e t h e s e two e x p r e s s i o n s ( ( 1 7 ) and ( 1 8 ) ) a r e i d e n t i c a l , w e have proved t h e Lemma. Thus we h a v e p r o v e d t h e f o l l o w i n g c o r o l l a r y t o t h e Lemma, and much t h a t precedes i t : THEOREM.
No i s an o r d e r e d commutative r i n g , w i t h 1 # 0.
Note, i n p a s s i n g , t h a t t h e r e i s a s i m p l e r u l e f o r d e s c r i b i n g t h e v a r i o u s o p t i o n s of xyz i n ( 1 7 ) and ( 1 8 ) ; namely t h e f o l l o w i n g . ( i ) A l l 1 6 c h o i c e s of
i,j,lE{L,R}.
(i,j,k)
o c c u r a s o p t i o n s of x y z , where
( i i ) 8 of t h e s e a r e l e f t o p t i o n s and 8 a r e r i g h t o p t i o n s .
( i i i ) ( i , j , k ) i s a l e f t o p t i o n i f f i t has an odd ( r e s p . even) number of L ' s
(resp. R ' s ) i n it.
( i v ) ( i , j , k ) i s a r i g h t o p t i o n i f f i t h a s a:) even
( r e s p . odd) number of L ' s ( r e s p . R ' s ) i n i t . The f o l l o w i n g was noted by Conway C24, p . 191.
(C.f.,
(4.06:41.)
4.08
Norman L . A l l i n g
152
xy
(19)
=
- x
Ixy - ( x L
xy - ( x - x ) ( Y 4.09
-
L
-
L
R
y ) , xy - ( x - x ) ( Y - y R R L y 1, xy - (x - x ) ( Y - y 1 ) . )(y
R
)I
ON NUMBERS G I V E N BY REFINEMENTS OF ( T I M E L Y ) CONWAY C U T S
We have s e e n i n S e c t i o n 4.02 t h a t e a c h XENO e q u a l s (LIR], where ( L , R ) i s a t i m e l y Conway c u t i n No: i . e . ,
( L , R ) i s a Conway c u t i n O b ( x ) .
Let L ' a n d R' be s u b s e t s of No s u c h t h a t ( i ) L ' < ( X I < R ' ;
(ii)for
a l l xL EL t h e r e e x i s t s x L ' E L ' s u c h t h a t xL 6 x L ' ; a n d ( i i i ) f o r a l l R R x E R t h e r e e x i s t s x ~ ' E R ' s u c h t h a t xR' 5 x
.
Then ( L ' , R ' )
w i l l be
c a l l e d a r e f i n e m e n t of ( L , R ) . be a r e f i n e m e n t of ( L , R ) ;
Let (L',R')
PROOF.
Let I
L
< (y} <
a n d I' = ( Y E N O :
R},
=
(LIR}.
L'
<
(y]
<
R'}.
we s e e t h a t X E I 'a n d t h a t I ' i s a s u b i n t e r v a l of t h e
By ( O , ( i ) - ( i i i ) ) i n t e r v a l I.
= (YENO:
then ( L ' I R ' )
Since x i s t h e s i m p l e s t element i n I , i t is c e r t a i n l y t h e
s i m p l e s t element of 1'. be a r e f i n a n e n t of ( L 1 , R 1 ) ,
Let (Ltl,R1')
ments of ( L , R ) ; (L',R')
and l e t ( L ' , R ' )
be refine-
t h e n ( L f f , R t l ) i s a r e f i n e m e n t of ( L , R ) .
i s p o s s i b l y an u n t i m e l y , r e p r e s e n t a t i o n of x ; w h e r e b y a
r e p r e s e n t a t i o n t h a t i s p o s s i b l y u n t i m e l y ( r e s p . i s u n t i m e l y ) we mean a n o t b e s u b s e t s of O b ( x ) Ob(x)).
x , b u t f o r which L ' a n d R ' may ( r e s p . t h e u n i o n of L ' and R' is n o t a s u b s e t of
i n No s u c h t h a t ( L ' I R ' )
Conway c u t ( L ' , R ' )
=
( S e e S e c t i o n 4.02 f o r d e t a i l s . )
We have s e e n t h a t 0
EXAMPLE 0 .
of No f o r which L ' s e n t a t i o n of 0.
<
(0) < R ' ;
-
[
then ( L ' , R ' )
I f t h e u n t o n of L ' and R'
u n t i m e l y r e p r e s e n t a t i o n of 0. n o t an i s o l a t e d o c c u r r e n c e .
I}.
Let L ' and R ' be any s u b s e t s
is possibly an untimely r e p r e is non-empty,
then (L',R')
is a n
The r e a d e r c a n e a s i l y v e r i t y t h a t t h i s i n
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.09
(2)
153
Let y
=
L R L R { y Iy ]&NO, a n d l e t IyL',yR') b e a r e f i n e m e n t of f y , y 1.
x
=
(x
+
y
L' +
Y, x
+
x
YL'I
R'
+ y, x
+ yR'].
By c o n d i t i o n ( 0 1 , a n d t h e f a c t t h a t No i s a n o r d e r e d g r o u p
PROOF.
( 4 . 0 8 ) , we know t h a t f o r each xL a n d x R t h e r e e x i s t s x L ' a n d xR' s u c h t h a t
<
xL + y 5 xL' + y
x
y
+
<
xR'
+
y S xR
+
y ; a n d we know t h a t f o r each y L
and yR t h e r e e x i s t s y L ' a n d y R ' s u c h t h a t x + yL 5 x x + y R ' I x + yR .
(31
+
yL'
< x
+
y
<
0
L' L' L' L' R' xy=(x y + x y - x y , x y + x y - x L' R' L' R' R' L' R' L' x Y + X Y - x y , x y + x y - x y }. PROOF of ( 3 ) .
U s i n g ( 0 ) a n d t h e f a c t t h a t No i s a n o r d e r e d r i n g
( 4 . 0 8 1 , we know t h a t t h e f o l l o w i n g h o l d .
(4)
F o r each xL a n d yL t h e r e e x i s t x L ' a n d y L ' f o r w h i c h
(i)
L
x y
+
xyL
-
xLyL 6 xL'y
+
xy
L' - x L' y L' < x y .
( i i ) F o r each xR a n d y R t h e r e e x i s t x R ' a n d y R ' f o r w h i c h
R x y
+
xyR
-
xRyR 5 x R ' y + xy
R'
( i i i ) For each xL a n d y R t h e r e exist
L
X Y
(iv)
+
R
PROOF of
L'
+
+
x y R'
- xL' y R' >
xy.
xyL
-
xRyL 2 xR'y + xyL'
-
xR'yL'
> xy.
4).
b y
consequence, 0 t h u s xL y
xLyR h xL'y
xL' a n d y R ' f o r w h i c h
f o r e a c h xL, t h e r e e x i s t s xL' 2 x L ; a n d f o r each y L t h e r e
By ( O ) ,
exists y
-
xy.
F o r each xR a n d yL t h e r e e x i s t x R ' a n d y L ' f o r w h i c h X Y +
(i)
xyR
- x R' y R ' <
x yL
L
< -
.
Thus, 0
xy - x
L'
< (x - xL')(y - y
y - xy
L'
+
xLyL 5 xL'y + x y L '
xL'
-
Y
L'
xL'yL'
L'
L (x - x )(y - yL).
AS a L L L L 6 X Y - X Y - x y + x y , a n d
<
) 5
xy.
Norman L. A l l i n g
154
4.09
x R , there e x i s t s x R ' 5 xR : a n d f o r each y R t h e r e R' R R Thus 0 < ( x - x R ' ) ( y - y ) 2 ( x - x ) ( y - y 1. As a R' R' R' R' R R R R
( i i ) By ( 0 1 , f o r e a c h
exists y
R'
4 yR.
<
consequence, 0 R
t h u s x y + xyR
-
xy
-
x
y - xy
xRyR 5 x
R'
+ x
-
y + xy R'
y
xR'yR'
( i i i ) By ( 0 1 , f o r e a c h xL, t h e r e e x i s t s x
R'
L'
2
x
L
L R xL'yR' 5 x y + xy
-
L R x y
By ( 0 ) . f o r each x R , t h e r e e x i s t s x
R'
4 x
Thus 0
- xL ' y
consequence, 0 > xy t h u s xy
< xL ' y
+
xyR'
-
- xy
R'
xL')(y +
+ x y ,and
; and f o r each yR t h e r e
L R x ) ( y - y 1. A s a x L ' y R' t x y - x Ly - x y R + x Ly R, a n d R'
y
5 yR.
x y - x y
< xy.
-
exists y
(iv)
> (x -
-
2 xy
) 2 (x
-
.
R
; a n d f o r e a c h yL t h e r e
L
.
consequence, 0
>
R' x y
+
R ' L' R xy - x y - x y L ' + x y L xy x y - xyL + x R y L , a n d L' R ' L' R L R L xy x y 5 x y + xy - x y i p r o v i n g ( 4 ) .
exists Y
t h u s xy
L'
<
2 Y
Thus 0
> (x - x
R'
)(y
-
y
L'
1 L (x
-
R
L
x ) ( y - y 1.
AS
a
-
R'
-
To c o m p l e t e t h e proof of ( 3 ) , we c a n r e a s o n as we d i d i n t h e proof of (1).
a
PROPERTIES OF DIVISION I N No
4.10
THEOREM. t h a t xy = 1 .
-
Were x then -x(z)
Let x b e i n No, w i t h x n o t 0.
T h e r e e x i s t s y i n No s u c h
Thus, No i s a n o r d e r e d f i e l d .
<
0 , a n d were -x t o h a v e a m u l t i p l i c a t i v e i n v e r s e z i n No,
1 = x(-z),
showing t h a t x h a s a m u l t i p l i c a t i v e i n v e r s e i n No.
Thus t o p r o v e t h e Theoren i t s u f f i c e s t o p r o v e t h e f o l l o w i n g . (0)
>
Assume t h a t x
0 in
No.
T h e r e e x i s t s a YENO s u c h t h a t xy = 1 .
I n t h i s s e c t i o n , t h e f o l l o w i n g r e s u l t w i l l a l s o b e of u s e . (1)
T h e r e e x i s t s u b s e t s L and R of t h e s e t of p o s i t i v e elements of No such t h a t L
<
t i m e l y (4.02).
R , w i t h x = { O , L I R ) , f o r which t h i s r e p r e s e n t a t i o n i s
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.10
PROOF of ( 1 ) . timely (4.02).
By d e f i n i t i o n x
>
Since x
t h a t 0 5 xL ( 4 . 0 2 : 1 4 ) .
=
[x
L
0 , and s i n c e 0
Let L
=
L
[ x : xL
R x 1 , the r e p r e s e n t a t i o n being
I
[I},
=
155
t h e r e must e x i s t a n xL s u c h
> 01, and
let R
=
[x
R
1.
Then x
=
[O,LIR}; e s t a b l i s h i n g ( 1 ) .
( 1 ) We may write x a s [ O , x L l x R } , w i t h e a c h
Conway d e f i n e d t h e m u l t i p l i c a t i v e i n v e r s e y of x as f o l l o w s
being timely. [24,
xL > 0 , t h e r e p r e s e n t a t i o n
p . 211.
(1
+
(xL
L L x)y ) / x , ( 1
-
+
(xR
-
R R x)y ) / x I.
Conway writes t h e f o l l o w i n g a f t e r t h i s d e f i n i t i o n .
"Note t h a t ex-
p r e s s i o n s i n v o l v i n g yL and yR a p p e a r i n t h e d e f i n i t i o n o f y . t h a t r e q u i r e s u s t o ttexplaintt t h e d e f i n i t i o n .
It is this
The e x p l a n a t i o n i s t h a t we
r e g a r d t h e s e p a r t s of t h e d e f i n i t i o n as d e f i n i n g new o p t i o n s f o r y i n terms So even t h e d e f i n i t i o n of t h i s y is an i n d u c t i v e one. t [ T h i s
of o l d o n e s .
i s i n a d d i t i o n t o t h e t t o t h e r t l i n d u c t i o n s by which we s u p p o s e t h a t i n v e r s e s L
for the x
and x
R
have a l r e a d y been f 0 u n d . 1 ~ ~C24, p . 21.1
In a footnote t
[24, p . 211 Conway g i v e s t h e f o l l o w i n g v e r y i n s t r u c t i v e example. Example 0. as [0,21}.
Let x R
=
3.
Thus x c a n be w r i t t e n , as i t is above i n ( l ) ,
.
Hence [ x ) i s empty, and xL
=
2.
By (21, t h e g e n e r a l f o r m u l a
for y is
Let yoL
=
0.
Note t h a t yoL
BY ( 3 ) , we s e e t h a t y I R
=
Let us make no c h o i c e of a y o
S i n c e t h e r e i s no y o R , t h e r e i s n o y 1
1/2.
U s i n g ( 3 ) , we s e e t h a t y Z L = 1 / 4 . y S R = 11/32, y G L = 21/64, y,R
< 1/3.
=
T h i s t h e n y i e l d s y,R = 318, y,L
43/128,
...
=
R
L
. .
5/16,
Writing t h i s i n t h e customary
Norman L. A l l i n g
156
way g i v e s y = ( 0 , 1/4, 5/16, 21/64,
... I
4.10
1/2,
3/8,
11/32,
The decimal v e r s i o n of t h i s i s even more i n s t r u c t i v e . {O,
0.25,
0.3125,
0,328125,
... I
0.5,
0.375,
0.34375,
431128,
...
I.
It is t h e following: 0.3359375,
... ) .
I n t r y i n g t o u n d e r s t a n d ( 2 ) more f u l l y , n o t e t h a t e x c e p t f o r 0 t h e o p t i o n s g i v e n i n (2) a r e d e t e r m i n e d i n p a r t by t h e c h o i c e of w h i c h o p t i o n o n e c h o o s e s t o c o n s i d e r , r i g h t or l e f t .
Let u s make t h e f o l l o w i n g
def i ni t i o n s .
(4)
(i)
R
-
x)y ) / x ,
be d e f i n e d t o be ( 1 + ( x L
-
x)y ) / x ,
Let [R:Ll(y) b e d e f i n e d t o be ( 1 + ( x
( i i ) l e t [L:R](y)
L
R
R
L
( i i i ) l e t [ L : L l ( y ) b e d e f i n e d t o be ( 1
+
( x L - x ) y L ) / x L , and
l e t [R:Rl(y) b e d e f i n e d t o be ( 1
+
(xR
(iv)
-
R R x)y )/x
.
Then, a c c o r d i n g t o (21,
PROOF of ( 0 ) .
Note t h a t (0) i s t r u e i f x
<
are no s i m p l e r p o s i t i v e e l e m e n t s ; t h u s f o r b ( x ) f o r which ( 0 ) i s t r u e f o r a l l X E N O , w i t h 0
< x
-
1.
Note a l s o t h a t t h e r e
2, (0)i s t r u e .
and b ( x )
<
Let a e o n
a.
We w i l l s t a r t o f f t h e f i n i t e i n d u c t i o n , as i n t h e e x a m p l e a b o v e , by d e f i n i n g y o t o b e 1.
(6)
Recall t h a t 1 = {Ol}: t h u s
yoL = 0, and there are no Y, R . Consider t h e f o l l o w i n g i n e q u a l i t y :
(7,n) ( i ) x y k <
1 , f o r a l l y,
L : and ( i l l 1
N o t e t h a t (7,O) i s t r u e , y I R be d e f i n e d t o be [L:L](y,); hypothesis,
<
xyn R , a l l yn R
.
Let u s d e f i n e y I L t o b e [R:L](y,),
and l e t
then, using our ( t r a n s f i n i t e ) induction
I n t r o d u c t i o n t o t h e s u r r e a l number F i e l d No
4.10
S i n c e No i s a n o r d e r e d r i n g , (7,l) h o l d s .
Let ncN such t h a t y n L a n d
Let y n + l L and y n+ 1
ynR a r e d e f i n e d , For which ( 7 . n ) holds.
= (1 +
(xR
-
x)yk)/x R , o r
(1
+
(xL
-
x)ynR)/xL; and
( i i i ) y n + l R = C L : L I ( ~ ~=) ( 1
+
(xL
-
x)y:)/xL,
+
(xR
-
x)yn )/x
=
Y,+~
(i)
(ii) Y
(iv)
and t h e ex-
be d e f i n e d a s f o l l o w s :
p r e s s i o n [*:.](.)
(9)
157
[R:L](y,)
~ += [L:R](yn) ~
=
Yn+l
=
[R:R](yn)
Now l e t t h e e x p r e s s i o n
(1
=
or
R
R
.
be d e f i n e d as f 011ow9 :
{ a : - ) ( . )
( i i i ) xyn+lR
-
1 = ((x
-
x )(I
L
-
xy,
(iv)
XY,+~
-
1 = ((x
-
R x )(1
-
'Yn
Using
(lo),
L R
we see t h a t t o prove t h a t (7,n+l) h o l d s , i t s u f f i c e s t o
prove t h e f o l l o w i n g :
PROOF of (11). that x
-
x
R
<
0,
Since
1
-
x < x
xynL
>
R
,
a n d s i n c e xynL
0, a n d h e n c e ((x
-
<
1 ( 7 , n , i ) , we s e e
R x )(1
-
xynL))/xR
-
158
Norman L. A l l i n g
{R:L](yn)
<
0; proving ( i ) .
>
s e e t h a t x - xL
<
0; proving ( i i ) .
x
-
x
R
<
0,
L
Since
-
1
xy
<
xynR
Since
x - xL > 0 , 1 - xy n proving ( i i i ) .
-
0, 1
Since
xL
<
xL
a n d since 1
((x - x
0, a n d
L
)(1
x, and since x y t
>
0, and ( ( x
x
< xR ,
<
0, and ( ( x
n
< x,
4.10
-
L x )(1
and s i n c e 1
-
<
xy
R x )(1
<
-
<
xynR ( 7 , n , i i ) , we
xynR))/xL
=
{L:R}(yn)
1 , ( 7 , n , i ) , w e see t h a t
x y t ) ) / x L = {L:Ll(yn) n
-
>
0;
( 7 , n , i i ) , we see t h a t xynR))/x
R
=
(R:R}(yn)
>
0;
p r o v i n g ( i v ) , a n d h e n c e a l l of ( 1 1 ) . T h u s we s e e t h a t ( 7 , n + l ) h o l d s .
Hence, by i n d u c t i o n , (7,111 holds f o r
all ncN.
(12)
F o r a l l m , ncN, y,
PROOF of ( 1 2 ) .
Since x
>
L
<
yn
R
. By (7.m) a n d (7,111. x y m L
L e t m , ncN.
0 , we see t h a t (12) f o l l o w s .
o
L R Let y b e d e f i n e d t o b e ( y n I y n 1 .
(13)
( i ) (xyIL
-
1
<
0, and ( i i ) 0
<
The f o l l o w i n g h o l d s :
(xyIR
-
1.
I n order t o p r o v e ( 1 3 1 , l e t u s e s t a b l i s h t h e f o l l o w i n g .
(14)
R
1
+
xR (y - [R:L](yn)),
- xLynR = 1
+
L x (y
-
[L:R](yn)),
L
-
xLyk
1
+
L x (y
-
[L:L](yn)),
xynR
-
xRyt = 1
+
xR ( y
-
[R:R](yn)).
(i)
X
Y
+
xy
(ii)
L x y
+
xy n
L (iii) x y
+
xy
xR y
+
(iv)
-
n
xRyk
=
=
and
<
1
<
xyn
R
.
I n t r o d u c t i o n t o t h e s ur r ea l number f i e l d No
4.10
R x y
PROOF of ( 1 4 ) . =
1
f
x R (y
xLy + x y R
n
1
+
-
(1
L
L 1 + x (y R
X Y f X Y
( x R - x ) y n L ) / xR )
f
(xL - x)ynR)/xL)
f
L
- x Ly,
-
(1
+
R
- x
R
L
1
=
Yn
R
1
=
-
L
XYn
+
L
- x
1
=
f
xL(y
( x Ly - 1 - x Ly, L
f
(xL - x ) y t ) / x R (x y
f
L
= 1 +
-
- x
1
x ) y n R ) / xR )
-
-
1
- x Ry nL
xynL))
f
[R:L](yn)), proving ( i ) .
R
+
[L:RI(y,)).
proving ( i i ) .
xynL)) =
L
x (y - C L : L I ( ~ , ) ) , proving ( i i i ) .
yn
R
+ xynR)) =
-
1 + xR (y
=
-
[R:RI(y,)),
proving ( i v ) .
o
Note t h a t by ( 1 4 , i i i ) a n d ( g , i i i ) , we h a v e
PROOF of ( 1 3 ) .
x Y
1 + xR ( y
=
R (x y
f
n
- ( 1 + (xR
1 +X(Y
1
=
R = l + ( xL y - 1 - x Ly R + x y n R ) ) =
-"n
x L (y - (1
X Y f X Y
xynL - xRynL
f
159
-
Y,
-
1 = xL ( y
[L:L](yn))
xL ( y
=
-
yn+lR)
<
0.
By
( 1 4 , i v ) , ( 9 , i v ) , a n d t h e d e f i n i t i o n o f y , we h a v e
R x Y
- x
XY,
f
-
-
1 = xR ( y
C R : R 1 ( y n ) ) = xR ( y
-
+ xynR
-
xLynR
-
-
1 = xL ( y
[L:R](yn))
=
xL ( y - y n + l L )
R ( g , i ) , a n d t h e d e f i n i t i o n of y , we h a v e x y + x y
R
C R : L I ( ~ , ) ) = x (y
PROOF of
R
t h e r e are no 1
-
yn+,L)
(15).
.
>
0 ; proving ( 1 3 , i i ) .
-
Let xy
By (131, z
z.
L < 1 < zR
n
<
0 ; proving
Were z
<
=
1 S z
L
Since 1 5
f o r sane z
z S. 1 , z
L
-,
0.
By ( 1 4 , i ) ,
L - l = x R( y -
(Ol}; t h u s l L
=
1 t h e n by ( 4 . 0 2 : 1 4 ) ,
R
(4.02:14),
>
o
Recall t h a t 1
.
-'Yn
0 o r zR S 1 f o r some z ; b o t h o f w h i c h a r e a b s u r d .
absurd.
yntlR)
By ( 1 4 , i i ) , ( 9 , i i ) , a n d t h e d e f i n i t i o n of y , we h a v e
(13,i).
x Ly
*
Y,
Were 1
< z
0 and z 5
t h e n by
o r l R 5 z f o r some l R ; b o t h o f w h i c h a r e
1.
By ( t r a n s f i n i t e ) i n d u c t i o n we have p r o v e d (0).
4.10
Norman L. A l l i n g
160
S i n c e ( 0 ) i m p l i e s t h e Theorem, t h e Theorem i s p r o v e d .
Having made a l l t h e s e c a l c u l a t i o n s , we can now s e e more
CONCLUSION.
2 ) comes a b o u t .
c l e a r l y how t h e e x p r e s s i o n define y ( =
o
I Y L I Y R 1 ) and prove t h a t xy
must know t h a t ( x y )
L
<
1
<
=
I n o r d e r t o prove ( 0 ) we m u s t 1.
I n o r d e r f o r xy t o be 1 , we
(13) must hold.
xyIR: i . e . ,
I n checking t o s e e
t h a t (13) does i n d e e d h o l d , e x p r e s s i o n s of t h e t y p e t h a t o c c u r on t h e l e f t hand s i d e of ( 1 4 ) m u s t be c o n s i d e r e d .
I n order t o study these expressions,
q u a n t i t i e s of t h e t y p e t h a t occur o n t h e r i g h t - h a n d s i d e of ( 9 ) a r i s e . These elements of No engender ( 2 ) ' and p r o v i d e t h e f o r m u l a e t h a t move t h e f i n i t e i n d u c t i o n frcm s t a g e n t o s t a g e n
1.
+
Even though t h i s l i n e o f r e a s o n i n g p r o v i d e s a m o t i v a t i o n f o r ( 2 1 , d o e s n o t r e d u c e t h e a u t h o r ' s admiration of Conway's i n s i g h t .
see t h a t t h e r i n g No i s a f i e l d seems r e m a r k a b l e i n d e e d .
it
Indeed, t o
To p r o v e i t i n
t h e way t h a t Conway d i d seems t o t h e a u t h o r l i t t l e s h o r t of i n s p i r e d . 4.20
DISTINGUISHED SUBCLASSES OF No
(1 ]
We h a v e shown t h a t element 1 i n No,
Let
l e t (n
n.1
+
11.1
=
nEN
i s t h e e l e m e n t 0 i n No, and t h a t
f o r which no1
+ 1 = {0,1,2,
{0,1,2,
=
...,n -
...
l,nl].
{Ol
is t h e
, n - 1 1 ) i n No; t h e n
Thus, by f i n i t e induc-
t i o n , t h e f o l l o w i n g i s proved:
(0)
For a l l n i n N , n.1
=
(O,l,2,
... , n
-
11 )EN,
I t i s convenient t o i d e n t i f y ncN w i t h n.lENo, s u b s e m i - g r o u p of No.
-1
(n)).
and t h u s r e g a r d N as a
We can a l s o c o n s i d e r t h e element I N [ ] , and c a l l i t w
as Conway d o e s [24, p . 121. (1)
(= b
Clearly n
<
w f o r a l l nEN, t h u s
No is a non-Archimedean f i e l d . The class On of von Neumann o r d i n a l s w a s d e s c r i b e d i n s e c t i o n 1 . 0 2 .
R e c a l l t h a t 0 i s t h e empty s e t , t h a t 1
=
{O}, etc.
Recall a l s o t h a t i f a
i s i n On, t h e n i t s s u c c e s s o r i s a u n i o n {a]. I t i s n a t u r a l t o a s s o c i a t e
4.20
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
OEOn w i t h f ( 0 )
=
0
t o let f(1)
[]]ENo,
=
f ( 6 ) = { ( f ( u ) ) a < B 11.
=
1
=
{O]]eNo,
Let BEOn and l e t
Then, one can e a s i l y see t h a t
f is a n o r d e r - p r e s e r v i n g map of On i n t o No.
b(f(8))
161
F u r t h e r , f o r a l l $€On,
B.
=
Fran t h i s we s e e , among o t h e r t h i n g s , t h a t
No i s a p r o p e r class. On o c c a s i o n , we w i l l i d e n t i f y BEOn w i t h f(B)cNo, even though t h e y a r e (To s e e t h i s t a k e No f o r example t o
q u i t e d i f f e r e n t objects i n s e t theory.
be g i v e n by t h e C u e s t a D u t a r i c o n s t r u c t i o n , g i v e n i n S e c t i o n 4.02.) ELEMENTS OF No H A V I N G F I N I T E BIRTHDAY
4.21
Let nEN, and c o n s i d e r t h e f o l l o w i n g a s s e r t i o n s .
(0,n) lNnl
= 2
n
,
10nl
=
2
n
-
1 , and 1M
n
I
n+ 1
=
2
...
,
PROOF.
C l e a r l y (0,O)
For each x
holds.
one p o i n t i n M
,
, we
(LIRIeN n+
=
( = 0 n+l
1.
(O,l),
hold.
that (Mn[
=
2n+’
-
F u r t h e r , i f b o t h L and R a r e non-empty t h e n x
Clearly Nn+l
1.
t h a t has a s u c c e s s o r i n M
l e a s t e l e m e n t of M n ) “+I
I
=
2”+l- 1
-
1
+
Let neN f o r which ( 0 , n )
can t a k e L and R t o c o n s i s t of a t m o s t
can be t a k e n t o be t h e immediate s u c c e s s o r of x
n
1.
For a l l neN, ( 0 , n ) h o l d s
(1)
M
-
*
n’
L
in M
n‘
By ( 0 , n ) we know
has one p o i n t i n i t f o r e a c h e l e m e n t of
p l u s t w o more p o i n t s ,
{ ] u ) ( u being t h e
and { v l ] ( v b e i n g t h e g r e a t e s t element i n M , ) . 2
=
2
n+l
R
., e s t a b l i s h i n g
(O,n+l).
Thus
162
Norman L. A l l i n g
4.21
S i n c e No i s an o r d e r e d f i e l d ( 4 . 1 0 ) , i t i s a f i e l d of c h a r a c t e r i s t i c 0, thus
(2)
t h e prime f i e l d of No i s t h e f i e l d Q of r a t i o n a l numbers. Any r a t i o n a l number c a n , of c o u r s e , be w r i t t e n as a / b , f o r acZ and
bsN.
a / b w i l l be s a i d t o b e i n r e d u c e d f o r m i f a a n d b a r e r e l a t i v e l y C l e a r l y e v e r y r a t i o n a l number x may be w r i t t e n i n r e d u c e d form.
prime.
F u r t h e r , r e d u c e d forms a r e u n i q u e .
Let a / b b e t h e r e d u c e d form f o r x .
x w i l l be c a l l e d d y a d i c i f t h e r e e x i s t s neZ+ for which b = 2 b e t h e s e t of a l l d y a d i c numbers.
.
Let D
C l e a r l y D i s a s u b r i n g of Q t h a t con-
Further,
t a i n s 1/2.
(3)
n
D i s t h e s m a l l e s t s u b r i n g of Q t h a t c o n t a i n s 1 / 2 .
(1) If b = 2n,
Let XED be w r i t t e n as a / b i n r e d u c e d form.
LEMMA 1 .
f o r sane ncZ+, t h e n ( i i ) x = { x
-
2-"1x
2-")
+
i n No.
Note t h i s r e p r e -
s e n t a t i o n may n o t be t i m e l y .
Let n
PROOF.
t h e n x is i n Z , x i s i n N
= 0;
1x1 '
d e f i n e t o b e m ) i s of t h e form [ m - 1 1 } (4.20:O). s i m p l e s t e l e m e n t of No b e t w e e n m - 1 a n d m + 1 .
-
s i m p l e s t e l e m e n t b e t w e e n -m - 1 a n d -m ( i i ) , provided n
Note t h a t I m
0.
-
>
is.
(4)
D e f i n e z t o be [x
(5)
22 = z + z =
(6)
z
+
x
-
2-"
<
22
F u r t h e r , -m
+
1 ) ( r e s p . [-m
whereas {m
-
-
+
-
2-"Ix
x
-
< z
2-"12
Then
+ 2-"].
+
+ x + 2-".
x
+
2-"];
thus,
Fran ( 4 ) we see t h a t
is the
x h a s t h e form 1 ) -m
1 ) ) is
+
11) ( r e s p .
0 and assume t h a t ( i ) and ( i i ) h o l d s f o r n
+ 1))
IZ
Clearly m is then t h e
+ 1 , showing t h a t
11 m
n o t a t i m e l y r e p r e s e n t a t i o n of m ( r e s p . -m), Now l e t n
a n d 1x1 ( w h i c h we
-
1.
[I
-m
I n t r o d u c t i o n t o t h e surreal number f i e l d No
x - 2
-n
<
163
z < x+2-".
On combining ( 7 ) a n d (61, w e f i n d t h a t 2x - 2 -(!l-l)
< x +
ZX - 2-(n-1)
<
- 2 -n <
z
< zx
+
22
< x +
z
+ 2-" < 2x + P - 1 )
; hence,
2-(n-1)*
A p p l y i n g t h e i n d u c t i o n h y p o t h e s i s t o 2 x , we know t h a t 2x i s t h e s i m p l e s t element i n No s u c h t h a t t h e f o l l o w i n g h o l d s :
-
2x
(10)
<
2- ( I P 1 )
2x
<
2x + 2
-(n-l)
F r a n ( 9 ) we know t h a t 22 i s i n t h e f o l l o w i n g i n t e r v a l :
(ZX
-
2 - ( n - l ) ,2x + 2 - ( n - 1 ) ) ,
(11)
t h e same o p e n i n t e r v a l i n No t h a t c o n t a i n s 2x
F r a n ( 7 ) we know t h a t
(10).
2
-
<
2-n
x
<
z + 2-".
Adding x t o b o t h s i d e s of ( 1 1 ) g i v e s u s
(12)
- 2 -n <
x + 2
< x
2x
+ 2
+
2-".
U s i n g ( 8 ) a n d ( 1 0 ) we s e e t h a t 2x i s t h e s i m p l e s t element i n t h e interval I
=
(x
+
z
-
x + z
2-",
+
2-").
U s i n g ( 5 ) we know t h a t 22 i s t h e
s i m p l e s t e l e m e n t i n I , a n d t h u s we s e e t h a t 2x dered f i e l d , x
=
z , p r o v i n g Lemma 1.
=
22.
Since No i s an or-
o
A s u b c l a s s S o f No w i l l be c a l l e d s y m m e t r i c i f
XES implies - x d .
U s i n g i n d u c t i o n , one e a s i l y sees t h a t t h e f o l l o w i n g i s t r u e : (13)
For a l l acOn, O a ,
M
a
and N
are s y m m e t r i c .
4.21
Norman L. A l l i n g
164
I f any of t h e s e sets h a s a g r e a t e s t element x , t h e n i t w i l l be c a l l e d t h e r a d i u s of t h e s e t i n q u e s t i o n .
Thus
f o r a l l a e o n , t h e r a d i u s of N a , M a ,
(14)
PROOF.
is a.
Ow i s a s u b s e t of D .
LEMMA 2.
0 , = 1-2,
and O a t ,
W e know t h a t 0 , is empty, t h a t 0 , = {O], t h a t 0 ,
-1,
-1/2,
1 , 21, e t c .
1/2,
0,
[-l,O,l],
=
Let n d be s u c h t h a t ( i ) O n i s a
s u b s e t of D , and ( i i ) t h e d i s t a n c e b e t w e e n s u c c e s s i v e e l e m e n t s i n 0 n i s 2
-k
,
f o r some k s Z + . S i n c e Oa and D are b o t h symmetric ( 1 3 ) . i t s u f f i c e s t o show t h a t O n +
i s a s u b s e t of t h e s e t of D t , f o r a l l nrN. g r e a t e s t element of 0 i n D.
n'
t h e n u i s i n N , and { u l } = u
v
induction hypothesis C
,
-(k+l)
.
t
.
If u is t h e
1 (4.20:0),
n
+
. As
-
2-k
-
which i s
n o t e d above
Let v be t h e immediate s u c c e s s o r of u i n On.
v ) / 2 = x , t h e n x i s i n D.
(15)
+
Assume t h a t u is n o t t h e g r e a t e s t element i n 0
( l ) , On i s f i n i t e .
x + 2
Let u be i n On
Let ( u
+
u , which we w i l l c a l l c, i s i n D and by t h e + Thus u = x - 2 - ( k t l ) , and v = f o r sane kcz
.
By Lemma 1 , ( u l v ) = x .
Summarizing what we have shown t h a t
if v is t h e immediate s u c c e s s o r of u i n O n t ,
then {ulv)
=
(u
+
v)/2,
and t h a t i t i s i n N n .
From t h i s we see t h a t On+l s a t i s f i e s ( i ) and ( i i ) a b o v e , a n d hence we have proved Lemma 2.
LEMMA 3 .
PROOF.
Ow is a s u b r i n g of
Recall t h a t 2
R
make no c h o i c e at a l l f o r 2
.
-
No t h a t c o n t a i n s 112.
I),
t h u s we may take 2L t o be 1 and may R 2h Let h = (01 l } ; t h u s hL 0 , and h = 1 (1
-
.
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.21
2hL - 2 L h L , 2Rh
=
{2Lh
+
=
{h
2hL - hLI h
h
<
+
1
<
h
2hR
-
h
R
-
2RhR
=
{h
I
. w
2Lh + 2hR
h
11.
+
-
Let x a n d y b e i n 0
und er s u b t r a c t i o n .
w
c o n s t r u c t i o n of No ( 4 . 0 2 1 ,
.
+
2hL
We know t h a t 0
<
h
=
1 ; showing
L
a n d Iy 1, a n d ( y
R
T h u s e a c h o p t i o n of x
o p t i o n s of
x
+
+
1 a r e f i n i t e s u b s e t s of 0
y a n d of x y i s i n 0
y a n d xy a r e f i n i t e i n number, x
THEOREM.
Ow =
+
w
.
Assume
b(y)'
w h e r e b(x)
+
b(y)
S i n c e t h e s e t of
y a n d xy a r e i n 0
w
.
0
D.
By Lemma 2, Ow i s a s u b s e t of D .
PROOF.
is closed
L R As s u c h , {x 1 , (x ) a r e f i n i t e
w'
k.
1 ; thus
we may r e g a r d x a s a n element of
t h a t t h e sum an d p r o d u c t of e a c h e l e m e n t o n Ok i s i n 0 =
2h
<
2 Rh L]
T h i n k i n g of No a s t h e C u e s t a
CD(Ob(x)) a n d y as a n e l e m e n t of CD(Ob(y)). s u b s e t s of O b ( x l ,
-
2 L h R , 2Rh
F i r s t n o t e t h a t , by ( 1 4 1 , 0
t h a t 1 / 2 cN2, a n d hence 1 1 2 ~ 0
Dutari
I
Since 1 is t h e simplest p o s i t i v e element,
1.
+
2hR
+
+
165
of Q t h a t c o n t a i n s 112.
By Lemma 3, Ow is a s u b r i n g
By (31, D is t h e smallest s u b r i n g of Q t h a t con-
t a i n s 1 / 2 ; t h u s D = Ow.
COMMENT.
The number
3 is i n Ow, b u t 1 / 3 is n o t ; t h u s 0 is n o t a w
field.
Mw
4.30
(0)
A number x i n
-n < x x
+
< n,
1, x
+
No w i l l be c a l l e d
and x = 112,
x
(x - 1 , x
114,
+ 114, x + 1 / 8 ,
THEOREM 0 ( C 2 4 , p p . 2 4 - 2 5 ] ) .
x
-
i f t h e r e e x i s t s ncN f o r w h i c h
x
-
... , x
... , x - 1/2", ... I 1/2", ... 1, [24, p . 241,
1/8, +
(1) Each deD i s a real number in No.
i n No, t h e n so are - x , x + y, a n d x y . ( i i i ) For e a c h real number x in No, l e t L = {qsQ: q < x ) a n d l e t R = {qcQ: q > XI. Then x = {L I R ) . (ii) If
a n d y a r e r e a l numbers
Norman L . A l l i n g
166
Given any g a p ( L , R ) i n Q , t h e n {LIR) i s a real number i n No.
(iv)
( i ) By Lemma 1 of S e c t i o n 4.21,
PROOF.
-
{x
2-"1
x + 2-n].
={-x - 2
-
-n
2-"
I
-x
+
y, x + y
-
-
-
(x - x ) ( y
x
2-")
+
2-m) x
2-"
+
+
- ( x - x R )(Y -
XY
L
y 1, xy
- (x -
(x
-
2-n))(y
-
(y
-
-
(x
-
2-"))(y
-
( y + 2 - 7 , xy
2 - 9 , xy
{xy - (=
I X Y - ,-(n+m)
I
xy
+
-
2-ml y x
+
+
2-m).
y
=
-x
-
Y
R
R y 11
- (x
2 T , xy
)I
-
= +
2 - 9 1 (y
(x + 2-"))(y
( 2 - n ) ( 2 - m ) l xy
-
(-
I
-
(y
+
2-9
-
(y
-
2-9
2 - n ) ( 2 - m ) , xy
-
=
(2-")(2-91
2 - ( n + m ) 1 ; s h o w i n g t h a t xy i s a r e a l number i n No.
< x ) and l e t R = { q E Q : q > X I . S i n c e by d e f i n i t i o n t h e r e e x i s t s nEN s u c h t h a t -n < x < n , L and R a r e non-empty. Clearly
( i i i ) Let L = { q E Q : (0)
-
-
R x )(Y
-
(x
and y = { y
y , x + y + 2 - m ) ; showing t h a t x + y i s
{xy - ( x {xy
( i i ) Let x a n d y b e r e a l
Using (4.08:19) we know t h a t xy =
L L x ) ( y - y 1,
-
L
-
xy
(x
2-"1
may be w r i t t e n as
a n d t h u s - x i s a r e a l number i n No.
a r e a l number i n No. {xy
-
(x
=
2-"),
+
dED
By ( 4 . 0 9 : 1 ) , x i s r e a l .
n u m b e r s i n No, w i t h x
(X
4.30
L and {x
coinitial. (4.02:16),
-
2-"]
q
a r e m u t u a l l y c o f i n a l a n d R and { x + 2-"]
By ( 4 . 0 2 : 1 6 ) , x { L I R ) is real.
As we have
=
(LIR].
are mutually
( i v ) Let ( L , R ) b e a g a p i n Q .
o
see i n S e c t i o n 4.21, 0
w
i s t h e r i n g D of d y a d i c n u m b e r s .
- D
S i n c e D i s d e n s e i n t h e f i e l d of real numbers R , a number r i n R associated w i t h s u b s e t s L
Clearly L < R.
=
{acD: a
< r)
a n d R = {bED: b
>
i s a t i m e l y r e p r e s e n t a t i o n of x.
Let x
=
= w.
c a n be
r ] of O w .
Let x = { L I R ] , a n d n o t e t h a t x i s n o t i n 0
{No,<,bl i s a f u l l class of s u r r e a l numbers ( 4 . 0 3 : 2 ) , b ( x ) on D.
By
w
.
Since
Thus ( L , R )
f ( r ) . Let f be t h e i d e n t i t y map
Clearly f is a n o r d e r - p r e s e r v i n g map of R i n t o M
w'
The d e f i n i t i o n s
o f a d d i t i o n a n d m u l t i p l i c a t i o n i n R , o n t h e o n e h a n d , a n d i n No o n t h e o t h e r a r e s u c h t h a t f i s a n i n j e c t i v e homomorphism of R i n t o M U . e.g.,
(See
S e c t i o n 1 . 1 6 , w h e r e much of t h i s i s c a r r i e d o u t i n t h e c a t e g o r y of
ordered g r o u p s . )
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.30 (1)
167
C l e a r l y f is a n isomorphism of R o n t o (XENO: x is a r e a l number). Let u s i d e n t i f y R w i t h {XENO: x r e a l ) , by means of f . M
u
t o be ( N ( } ( 4 . 2 0 ) .
C l e a r l y w i s t h e greatest e l e m e n t i n M
is t h e least element i n M
(2)
<
C l e a r l y , d-
d-,
d + E Mu
<
d
- R,
w
.
I
d e f i n e d + t o be {d d e f i n e d - t o b e {d
(3)
For example, r e c a l l t h a t w was d e f i n e d
contains other elements.
-
and
-0 =
{IN]
Given d E D we c a n
d + 1 , d + 1/2,
-
1, d
1/2,
... , d
... , d -
d + , f o r a l l d i n D.
and d
w'
-
<
2-"
d-
<
d
+
l/n,
... ]
... [
l/n,
and
d].
Note t h a t
< d+ <
d
+
2-" f o r a l l n d .
F i n a l l y , we c a n g i v e t h e f o l l o w i n g d e s c r i p t i o n of Mu.
(4)
M
w
is t h e u n i o n o f t h e f o l l o w i n g d i s j o i n t s u b s e t s : ( i ) R, ( i i ) { + w ) ,
a n d ( i i i ) id*: dsD). PROOF.
We h a v e s e e n t h a t M
w
contains all set d e f i n e d i n ( i ) - ( i i i ) .
I t i s a l s o clear t h a t t h e s e t s described i n ( i ) - ( i i i ) are d i s j o i n t . X E M ~ ;then b(x) 5 w.
-
t h a t b ( x ) = w, l e t L
If b ( x ) < w, t h e n X E O ~ , a n d h e n c e i s i n R . (dEOw: d
< XI, and l e t
i s a C u e s t a D u t a r i c u t i n Ow, a n d x i s empty t h e n x
= -w.
=
{L[R).
R = {dcOw: d
Assume t h a t L and R a r e non-empty;
If (L,R) is a gap i n D, t h e n XER
Dedekind-cut i n D.
( L , R ) is n o t a gap i n D; t h e n i t h a s a c u t p o i n t dED.
If doR, t h e n x = d
-
.
0
> XI; then
If R Is e m p t y , x
- D.
-
Let
Assume
w.
(L,R)
If L
thus (L,R) is a
Assume now t h a t
If dcL, t h e n x = d'.
168
Norman L. A l l i n g Conway [ 2 4 , p . 2 4 1 d e f i n e s x t o b e
B I B L I O G R A P H I C NOTE.
number i f and o n l y if - n (X
-
- 1, x
1/2, x
-
<
1/3,
4.30
x
< n
... I
f o r sane i n t e g e r n , a n d x
x
+
1 , x + 1/2, x + 113,
We h a v e c h o s e n t o u s e s i m p l e r numbers, 1/2",
Ira
=
...1".
e a c h of which i s i n D ,
r a t h e r t h a t t h e numbers l / n , as Conway d o e s [ 2 4 , p . 2 4 1 , s i n c e some of
.
t h e s e numbers a r e i n a r e i n N
Note t h a t t h u s t h e r e p r e s e n t a t i o n ( x
1/3, 1 / 5 , e t c . a r e i n Nu.
-
is a r e f i n e m e n t of ( x
1/2n)
For example 1/2 a n d 114 a r e in D , w h i l e
l/n, x
+
l/n).
-
By ( 4 . 0 9 : l )
1/2",
X
+
they are
e q u i v a l e n t ; t h u s t h e q u e s t i o n of which r e p r e s e n t a t i o n o n e uses s e e m s n o t t o be of major i m p o r t a n c e .
THE MAP XENO
4.40
+
w
x
ENO
+
F o l l o w i n g t h e usual p r a c t i c e s i n an o r d e r e d f i e l d K , f o r XEK l e t 1x1 =
max.(x,
-XI.
For x a n d y i n K , write x
1x1 5 n l y l a n d ( y I 4 n l x l . s u r a t e iff x (0)
a
a
y.
a
y i f t h e r e e x i s t s ncN s u c h t h a t
F u r t h e r , we w i l l s a y t h a t x a n d y a r e commen-
Clearly
i s an e q u i v a l e n c e r e l a t i o n o n K .
Let y / ( a ) d e n o t e t h e e q u i v a l e n c e c l a s s e s mod Clearly
O/(a)
i s {Ol.
a
t o which y belongs.
C l e a r l y K is Archimedean i f and o n l y i f K h a s two
e q u i v a l e n c e classes: ( 0 ) a n d K* ( = {xcK: x # 0 1 ) .
Let y b e i n No*.
<
x
01).
<
nlyl}.
Let J ( y )
=
{xcNo: t h e r e e x i s t s ncN s u c h t h a t l y l / n
C l e a r l y J ( y ) i s a n o n - e m p t y i n t e r v a l i n No'
Using (4.02:7),
( = (XENO:
x >
we know t h a t t h e r e e x i s t s a s i m p l e s t e l e m e n t i n J(y),
which Conway c a l l s t h e l e a d e r of y [24, p . 311. T h u s t h e l e a d e r of YENO*
is t h e s i m p l e s t p o s i t i v e element i n y / ( p ) .
The s i m p l e s t l e a d e r of a l l i s 1 , t h e o n l y p o s i t i v e n u m b e r b o r n o n d a y 1 .
Conway d e f i n e s wo t o be 1 .
He l e t s w 1 be t h e s i m p l e s t l e a d e r t o t h e r i g h t
4.40
I n t r o d u c t i o n t o t h e surreal number f i e l d No
of 1 i n No, namely w, which i s i n N
w
(4.30).
169
I n g e n e r a l , wx w a s d e f i n e d by
+
Conway as f o l l o w s , where r and s r a n g e o v e r R :
(1)
wx
=
(0,rw
xL
Isw
xR
1 C24, p . 311.
I t w i l l be c o n v e n i e n t , o n o c c a s i o n , t o r e s t r i c t r a n d s i n ( 1 ) t o be
S i n c e Qt and R t are m u t u a l l y c o f i n a l a n d m u t u a l l y c o i n i t i a l , t h e
i n .'Q
v a l u e of w x g i v e n by t h e r e s u l t i n g d e f i n i t i o n (which we w i l l c a l l (1,Q')) i s t h e same as t h a t g i v e n by ( 1 ) .
L e t u s m a k e s u r e t h a t t h e d e f i n i t i o n , as g i v e n i n ( I ) , agrees w i t h those g i v e n i n t h e p a r a g r a p h p r e c e d i n g ( 1 ) . 0 =
[Ol], UJ
-1
which i s 1 .
[Ol],
1 =
.
is, a c c o r d i n g t o ( 1 ) .
thus w -1
=
1
=
[lo],
{I], =
0
=
Let u s d e t e r m i n e what
(O,rll = w. thus w-l
t h u s by ( 1 1 , w
[Olrl, a n e l e m e n t we
d e f i n e d i n (4.30) t o be 0'.
For x , y , a n d z i n K (2)
<<
x
y i f nx
Thus x
(3)
0
<
wu
, we
w i l l write
y f o r a l l nEN.
<
y iff x
y a n d x a n d y a r e incommensurate.
Clearly x
<<
LEMMA 0.
Let u a n d v be i n No.
y and y
<<
z implies x
<<
(1) u
z.
<
v i f and o n l y if ( i i )
<< wv. PROOF.
2b(v)). (1).
<<
<
+
Let u s p r o c e e d by i n d u c t i o n o n m a x . ( b ( u )
Assume t h a t ( i ) h o l d s .
<
u
R
6 v,
or
b(v), 2b(u),
Using i n d u c t i o n a n d t h e d e f i n i t i o n of w
i t i s e v i d e n t t h a t a l l wu a r e p o s i t i v e .
e x i s t s a uR w i t h u
+
By ( 4 . 0 2 : 1 4 ) ,
( b ) there e x i s t s v
L
U
( a ) there
w i t h u S vL
<
v.
If
Norman L. A l l i n g
170
( a ) h o l d s then wu << w' induction, w
U
R
<<
R
.
I f uR
-
v then w
w v , and u s i n g (31, w
U
approximately a s a b o v e , we s e e t h a t w ( i i ) . Now assume t h a t ( i i ) h o l d s .
v t h e n of c o u r s e mu
=
U
<<
4.40
<<
U
R
< v t h e n , by
V
<<
w ; proving t h a t ( i ) implies
Assume f o r a moment t h a t u L v .
>
If u
w v , which i s a b s u r d .
If u =
v t h e n we may u s e t h e
> > w V , which a g a i n i s
W'
Thus we have shown t h a t ( i i ) I m p l i e s ( i ) . o
LEMMA 1 . uy.
If u
If ( b ) h o l d s , t h e n , a r g u i n g
wV.
f a c t t h a t ( i ) implies ( i i ) t o conclude t h a t absurd.
wV.
( i ) For a l l xcNo+ t h e r e e x i s t s a unique ycNo s u c h t h a t x
a
F u r t h e r , ( i i ) b(wy) 5 b ( x ) . L R L x may be w r i t t e n i n t h e form {O,x ( x 1 w i t h x
By ( 4 . 1 0 : l ) .
PROOF'.
P r o c e e d i n g by i n d u c t i o n o n
and xR p o s i t i v e , and chosen t o b e i n Ob(x).
R b ( x ) , we see t h a t f o r e a c h xL ( r e s p . x ) t h e r e e x i s t s a unique yL ( r e s p . y
R
such t h a t xL = uy
L
(resp. xR
R s i m p l e a s xL ( r e s p . x 1.
X I , then ( i ) is proved.
R a
wy
1, w i t h wy
L
R ( r e s p . wy ) a t l e a s t a s
I f x i s commensurate w i t h o n e of i t s o p t i o n s , s a y Note a l s o t h a t b ( u y ' ) 2 b ( x ) , s h o w i n g t h a t ( i i )
Assume t h a t x i s c o m m e n s u r a t e w i t h none of i t s o p t i o n s ; then by
holds.
Lemma 0 , r w y {yLlyR].
L
<< x <<
R SW'
Then y i s i n No.
,
f o r a l l r , SER'.
Let y b e d e f i n e d t o b e
Let I be d e f i n e d t o be {ZENO: xL
L
l e t J be d e f i n e d t o be (zcNo: rwy
<<
z
<<
<
z
<
R swY , f o r a l l r , s E R + } .
i s commensurate w i t h n o n e of i t s o p t i o n s , I
=
J.
x
R
1 , and
Since x
Since x i s , by d e f i n i -
t i o n , t h e s i m p l e s t element i n I and since wy is d e f i n e d t o be t h e s i m p l e s t e l e m e n t i n J , we s e e t h a t x
= w'.
I n t h i s c a s e , of c o u r s e , b ( x )
A s t o t h e uniqueness of y , t h a t f o l l o w s frcm Lemma 0 .
THEOREM.
For a l l x and y i n No, wX+'
= wXwy.
=
b(uy).
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.40
Proceeding by i n d u c t i o n on b ( x )
PROOF. =
Then by Lemma 0, X > 0 and Y > 0 .
uy.
(4)
X
=
{O,aw
X
L bw
X
R
+
b(y), let X
171
=
w
X
and l e t Y
By d e f i n i t i o n ,
L
R
] and Y = (0,cwy IdaY ) , where a , b , c , d a r e i n R t .
Then XY =
L
(5)
10. awx
(6)
L (0, a w x ty
+'
+
+
cw '+Y
cw '+Y
L
L
-
acw
-
acw
L
L
ty
ty
L ~
L
bw
, bw
R
R
ty
+'
t
dw"Y
t
dw"y
R
-
bdwX ty
R
R
I
R
-
R R bdwX ty
I
S i n c e ( 5 ) and ( 6 ) a r e i d e n t i c a l , t h e Theorem is proved. Applying t h e Theorem, we o b t a i n t h e f o l l o w i n g c o r o l l a r y :
(7)
For a l l XENO, w
-X
=
l/w
X
.
F u r t h e r , t h e map t h a t takes XENO t o w x i s
a n o r d e r - p r e s e r v i n g homomorphism o n t o t h e class A of a l l l e a d e r s i n
No; t h u s A i s an o r d e r e d group, under m u l t i p l i c a t i o n , t h a t is o r d e r l s u n o r p h i c t o t h e a d d i t i v e group (No,+) of No.
4.41
FINITE LINEAR COMBINATIONS OF w - x ( 1) ,
... , w - x ( n )
OVER R
We w i l l be concerned i n t h i s s e c t i o n w i t h f i n i t e l i n e a r combinations
Norman L . A l l i n g
172
of elements of t h e form w Y o v e r R.
4.41
Let aeR, and l e t Y E N O .
The f o l l o w i n g
a r e obvious.
(0)
(i)b 0.~')
= 0,
0
f o r all YENO, and ( i i ) b ( a w 1 = b ( a ) , f o r all aeR.
Since b ( - x ) = b(x), and having c o n s i d e r e d ( O , ( i ) ) ,
L
R
Let a = { a la J E R .
f i e l d of Mu.
Since b ( a ) 4
W,
w e know ( b y c o n v e n t i o n
R L t h a t ( a L , a ) i s a t i m e l y r e p r e s e n t a t i o n f o r a: t h u s each a
(4.02:15)),
and each a
F u r t h e r , w e saw i n (4.30) t h a t R i s a sub-
Ow.
=
0.
A s we
As b e f o r e , l e t D denote t h e r i n g of a l l dyadic numbers ( 4 . 2 1 ) .
saw i n S e c t i o n 4 . 2 1 , D
>
assume t h a t a
R . is i n D.
Since a { aL wY
LEMMA.
awy
PROOF.
Let {a
=
L'
I
>
0 , we may a l s o assume t h a t each aL
0.
aR wY I . {dcD: 0
=
>
<
d
<
a) and l e t [ a
R'
) = (dED: d
>
a].
S i n c e D i s dense i n R we know t h a t L R ( a L ' , a R ' ) i s a refinement of ( a ,a ) (4.09:O).
- D.)
( C l e a r l y ( a L 1 , a R ' ) i s t i m e l y i f and o n l y i f a d a
= (3''
I
By ( 4 . 0 9 : 1 ) ,
aR').
By d e f i n i t i o n ,
aJ
{aLwY
=
aLwy
R
+
awy
+
-
L awy
R
aLJ
,
L
aLw Y , a Rwy + a wY aRwY
+
au'
L
-
R
-
R
aRwY
I
L aRwY I .
There a r e f o u r e x p r e s s i o n s f o l l o w i n g t h e e q u a l i t y s i g n i n ( 3 1 , w h i c h we w i l l c o n s i d e r i n o r d e r .
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.41 L
aLwY + y
L
-
L
<
y,
wy
aLwY
<<
L
-
aLwY + ( a
=
aL)WY
L
Since aL < a ,
L
o <
a - a
S i n c e t h e r e e x i s t s aL’
( 4 . 4 0 , Lemma 1 ) .
my
.
173
>
.
Since
a L , we s e e
t h a t t h e following holds.
R
Consider now t h e second e x p r e s s i o n i n ( 3 ) : aRwY
(a
+
-
aR)wY
(4.40, Lemma 1 ) . (5)
a R J+ a J
R
.
< aR, a
Since a
-
aR
<
(a
-
- a w
<
YR --
<<
y R , wY
wy
R
Thus R
R
-
L YR a )w
Lemma 1 ) .
(4.40,
Since y
0.
aRJ
< o <
aL’J
<
am’.
R A aLwY + a u y - aL wY
Consider now t h e t h i r d e x p r e s s i o n i n ( 3 ) : aLwY
R
aRwY+ aw
.
o <
Since aL < a ,
a
- aL .
Since y
<
y R , ,Y
L
-
aRwY
=
<<
J
R
Thus R
(6)
auy
< a R ’ u Y < aLwY + aWy - aL w YR .
Consider now t h e l a s t e x p r e s s i o n i n ( 3 ) : aRwY
(a
+
-
aR)wY
( 4 . 4 0 , Lemma 1 ) .
(7)
L
.
Since a
+
COROLLARY 0. =
<
Since y
0.
L
<
y , Y.
L
<<
UY
L aWY - aR W yL .
that t h e following h o l d s : awy
C
aR
=
There e x i s t s aR’ < a ; t h u s ,
a J < aR’mY < aRwY
+
-
L
R
Hence, b y ( 4 ) - ( 7 ) , auy
Then A
< aR, a
aRwY+ a0y
{aLw-‘
{aL’WyI a R ’ u Y ) . By ( 4 . 0 9 : 1 )
= =
iaLwyI a R u Y ) ; proving t h e Lemma.
Let a , ccR, x +
cw-y,aw-x
, we c o n c l u d e
+
<
y be i n No, A
=
aw- X and l e t C
cL~-YlaRw-X +cW-y,aW-x + cRu-Y].
=
c w- Y
.
Norman L. A l l i n g
174
By d e f i n i t i o n ,
PROOF.
(8)
A
C = {AL
t
4.41
+
C, A
CLI AR
+
c
C, A +
+
R
}.
Using Lemma 0, t h e r i g h t hand s i d e of ( 8 ) i s s e e n t o equal
T h i s may a l s o be g e n e r a l i z e d as f o l l o w s .
COROLLARY 1 .
L e t nEN l e t , a i and ci be i n R, f o r i
<...<
-x(l) + Let A = alw
l e t x(1)
c = c w- x ( l )
t
1
(10)
“alL (a, (a,
+
+
R +
x ( n ) be i n No.
...
t
c
-x(n).
n
c , ) w-x(l)
+
clL).w-x(l)
+
cl)-w - x ( l )
+
( a , + c1R ~ * w - ~ ( ’ )t
PROOF.
Let
x ( n ) be i n No.
... ... ...
+
(a
t
(an
+
( a nR
t
(an
+
n
m i s not equal t o 0.
a
4 . 4 0 h o l d s , w- x ( ’ )
>>
>
< ,
cn)*w- x ( n ) , cnL)-w-x(n)~
+
+
+
cn)*w- x ( n )
#
c:)-w-~(n)].
alw-’(’)
0 or A
... , <
... >> w - x ( n ) ,
BIBLIOGRAPHIC NOTE.
and l e t
anw- x ( n ) ,
+
= 1,.
<
o
..,n, and l e t x(l)
... + anw- x ( n ) .
Let m be t h e l e a s t element i n [ l ,
Then A
Since x(1)
=
1 ,...,n, and
=
l e t aicR, f o r i
nEN,
Let A
a l l of t h e a i l s a r e 0.
PROOF.
...
+
Apply t h e d e f i n i t i o n of a d d i t i o n and Lemma 0.
PROPOSITION.
... , <
Then A + C
...
=
0 a c c o r d i n g as am
< ,
Assume t h a t not
...*n) >
such t h a t
0 o r am < 0.
x ( n ) , a n d since Lemma 0
Of
Section
Sane of t h e r e s u l t s i n t h i s s e c t i o n may be new.
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.50 4.50
THE SIGN-EXPANSION
F o l l o w i n g Conway [24, p p . 30-311, l e t a
Let X E N O , w i t h b ( x ) = B .
B, l e t
175
L ( x ) = [yEOa: y
a
x is not i n 0
a'
< X I , and l e t Ra(x)
= ( ~ € 0 y~ >:
XI.
t h u s ( L ( x ) , R ( x ) ) is a Cuesta Dutari c u t i n 0 a
.
[L,(x)] R a ( x ) ] , and n o t e t h a t x EN a a a 1 B, and let x
>
.
Let x
a, a
=
Now a s s u m e t h a t
a , x f xa.
ath a p p r o x i m a t i o n
t o x [24, p .
( x ) ) i s a s u b s e t of L
(x) ( r e s p .
Conway c a l l s x
x.
=
Since B
>
S i n c e f3
<
the
291. (0)
If a. ,< a , I 6, t h e n ( x a
PROOF. R
a1
Since L
( x ) ) , x and x
a1
a0
a0
1
(x) (resp. R
b o t h f i l l (4.02)
we s e e t h a t L a o ( x a l )
=
[Lao(xa,)I R
{L ( X I / R
a0
(xa
=
= 1
La
(X) 0
a0
x
a0
a0
t h e Conway c u t ( L
a n d Ra a0
.
(XI]
0
(x
=
a1
xa
0
= R
.
a0
(x), R
a1
a0
( X I ) i n No,
Thus
(XI. a0
(Xa 1
1
=
a0
Let u s c a l l ycNo a p r e d e c e s s o r of x, a n d write y < t x i f t h e r e e x i s t s a
<
B such that y
a s u c c e s s o r of y.
=
x
.
I f y i s a n p r e d e c e s s o r of x we w i l l refer t o x as
x i s t h e immediate s u c c e s s o r of y i f y
i s no z ~ N of o r which y
<
z
<
x.
<
x, and t h e r e
C l e a r l y 0 i s t h e o n l y e l e m e n t i n No t h a t
has no p r e d e c e s s o r . (1)
Under < t , No i s a p a r t i a l o r d e r c l a s s .
PROOF.
Assume t h a t x o c t x 1 a n d x 1
ct
y ; t h e n by ( 0 1 , x o
p r o v i n g t h a t No i s a p a r t i a l l y - o r d e r e d c l a s s u n d e r
Let t h e s e t of p r e d e c e s s o r s of x, p r N o ( x )
(2)
=
ct XI.
p r ( x ) i s a w e l l - o r d e r e d s e t u n d e r < t . Under < t , No i s a tree.
c t y;
176
Norman L . A l l i n g PROOF.
C l e a r l y acB
We w i l l r e f e r t o
4.50
xacprNo ( x ) i s an o r d e r - p r e s e r v i n g b i j e c t i o n .
+
a s t h e t r e e o r d e r on No.
See e . g . ,
c57, pp. 292-
3201 o r [55, pp. 315-3263 f o r r e f e r e n c e s t o t h e t h e o r y o f t r e e s .
Since 0
i s t h e element of No t h a t h a s no p r e d e c e s s o r , 0 i s t h e r o o t of No. Let G be a n o r d e r e d group G .
(3)
For ycG l e t
s i g n ( y ) be + i f y
>
0,
( i i ) s i g n ( y ) be 0 i f y
=
0 , and l e t
<
0.
(i)
( i i i ) s i g n ( y ) be
-
if y
Returning our a t t e n t i o n a g a i n t o No,
(4)
l e t o ( x ) ( a ) be d e f i n e d t o be s i g n ( x
-
xa), for a l l a
B U ( X ) E [ + } w i l l be c a l l e d t h e Sign e x p a n s i o n of x. and
as follows: l e t
[ * I Bis a n o r d e r e d s e t .
-<
0
<
+.
bl(Z Y
= Y,
Y
b(x).
Let u s o r d e r
6
Next l e t C be t h e u n i o n of ( { k } )Beon,
is an o r d e r e d s e t .
and b ' ( E ) =On.
-,
0,
Then, under t h e l e x i c o g r a p h i c o r d e r i n g ,
C i n t o O n by t a k i n g e a c h p o i n t of [ + I B t o 6.
( I + ) B ) B E Y ; then E
<
L e t b' map
Let L y be t h e u n i o n o f
Clearly E be t h e u n i o n of ( 2 y ) y e 0 n ,
F u r t h e r t h e f o l l o w i n g h o l d , f o r a l l BcOn.
B I B L I O G R A P H I C NOTE.
L y i s d e f i n e d [55, p p . 3 1 6 1 t o b e t h e f u l l
b i n a r y t r e e of h e i g h t Y .
E h r l i c h [ 2 8 1 c a l l s L t h e f u l l b i n a r y t r e e of
h e i g h t On.
The h e i g h t f u n c t i o n of t r e e t h e o r y [55, p p . 3151 i s t h e f u n c -
tion bl.
(5)
(i)
u maps No i n t o
(ii)
b'.o
z,
= b,
( i i i ) o ( N ) i s c o n t a i n e d i n (*}',
B
(iv)
d o B ) i s c o n t a i n e d i n E 8'
and
177
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.50
One c a n d e f i n e t h e r e l e v a n t o r d e r s o n Z Y and o n I d i r e c t l y .
However,
many fewer s p e c i a l cases must be c o n s i d e r e d i f we p r o c e e d a s f o l l o w s .
L e t Z y A = { t ~ ( f , O ] ~ :t h e r e e x i s t s 6
< bY - ( s ) ,
t(a)
Clearly b
zY1
= f,
maps 1 Y
Y
-
b Y ^ ( t )< Y s u c h t h a t f o r a l l a
a n d f o r a l l a f o r which b - ( s ) 2 a Y o n t o Y.
<
Y , t h e n t ( a ) = 01.
Let Eye b e l e x i c o g r a p h i c a l l y o r d e r e d .
Then
is a n o r d e r e d s e t .
For seZy, w i t h b ( s )
< 6, a n d (6)
s o cZy
B , l e t s A c Z y As u c h t h a t s * ( u ) =
=
s - ( a ) = 0 , f o r a l l a f o r which 6 6 a
Let s o ~ E ya,n d l e t B o
1
, and
that so
+
-
Now l e t
byA(soA) BO.
sl.
-
=
=
b y A ( s A ) ,f o r a l l s c Z y .
By d e f i n i t i o n , B o
b(so).
S1€Ey,
for a l l a
S(u),
< Y.
seZy + s n e Z Y A i s a b i j e c t i o n s u c h t h a t b ( s )
PROOF. -
=
<
and l e t B1 = b ( s l ) .
Y.
Then Assume
I f B o = B = B 1 , t h e n i t i s clear t h a t s o A f s l A . Assume
< E l , t h e n s o ^ ( B o ) 0 a n d s l A ( B 0b) f ; t h u s s o A C s l A . Assume t h a t b 0 > B l r t h e n s l n ( B 1 ) 0 a n d S O A ( B 1 ) t f; t h u s s o A f s l A . Hence t h e mapping s + s A is i n j e c t i v e . Let t E E y n , a n d l e t b y A ( t ) = B ; t h e n 8 < Y . t h a t B,
Let s =
<
6.
tls:
-
i . e . , l e t s be t h e r e s t r i c t i o n of t t o t h e s e t of a l l o r d i n a l s
Then c l e a r l y seZy a n d s- = t .
Let Y 6 6 6
c , a n d l e t t Y E EY
o
6
A
b e e x t e n d e d t o i y ( t y )=
d e f i n i n g t ( a ) t o b e 0, f o r a l l a f o r which Y 5 a
Y
f ol 1owing
(7)
<
6.
Y i s a n o r d e r - p r e s e r v i n g map of Z Y
i6
(ii)
i5 6 - i 6y = i y 5,
and
( i i i ) i Yy is t h e i d e n t i t y mapping of EyA.
into
by
C l e a r l y we h a v e t h e
.
(i)
t6EEgA,
Z6*,
4.50
Norman L. A l l i n g
178
Using ( 7 ) we s e e t h e f o l l o w i n g .
(8)
The d e f i n i t i o n of o r d e r on C i s independent of t h e c h o i c e o f Y, a n d i t r e n d e r s Z ( r e s p . C ) i n t o an ordered c l a s s ( r e s p . an o r d e r e d s e t ) .
Y
< {Yj.
Given any s u b s e t S of C , t h e r e e x i s t Y E O n s u c h t h a t b ' ( S )
a well-defined order-preserving i n j e c t i o n , for
mapping SES
+ s " E i~s ~t h e n
a l l such Y .
Let S" be t h e image of S I n C Y under t h i s mapping.
(9)
The
I n t h e f o l l o w i n g pages, " w i l l always have t h i s meaning. Following Conway C24, p.301, we have t h e f o l l o w i n g . LEMMA.
a is an o r d e r - p r e s e r v i n g map of No i n t o E .
PROOF.
Let x and y be i n No, w i t h x
t i o n on b(x)
+
f o r which x 5 yL
Let u s p r o c e e d by i n d u c -
By ( 4 . 0 2 : 1 4 ) w e know t h a t ( i ) there exists sane y
b(y).
<
< y.
xR f o r which
y , o r ( i i ) t h e r e e x i s t s some
x
<
x
R
L
5 y.
If ( i ) holds then we may use t h e i n d u c t i o n h y p o t h e s i s t o conclude t h a t a ( x ) S
L
< a ( y ) ; which shows t h a t
o ( y L ) and t h a t a ( y )
<
u(x)
<
w e may use t h e i n d u c t i o n h y p o t h e s i s t o conclude t h a t u ( x ) 5 a ( y ) ; which shows t h a t o ( x )
<
I f ( i i ) holds
a(y).
a(x
R
R
and a(X )
o(y).
THE STRUCTURE OF E AND THE SIGN-EXPANSION
4.51
The o r d e r e d c l a s s E w a s d e f i n e d and s t u d i e d i n S e c t i o n 4.50.
operator 0.
then
a s follows: l e t - ( + I
o p e r a t e on
For each S E E , l e t
b'(s);
(0)
-
-9
i s i n Z.
_ _
( s) =
-3
9,
=
-, - ( - I
= +,
Let t h e
and -(O) =
be d e f i n e d a s f o l l o w s : - s ( a ) = - s ( a ) , f o r a l l a Clearly
f o r a l l SEE, and f o r a l l XENO, a ( - x )
=
a(X).
<
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.51
Henceforth we w i l l write -s a s
179
Let 0 b e t h e e l e m e n t i n { + l o :
-9.
i . e . , 0 i s t h e empty mapping; t h a t i s i t i s j u s t t h e empty s e t .
Conway d e f i n e d x
a
124, p . 291, c a l l i n g i t t h e a% approximation o f x.
We d e a l t w i t h t h i s n o t i o n , u s i n g t h e same n o t a t i o n and terminology, i n F u r t h e r , u s i n g i t we i n t r o d u c e d t h e idea of t h e o r d e r
S e c t i o n 4.50.
No, under which No is a t r e e .
(1)
o f s t o t h e set a.
If a, 6 a 1 5 8 , t h e n (sa
PROOF.
(sa )
1 010
-
On
Let us now c o n s i d e r t h e same i d e a s f o r E.
Let S E E , and l e t B = b ’ ( s ) . the restriction
1 a0
I f a 5 B , l e t sa Clearly s
=
s
a0
= s.
= s(a:
i . e . , l e t su be
Further,
.
(slal)lao,which i s j u s t s l a o .
L e t us c a l l teE a p r e d e c e s s o r of s, and write t < t s, i f there e x i s t s
a < 8 such t h a t t
=
s
a
(cf. (4.50)).
If t i s an p r e d e c e s s o r of s we w i l l
refer t o s as a s u c c e s s o r of t ( c f . ( 4 . 5 0 ) ) .
s i s t h e immediate s u c c e s s o r
of t i f t < t s, and i f there is no ueL f o r which t < t u < t s. C l e a r l y 0 i s t h e o n l y element i n L t h a t has no p r e d e c e s s o r . (2)
Under < t , E is a p a r t i a l order class. PROOF.
Assume t h a t so
8,
and
9,
t; then by (11,
9,
C t t.
o
Let t h e set of p r e d e c e s s o r s of s, p r E ( s ) , be {teE: t < t s}.
(3)
p r ( s ) i s a well-ordered set under < t . Under < t , 1 is a tree.
PROOF.
C l e a r l y aEB
+
s E p r z ( s ) i s an o r d e r - p r e s e r v i n g b i j e c t i o n . a
Let < t be c a l l e d tree o r d e r on L.
Note: 0 i s t h e root of t h e t r e e E.
F o l l ~ i n gConway c24, Theorem 181, we have t h e f o l l o w i n g .
Norman L. A l l i n g
180
THEOREM.
is an order-preserving map of No onto
Q
PROOF.
Q
4.51
r,
w i t h b'
s u f f i c e s t o prove t h a t Let
Y
S E ( ~ }
b.
was d e f i n e d i n (4.501, where i t w a s shown t o be an order-
preserving map i n t o Z (Lemma 4.50) f o r which b t * o = b ( 4 . 5 0 : 5 ) .
Cy.
-0 =
.
(I
is s u r j e c t i v e .
For each B
< Y,
Thus, i t
Let YEOn and assume t h a t o ( O y )
t h e r e e x i s t s a unique y E N
5
=
such t h a t
0
o(yg) = SB.
Let x = { ( y 8 : s
(4)
8
< s and
B
-
<
Yll ( y B : s8
>
s and B
< Yl).
Since (No,<,bl is a f u l l c l a s s of surreal numbers (4.03:2), b ( x ) 6 Y.
< Y.
Let s
>
s
B
Note t h a t a(x)(B)
(resp. s
<
6 Y , b t ( o ( x ) ) 6 Y.
COROLLARY.
4.52
+
-
s ) i f f s(B) B
Hence
Q(X) =
(resp. - ) i f f x
> y B (resp.
thus a ( x ) l Y
+ (resp. -);
=
x
<
s.
iff 8 Since b ( x ) y
s , and b ( x ) = Y .
( r , < , b ' } is a f u l l c l a s s of s u r r e a l numbers.
THE NEAREST COMElON PREDECESSOR OF A SUBCLASS OF C
Let S be a non-empty s u b c l a s s of C . predecessor of S i f a S t s , f o r each SES.
aEE w i l l be c a l l e d a common
A cmmon predecessor c of S w i l l
be c a l l e d t h e n e a r e s t common predecessor of S, i n symbols c = ncp(S), i f given any common predecessor a of S t h e n a S t c .
(0)
S has a nearest common predecessor. PROOF.
Let m i n . ( b t ( S ) )
of t h e c h o i c e of seS,
=
Y , and l e t
for a l l a 6 61.
Further, i t is unique.
r
=
{a <
Clearly
Y:
r
S(a)
is independent
i s a lower-saturated
I,) I ) . subset of Y (1.021, which may b e e m p t y ( f o r example i f S = { ( +( Let 8 be t h e l e a s t ordinal t h a t i s g r e a t e r than a l l of t h e elements of Let S ~ E S ,and l e t c ( a ) a l l S E S and a l l a
<
=
s , ( a ) , f o r a l l ae[O,B).
B, c(a)
=
c is a common predecessor of S.
r.
By t h e c h o i c e of B , f o r
s ( a ) ; t h u s f o r a l l SES,
sB
-
c ; showing t h a t
Now l e t a be a common p r e d e c e s s o r of S ,
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.52 and l e t B '
=
c
.
=
ncp(S)
Since a
b'(a).
sBl,
=
181
f o r a l l SES, B ' 5 6, a n d a
=
c
Thus
B"
Now l e t d b e a n e a r e s t common p r e d e c e s s o r of S; t h e n d S t c A s a consequence c = d .
and c St d .
EXAMPLES.
( i ) ncp([s,l)
=
so.
( i i ) I f so is a common p r e d e c e s s o r of
S o , a n d S is t h e u n i o n S o and [ s o ) , t h e n s o = n c p ( S ) .
Let c = n c p ( S ) .
(1)
S 1 ~ Ss u c h
t h a t so 5 c 5
Assume f o r a moment t h a t e i t h e r ( I ) s
PROOF. (ii) s
< c
holds.
Let 6
=
or 0, f o r a l l SCS.
c ; which i s a b s u r d .
f o r a l l S E S , or
S i n c e c A ( B ) = 0 , a n d since c is a common p r e d e c e s -
bl(c). = +
> c
9,.
W i t h o u t l o s s of g e n e r a l i t y we may assume t h a t ( i )
f o r a l l SES.
sor of S, s^(B) =
There e x i s t s o ,
B)
Thus s
=
+,
Were t h e r e
8,hS
f o r a l l SES.
w i t h s,^(B) = 0 , s,
Hence
BEr,
which i s
absurd.
PROOF. so S c 6
bt(c)
<
8,.
Let
8,
<
8,
be i n {+IY, and l e t c
Since b'(s,)
= Y =
Y ; thus so f c f sl.
LEMMA.
b ' ( s l ) , and s i n c e
Hence s o
< c<
3,
<
By ( 2 1 , j ,
j,
and
3,
are unequal,
9,.
There exists a unique
- [Jol). A s s u m e , f o r a moment
Let Y b e t h e l e a s t e l e m e n t i n b ' ( J ) .
that there exists j,
ncp((jo,jl)).
n c p ( ~ s , , s , ) ) . By ( 1 )
Let J be a non-empty i n t e r v a l i n Z.
j o c J such that b f ( j o ) < b ' ( J PROOF.
=
i n J , for which b * ( j o ) = Y = b t ( j l ) .
< c <
j , , and b ' ( c )
<
Y.
Let c =
S i n c e J is a n i n t e r v a l ,
c i s i n J ; which i s a b s u r d .
THEOREM. joEJ
such t h a t b F ( j o )< b ' ( J PROOF.
(1)
Let J be a non-empty i n t e r v a l i n 2 .
-
The u n i q u e e l e m e n t
{ j , ) )is ncp(J).
Let j , be as d e f i n e d i n t h e Lemma, a n d l e t c
a n d t h e f a c t t h a t J is a n i n t e r v a l , we see t h a t
CEJ.
=
ncp(J).
Using
By c o n s t r u c t i o n
Norman L. A l l i n g
182 bl(j,) 5 bt(c).
that bt(c)
=
t h a t c = j,.
Since c
bt(j,,).
=
n c p f J ) , c St j,; t h u s b l ( c ) 5 b t ( j o ) ; and we see
Since
i s unique having t h e s e p r o p e r t i e s , we s e e
j,
o
B I B L I O G R A P H I C NOTE.
i n Ey has been d e f i n e d . 4.53
4.52
The n e a r e s t common p r e d e c e s s o r of two e l e m e n t s
See, e.g.,
[55, pp. 316-3171.
THE TREE STRUCTURE OF A FULL CLASS OF SURREAL NUMBERS
Let { F , < , b , Y ) b e a c l a s s of s u r r e a l numbers of h e i g h t Y ( 4 . 0 3 ) .
-
P r o c e e d i n g v e r y much a s we d i d f o r N o i n S e c t i o n 4.50, l e t XEF, and l e t {YEF: y < x a n d b ( y ) < a ] , b ( x ) = B ; t h e n B < Y. Let a < 8, l e t La(x) a n d l e t R,(x)
= {YEF:
y
F ( < , a ) , t h u s ( L a ( x ) , R,(x))
> x and b(y) <
a].
S i n c e B > a , x is not i n
is a Cuesta Dutari c u t i n F ( < , a ) .
{La(x)l Ra(x)}, and n o t e that x a c F ( = , a ) .
Let xu
S i n c e B > a , x f x a'
=
Recall
(4.50) t h a t Conway C24, p.291 c a l l s xa t h e u t h a p p r o x i m a t i o n t o x.
Let us c a l l yeF a p r e d e c e s s o r ( c f . ( 4 . 5 0 ) ) of x , a n d write y
( c f . ( 4 . 5 1 ) ) i f there e x i s t s a
<
s u c h t h a t y = xa
.
I f y is a n predeces-
sor of x we w i l l r e f e r t o x as a s u c c e s s o r ( c f . (4.50)) o f y .
x is t h e
immediate s u c c e s s o r of y i f y c t x , and i f t h e r e is no zcF f o r which y
(1)
C l e a r l y 0 is t h e o n l y element i n F t h a t has no p r e d e c e s s o r . Under < t , F is a p a r t i a l o r d e r c l a s s . Let t h e s e t of p r e d e c e s s o r s of x, pr,(x)
(2)
= (YEF: y < t x ) .
p r ( x ) i s a w e l l - o r d e r e d set under < t . Under < t , F is a tree. We w i l l refer t o
< t as t h e tree order on ( F , < , b , Y l ( c f . 4.53). --
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.53
I n S e c t i o n 4.51 we a l s o d e f i n e d t h e same terms o n Z ,
183 h o w e v e r i t was
To prove t h a t t h e two i d e a s are t h e same,
based o n a d i f f e r e n t d e f i n i t i o n .
i t s u f f i c e s t o prove t h e f o l l o w i n g . LEMMA.
Let X E N O , b ( x ) = Y , and l e t u ( x )
PROOF.
Let u s proceed by i n d u c t i o n o n Y .
b ' ( a ( x 1) %
x s
% %'
8
=
Let a
b'(sg).
<
6.
x ) i f f s > s (resp. s a % a %
<
If %
<
Y , a(x
%
= S
% a
+
(resp. -1 i f f x
sa) i f f s ( a ) =
%
%'
A s we have s e e n ( 4 . 5 0 : 5 ) ,
Using t h e f a c t t h a t ( x )
we see t h e f o l l o w i n g : a ( x B ) ( a ) =
(4.50:0),
<
=
= 3.
+
%
( r e s p . -1;
>
x
a
= x
a
(resp.
thus u(x ) %
=
o
u i s an i s a n o r p h i s m of { N o , < t ) o n t o {E,<,}.
COROLLARY.
THEOREM.
Let { F , < , b , Y ) and { F f , b q , < ' , Y }be f u l l c l a s s e s of s u r r e a l
n u m b e r s of h e i g h t 7 .
Each h a s a t r e e o r d e r , C t and
respectively.
The
unique s u r r e a l monomorphism f of F o n t o F ' ( T h e o r a 4.03) i s a n isomorphism of { F , < t I o n t o { F q , < t l ) .
THE PREDECESSOR CUT REPRESENTATION OF A SURREAL NUMBER
4.54
Let { F , < , b , Y } be a class of s u r r e a l numbers of h e i g h t Y ( 4 . 0 3 ) . L e t X E F ( = , B ) ;then %
B. (0)
<
Y.
I n S e c t i o n 4.54 we d e f i n e d x a , f o r a l l a
<
Thus we have t h e f o l l o w i n g .
x
=
{{x : x
PROOF.
a
a
<
x and a
<
{xu: x
>
x and a
< %I}.
S i n c e { F , < , b , Y } and {Z , < , b ' , Y ) a r e i s o m o r p h i c , i t s u f f i c e s
Y
t o e s t a b l i s h (0) f o r { Z y , < , b T , Y ) . To prove t h i s we need only prove ( 0 ) f o r ( r y , < , b f ) . Let y be t h e number d e f i n e d o n t h e r i g h t hand s i d e of ( 0 ) . S i n c e { Z , < , b ' } i s a f u l l c l a s s of surreal numbers ( 4 . 0 3 : 2 ) , b ' ( y ) S. b ' ( x ) =
184 Let a
B.
>
< 8.
Note t h a t x ( a ) =
+
(resp. -1 i f f x
(resp. y
<
x a ) i f f y ( a ) = + ( r e s p . -1;
B, we see t h a t x
=
y.
y
4.54
Norman L. A l l i n g
xa
>
xu ( r e s p . x
thus y[B = x.
The f o l l o w i n g w i l l be c a l l e d t h e p r e d e c e s s o r
< x and a <
>
x and a
(2)
Let (L,R) be t h e p r e d e c e s s o r c u t r e p r e s e n t a t i o n of x.
a
iff
Since b ' ( y ) 5
of x:
< 61).
((xu: xa
(xa: x
x
cut r e p r e s e n t a t i o n
(1)
B),
<
Then I L 1
+
IS(
= Ib(X)I.
4.60
ALTERNATIVE AXIOMS FOR A FULL CLASS OF SURREAL NUMBERS
W e w i l l now g i v e a n a l t e r n a t i v e s e t of a x i o m s f o r a f u l l c l a s s of
surreal numbers of h e i g h t B, t h e f i r s t set of axioms b e i n g g i v e n i n S e c t i o n
4.03.
If B = On, l e t [O,B)
d e n o t e On.
F i r s t we have t h e f o l l o w i n g O R D E R
AXIOM: (0)
Assume t h a t S is a n o r d e r e d class. We w i l l c a l l t h e f o l l o w i n g t h e BIRTH-ORDER AXIOM:
(B)
Assume t h a t t h e r e e x i s t s a map b of S o n t o [ o , ~ ) . S , < , b , B ) s a t i s f i e s (0) a n d (9). For a
-
b-'([O,aI),
<
8, l e t S ( < , a ) =
and l e t S ( - , a ) = b-' ( a ) .
S i n c e Conway o f t e n c a l l s XENO f q s i m p l e r f tt h a n YENO, i f b ( x )
<
b ( y ) , we
w i l l call t h e n e x t axiom a b o u t [ S , <, b ) t h e SIMPLE DENSITY AXIOM:
<
(SD) F o r a l l a t h a t x,
<
y
B , a n d a l l x,
<
x,.
<
x I E S ( = , a ) , t h e r e exists y c S ( < , a ) s u c h
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.60
105
S i n c e t h e f o l l o w i n g axiom resembles t h e TI -set c o n d i t i o n , a n d i t h a s
5
t h e f u l l n e s s c o n d i t i o n of
i n i t , we w i l l c a l l i t t h e FULL ETA
(4.03:2)
AXIOM:
(FE)
< 6 , and s u b s e t s
Given a L
< (y) <
<
L
R of S ( < , a ) , t h e r e e x i s t s YESs u c h t h a t
R , f o r which b ( y ) S u .
L e t [ S , < , b , B ) s a t i s f y (SD) a n d (FE) ( i n a d d i t i o n of ( 0 )a n d (B)), a n d l e t i t b e c a l l e d a new f u l l class of s u r r e a l numbers of h e i g h t Let J b e a non-empty i n t e r v a l i n S.
LEMMA 0.
ment x o ~ Js u c h t h a t ( b ( x , ) l
<
b(J
-
C l e a r l y t h e r e e x i s t s X,EJ f o r w h i c h b ( x , )
which b ( x , )
T h e r e e x i s t s a n ele-
C l e a r l y x, is u n i q u e .
{x,,]).
-
i t has a least
S i n c e b ( J ) i s a n o n - e m p t y s u b s e t of [ O , B )
PROOF. element a .
=
u.
< x,.
<
Let X , E J f o r
With o u t l o s s of
By t h e T h e S i m p l e D e n s i t y Axiom
(SD), t h e r e e x i s t s y E S ( < , a ) s u c h t h a t x, YEJ. S i n c e b ( y )
u.
Assume, f o r a m o m e n t , t h a t x, b x l .
g e n e r a l i t y , we may assume t h a t x,
B.
<
< x,.
y
S i n c e J is a n i n t e r v a l ,
a , we see t h a t u i s n o t t h e l e a s t e l e m e n t of b ( J ) ; which
i s absurd.
CONWAY'S SIMPLICITY THEOREM. S ( < , a ) be
given.
<
Let a
8,
and let subsets L
There e x i s t s a u n i q u e XES s u c h t h a t L
< (XI <
<
R of
R , for
which b ( x ) i s m i n i m a l .
PROOF. non-empty.
Let J
=
(YES: L
<
[y]
<
R).
By t h e F u l l Eta Axiom ( F E ) , J is
By Lemma 0, J has a u n i q u e e l e m e n t x h a v i n g m i n i m a l b i r t h d a y .
0
Note t h a t s i n c e C o n w a y ' s S i m p l i c i t y Theorem h o l d s , x i s u n i q u e l y
d e t e r m i n e d by L a n d R . d e n o t e x.
F o l l o w i n g Conway, l e t t h e symbol ( L I R I b e u s e d t o
C o n t i n u i n g t o u s e Conway's n o t a t i o n a n d c o n v e n t i o n s C24, p . 41, L
R
L
we w i l l w r i t e x a s ( x ( x 1 , w h e r e [ x 1 L R w i t h y = Iy Iy 1 .
=
R L and [x 1
=
R.
Let y b e i n S,
186
(4)
4.60
Norman L. A l l i n g
x 5 y i f and o n l y i f
(i)
(ii) x
<
Assume t h a t ( i ) h o l d s .
PROOF.
< x
S i m i l a r l y xL
Then x 4 y
5 y , and hence xL
T h u s we s e e t h e f o l l o w i n g : ( * ) yL
L
y for a l l x
<
L
<
. yR, and hence x
y; t h u s ( i i ) h o l d s .
<
y
R
.
Assume now t h a t
Also assume f o r a moment, t h a t (1) i s f a l s e : i . e . , t h a t y < x.
( i i ) holds.
for all x
<
yR f o r a l l y R , and xL
, x R , y L , and y R .
we s e e t h a t b ( y )
<
see t h a t b ( x )
<
b(x).
<
y
< x <
y
R
,
<
a n d (**I xL
y
< x < xR ,
Using t h e Conway S i m p l i c i t y Theorem a n d ( * ) ,
Using t h e Conway S i m p l i c i t y Theorem and ( * * ) , we
b ( y ) ; which i s a b s u r d .
The c o n t r a p o s i t i v e of ( 4 ) i s t h e f o l l o w i n g :
(5)
(i)
x
<
y i f and o n l y i f
( i i ) x 4 yL, for sane y L , or
5 y, f o r sane x
S i n c e x i s t h e s i m p l e s t element
< [ y ) < R f o r which in S f o r which L < [ x < R , b ( x )
o
THEOREM 0.
Every new f u l l c l a s s o f s u r r e a l n u m b e r s of h e i g h t 6,
[ S , < , b , B ] i s a f u l l class of surreal numbers of h e i g h t PROOF.
.
By ( F E ) , t h e r e e x i s t s YESs u c h t h a t L
PROOF.
4 b(y) 5 a .
R
For a l l Conway c u t s (L.R) i n S ( < , a ) , x = [LIRIcS 5 , a ) .
LEMMA 1 .
b(y) 5 a.
xR
B (4.03).
Apply t h e Conway S i m p l i c i t y Theorem and Lemma 1 .
THEOREM 1 .
E v e r y f u l l c l a s s o f s u r r e a l n u m b e r s of h e i g h t
B,
[ S , < , b , B ] , i s a new f u l l class of surreal numbers of h e i g h t 6. PROOF.
-
C l e a r l y (0) a n d (8) h o l d .
of S ( < , a ) , and l e t y
(LIR].
As t o ( F E ) , l e t L
<
R be s u b s e t s
Since [ S , < , b , e ] i s a f u l l Class of surreal
n u m b e r s , y c S ( S , a ) ; t h u s (FE) h o l d s .
Let x
<
y be i n S ( - , a ) .
L R Let ( x ,x 1
and ( y L , y R ) b e ( t i m e l y ) Conway c u t r e p r e s e n t a t i o n s f o r x a n d y ( 4 . 0 2 ) .
BY
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.60
( 5 ) , ( a ) t h e r e e x i s t s yL s u c h t h a t x S y
x
R
5 y.
L b(y )
<
L
, o r ( b ) t h e r e e x i s t s xR s u c h t h a t
Assume t h a t ( a ) h o l d s , t h e n x 5 y
b(y)
holds, then x
<
L
<
b(x), x f yL; and hence x
= a =
L
xR f y; and hence x
<
x
L
R
< y.
< y
L
<
R
Thus (SD) h o l d s .
COMMENTARY ON THE NEW SET OF A X I O M S . l i k e ( 0 ) and ( B ) seem f o r c e d .
R
S i n c e ( y Iy ) i s t i m e l y ,
y.
Assume t h a t ( b
y. R
<
S i n c e ( x Ix ) i s t i m e l y , b ( x )
xR S y.
187
b(x) = a
=
b(y)
o
C l e a r l y s a n e t h i n g v e r y much
The F u l l E t a Axiom ( F E ) , i s a n a p p a r e n t l y
s l i g h t l y weaker v e r s i o n of the o l d f u l l n e s s axiom (4.03:2).
If a n y t h i n g i s
novel about t h i s new s e t of axioms i t would seem t o b e t h e S i m p l e D e n s i t y Axiom, ( S D ) .
I n s e v e r a l of t h e e x p l i c i t c o n s t r u c t i o n s , t h i s c o n d i t i o n
arises very n a t u r a l l y .
tion (4.02:3,ii),
For example, i t does i n t h e C u e s t a D u t a r i c o n s t r u c -
and i n t h e sign-expansion construction (4.52:2).
One c o u l d make a s l i g h t g e n e r a l i z a t i o n of ( B ) by a l l o w i n g b ( S ) t o b e a n y s u b c l a s s C of On.
T h i s c o u l d be c o n v e r t e d t o (81, b y l e t t i n g g be t h e
o r d e r - p r e s e r v i n g map of C o n t o [O, f3)
, and
by d e f i n i n g a new b i r t h - o r d e r map
b f t o be g - b .
S i n c e t h e F u l l Eta Axiom ( F E ) , d e a l s w i t h t w o t h i n g s , i t c o u l d b e b r o k e n u p i n t o t h e f o l l o w i n g two t h e f o l l o w i n g two axioms: the ETA AXIOM ( E ) t h a t asserts t h a t s u c h a n element y e x i s t s ; and t h e FULLNESS A X I O M ( F )
(E) i s a d e n s i t y ( F ) t e l l s u s some-
t h a t asserts t h a t s u c h a y e x i s t s f o r which b ( y ) 5 a.
c o n d i t i o n and is a n a l o g u e s w i t h a n rl -set c o n d i t i o n . 5 t h i n g a b o u t e l e m e n t s of h e i g h t 2 a i n t h e t r e e s t r u c t u r e of [ S , < , b , B } . a l s o c o n t r o l s how l a r g e b ( S(<,a).
{ - I- 1 )
<
On t h e o t h e r hand, (SD) d e a l s w i t h t h e p o i n t s of h e i g h t
between x, and x , .
C l e a r l y (FE)
=
=
1.
L e t u s s e e i f i t s a t i s f i e s (FE).
t h e empty s e t , L
=
0
=
R.
Let b map
Clearly {S,<,b,f3} satisfies ( 0 ) a n d ( B ) . Let a
Let YES; t h e n L
s e e t h a t {S,<,b,B] s a t i s f i e s (FE).
a in S
(F) + ( E ) .
Let S be any o r d e r e d c l a s s , s u c h t h a t IS1 2 2.
EXAMPLE 0.
a l l of S t o OEOn, and l e t 6
It
is, when a p p l i e d t o a Conway c u t i n
<
f3;
<
then a
{y]
<
R.
=
0.
S i n c e S(<,O)
Since S
By a s s u m p t i o n , I S 1 2 2.
=
S(S,O),
Let x,
<
is
we
x 1 be
188
Norman L. A l l i n g
i n S(=,O). (B)
t
Since t h e r e i s no p o i n t y i n S ( < , O ) , (SD) f a i l s .
PROPOSITION 0 .
If { S , < , b , B } s a t i s f i e s
IS(=,O)l Let S
EXAMPLE 1 .
B = 2.
+
-
(B),
(O),
a n d (SD); a n d i f S
= 1.
[ c , d , e, f , g ) , ordered l e x i c o g r a p h i c a l l y .
Let
Let b ( e ) = 0 , and l e t t h e b i r t h d a y map b map c , d , f , and g t o 1 .
Clearly { S , < , b , 2 ] s a t i s f i e s ( 0 ) and (B).
<
(y]
Note (FE) t a l k s a b o u t e x i s t e n c e
L e t a ( i n axiom ( F E ) ) be 0; t h e n L and R are empty,
not uniqueness.
<
Hence ( 0 )
(FE) h o l d and (SD) f a i l s i n t h i s Example.
is non-empty, t h e n
L
4.60
R f o r a l l YES, (FE) h o l d s a t t h i s l e v e l .
Now l e t a
=
1.
Since Having
d e a l t w i t h t h e Conway c u t ( 0 , 0 ) , we h a v e o n l y t w o more Conway c u t s of S(<,1)
t o investigate:
( i ) ( 0 , i e ) ) and ( i i ) ( [ e l , 0 ) .
p r e s e n c e of c and d s u f f i c e t o s a t i s f y (FE). f and g s u f f i c e t o s a t i s f y (FE); t h u s {S,<,b,2)
I n case ( i i
:n c a s e ( I ) , t h e
, the
satisfies (0)
S i n c e t h e r e i s no element i n S between c and d c S ( - , l ) ,
+
p r e s e n c e of
fB)
+
(FE).
S , < , b , B ] does not
s a t i s f y (SD). Thus from Examples 0 and 1 we see t h a t we have proved t h e f o l l o w i n g :
(SD) i s independent of (0) t (B) + ( F E ) .
THEOREM 2.
Let S be any o r d e r e d c l a s s .
EXAMPLE 2 .
C l e a r l y ( S , < , b , B ] s a t i s f i e s (0) + (B).
o n t o s m o C0,B). Since
Let b be any i n j e c t i o n of S
IS(=,a)l
1 , for all a
=
< 6;
(SD) i s v a c u o u s ,
t h u s ( 0 ) + (B) + (SD) h o l d , f o r t h i s
example. EXAMPLE 3 .
on S.
Let
B
2
3, let S
-
[O,B),
<
B.
and l e t b be t h e i d e n t i t y map
-
A s we saw above, [ S , < , b , B J s a t i s f i e s (0) + (B) + (SD). S i n c e B 2 3,
there exist a
< B
such t h a t a
+
2
Let L
(FE) f a i l s f o r t h i s Conway c u t i n S(<,a + 2 ) . THEOREM 3.
{ a ] and l e t R = { a + 11.
Thus we have t h e f o l l o w i n g :
(FE) i s independent of (0) + (B) + (SD).
BIBLIOGRAPHIC NOTE.
The m o t i v a t i o n f o r t h i s S e c t i o n is d e s c r i b e d a t
t h e e n d of S e c t i o n 4.03.
The Simple Density Axiom grew out of s u c h t h i n g s
as the proof of (5) i n C6, p. 2441.
The i d e a o f u s i n g i t as a n axiom is
4.60
189
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
d u e t o Timothy A . S w a r t z , who U n i v e r s i t y of R o c h e s t e r .
- at
- was
t h e time
Using (0) + (B)
an u n d e r g r a d u a t e a t t h e
(SD), S w a r t z w a s a b l e t o p r o v e
+
an i n t e r e s t i n g embedding theorem f o r s u c h s y s t e m s ( d u r i n g t h e f a l l of
1985).
When h e a d d e d axioms ( 0 ) a n d ( 2 ) of [71 t o h i s s y s t e m s , i t became
e q u i v a l e n t t o (0) + ( 8 ) 4.61
+
(SD)
+
(FE).
CONWAY CUTS, ORDERED BY EXTENSION, AND CUESTA DUTARI CUTS
L e t X be a n o r d e r e d s e t .
Recall (4.02) t h a t C D ( X ) d e n o t e s t h e s e t of Recall a l s o t h a t we p u t a n o r d e r o n
a l l Cuesta Dutari c u t s i n X (1.20).
C D ( X ) i n S e c t i o n 4.02, u n d e r which i t i s a n o r d e r e d s e t ( 4 . 0 2 : 2 ) .
L e t CC(X) b e t h e set of a l l Conway c u t s i n X (1.20). be i n C C ( X ) .
(L,,R,)
L e t (L,,R,)
( L 1 , R I ) w i l l be c a l l e d a n e x t e n s i o n o f ( L , , R , ) ,
and if Lo
is a s u b s e t of L , and R,
i s a s u b s e t of R , .
o r d e r s e t under e x t e n s i o n .
C l e a r l y t h e u n i o n of any non-empty s u b s e t T o f
Clearly
C C ( X ) is a p a r t i a l
C C ( X ) , w h i c h i s o r d e r e d by e x t e n s i o n , is a Conway c u t i n X t h a t i s a n
By Z o r n ' s Lemma ( s e e e . g . ,
e x t e n s i o n o f a l l t h e Conway c u t s i n T .
we have p r o v e d p a r t ( i ) of t h e f o l l o w i n g :
p.33]), (0)
C50,
(1)
Every (L,RlsCC(X) h a s a maximal e x t e n s i o n (L*,R*) i n CC(X).
( i i ) C D ( X ) i s t h e set of a l l maximal e l e m e n t (L*,R*) of C C ( X ) .
PROOF.
I n o r d e r t o p r o v e ( i i ) , assume f o r a moment, t h a t ( L * , R * )
not a Cuesta Dutari cut i n X. u n i o n of L* and R*.
x
>
y , t h e n l e t R**
(L**,R**)
-
If ( i ) t h e r e exists YEL s u c h t h a t x
b e t h e u n i o n of L* and {x} a n d l e t R** e x t e n s i o n of ( L * , R * ) ,
which i s a b s u r d .
R*.
<
y, t h e n l e t L**
C l e a r l y (L**,R**)
is a proper extension of (L*,R*),
t h a t ( i ) a n d ( i i ) d o n o t h o l d ; t h e n L*
-
i s a proper
If ( i i ) t h e r e e x i s t s Y E R s u c h t h a t
b e t h e u n i o n of R* a n d { X I a n d l e t L**
-
is
Then t h e r e e x i s t s x i n X which i s n o t i n t h e
<
{x}
which i s a b s u r d .
< R*.
L*.
Clearly
Assume now
Let L** be t h e u n i o n of
L* and { x } a n d l e t R** R*. Then (L**,R**) i s a p r o p e r e x t e n s i o n of (L*,R*), which i s a b s u r d . C o n v e r s e l y , c l e a r l y e v e r y C u e s t a D u t a r i c u t i n X i s a maximal Conway c u t in X .
Norman L. A l l i n g
190
Let (L,R)cCC(X). a n d l e t R-
letL
(1)
+
=
-
X
Let L- = {xEX: t h e r e exists Y E L s u c h t h a t x 6 y } ,
Let R +
L-.
4.61
(xcX:
=
t h e r e e x i s t s ycR s u c h t h a t x L y } , a n d
+
= X - R .
- -
(i) (L ,R
+
a n d (L , R
( i i ) Let ( L * , R * )
+
a r e maximal extensions of (L,R) i n C C ( X ) .
b e a maximal e x t e n s i o n of (L,R) i n C C ( X ) .
t h e l i n e a l o r d e r i n g o n CD(X), (L-,R-)
Then, i n
<, ( L t , R + ) .
I (L*,R*)
PROOF.
C l e a r l y L- is t h e smallest l o w e r - s a t u r a t e d s u b s e t of X t h a t
contains L.
S i n c e L* is a l o w e r - s a t u r a t e d s u b s e t of X t h a t c o n t a i n s L , L-
is a s u b s e t of L*; hence (L-,R-) smallest u p p e r - s a t u r a t e d
5 (L*,R*) ( 4 . 0 2 : 1 , i i i ) .
C l e a r l y R + is t h e
s u b s e t of X t h a t c o n t a i n s R .
u p p e r - s a t u r a t e d s u b s e t of X t h a t contains R , (L*,R*) S (L+,R+) ( 4 . 0 2 : 1 , i i I ) .
Rt
S i n c e R* i s a n
i s a s u b s e t of R * ;
hence
o
Let ( S , < , b , B ) s a t i s f y ( 0 ) + ( B )
+
( F E ) (4.60), l e t a
<
6 and l e t
(L*,R*) b e I n CD(S(< , a ) ) . (2)
<
If yES(S,a) s u c h t h a t L*
PROOF.
R* t h e n b ( y )
a.
o
For a l l ( L , R ) E C C ( S ( < , a ) t h e r e e x i s t s y s S ( - , a )
PROOF.
=
S i n c e (L*,R*) i s a Cuesta Dutari c u t i n S ( < , a ) , t h e u n i o n of
L* and R* is S ( < , u ) ; t h u s b ( y ) L a .
(3)
<
{y]
such t h a t L
By ( 0 1 , ( L , R ) h a s a maximal e x t e n s i o n ( L * , R * )
w h i c h i s a n e l e m e n t of C D ( X ) . which t h e f o l l o w i n g h o l d s : L*
i s a s u b s e t of R*,
<
<
{y}
<
R.
i n CC(S(<,a)),
By (FE) a n d ( 2 ) , t h e r e e x i s t s y ~ S ( = , a )f o r {y]
<
R*.
we c o n c l u d e t h a t L < { y )
S i n c e L is a s u b s e t of L * a n d R
<
R.
I N T R O D U C T I O N T O THE SURREAL NUMBER F I E L D No
4.00
SURREAL NUMBERS
In J . H . Conway's book, On Numbers and Games C241, t h e b a s i c c o n s t r u c t i o n o f numbers i s t h e f o l l o w i n g : (0)
I f L a n d R a r e two s e t s of n u m b e r s , a n d i f no member of L is t any
member of R , t h e n ( L I R } i s a number.
A l l numbers are c o n s t r u c t e d i n
t h i s way [ 2 4 , p . 41.
How t h e n d o e s o n e g e t s t a r t e d c o n s t r u c t i n g n u m b e r s u s i n g C o n w a y ' s
construction?
The empty s e t is a s e t of numbers which we know e ists.
L and R be empty. (01,
Note t h a t no member of L is 2 any member of R
[LIR] i s a number.
Let u s c a l l t h i s number 0 .
Let
t h u s , by
Conway C24, p . 41
adopted t h e following n o t a t i o n a l convention:
If x
=
( L I R } w e w r i t e xL f o r a t y p i c a l member of L , a n d
t y p i c a l member of R ; t h u s x e, f , e, f ,
... ) , ... 1 .
option _---
of x.
we mean t h a t x
L R {x Ix 1 .
= =
If we write x = ( a , b , c ,
... I d ,
and R
=
Id,
x L i s c a l l e d a l e f t o p t i o n of x , a n d x R i s c a l l e d a r i g h t If L ( r e s p . R ) i s empty, we may i n d i c a t e t h i s by l e a v i n g t h e
p l a c e where L ( r e s p . R ) would a p p e a r b l a n k . =
... }
( L I R ) , where L = [ a , b , c ,
xR for a
[I).
Hence ( ( 0 1 l a }
=
([O}
I],
and 0
I n K n u t h ' s m a t h e m a t i c a l n o v e l l a o n s u r r e a l n u m b e r s [52] he u s e s s l i g h t l y d i f f e r e n t n o t a t i o n i n t h e body of t h e t e x t . writes x
=
(X
X 1. L' R
For example, Knuth
We have c h o s e n t o a d o p t most of Conway's n o t a t i o n .
is n o t o n l y v e r y compact a n d e a s y t o u s e , b u t i t s u g g e s t s
feels
-
t h e r i g h t way t o t h i n k a b o u t t h e s u b j e c t .
-
It
t h e author
118
Norman L . A l l i n g
4.00
Conway t h e n d e f i n e s o r d e r between numbers a s f o l l o w s :
(1)
( 1 ) x 6 y i f and O n l y i f ( i i ) no y R 2 x and x S no x
L
.
Note t h a t ( 1 , i ) is a s t a t e m e n t about n u m b e r s , a n d t h a t ( 1 , i i ) i s a s t a t e m e n t a b o u t s e t s of n u m b e r s .
Conway d e s c r i b e s 0 as t h e t l s i m p l e s t t t
number t h a t was ttborn on day 0" [24, p.
111.
T h i s seems f i t t i n g i n d e e d ,
{*I.]. The numbers 1 = Conway s a y s of them t h a t
s i n c e i t i s b u i l t up fran t h e empty s e t u s i n g o n l y
{Ol) and -1
-
(10) a r e a l i t t l e more c o m p l e x .
t h e y were e a c h " b o r n o n d a y 1 " [ 2 4 , verify t h a t ( 1 , i i ) holds. and t h a t ( b ) 0 < j l ) ,
p . 111.
To s e e t h a t 0 2 1, w e m u s t
To do t h a t i t s u f f i c e s t o show t h a t ( a ) lo)
<
0,
S i n c e ( a ) and ( b ) a r e both t r u e , we see t h a t 0 2 1 .
Conway goes on t o make t h e f o l l o w i n g d e f i n i t i o n s : (2)
(i)
y 2 x i f and o n l y i f x 6 y ,
x = y i f and o n l y i f x 6 y and y S x , x < y i f and o n l y i f x 6 y and i f x + y , and ( i v ) y > x i f and o n l y i f x < y.
(ii)
(iii)
Perhaps t h e o n l y s u r p r i s e i s t h a t ( 2 , i i ) is a definition.
Conway
y, -x,
and xy
ends h i s s h o r t l i s t of remarkable s t a t e m e n t s by d e f i n i n g x
+
i n d u c t i v e l y f o r all numbers x and y as f o l l o w s . L R R I x L + y , x + y Ix + y , x + y 1 .
(3)
x + Y
(4)
-x = (-x
(5)
x y - ( x y + x y
=
R
L
1-x I .
L
L x Y
+
L L R R R R - x y , x y + x y - x y J R L R R L X Y - x y , x y + xy - x Ry L ) *
L
A t f i r s t g l a n c e t h e s e d e f i n i t i o n s may l o o k c i r c u l a r .
Note, f o r
example, i n ( 4 ) i f we know how t o form t h e n e g a t i v e of a l l t h e o p t i o n s of x used t o d e f i n e x , t h e n (4) i s n o n - c i r c u l a r .
S i m i l a r l y , i n ( 3 ) i f we c a n
p r e f o r m a l l t h e i n d i c a t e d a d d i t i o n s among o p t i o n s of x and y and y and x
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.00
t h e n w e c a n compute t h e s e t s on t h e l e f t i n ( 3 ) .
119
The same may b e s a i d of
(5). Conway a l s o showed C24, p p . 16-17] t h a t , i f x
(6)
L
<
[XI
4.01
<
=
[LIR] then
R.
CONWAY'S CONSTRUCTION
C o n w a y ' s c o n s t r u c t i o n a n d most of t h e p r o o f s h e g i v e s , a r e b y induction.
One of Conway's v e r y u s e f u l i d e a s i s t h a t of t h e b i r t h o r d e r of
s u r r e a l numbers.
As we w i l l s e e t h i s i s o n e of t h e most i m p o r t a n t Given aEOn, Conway d e f i n e s
p r o p e r t i e s of surreal numbers. (0)
0 as t h e s e t of a l l numbers b o r n before day a , a M
as t h e s e t of a l l numbers b o r n o n o r b e f o r e day a t a n d
N
as t h e s e t of a l l numbers b o r n o n d a y a C24, p . 291.
<
S i n c e t h e r e are no o r d i n a l s a
0, 0 ,
=
The s e t s L a n d R which
0.
a r e a v a i l a b l e t o make n u m b e r s o n day 0 a r e v e r y few: o n l y t h e empty s e t . Thus t h e o n l y Conway c u t ( 1 . 2 0 ) i n 0 is ( 0 , 0 ) .
We c a n t h i n k of a number as
a n e q u i v a l e n c e c l a s s o f Conway c u t s ( L , R ) i n NO, u n d e r t h e e q u i v a l e n c e relation (4.00:2,ii). (4.00:O).
Thus we s e e t h a t M ,
=
No
=
(01,
0 being
Ill
Now t h a t t h e n u m b e r s o n day 0 h a v e been c r e a t e d , t h e c a l e n d a r
a d v a n c e s , as i t were, a day t o day 1 .
On d a y 1 t h e r e a r e two s e t s o f n u m b e r s :
t h e e m p t y s e t 0 and [ O } .
Thus t h e r e a r e two Conway c u t s i n 0 , , ({01,0) a n d ( 0 , { 0 1 ) .
we w i l l d e f i n e t o be [Ol], and - 1 , t h e elements i n N , .
C l e a r l y 0,
=
Thus 1 , which
which we w i l l d e f i n e t o b e
M,
= [0,+11; thus
(lo],
are a l l
we a r e r e a d y t o b e g i n
t o c o n s i d e r t h e numbers c r e a t e d o n day 2. Conway d e f i n e s t h e c l a s s of a l l n u m b e r s c r e a t e d i n t h i s way a s No [24,
(I)
p . 41.
He shows [ 2 4 , p . 301 t h a t
g i v e n any xcNo t h e r e e x i s t s a u n i q u e aEOn s u c h t h a t X E N ~ .
Norman L. A l l i n g
120
4.01
Let a be c a l l e d t h e b i r t h d a y of x , and l e t i t be denoted by b ( x ) . w i l l c a l l b the b i r t h order function.
y if b(x)
<
Conway writes t h a t x i s s i m p l e r t h a n
Since On i s well-ordered t h e p h r a s e t h a t sane element i s
b(y).
" t h e s i m p l e s t element such t h a t
..
.I1
makes s e n s e .
Conway g i v e s t h e follow-
i n g very i l l u m i n a t i n g d e s c r i p t i o n of t h e c r e a t i o n p r o c e s s . numbers w i t h L
<
No, I L I R I
R in
=
G i v e n s e t s of
x
i s t h e s i m p l e s t element of No s u c h t h a t L
(2)
We
< {XI <
R.
Conway r e f e r s t o t h i s as "The S i m p l i c i t y Theorem" C24, Theorem 1 1 , p . 231.
Henceforth we w i l l r e f e r t o (2) as ttConwayls S i m p l i c i t y Theorem".
is a v i t a l i n g r e d i e n t i n many of
we w i l l s e e , Conway's S i m p l i c i t y Theorem
our c o n s i d e r a t i o n s .
Note a l s o t h a t P =
As
( Na
aeon
i s a p a r t i t i o n of No.
T h i s p a r t i t i o n can a l s o be g i v e n by g i v e n a map b , which maps each element
t o t h e index a .
in N
Given b , t h e n N
=
b
-1
( a ) . b c a n b e t h o u g h t of a s
a s s i g n i n g t h e b i r t h o r d e r t o t h e e l e m e n t s of No. more d e t a i l s . ) day 1 .
-2,
Thus 0 i s born f i r s t , on day 0.
-1/2,
( S e e [5, 384-3851 f o r
1 and -1 are born n e x t , on
1/2, and 2 are born n e x t , on d a y 2, e t c .
One of t h e t h i n g s t h a t Conway had t o d e a l w i t h was t h e f o l l o w i n g :
"A
m o s t i m p o r t a n t comment whose l o g i c a l e f f e c t s w i l l be d i s c u s s e d l a t e r i s that
the n o t a t i o n of
equality
is a
defined relation.
Thus a p p a r e n t l y
d i f f e r e n t d e f i n i t i o n s w i l l produce t h e same number, and we m u s t d i s t i n g u i s h
form
{LIR] of a number a n d t h e number i t s e l f . I t C24, p . 51 U s i n g ( 2 ) we c a n g i v e a d r a m a t i c i l l u s t r a t i o n of t h i s , n a m e l y t h e
between t h e following:
(3)
Let L and R be s u b s e t s of No s u c h t h a t L
< {O] <
R ; then
0 = ILIRf.
S i n c e t h e class of a l l o r d i n a l numbers On i s , i n a very n a t u r a l way, a s u b c l a s s of No C24, pp. 27-281, we see t h a t (4)
No i s a proper class.
One of t h e n a t u r a l t h i n g s t o d o , in t r y i n g t o c o n s t r u c t No i n a more c l a s s i c a l manner w i t h i n a conventional s e t t h e o r y , would be t o c o n s i d e r t h e
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.01
121
v a r i o u s Conway c u t s f r m which x c a n be d e f i n e d , s a y a s o r d e r e d p a i r s , a n d then pass t o equivalence c l a s s e s .
The d i f f i c u l t y of d o i n g t h i s can be s e e n
i n ( 3 1 , s i n c e 0 h a s a p r o p e r c l a s s of Conway c u t s ( L , R ) s u c h t h a t 0
F u r t h e r , ( 3 ) i s n o t a n i s o l a t e d o c c u r r e n c e , as we w i l l now show.
{LIR).
Let x = I L ( R 1 , where (L,R) i s
>
t h a t f o r a l l f3cOn w i t h 8 ( r e s p . R ) union [ 8 } .
(5)
=
x
=
[L (R
$
6
a Conway c u t i n No.
a, (-6)
<
L and R
<
There exists acOn s u c h
(8).
Let L
8
(resp R ) be L
B
Then
1, f o r each B > a .
Hence. we see ( 5 ) t h a t e a c h XENO has a p r o p e r c l a s s o f Conway c u t s (L',R')
such t h a t {L'IR') 4.02
=
x.
THE CUESTA DUTARI CONSTRUCTION OF No
Let T be an o r d e r e d s e t .
R e c a l l (1.20) t h a t a C u e s t a Dutari c u t i n T
is a p a i r of s u b s e t s (L,R) of T , s u c h t h a t ( i ) L L and R i s T .
<
R and ( i i ) t h e u n i o n o f
Let CD(T) = { C u e s t a D u t a r i c u t s i n T I .
S i n c e ( 0 , T ) and
(T, 0) a r e Cuesta Dutari cuts, (0)
C D ( T ) i s never empty.
Assume t h a t M is an o r d e r e d set which c o n t a i n s T , s u c h t h a t t h e o r d e r
on M i n d u c e s t h e o r i g i n a l o r d e r on T: i . e . , (M,6) (1.10).
Let (L,R)ECD(T). X E M w i l l be s a i d t o
i s a n e x t e n s i o n of (T,S)
rill (L,R)
in M if L
< [XI <
R.
Let x ( T ) , t h e C u e s t a D u t a r i c o m p l e t i o n o f T , b e t h e u n i o n of T a n d CD(T), o r d e r e d a s f o l l o w s : (1)
(i)
i f x and y a r e i n T , l e t them be o r d e r e d as t h e y were i n T ;
( i i ) i f XET and y (iii) i f x
=
=
(L,R), y
(L,R)€CD(T). =
s u b s e t of L ' . (2)
x ( T ) i s an o r d e r e d s e t .
x
<
y i f X E L , and y
(L',R')&CD(T), t h e n x
<
< x
i f XER;
y i f L is a p r o p e r
Norman L. A l l i n g
122
Let x , y , and z be i n x ( T ) , w i t h x < y a n d y
PROOF.
that x
(3)
< z
4.02
< z.
c o n s i d e r t h e e i g h t e a s i l y proven c a s e s s e p a r a t e l y .
< t, in < c, i n
For all t ,
(i)
( i i ) For a l l c,
To show
o
T , t h e r e e x i s t s CECD(T) with t o < c C D ( T ) , t h e r e exists
tET
< t,. < t < C,.
with c,
( i i i ) ( 0 , T ) i s t h e l e a s t and ( T , 0 ) i s a g r e a t e s t element of x ( T ) .
PROOF.
Let t o
< c <
then t o
Let t c L ,
-
be elements i n T.
< c,.
w i t h c,
(L,,R,)&CD(T),
L,.
< tl,
t,, establishing L o ; then c ,
(i).
Let c
Let c,
=
= ((-m,to],(to,+-));
and c,
(L,,Ro)
=
Then, by d e f i n i t i o n , Lo is a proper s u b s e t of
< t <
c,, establishing (ii).
I f T i s empty
t h e n x ( T ) h a s o n l y o n e p o i n t i n i t , namely ( 0 , 0 ) ; e s t a b l i s h i n g ( i i i ) i n case T
=
0.
Assume now t h a t T is non-empty.
For a n y ~ E T ,( 0 , T )
< t <
(T,0). (4)
c
=
f i l l s t h e Cuesta D u t a r i c u t ( L , R ) i n x ( T ) .
(L,R)ECD(T),
PROOF. NOTE.
L
R
By d e f i n i t i o n , f o r a l l x EL and a l l x ER, xL
<
<
c
x
R
.
o
Even though each Dedekind c u t i n T i s a C u e s t a D u t a r i c u t i n
T , t h e C u e s t a Dutari completion x ( T ) of T p l a y s a very d i f f e r e n t r o l e t h a n
Dedekind used gaps i n t h e r a t i o n a l
does t h e Dedekind completion 6(T) of T .
numbers Q t o d e f i n e i r r a t i o n a l n u m b e r s , a n d t h u s d e f i n e R.
Since R is
Dedekind-complete i t has no gaps; t h u s t h e Dedekind completion of R, i s R. S i n c e C D ( T ) i s n e v e r empty (01, t h e C u e s t a D u t a r i completion x ( T ) of T always c o n t a i n s T as a proper s u b s e t .
I n p a r t i c u l a r , R is a p r o p e r s u b s e t
of x ( R ) . Let T o be t h e empty s e t .
d e f i n e d T , t o be x(T,,),
Cuesta D u t a r i [251 a n d H a r z h e i m C431 t h e n
and noted t h a t T , = [ ( 0 , 0 ) ) .
Assume t h a t f o r sane
BEOn t h a t a f a m i l y (Ta)a
then define T
d i n a l , than d e f i n e T
set.
t o be x ( T , ) ; 8 t o be t h e union of
6 F u r t h e r , Harzheim proved t h a t
and i f 8 is a non-zero l i m i t o r -
Note t h a t T
WO
is a n n o -
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.02
(5)
T
is a n
IRI
+
<
-set 143, p.1191.
5
Let L a n d R be s u b s e t s of T
PROOF. ILI
q
WE'
=
(avcT
6:
a r e s u b s e t s of T
5
C'E
and ( i i )
s u c h t h a t L and R are s u b s e t s of T Clearly L'
Let R' be d e f i n e d t o be T
6'
such t h a t L Then
(L',Rf)eCD(T6).
(6)
< w5
< R,
is r e g u l a r and s i n c e
5
t h e r e e x i s t s acL s u c h t h a t a t 5 a } .
s a t u r a t e d s u b s e t of T
i n Tw 5
f o r which ( i ) L
S i n c e , by a s s u m p t i o n ( 1 . O O : O ) , w
( i i ) holds, there e x i s t s 6
L'
123
<
R,
T6+,, and L
6
-
L'.
is a lower
S i n c e L and R
R i s a s u b s e t of R I .
< {cl} <
Let
6'
Let c '
=
By c o n s t r u c t i o n , c' is
R.
. By i n d u c t i o n
is defined.
F o l l o w i n g Conway [24,
p . 291, l e t Oa ( t h e s e t o f " o l d n u m b e r s " ) b e
d e f i n e d t o be Ta, l e t N
a
( t h e set of Ifnew numbers") b e T
a+l
-
Ta, a n d l e t
Mu ( t h e s e t of "made numbersTf)b e Ta + l ' N o t e t h a t Ma i s t h e u n i o n o f 0 a a n d Nu C24, p . 291. F i n a l l y , l e t No be d e f i n e d t o be t h e u n i o n of (Oa)aEOn [24, p . 41.
See a l s o C60, p . 491 o n s u c h u n i o n s .
N o t e t h a t No i s a l s o t h e u n i o n o f (M M
a
and t h a t i f a
a adn'
<
6, t h e n
i s a p r o p e r s u b s e t of M6; t.hus g i v e n X E N O , t h e r e e x i s t s a l e a s t BEOn
s u c h t h a t XEM
6
(= T6+l).
t h e b i r t h d a y of x.
Let b ( x ) d e f i n e d t o be f3, a n d l e t i t be c a l l e d
b w i l l be c a l l e d t h e b i r t h - o r d e r
t h a t b ( x ) is t h e u n i q u e e l e m e n t i n On s u c h t h a t X E N , , ( ~ ) ( - Tg+l C l e a r l y f o r each acOn, b - l ( [ O , a ) ) Note t h a t (NaIaeOn
=
Oa, b - l ( [ O , a ] )
i s a p a r t i t i o n o f No.
=
Note
f u n c t i o n o n No.
Ma, a n d b - l ( { a } )
-
T6). =
N
a'
I n [5, p p . 384-3851 we c a l l e d
s u c h a p a r t i t i o n a Conway p a r t i t i o n , a n d c o n s i d e r e d s o m e o f t h e i r p r o p e r ties.
The f o l l o w i n g i s of g r e a t i m p o r t a n c e .
Norman L . A l l i n g
124
CONWAY'S SIMPLICITY THEOREM. L
<
Let I
R.
=
Let L and R be s u b s e t s of No f o r which
< {y) <
(YENO: L
4.02
R).
Then ( i ) I is non-empty, a n d ( i i )
t h e r e e x i s t s a u n i q u e x c I s u c h t h a t b ( x ) < b ( y ) , f o r all Y E I[24,
is an
11
5
i s r e g u l a r , a n d f o r which U is a s u b s e t of 0
-set (51, I i n t e r s e c t Ow
5
-
least element.
b ( J ) i s a n o n - e m p t y s u b c l a s s of On; l e t 5 be i t s
Let x a n d x' be i n J s u c h t h a t b ( x ) = B
f o r a moment, t h a t x
t <
XI.
T h e r e e x i s t s a u n i q u e XEJs u c h
{XI implies b(x) < b(y).
PROOF ( o f ( 7 ) .
T5.
Since
is a non-empty.
Let J be a n o n - m p t y i n t e r v a l i n No,
-
.
5
that y d
T6+,
Since U i s a s e t t h e r e e x i s t s
Let U be t h e u n i o n of L and R .
a CEOn s u c h t h a t w
(7)
(Cf.
Theorem 1 1 , p . 231.) PROOF.
0
{XI.
<
=
b(t)
6'
<
Assume
S i n c e b ( x ) = 6 = b ( x ' ) , x a n d x ' are i n N
X I .
By ( 3 , i i ) we know t h a t t h e r e i s a n e l e m e n t t E T
Since tcT
b(x').
6'
6
=
such t h a t x
<
S i n c e J is a n i n t e r v a l , t i s i n J ; which i s
6.
absurd, proving ( 7 ) . Apply ( 7 ) t o I , a n d t h e Conway S i m p l i c i t y Theoran i s p r o v e d .
o
Let ( L , R ) b e a Conway c u t i n No ( 1 . 2 0 ) ; i . e . , l e t L and R b e s u b s e t s of No f o r w h i c h L
<
R.
By C o n w a y ' s S i m p l i c i t y T h e o r e m , t h e r e e x i s t s a
u n i q u e X E N O of minimal b i r t h d a y s u c h t h a t L
<
{x)
<
R.
Since x is uniquely
d e t e r m i n e d by L a n d R , l e t us f o l l o w Conway [24, p . 41 a n d use t h e symbol (LIR) t o d e n o t e x .
We w i l l a l s o s a y t h a t t h e Conway c u t ( L , R ) r e p r e s e n t s
x , a n d t h a t ( L , R ) i s a Conway cut r e p r e s e n t a t i o n of x.
(8)
Given any Conway c u t ( L , R ) a n d ( L ' , R ' )
< R)
=
PROOF,
(yeNo: L'
<
(y)
<
R'),
-
of No f o r which (YENO: L
t h e n (LIR)
Apply Conway's S i m p l i c i t y Theorem.
-
x
< (y}
(L'IR').
o
We w i l l s a y t h a t L a n d L ' ( r e s p . R and R ' ) a r e m u t u a l l y c o f i n a l
4.02
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
125
( r e s p . m u t u a l l y c o i n i t i a l ) i f f o r a l l aEL t h e r e e x i s t s a'EL' s u c h t h a t a 5 a ' , a n d f o r a l l a f E L * t h e r e e x i s t s acL s u c h t h a t a' 5 a ( r e s p . i f f o r a l l CER t h e r e exists c ' E R '
such t h a t c 5 c').
such t h a t c' 2 c , and f o r a l l c'ER'
t h e r e e x i s t s CER
We w i l l s a y t h a t t h e Conway c u t s ( L , R ) a n d ( L f , R ' ) a r e
equivalent i f L and L ' are m u t u a l l y c o f i n a l and R and R ' Clearly (8) implies t h e following:
coinitial. (9)
If ( L , R ) a n d ( L ' , R ' )
EXAMPLE 0 . and R '
=
are mutually
Let L
a r e equivalent, then {L'IR'] =
and R
{-I}
11/21; t h e n 0 = { L f l R f ] .
cofinal.
=
[ l ] ; then 0
=
=
[LIR}.
(LIR].
C l e a r l y L and L '
Let L '
[-21
are not mutually
are not mutually c o i n i t i a l .
F u r t h e r , R and R '
=
Thus [LIR]
=
[ L f l R t } , and yet ( L , R ) and ( L f , R ' ) are not equivalent. A Conway c u t r e p r e s e n t a t i o n ( L , R ) of x w i l l be c a l l e d t i m e l y i f L and
R are s u b s e t s of 0
b ( x ) ( = Tb(x))*
Note t h a t t h e Conway c u t r e p r e s e n t a t i o n s
of 0 i n E x a m p l e 0 a r e n o t t i m e l y .
o
s e n t a t i o n of (10)
Also n o t e t h a t t h e o n l y t i m e l y r e p r e -
i s 10101.
Each XENO has a t i m e l y r e p r e s e n t a t i o n . PROOF.
Let b ( x ) = 8 ; t h e n XEN
a n element ( L , R ) i n CD(T ) . 0
8'
i.e.,
x is
in T
B+1
- TO.
Hence x i s
(L,R) i s t h e n a t i m e l y r e p r e s e n t a t i o n of x.
Another way t o d e s c r i b e s u c h a Conway c u t ( L , R ) i s t o n o t e t h a t L [ ~ E O ~ ( y~ <) x] : and that R ( ( Y E O ~ ( ~ ) y:
s e n t a t i o-n _------
=
[y€Ob(,):
y
>
XI [ 2 4 , p . 291.
x [24, p . 291.
=
Let us d e f i n e
< X ) , [ Y E O ~ ( ~y ) >: X I ) t o b e t h e C u e s t a D u t a r i
of
o
cut
repre-
Note also t h a t t h e Cuesta Dutari cut
r e p r e s e n t a t i o n of x is a C u e s t a D u t a r i c u t i n 0
S i n c e we h a v e b u i l t b(x)' up No, i n t h i s s e c t i o n , u s i n g C u e s t a Dutari c u t s have t h e f o l l o w i n g r e s u l t .
(12)
Let ( L , R ) a n d ( L ' , R f 1 b e t i m e l y Conway c u t r e p r e s e n t a t i o n s i n No,
Norman L. A l l i n g
126
4.02
such t h a t [LIR} = [ L t l R f } ; t h e n ( L , R ) and ( L ' , R ' )
a r e equivalent.
Assume f i r s t t h a t ( L , R ) i s t h e C u e s t a D u t a r i c u t r e p r e -
PROOF. s e n t a t i o n of x .
Let b ( x )
> x}.
6; then XECD(O 1, L
=
B
<
[x)
<
{ Y E O ~ ( ~y )<:
X I and
R
=
and s i n c e ( L ' , R ' )
i s t i m e l y , L ' is
a s u b s e t of L and R ' is a s u b s e t of R ; t h u s f o r a l l a ' c L '
t h e r e e x i s t s acL
( Y E O ~ ( ~ )y:
Since L '
=
R',
s u c h t h a t a ' 5 a , and f o r a l l c ' E R ' Since x L and R '
=
[L'IR'),
t h e r e e x i s t s C E R such t h a t c S c ' .
< [ y } < R') is empty;
{ Y E O ~ ( ~ )L: '
i s c o i n i t i a l i n R : i , e . , f o r a l l acL t h e r e e x i s t s a ' E L '
such t h a t
such t h a t c' 5 c .
a 5 a ' , and f o r a l l CER t h e r e e x i s t s c ' E R ' these
t h u s L ' is c o f i n a l i n
Combining
f i n d i n g s , we see t h a t L and L ' are m u t u a l l y c o f i n a l and R and R' are Now d r o p t h e a s s u m p t i o n t h a t [ L I R ] i s t h e C u e s t a
mutually c o i n i t i a l .
D u t a r i c u t r e p r e s e n t a t i o n of x .
Fran t h e above argument we see t h a t L and
L ' are both m u t u a l l y c o f i n a l w i t h {yEOb(x): y
m u t u a l l y c o f i n a l with each other.
< XI,
and hence t h e y a r e
S i m i l a r l y , one can show t h a t R and R '
a r e mutually coinitial.
(13)
(i) x S y iff ( i i ) x
<
R
f o r a l l y R , and xL
y
Assume f i r s t t h a t ( i ) h o l d s .
PROOF.
<
y for all x
Then x 5 y
<
y
R
,
L
.
and xL
< x
S
L
y , f o r a l l yR a n d x ; e s t a b l i s h i n g ( i i ) . C o n v e r s e l y , assume t h a t ( i i )
holds.
Assume, f o r t h e moment, t h a t ( i ) i s f a l s e : i . e . , t h a t y
we s e e t h a t t h e f o l l o w i n g hold: ( a ) y
.
x R , f o r a l l x L , x R , y L , and y R ( a ) we s e e t h a t b ( y ) we see t h a t b ( x )
<
<
L
<
y
< x <
y R , and ( b ) x
<
x.
L
<
Thus y
<
x
<
Using t h e Conway S i m p l i c i t y Theorem a n d Using t h e Conway S i m p l i c i t y Theorem and ( b )
b(r).
b ( y ) ; which i s a b s u r d .
The c o n t r a p o s i t i v e of (13) i s t h e f o l l o w i n g :
<
L
y iff (ii)x 5 y
, for
sane y L ,
or xR 6 y , f o r sane x
R
.
(14)
(i)x
(15)
U n l e s s s t a t e d t o t h e c o n t r a r y , w e w i l l assume h e n c e f o r t h t h a t a l l r e p r e s e n t a t i ons under c o n s i d e r a t i on are t i me1y
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.02
127
Ccmbining ( 9 ) and ( 1 2 ) we have t h e f o l l o w i n g : (16)
L e t ( L , R ) a n d (L',R')
b e Conway c u t s i n No.
e q u i v a l e n t i f and o n l y i f { L ' I R ' )
=
( L , R ) and ( L ' , R ' ) a r e
(LIRJ.
Note t h a t we c a n d e s c r i b e t h e C u e s t a Dutari c o n s t r u c t i o n No of t h e
s u r r e a l numbers as f o l l o w s . (17)
XENO i f and o n l y i f x i s t h e f o l l o w i n g o r d e r e d p a i r of sets: ((yENo(<,b(x)): y
< XI, (yENo(<,b(x)):
B I B L I O G R A P H I C NOTE.
y
>
XI).
Much of what was g i v e n i n f i r s t p a r t of t h i s
s e c t i o n i s at most a s l i g h t r e w o r k i n g of C u e s t a D u t a r i ' s p a p e r [ 2 5 1 of 1954.
( 5 ) c a n b e f o u n d i n Harzheim's e l e g a n t paper 143, pp. 116-1201, of
1964.
These i d e a s were t h e n a p p l i e d t o i d e a s developed by Conway C24, p p .
4-29],
i n 1976.
Cuesta-Dutari
N o t e a l s o how many s i m i l a r i t i e s t h e r e a r e between t h e
c o n s t r u c t i o n and von Neumann's c o n s t r u c t i o n of t h e c l a s s On
of a l l o r d i n a l n u m b e r s , ( 1 . 0 2 ) .
The new r e s u l t s of t h i s s e c t i o n a r e a
r e s u l t of j o i n t work w i t h P h i l i p E h r l i c h , much of w h i c h a p p e a r e d , w i t h o u t many of t h e p r o o f s , i n [6].
4.03
AN ABSTRACT CHARACTERIZATION OF A FULL CLASS OF SURREAL NUMBERS
Thus f a r i n t h i s monograph we have d i s c u s s e d two s l i g h t l y d i f f e r e n t c o n s t r u c t i o n s of c l a s s e s of s u r r e a l numbers: Conway's [241 ( S e c t i o n s (4.00) and ( 4 . 0 1 ) ) , and t h e one g i v e n i n (4.02) u s i n g Cuesta Dutari c u t s .
In
Chapter 5 we w i l l g i v e a n o t h e r o n e , u s i n g Conway's n o t i o n of t h e s i g n e x p a n s i o n of a surreal number as i t s d e f i n i t i o n .
We have yet t o show t h a t
Conway's surreal numbers a r e e s s e n t i a l l y t h e same as t h o s e c o n s t r u c t e d i n (4.02).
This s t a t e of a f f a i r s resembles, t o some d e g r e e , t h e s t a t u s of t h e
v a r i o u s c o n s t r u c t i o n s of t h e real numbers; as Dedekind cuts i n t h e r a t i o n a l n u m b e r s , a s l e a s t upper bounds of bounded s u b s e t s of t h e r a t i o n a l numbers, a s e q u i v a l e n c e c l a s s e s of Cauchy sequences of r a t i o n a l numbers, as i n f i n i t e decimals, e t c .
Each of t h e s e c o n s t r u c t i o n s has i t s u s e s .
Once i t h a s been
shown t h a t t h e f i e l d of a l l real numbers is, u p t o i s o m o r p h i s m , t h e o n l y c o m p l e t e o r d e r e d f i e l d , t h e n o n e c a n p a s s f r e e l y back and f o r t h between t h e s e v a r i o u s c o n s t r u c t i o n s , as need o r i n c l i n a t i o n s u g g e s t s .
Norman L. A l l i n g
1 28
4.03
I n t h i s s e c t i o n we w i l l d e v e l o p a n a b s t r a c t d e s c r i p t i o n o f w h a t we w i l l c a l l a f u l l c l a s s F of s u r r e a l n u m b e r s , a n d show t h a t any two s u c h
classes have an o r d e r - p r e s e r v i n g i s a n o r p h i s m between them t h a t p r e s e r v e s T h i s shows t h a t t h e two c o n s t r u c t i o n s g i v e n s o f a r of
birth-order.
No a r e
indeed isomorphic. I t p r o v e s u s e f u l t o g e n e r a l i z e t h i s d i s c u s s i o n a l i t t l e f u r t h e r by
c h o o s i n g a n i n i t i a l segment S of On. S = [O,B)
If S i s a p r o p e r s u b c l a s s of On, t h e n
f o r a u n i q u e 8 ~ O n . If n o t t h e n l e t
L e t ( a ) F be a n ordered class.
On; t h e n a g a i n S
=
[O,B).
Assume ( b ) t h a t a f u n c t i o n b h a s been
d e f i n e d which maps F o n t o t h e s u b c l a s s [O,B) birth-order function. -___
=
of On, t h a t w i l l be c a l l e d a
Assume ( c ) t h a t , g i v e n any Conway c u t ( L , R ) i n F f o r
< a < 6, t h e n t h e r e e x i s t s a u n i q u e XEF f o r which b ( x ) i s m i n i m a l , s u c h t h a t L < ( X I < R . Such a t r i p l e { F , < , b . f i ] w i l l b e c a l l e d a which b ( L ) , b(R)
class ___-
of s-u r r e a l -____ n u m b e r s -of ---h e i g h t 8.
I
If B = O n , l e t u s a l s o d e n o t e
I F , < , b , B ] by ( F , < , b } , a n d l e t u s c a l l i t a c l a s s of. s u r r e a l n u m b e r s . C o n d i t i o n ( c ) w i l l a l s o b e referred t o as Conway's S i m p l i c i t y Theorem. Let ( F , < , b , B } b e a class of surreal numbers of h e i g h t B.
t h a t s i n c e b maps F o n t o [ O , B ) , t h e n F is a proper class.
Next, n o t e t h a t s i n c e Conway's S i m p l i c i t y
Theorem h o l d s , x i s u n i q u e l y d e t e r m i n e d by L and R . t h e symbol ( L I R I be used
Note f i r s t
t h a t i f BEOn, t h e n 161 2 I F I ; and i f B = On
t o denote x.
F o l l o w i n g Conway, l e t
C o n t i n u i n g t o u s e Conway's n o t a t i o n L
a n d c o n v e n t i o n s , ( 4 . 0 0 ) o r C24, p . 41, we w i l l write x as { x Ix L (x
(0)
=
R
L and ( x 1
R.
Let y be i n No, w i t h y
( i ) x 2 y i f f (ii) x
PROOF. Similarly x ( i i ) holds.
x.
=
R
1 , where
L R ( y ( y 1.
=
< yR f o r a l l yR, a n d xL < y for a l l xL .
Assume t h a t ( i ) h o l d s .
Then x 2 y
<
yR, and hence x
L
< yR
.
< x S y , a n d h e n c e xL < y ; t h u s ( i i ) h o l d s . Assume now t h a t Also assume, f o r a moment, t h a t ( i ) i s f a l s e : i . e . , t h a t y <
Thus we s e e t h a t t h e f o l l o w i n g h o l d : ( f ) yL
< y < x < y R , a n d (**I
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.03
xL < y
< x <
R
xR, for a l l xL, x
L R y , and y
,
<
Theorem a n d ( * ) we s e e t h a t b ( y )
<
Theorem and ( * * ) we see t h a t b ( x )
.
b(x).
129
Using t h e Conway S i m p l i c i t y U s i n g t h e Conway S i m p l i c i t y
b ( y ) ; which i s a b s u r d .
n
The c o n t r a p o s i t i v e o f ( 0 ) i s t h e f o l l o w i n g :
(i) x
(1)
<
y i f f ( i i ) x 6 y L , f o r sane y L , o r x
F o r e a c h a ~ [ O , 8 ) ,l e t and l e t F ( = , a ) (2)
=
F(<,a)
=
b-'([O,a)),
R
2 y , f o r sane x
let F(6,a)
=
b
-1
R
.
([O,a]),
b-'([a)).
{F,<,b,B] is
full
i f f o r a l l Conway c u t s (L,R) i n F ( < , a ) , [LIR) i s i n
F ( < ,a).
i s a f u l l class of s u r r e a l numbers of
{ N o , < , b } , as d e f i n e d i n (4.021, h e i g h t On.
i s a l s o a f u l l c l a s s of s u r r e a l numbers of
C o n w a y ' s c l a s s No [ 2 4 ] h e i g h t On. p.
T h i s may be s e e n by c o n s u l t i n g Conway's c o n s t r u c t i o n o f No C24,
41, h i s d e f i n i t i o n o f o r d e r a n d e q u a l i t y o n No 1 2 4 , p . 41 ( w h i c h i n -
spired (0) and (4.02:13), c o n s u l t i n g [24,
pp.
a n d i s i n c o m p l e t e a g r e e m e n t w i t h e a c h ) , by
15-17],
by n o t i n g t h a t Theorem 1 1 1 2 4 , p . 231 is what
we h a v e c a l l e d Conway's S i m p l i c i t y Theorem, a n d by r e a d i n g [24, pp. 29-30]. Let [ S , < , b ] d e n o t e t h e o r d e r e d c l a s s of a l l s i g n - e x p a n s i o n s , as g i v e n by Conway [24, pp. 30-311.
See a l s o ( 5 . 3 0 ) .
t h a t t h e s i g n - e x p a n s i o n map X E N O map t h a t p r e s e r v e s b i r t h - o r d e r ; n u m b e r s of h e i g h t O n .
+
Conway s h o w e d C24, p . 301
(x)ES is a s u r j e c t i v e order-preserving
t h u s ( S , < , b ] i s a f u l l c l a s s of s u r r e a l
T h i s c a n a l s o b e s h o w n d i r e c t l y by w o r k i n g o n
is,<, b } . Let { F , < , b , B ) be a full c l a s s of s u r r e a l numbers.
= a
<
8.
then L(x)
Let L ( x )
< {XI <
=
ItcF(<,a): t
R(x).
d e f i n e d t o be [ L ( x ) ( R ( x ) ] .
< XI
and l e t R(x)
=
Let X E F , w i t h b ( x ) itsF(<,a):
Assume t h a t L ( x ) a n d R ( x ) a r e s e t s . By ( 2 1 , z i s i n F ( S , a ) .
~ ( x )a n d R ( x ) i s F ( < , a ) , b ( z )
= a.
t
>
x);
Let z be
S i n c e t h e u n i o n of
S i n c e C o n w a y ' s S i m p l i c i t y Theorem
Norman L. A l l i n g
130
holds, z
=
4.03
f o r No, t h e
Let ( L ( x ) , R ( x ) ) b e c a l l e d , as i t w a s i n ( 4 . 0 2 )
x.
Cuesta D-u t a r i --cut -~
r e p r e s e n t a t i o n of x i n I F , < , b , f 3 } .
1 _ 1
T h u s we h a v e p r o v e d
the following: Each XEF f o r which L ( x ) a n d R ( x ) a r e s e t s , x h a s a unique Cuesta
(3)
D u t a r i c u t r e p r e s e n t a t i o n i n { F , <, b , B } . Let [ F , < , b , B ) , { F ' , < , b * , B I , a n d [ F t t t , < , b l l , B l b e f u l l c l a s s e s of
L e t f be a map form F i n t o F' and l e t h be a
s u r r e a l numbers of h e i g h t B. m a p p i n g of F' i n t o F". YEF i m p l i e s f ( x )
f w i l l be c a l l e d a s u r r e a l monomorphism i f ( i ) x
< f ( y ) h o l d s , and
<
f o r a l l XEF.
(ii) i f b(x) = b ' ( f ( x ) ) ,
C l e a r l y we have t h e f o l l o w i n g . If f and h a r e s u r r e a l monomorphisms, t h e n h - f is a surreal monomor-
(4)
phi s m
.
( i ) There e x i s t s a s u r r e a l monomorphism of F i n t o F'.
LEMMA.
Let f a n d g b e a s u r r e a l monomorphism of F i n t o F ' , t h e n f = g . s u r r e a l monomorphism f of F i n t o F ' , m a p s F o n t o F ' .
Let
aE[O,e)
( i i i ) Any
( i v ) F o r a l l aEOn
is a s e t .
s u c h t h a t a 5 8, b-'([O,a)) PROOF.
(ii)
a n d assume t h a t t h e r e e x i s t s a s u r r e a l monomor-
p h i s m f of F ( < a ) o n t o F ' ( < & ) , a n d t h a t F ( < a ) a n d F ' ( < a ) a r e s e t s . t h a t 0 has t h i s p r o p e r t y . )
Let x , ~ F ( = a ) , a n d l e t ( L , , R , )
Dutari cut representation.
Clearly (f(L,),f(R,))
F'.
Let x,'
n u m b e r s of h e i g h t 6 ( 2 ) , b ' ( x , ' ) x,;
be its Cuesta
is a Cuesta Dutari cut i n
S i n c e [ F t , < , b l , B } i s a f u l l c l a s s of s u r r e a l
= {f(Lo)lf(Ro)].
i s equal t o F ' ( < a ) , b ' ( x , ' )
I a.
S i n c e t h e u n i o n of f ( L , ) a n d f ( R , )
2 a ; thus b'(x,')
=
Let x , E F ( = a ) , w i t h x o
a.
a n d l e t ( L , , R , ) b e t h e Cuesta D u t a r i c u t r e p r e s e n t a t i o n of x , .
( f ( L , ) , f ( R , ) ) i s a C u e s t a Dutar i c u t i n F 1 . S i n c e xo
< x,,
R
f(xl)
>
< xl,
<
Clearly
Let x l t = [ f ( L , ) l f ( R , ) l .
s i n c e x, and x , t F ( = a ) , a n d hence a l l x o R and a l l x l L are i n
F ( < a ) , we may i n v o k e ( 1 ) t o c o n c l u d e t h a t ( a ) x p ( b ) x,
(Note
f o r some x,
L f ( x l )cf(R,).
R
.
< xlL, for
some x l L , o r L
Assume t h a t ( a ) h o l d s ; t h e n x i ER,,
Since f ( R , ) R
and thus
> ( f ( x , ) ] , we s e e t h a t f ( x , ) > f ( x , ) .
Assume t h a t ( b ) h o l d s ; t h e n x, E L , , a n d t h u s f ( x , )
<
R f ( x , )cP(L,).
Since
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.03
<
f(L,)
{ f ( x , ) } , we s e e t h a t f ( x o )
<
131
Since each x , ~ F ( = c r ) is
f(x,).
u n i q u e l y d e t e r m i n e d by a C u e s t a D u t a r i c u t i n t h e s e t F ( < a ) , F(6a) i s a g a i n
a set.
Let z ' E F ' ( = a ) , w i t h z '
cut i n F'(
Let z
F(
=
=
{L'IR'),
where ( L ' , R ' )
i s a Cuesta D u t a r i
i s a Cuesta D u t a r i cut (L.R) i n
Clearly (f-'(L*),f-'(R'))
{LIR}; t h e n z EF ( =a ) a n d f ( z ) =
Since each z'cF'(=a) i s
2'.
u n i q u e l y d e t e r m i n e d by a C u e s t a D u t a r i c u t i n t h e s e t F ' ( < a ) , F'(Sa) i s again a s e t .
As t o ( i i ) , a s s u m e t h a t t h e r e s t r i c t i o n of f t o F ( < a ) ) ,
f l F ( < a ) , is equal t o glF(
=
( g ( L o ) , g ( R , ) ) i s a Cuesta Dutari c u t i n F'(
{ f ( L o ) ( f ( R o ) )= ( g ( L o ) ( g ( R o ) ) = g ( x o ) . THEOREM.
Thus f ( X o )
0
Assume t h a t { F , < , b , B ) a nd { F ' , < , b ' , B )
s u r r e a l numbers of h e i g h t B.
Clearly
a r e f u l l c l a s s e s of
The r e exists a u n i q u e surreal monomorphism f
of F o n t o F ' . COROLLARY.
T h e r e e x i s t s a u n i q u e s u r r e a l monomorphism, which maps
t h e c l a s s of Conway's s u r r e a l numbers C241 onto t h e c l a s s of s u r r e a l numb e r s c o n s t r u c t e d i n (4.02) BIBLIOGRAPHIC NOTE.
( a n d i n C71).
I n [ 7, ( 1 1 1 t h e a u t h o r a n d P h i l i p E h r l i c h m a d e ,
what h e r e i s a theorem ( 4 . 0 3 : 0 ) ,
into a condition.
I t w a s E h r l i c h who
f i r s t s u s p e c t e d t h a t r a t h e r t h a n b e i n g a c o n d i t i o n i t was, i n f a c t , a
theorem.
He w e n t o n t o show t h a t ( 0 , i ) i m p l i e d ( O , i i ) , w h i l e t h e a u t h o r
showed t h a t ( 0 , i i ) i m p l i e d ( 0 . i ) . [6],
b o t h authors - separately
-
s u r r e a l numbers of h e i g h t 6 a g a i n .
Having m i s s e d t h i s f a c t when we w r o t e c o n s i d e r e d t h e axioms f o r a f u l l class of The r e s u l t s of t h e a u t h o r s r e f l e c t i o n s
may be found i n S e c t i o n 4.60. 4.04
SUBTRACTION I N No
Assume t h a t No i s t h e c o n s t r u c t i o n of t h e s u r r e a l numbers by means of C u e s t a D u t a r i c u t s , t h a t we c o n s i d e r e d i n S e c t i o n 4.02. f u l l c l a s s of s u r r e a l n u m b e r s ( o f h e i g h t O n ) .
Then (N o , < , b } i s a
By T h e o re m 4 . 0 3 we can
t r a n s f e r a l l the or e m s t h a t a r e s t a t e d p u r e l y i n terms of t h e l i n e a r o r d e r and t h e b i r t h - o r d e r f u n c t i o n , t o any o t h e r f u l l c l a s s of s u r r e a l numbers.
132
Norman L . A l l i n g
4.04
Following Conway [ 2 4 , p . 51, we w i l l c o n s i d e r e x p r e s s i o n s { L I R ) , f o r which L and R a r e s u b s e t s of No, and c a l l s u c h an e x p r e s s i o n s games.
a game { L I R ) , t h e n i t i s a number i f and o n l y i f L Let x
( L I R } b e i n No.
=
<
Then, by assumption (4.02:15),
t i m e l y Conway c u t r e p r e s e n t a t i o n of x.
Given
R.
(L,R) is a
Following Conway, d e f i n e s u b t r a c -
t i o n as f o l l o w s : l e t -x be d e f i n e d t o be { - R l - L ] . (0)
(i)
-x i s an element i n No, which i s independent of r e p r e s e n t a t i o n ,
(ii)
-(-XI
=
x,
( i i i ) b(-x)
=
b ( x ) , and
x <
(iv)
y i f and o n l y i f -y
< -x. i s a game.
I n i t i a l l y a l l we know i s t h a t ( - R l - L )
PROOF.
We w i l l
p r o v e ( 0 ) by i n d u c t i o n , b y a s s u m i n g t h a t ( 0 ) h o l d s on Oa ( = b - ' ( [ O , a ) ) . Note t h a t i f a = 0 , t h e n Oa i s e m p t y , t h u s ( 0 ) h o l d s o n O a ; h e n c e o u r i n d u c t i o n i s launched.
-
a.
By ( O , i v ) , - x R
<
thus b(x)
(= b-'(a));
< xR.
To see t h a t (0) holds on Ma ( = b - l ( [ O , a l ) , F i r s t n o t e t h a t each x L
-x ; t h u s ( - R 1 - L )
L
let X E N ~
and xR i s i n Oa, and x
i s a number.
I n i t i a l l y we may
assume t h a t ( L , R ) i s t h e C u e s t a D u t a r i c u t r e p r e s e n t a t i o n o f x. ( L o , R o ) b e any Conway c u t i n 0
CL
t h a t r e p r e s e n t s x.
L o and L are m u t u a l l y c o f i n a l a n d R ,
(4.02:16),
coinitial.
Since ( 0 ) holds on Oa,
-(-x)
=
=
I-RI-L);
-{-RI-L)
=
. a
Thus by (4.02:16)
Since ( 0 , i i ) h o l d s on Oa,
{ L I R ] = x; showing t h a t ( 0 , i i ) h o l d s on Ma,
t h e element of No of minimal b i r t h d a y s u c h t h a t L and conclude t h a t t h e r e i s no element ~ b ( - x ) 2: b ( x ) .
and -R are m u t u a l l y
we know t h a t -R,
e s t a b l i s h i n g ( 0 , i ) on M
Let
Then, as we saw i n
and R are m u t u a l l y
c o f i n a l , and t h a t -Lo and -L are m u t u a l l y c o i n i t i a l . (-Rol-Lo}
L
<
x
<
€ such 0 ~t h a t -R
Since x is
R , we may a p p l y ( 0 )
<
(z}
<
-L.
Thus
S i n c e (No,<,b} i s a full class of surreal numbers (4.03:2),
b ( - x ) 5 a , showing t h a t b(-x) = b ( x ) , and hence showing t h a t ( 0 , i i i ) h o l d s o n Ma. on M
Ci
Now l e t ycM
, we
a'
w i t h y = ( L t l R t ] , and w i t h x
<
y.
Since ( 0 , i ) holds
may assume, without l o s s of g e n e r a l i t y , t h a t ( L t , R t )i s t h e
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.04
C u e s t a D u t a r i c u t r e p r e s e n t a t i o n of y. =
(L,R), y
=
F i r s t assume t h a t b ( y )
< y.
Since x
(L',R')ECD(T~).
is a p r o p e r s u b s e t of - R ;
y is i n R.
t h u s -y
<
a ; then x As
U s i n g ( 0 ) o n O a , we know t h a t Assume now t h a t b ( y )
-x.
and we s e e t h a t - y
Hence - y c ( - R ) ,
=
L i s a p r o p e r s u b s e t of L'.
a c o n s e q u e n c e , R ' is a p r o p e r s u b s e t of R . -R'
133
<
-x.
<
a ; then
T h u s we h a v e s h o w n
t h a t (0) h o l d s o n Ma.
A D D I T I O N I N No
4.05
Conway d e f i n e s a d d i t i o n i n No as f o l l o w s C24, p . 51:
x + y = Ix
(0)
L
L
R
R
+ y , x + y Ix
+ y , x t y }.
L e t u s s e e sane of t h e m o t i v a t i o n b e h i n d t h i s d e f i n i t i o n .
< x <
xL
x
R
and y
L
<
y
<
y
.
R
S i n c e we want No t o be a n o r d e r e d g r o u p u n d e r
<
a d d i t i o n , we m u s t h a v e ( i ) xL + y
<
<
xR + y , a n d ( i v ) x + y
x + y
R
.
b i r t h d a y s u c h t h a t i n e q u a l i t i e s (1)
[OL
-
(ii) 0
+
1 and 1 + 0 a r e b o t h t h e number 1 .
PROOF.
Since 0 = L
R
0 10
+
0, and there are no 1
L
1)
=
I0
+
0, 0
R
.
+
there are no 0 0
R
1
=
x + y, (iii) x + y
( i v ) hold.
0 i s t h e number 0, a n d
[I},
<
Thus x + y i s t h e e l e m e n t of smallest
+
+ 0, 0 +
I0 + 1
x + y , ( i i ) x + yL
0
(i)
(1)
Note t h a t
(1)
= 0.
L
R and no 0 ; t h u s 0
Recall t h a t 1
=
+
0 =
[Ol); t h u s l L
=
R
Thus 0 + 1 = [OL + 1 , 0 + lLIOR + 1 , 0 + 1 1 =
0 1 1 = I011 =1.
0
On o c c a s i o n w e w i l l p r o c e e d b y i n d u c t i o n o n t h e n a t u r a l ,
or
H e s s e n b e r g , sum of t h e o r d i n a l n u m b e r s b ( x ) a n d b ( y ) , b ( x ) + b ( y ) , e t c . Recall t h a t g i v e n a n o r d i n a l number a
>
0 i t has t h e following unique
e x p a n s i o n , which i s c a l l e d C a n t o r ' s normal form C18, p . 2371:
134
Norman L. A l l i n g
w
(2)
“1
B,
+
...
w
+
“n
Bn, w i t h
... > q n ,
>
q,
4.05
>
and w
... ,
B1,
Bn > 0.
R e g a r d i n g s u c h s u m s as f o r m a l power s e r i e s a l l o w s u s t o a d d a n d
Let u s a l s o r e g a r d t h e o r d i n a l number 0 as t h e empty expan-
m u l t i p l y them.
s i o n ( 0 1 , a n d l e t i t b e a d d e d a n d m u l t i p l i e d a s a formal power series. Such sums and p r o d u c t s g i v e r i s e t o t h e n a t u r a l numbers.
a n d p r o d u c t of o r d i n a l
C55, pp. 246-2611.)
(See e.g.,
Note t h a t we have y e t t o p r o v e t h a t x we know t h a t i t is a game ( 4 . 0 4 ) .
t i o n of a d d i t i o n between games.
+
y i s a l w a y s a number; however
F u r t h e r , we may r e g a r d (0) as a d e f i n i -
As s u c h we c a n e s t a b l i s h p r o p e r t i e s a b o u t
F i r s t note that
it.
(3)
0
+
x
=
PROOF.
x , f o r a l l XENO. Assume t h a t 0
+
u = u is t r u e f o r a l l UEO
U’
and l e t b ( x ) = a.
(Note that i f a
=
0, t h e i n d u c t i o n h y p o t h e s i s i s empty, x
noted i n ( 2 ) , 0
+
x
+
xR I .
Since 0
x.)
By d e f i n i t i o n , 0
+
x
(0
=
+
xLIO
L R h y p o t h e s i s ) i s {x Ix 1, which i s x .
x
(4)
+
y
=
PROOF.
+
x = {OL
+
x, 0
+
L R x 10
+
x, 0
( 1 1 , t h e r e a r e n o 0L , o r 0 R : i . e . , 0 h a s n o o p t i o n s
=
Thus 0
(4.00).
=
0 , a n d , as
=
+
R x 1 , which ( u s i n g o u r i n d u c t i o n
o
y + x , f o r a l l x , YENO. Let x a n d y be chosen s o t h a t f o r a l l U , V E N O ,
n a t u r a l sum, b ( u )
+
(Note t h a t i f
b ( v ) , i s less t h a n b ( x )
f o r which t h e
b(y) = a; then u
v
v
=
(Y
=
0, t h e n t h i s i s t h e empty i n d u c t i o n h y p o t h e s i s ,
x
y , and hence x
+
y
+
+
+
u.
= 0 =
x.)
S i n c e any o p t i o n z of x ( r e s p . y ) i s s i m p l e r
than x (resp. y): i.e., b ( z )
< b ( x ) ( r e s p . b ( z ) < b ( y ) ) , we c a n u s e t h e
=
y
+
i n d u c t i o n h y p o t h e s i s t o show t h a t x [y
t
L
x , y
L + x I y + x R , yR
+ XI
=
y
+
y
+
x.
=
IxL 0
+
y, x
+
yLlxR
+
y,
x
+
yR)
=
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.05
135
Let u s now c o n s i d e r two s t a t e m e n t s which we w i l l p r o v e f o r a l l u, v ,
a n d z i n No.
w,
The f i r s t o f t h e s e , we w i l l c a l l P ( u , v : w , z )
is the
following: ( i ) u 2 v a n d ( i i ) w 5 z, implies ( i i i ) u
(5)
w 2 v
+
(iv) Strict
z.
+
i n e q u a l i t y i n ( i ) or ( i i ) i m p l i e s s t r i c t i n e q u a l i t y i n ( i i i ) . O u r i n d u c t i o n w i l l be w i t h r e s p e c t t o rnax.(b(u) + b ( v ) , b(w) + b ( z ) ) , Let N ( x , y ) b e t h e s t a t e m e n t t h a t
which we w i l l d e f i n e t o be b ( P ( u , v : w , z ) ) .
x
(6)
+
L
y i s a number: i . e . , t h a t {x
+
Let b ( N ( x , y ) ) be d e f i n e d t o be b ( x )
y, x +
+
L
y 1
< {xR
t
R
y , x + y }.
b ( y ) , and consider t h e f o l l o w -
ing statement:. ( 5 ) a n d ( 6 ) h o l d , f o r a l l u, v , w , z , x , a n d y i n No.
(7)
PROOF.
To e s t a b l i s h ( 7 ) , we w i l l p r o c e e d b y i n d u c t i o n o n
max.(b(P(u,v:w,z)),
b ( N ( x , y ) ) ) , w h i c h we w i l l c a l l b ( Q ( u , v : w , z : x , y ) ) .
(Note t h a t i f b ( Q ( u , v : w . z : x , y ) ) (5) and ( 6 ) a r e t r u e . )
Let a
>
= 0,
then u
=
v = w
=
z =
x
=
y
=
0, and
0 , a n d assume t h a t f o r a l l u, v , w , z , x ,
<
a n d y i n No, f o r which b ( Q ( u , v : w , z : x , y ) )
a, t h e n (5) and ( 6 ) holds.
Now
l e t b ( Q ( u , v : w , z : x , y ) ) = a. L
<
<
x
u 5 v iff u
<
Note t h a t ( a ) u R
z < z , ( e ) xL (4.02:13),
x iff w
<
R
x
<
x
a n d wL
u R
< u R , ( b ) v L < v < v R , ( c ) w L < w < w R , ( d ) zL <
, and
v
R
(f) y
x
<
y
<
y
R
.
Note a l s o t h a t , b y
< v , f o r a l l v R a n d a l l uL ; and
a n d uL
< x, for a l l
L
R
and w
L
.
that w 2
S i n c e ( e ) a n d ( f ) h o l d , and
s i n c e t h e r e l e v a n t i n d u c t i o n h y p o t h e s i s h o l d s , we see by (51, t h a t x
xR (8)
t
y, x x
+
t
yL< x
t
y R , xL + y
<
x
t
y R , and x + y L
<
y is a number, as a l s o are u + w a n d v + x .
xR + y: t h u s
L
+
y
<
Norman L . A l l i n g
136
t e l l s us that u
Now n o t e t h a t (4.02:13)
<
u + w
( v + zIR and ( u
wIL < v
+
4.05
w 5 v
+
(v
+ z, for a l l
z if and o n l y i f
+
zIR, a n d a l l ( u
+
L
w) ,
+
But by ( 8 ) an d ( 0 1 , we h a v e
(9)
u
+
w 5 v + z iff u + w < v
u
+
wL < v
+ z, f o r a l l v
R
R
+
z, u + w
, zR , u L ,
<
v
and w
z R , uL
+
+
w < v
+
z, a n d
.
L
Then, u s i n g ( a ) - ( d ) , we know t h a t
Assume t h a t ( 5 , i ) a n d ( 5 , i i ) h o l d . t h e following holds:
u
(i)
(10)
L
(iii) u
< u
5 v
R
< v ,
< vR ,
(iv) w
and ( i i ) w
< zR ,
L
( v ) uL
< w
<
2 z
R
<
z ; which i m p l y t h a t
v, and ( v i ) w
L
< z.
Thus, we may i n v o k e t h e i n d u c t i o n h y p o t h e s i s , a p p l y i t t o ( l O , ( i i i ) ( v i ) ) u s i n g (51, t o e s t a b l i s h t h e r i g h t hand s i d e o f ( 9 ) .
With t h e a i d of
( 9 ) we h av e t h u s p r o v e d ( 5 , i i i ) .
Now assume t h a t s t r i c t i n e q u a l i t y h o l d s i n ( 5 , i ) o r ( 5 , i i ) .
that u
<
t h a t ( i ) uR S v , f o r s a n e u
see t h a t u
+
(4.02:14)
R
.
Assume f i r s t t h a t ( h ) h o l d s .
L
L
.
w 2 vL + z ; t h u s u + w 5 ( v + z I L , f o r some ( v + z ) w e see t h a t u + v < v + z . Assume, l a s t l y , t h a t ( i ) h o l d s .
w 5 v
t h u s ( u + w ) S~ v + z ,
+ z;
we s e e t h a t u
+
w
<
v
+ z,
,
or
U s i n g ( 5 ) we
+
(4.02:14),
(5), u R
we know t h a t ( h ) u 5 v L , f o r some v
Applying (4.02:14)
v.
Since
we may a s s u m e , w i t h o u t l o s s of g e n e r a l i t y ,
a d d i t i o n i s commutative ( 4 ) ,
f o r some ( u + w)
s h o wi n g t h a t ( 5 , i v ) h o l d s .
R
.
By By By
By i n d u c -
t i o n we h av e e s t a b l i s h e d ( 7 ) .
(x
(11)
+ y) + z =
PROOF.
“x
+
Y)
+ 2,
( y + z ) , f o r a l l x, y , an d z i n No.
+
Let u s p r o c e e d by i n d u c t i o n o n b ( x ) + b ( y ) + b ( z )
as we d i d ab o v e. L
x
(x
Thus (x + y ) + z = +
Y)
+
2
L )(x
+
y)
R
+
z, (x
+
y)
+
z
R
1
=
=
a , much
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.05
I(x L (xR
y) +
+
+
y)
+
L
137
LI
z, (x
+
y
z, (x
+
y)
+
z
z, (x
+
y ) + z, (x
+
y)
+
z R ] . We may now a p p l y t h e i n d u c t i o n
+
R
h y p o t h e s i s an d c o n c l u d e t h a t t h e l a s t number i s
(13)
x
+
(-x)
PROOF.
0, f o r a l l xcNo.
=
x
+
(-XI
(xL IxR
+
(-x R I-xL )
=
[xL
+
(-XI, x
+
(-XI L I
(-xR )1xR + (-XI, x + (-xL 1). Let u s p r o c e e d by i n d u c t i o n by a s s u m i n g t h a t u + (-u) = 0 , f o r a l l u s u c h t h a t R b ( u ) < a , w h e r e b ( x ) = a . We know t h a t ( i ) xL < x < x We h a v e s e e n R L L ( 4 . 0 4 : 0 , i v ) t h a t ( i ) i m p l i e s t h e f o l l o w i n g : ( i i ) -x < -x < -x , Adding x
xR
+
(-XI, x
+
(-XI
RI
-
=
(xL
+
(-XI,
x
+
.
t o t h e r i g h t - h a n d s i d e of ( i i ) , u s i n g t h e i n d u c t i o n h y p o t h e s i s , a n d ( 5 ) g i v e s us ( i i i ) xL
+
(-x)
<
0. Adding -xR t o t h e r i g h t - h a n d s i d e of
u s i n g t h e induction hypothesis, and (5) g i v e s us ( i v ) x
+
(i),
(-x R ) < 0.
Adding xR t o t h e l e f t - h a n d s i d e of ( i i ) , u s i n g t h e i n d u c t i o n h y p o t h e s i s , and (5) gives us (v) 0
< xR
+
(-XI.
F i n a l l y , a d d i n g -xL t o t h e l e f t hand
s i d e of ( i ) , u s i n g t h e i n d u c t i o n h y p o t h e s i s , a n d ( 5 ) g i v e s u s ( v i ) 0
L
(-x 1.
< x
+
Taken t o g e t h e r , ( i i i ) - ( v i ) show t h a t t h e f o l l o w i n g h o l d s :
S i n c e 0 i s t h e s i m p l e s t element i n No, ( 1 3 ) i s p r o v e d . Canbining ( 3 ) , proved t h e following:
( 4 1 , ( 7 1 , ( 8 ) , ( 1 1 1 , a n d ( 1 3 1 , we s e e t h a t we h a v e
4.05
Norman L. A l l i n g
138
Under a d d i t i o n No i s an o r d e r e d A b e l i a n g r o u p .
THEOREM.
The i d e a of u s i n g t h e n a t u r a l or H e s s e n b e r g sum
B I B L I O G R A P H I C NOTE.
i n t h e i n d u c t i o n , as we d i d a b o v e , c a n b e f o u n d i n G o n s h o r ' s b o o k o n t h e s u b j e c t [38].
Of c o u r s e t h e i d e a of n a t u r a l sums a n d p r o d u c t s p e r m e a t e s A f t e r a l l , Conway's normal form f o r a s u r r e a l number
Conway's work [24].
w h i c h we w i l l c o n s i d e r i n C h a p t e r 6 , i s a g e n e r a l i z a t i o n
[ 2 4 , pp. 32-33],
Normal f o r m s were a l s o u s e d by S i k o w s k i 1661, i n
of C a n t o r ' s normal form.
1 9 4 9 , t o c o n s t r u c t a n i n t e r e s t i n g f i e l d , which c a n be r e g a r d e d i n a v e r y n a t u r a l way as a s u b f i e l d of No. MULTIPLICATION I N No
4.06
Conway's d e f i n i t i o n of m u l t i p l i c a t i o n ([24, p . 51
Or
However, Conway g i v e s C24, p . 61
at first glance, a l i t t l e surprising.
sane more m o t i v a t i o n f o r i t , which we w i l l now expand o n .
<
i n No; t h e n xL
(0)
x - x
L
,
x
-
y
< xR
yL, xR
-
L
<
y
x , and yR
-
and y
<
y
R
( 4 . 0 0 : 5 ) ) seems,
.
Let x a n d y b e
Hence
y a r e a l l greater t h a n z e r o .
We w a n t t o d e f i n e a m u l t i p l i c a t i o n o n No u n d e r which i t w i l l be an o r d e r e d f i e l d ; t h u s w e must have t h e f o l l o w i n g :
(1)
L
L
L
(i)
O < ( x - x ) ( y - y ) = x y - x y - x y
(ii)
o <
(iii)0
(iv)
(xR
-
x)(y
R
-
R
-
-
R
R R
=
xy
< ( x - x L ) ( yR - y )
=
-xy
+
L x y
=
-xy
+
R x y +xyL
(xR - x ) ( y
-
yL
xy
L L x y ,
+
y)
o <
x y
L
+ x y
xyR
-
,
xLyR, a n d
- x Ry L ,
A s a consequence we s e e t h a t
(2)
(i)
L x y
( i i i ) xy
+
<
-
xyL
L
x y
+
L L x y
<
R xy, ( i i ) x y L R
xyR - x y
,
+
and ( i v ) xy
xy
<
R
R
R R x Y
x y + xy
< L
XY.
-
R L x y
.
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.06
139
Using ( 1 ) a n d ( 2 ) we s e e t h a t f o r No t o b e a n o r d e r e d f i e l d t h e f 011owing i nequal i t y must hol d: L L L L R R R R I x y + x y - x y , x y t x y - x y ) < x y < L R L R R Ix Y XY - x y , x y i. xyL - x R y L ) .
(3)
+
Hence by d e f i n i t i o n of x y , [24, p . 51, i s L
(4)
L L R R R R - x y , x y t x y - x y l R L R R L R L Y - x y , x y + x y - x y } .
I X Y + X Y
L
X Y + X
L
Thus xy is t h e s i m p l e s t element i n No s u c h t h a t (3) h o l d s . S i n c e ( 1 ) ( o r e q u i v a l e n t l y (2) o r ( 3 ) ) h a s y e t t o b e p r o v e d , we do n o t know t h a t xy i s a number, w e know o n l y t h a t i t i s a game.
As such t h e
f o l l o w i n g h o l d s f o r a l l x , YENO.
PROOF.
( xL0 + x o L
(1)
=
Recall t h a t 0
- x L0 L , x R 0
+
=
xOR
( 1 1;
-
t h u s t h e r e a r e n o O L o r a n y 0R
x R 0R
I
x L0
xOR
i.
-
0 , e s t a b l i s h i n g ( i ) . To prove ( i i ) , r e c a l l t h a t 1
R 0 , a n d t h e r e i s no 1
.
L {x 1
+
xlL
+
L (x 1
+
x0
-
-
R XLlL, x 1
L R x OIx 1
i n d u c t i o n on b ( x )
+
+
x0
-
xLOR, x R o + xoL
- {Ol 1 ;
.
x0 =
XROLJ
=
t h u s 1L
=
Proceeding by i n d u c t i o n on b ( x ) , we see t h a t x l = xlR
-
xRIRI xL1
- x R 01
=
(x
L
Ix
R
+
xlR
}
x.
-
-
x L I R , xR1 + x l L
-
x R1L 1 =
To prove ( i i i ) , proceed by
By d e f i n i t i o n ( 4 ) we know t h a t xy i s equal t o
b(y).
-
xRyR
I
x y
+
xy
R
-
-
R R y x
I
L y x
+
yxR
-
LR R R L x y , x y + xyL - x y I = L L L L R R R R R R L R L L R (yx + y x - y x , y x + y x - y x I y x L + y x - y x , y x + y x - y x } = L
+ xyL
-
xLyL, xRy
t y L x + yxL
-
yLxL, y R x + y x
{x y
yx.
+
xyR
R
y L x R , y R x + yxL
-
y Rx L 1
=
L a s t l y , t o e s t a b l i s h ( i v ) , l e t us proceed by i n d u c t i o n on b ( x ) + b ( y ) .
Norman L. A l l i n g
140
4.06 R
By t h e d e f i n i t i o n of s u b t r a c t i o n (4.041, I(-xR)y
-
(-x)yL
+
(-xR)yL, (-xL)y
+
L
L
( - x ) y = I-x I-x I I y Iy ( - x ) y R , - ( - x L ) yR I (-x ) y +
-
( - x R ) y R , ( - x L )Y
I-x R y - xy L -(xy).
which i s
(-x)y
R
+
=
(6)
(x
(-y)x
x Ry L , -x L y - xy R =
L
1(x (x
(x
y)z L
+
y z , f o r a l l x , y , and z i n No.
+
(x
+
-
y)2L
+
y)z
(xL
+
L
(x + y)zL
-
(x + y )z
y)z
+
(x
y)zL
-
(xR
R y )z
+
(x + y)zL
-
(x
+
+
y,
+
Rlz =
Y)ZL,
+
+
+ L Ix R
(xL + y ,
=
y )z
+ +
=
BY d e f i n i t i o n , ( x
t
+
R
L R R R R R x y I-x y - xy + x y , -xLy - xyL t xLyL} - ( y x ) ; t h u s u s i n g what we have j u s t p r o v e d , +
x(-y); establishing ( i v ) .
y)z = xz
+
PROOF.
(x
+
By ( i i i ) , - ( x y )
-(xy)
(7)
I
( - x ) y L - ( - x L ) yL 1. Using t h e i n d u c t i o n h y p o t h e s i s , and we s e e t h a t t h i s last game i s e q u a l t o t h e f o l l o w i n g game:
(4.05).
=
R
L
,
Y)ZL,
R
y )z
L
I.
We w i l l p r o v e ( 6 ) by i n d u c t i o n on b ( x )
+
b(y)
+ b(z).
Then, making
t h e usual i n d u c t i o n h y p o t h e s i s , ( 7 ) i s e q u a l t o t h e f o l l o w i n g :
(8)
I XL Z + Y Z + X Z ~ L- XL ,Z x z + yL z + y zL - y Lz L, y zR - y Rz R 1
xR z
+
yz
+ XZR
-
XRZR,
xz
+
y Rz
xL z
+
yz
+ XZR
-
XLZR,
xz
+
yL2 + y zR
-
yL z R ,
xR2
+
yz
+
xzL - xRzL, x z
+
y Rz
-
y R z L 1.
Now expanding x z
+
t
+
yzL
yz, we f i n d t h a t i t
is e q u a l t o t h e f o l l o w i n g :
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.06
L {x z
+
-
xzL
xLzLt xRz
x L z + x z R - xL zR , x R z + x z L - xR zL 1 +
R R xzR - x z
.+
L L L R R R R tyLz+yz - y z , y z + y z - y z
(9)
L Ix z
+
xz
yLz
+
L x z
xz
.+
.+
xzL .+
xzR
L y z
-
.+
-
L L x z
yzL
.+
R yz, x z
L L y z , xz
-
L R x z
+
.+
+
yLz + y z R
xzR
R y z
R y z , x z + xzL
yzR - yLzR, x z
.+
141
R y z
+
.+
-
y L zR , y R z
.+
y zL - y R zL l =
R R x z + yz,
yz
R
-
xRzL yzL
-
R R y z +
I
yz,
R L y z I.
S i n c e t h e games i n ( 8 ) a n d ( 9 ) a r e e q u a l , ( 5 ) i s p r o v e d .
4.07
ORDER A N D MULTIPLICATION I N No
L e t x l , x,,
y l , a n d y, be e l e m e n t s i n No.
Conway s t a t e s a n d p r o v e s
t h e f o l l o w i n g t h r e e r e s u l t s [ 2 4 , Theorem 8, p p . 19-20], THEOREM. xy is a number. (i)
x , S x, a n d ( i i ) y 1 5 y 2 i m p l y ( i i i ) x 1 y 2
.+
x,y,
( i v ) I f ( i ) and ( i i ) are strict t h e n ( i i i ) is strict I f xl
=
x, t h e n x l y
We w i l l p r o v e ( 0 )
-
=
5 xlyl
.
.+
x2y2.
x,y.
(2) t o g e t h e r , by i n d u c t i o n .
Conway r e f e r s t o t h e i n e q u a l i t y o f ( 1 , i i i ) a s P ( x l , x 2 : y 1 , y 2 ) C24, p 191.
I f ( 1 , i i i ) h o l d s t h e n we w i l l write t h a t t t P ( x , , x , : y l , y , )
holds".
( 1 , i i i ) i s a s t r i c t i n e q u a l i t y we w i l l w r i t e t h a t " P ( x , , x 2 : y 1 , y , )
strictly".
(3)
If
holds
Now assume t h a t x , 5 x, 6 x , , a n d t h a t y 1 5 y2 6 y B .
I f ( i ) P(x1,x2:y1,y2) and (ii) P(x2,x,:yl,y,) P(x,,x,:yl,y2)
holds.
hold, then ( i i i )
F u r t h e r , ( i v ) i f a t l e a s t o n e of ( i ) or ( i i )
holds s t r i c t l y , then ( i i i ) holds s t r i c t l y .
Norman L. A l l i n g
142
4.07
Assume t h a t ( 3 , i ) a n d ( 3 , 1 1 1 h o l d ; t h e n we h a v e t h e
PROOF of ( 3 ) .
following:
(4)
( i ) xlyZ
+
x2Y1 5 X , Y ,
x 2 y 2 , and ( i i ) x2y2
+
X,Y,
+
5 x2y1
x,y,.
+
Adding ( 4 . i ) a n d ( 4 , i i ) g i v e s u s
C a n c e l i n g t h e sum of t h e c e n t e r two terms o n e a c h s i d e of ( 5 ) shows that P(xI,x3:y1,y2) holds;
proving ( 3 , i i i ) .
I f ( 3 . i ) or ( 3 , i i ) is s t r i c t ,
t h e n s o is (5); thus ( 3 , i v ) h o l d s , hence (3) is proved.
(6)
If
( i ) P ( x , , x , : Y , , y 2 1 a n d ( i i ) P ( x , , x z : Y z , ~ ,) h o l d , t h e n ( i i i )
P(x1,x2:y1,y3) holds.
F u r t h e r , ( i v ) i f a t l e a s t o n e of ( i ) or ( i i )
holds s t r i c t l y , then ( i i i ) holds s t r i c t l y . PROOF of
(6).
Assume t h a t ( 6 , i ) a n d (6,111 h o l d ; t h e n we h a v e t h e
fo l l o w i n g :
(7)
(i)
xIy2
+
xZyl 5 xly,
+
x 2 y Z r and ( i i ) xlyB
t
x,y,
2 x1y2
t
x,y,.
Adding ( 7 , i ) a n d ( 7 , i i ) g i v e s us
C a n c e l i n g t h e two i d e n t i c a l terms i n ( 8 ) g i v e s
US
P(X,
,X2
:yl ,Y3),
e s t a b l i s h i n g (71, a n d h e n c e c o m p l e t i n g t h e p r o o f of (6).
PROOF of (O), ( 1 1 , a n d ( 2 ) . n a l 2 ( b ( x ) + b ( y ) ) t o prove ( 0 ) ;
Let u s p r o c e e d by i n d u c t i o n on t h e o r d i on b ( x l ) + b ( x , ) + b ( y , ) b(y,) t o prove
( 1 ) ; and on 2 ( b ( x , ) + b ( y ) ) ( w h i c h e q u a l s 2 ( b ( x , )
+
+
prove ( 2 ) .
To p r o v e ( O ) ,
we m u s t e s t a b l i s h t h e f o l l o w i n g :
b ( y ) ) i f X1
-
X,),
to
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.07
143
T h a t i s , t h e f o l l o w i n g i n e q u a l i t i e s m u s t be e s t a b l i s h e d :
Assume f i r s t t h a t x L ’ 5 x L 3 .
Then, u s i n g t h e i n d u c t i o n h y p o t h e s i s ,
we may a p p l y ( 1 ) a n d c o n c l u d e t h a t P ( x L 1 , x L 3 : y L 1 , y ) h o l d s a n d t h a t
P(xL3,x:yL1 , y R 3 ) holds s t r i c t l y .
Adding t h e two a s s o c i a t e d i n e q u a l i t i e s
together gives us t h e following:
On c a n c e l i n g o u t x L 3 y L 1 a n d s u b t r a c t i n g xL1yL1
+
x L 3 y R 3 f r a n ( 1 11, we
obtain (10,i).
Assume, o n t h e o t h e r h a n d , t h a t xL3 5 xL h y p o t h e s i s , we may a p p l y ( 1 )
’.
Then, u s i n g t h e i n d u c t i o n
and conclude t h a t P ( x L 1 , x : y L 1 , y R 3 ) holds
R s t r o n g l y and t h a t P ( x L 3 , x L 1 : y , y 3 , h o l d s .
A d d i n g t h e two a s s o c i a t e d i n -
e q u a l i t i e s together gives us t h e following:
(12)
xLlyR3 + xyLl
+
xL3yR3
+
xLl Y
<
XL’yL1 + xyR3 +
On c a n c e l i n g o u t xL1yR3 a n d s u b t r a c t i n g xL1yL1 obtain ( 1 0 , i ) again.
+
xL3y
+
,LlyR,.
x L 3 y R 3 fran (121, we
144
Norman L. A l l i n g
4.07
U s i n g t h e i n d u c t i o n h y p o t h e s i s , we m a y a p p l y ( 1 ) a n d c o n c l u d e t h a t P(xL1,x:yL1,y) and P(x,xR4:yL4,y) hold strongly.
Adding the two a s s o c i a t e d
i n e q u a l i t i e s t o g e t h e r gives us t h e f o l l o w i n g :
L (13) x 'y
+
xyL1 + x y
t
xR4yL4
<
xLiyLi
+
xy
+
xyL4
On c a n c e l i n g o u t xy a n d s u b t r a c t i n g x L 1 y L 1
+
+
xR4y.
x R 4 y L 4 from ( 1 3 ) ,
we
obtain (10,ii). Using t h e i n d u c t i o n h y p o t h e s i s , w e may a p p l y ( 1 ) a n d c o n c l u d e t h a t P(x,xR2:y,yR2) and P(xL3,x:y,yR3) hold strongly.
Adding t h e two a s s o c i a t e d
i n e q u a l i t i e s t o g e t h e r g i v e s us t h e f o l l o w i n g :
(14)
xyR2
+
x R 2 y + xL3yR3 + xy
< xy
+
,RzyR2
+
L
3y
+
xyR,*
On c a n c e l i n g o u t xy a n d s u b t r a c t i n g x L 3 y R 3 + x R z y R 2 from ( 1 4 ) ,
we
obtain (10,iii).
Assume t h a t x R z 4 x R 4 . apply
T h e n , u s i n g t h e i n d u c t i o n h y p o t h e s i s , we may
and conclude t h a t
(1)
P(x,xR2:yL4 ,yRz) h o l d s s t r o n g l y .
P(xR2,xRr:yL4,y) holds and t h a t
Adding t h e two a s s o c i a t e d i n e q u a l i t i e s
t o g e t h e r gives u s t h e f o l l o w i n g :
On c a n c e l i n g o u t x R z y L 4 a n d s u b t r a c t i n g x R 2 y R 2 + x R 4 y L 4 fran ( 1 5 1 , we obtain (10,iv).
Assume, on t h e o t h e r h a n d , t h a t x R 4 5 x R 2 .
the induction hypothesis,
Then, u s i n g
we m a y a p p l y ( 1 ) a n d c o n c l u d e t h a t
P ( x , x R 4 : y L 4 , y R z ) h o l d s s t r o n g l y a n d t h a t P ( x R 4, x R 2 : y , y R 2 ) h o l d s . t h e s e two i n e q u a l i t i e s t o g e t h e r gives us t h e f o l l o w i n g :
(16)
xyR2
+
xR4yL4 + xR4yR2
t xR2y
xyL4
+
,hyR2
R $y
+
xR2yR2.
Adding
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.07
On c a n c e l i n g o u t x R 4 y R 2a n d s u b t r a c t i n g x R z y R z + xR'yL' obtain (10,iv) again.
145
f r m ( 1 6 1 , we
T h i s c o m p l e t e s t h e i n d u c t i v e proof of ( 0 1 , t o t h e
n e x t h i g h e r l e v e l of t h e i n d u c t i o n . L e t u s now t u r n o u r a t t e n t i o n t o ( 1 ) .
(17)
Assume t h a t ( 1 . i ) a n d ( 1 , i i ) a r e b o t h s t r i c t : i . e . , ( 1 , i ) x,
t h a t we h a v e
< x, and ( 1 , i i ) y 1 < y,.
By ( 4 . 0 2 : 1 4 ) a n d ( 1 , i ) t h e r e i s a n x Z L w i t h x, 5 x z L o r t h e r e i s a n
x l R w i t h x l R 5 x,:
(18)
i.e.,
( i ) t h e r e i s a n x p L w i t h x i 5 x Z L < x,,
R
( i f )t h e r e i s a n x1
. with x I
Assume t h a t ( 1 8 , i ) h o l d s . f a c t t h a t y1
(19)
<
x,
R
or
6 x,.
We may u s e t h e i n d u c t i o n h y p o t h e s i s ; t h e
< y z ; a n d (3) t o c o n c l u d e t h e f o l l o w i n g :
P(x, ,x,:y, ,y,)
holds s t r o n g l y , s i n c e P(x, ,xZL:y, , y 2 ) h o l d s and
L
P(x, , x z : y l , y z ) h o l d s s t r i c t l y . Assume t h a t ( 1 8 , i f ) h o l d s . f a c t t h a t y1
(20)
We may u s e t h e i n d u c t i o n h y p o t h e s i s ;
the
< y,; and (3) t o c o n c l u d e t h a t R
P ( x I , x p : y l , y a ) h o l d s s t r o n g l y , since P ( x , , x l : y l , y z ) h o l d s s t r i c t l y
R and P(xl , x z : y 1 , y 2 ) holds. Now assume t h a t
(21)
( 1 , i ) i s s t r i c t and t h a t ( 1 , i i ) h o l d s , b u t i s n o t s t r i c t : i . e . , t h a t ( 1 , i ) x,
<
x, and ( 1 , l i ) y1 = y 2 .
146
Norman L. A l l i n g
By ( 4 . 0 2 : 1 4 )
a n d ( 1 , i ) there is a n x p L w i t h x , 5 x Z L or t h e r e i s a n
x I R w i t h x l R 5 x,:
(22)
i.e.,
t h e r e i s a n x Z L w i t h x , 4 x,
(i)
<
( i ' ) there i s a n x l R w i t h x,
Assume t h a t ( 2 2 , i ) h o l d s . f a c t t h a t y , 6 y,;
(23)
4.07
x1
L R
<
x,,
or
5 x,.
We may u s e t h e i n d u c t i o n h y p o t h e s i s t h e
and (3) t o conclude t h e f o l l o w i n g :
P(x, ,x,:yl , y 2 ) h o l d s , s i n c e P ( x , , x Z L : y , ,Y,) a n d P(x,Llx,:Yl
,Y,)
h o i d. Assume t h a t ( 2 2 , i ' ) h o l d s .
We may u s e t h e i n d u c t i o n h y p o t h e s i s ; t h e
f a c t t h a t y , 5 y 2 ; and (3) t o c o n c l u d e t h a t
Now a s s u m e t h a t (25)
( 1 , i ) a n d ( 1 , i i ) h o l d , b u t are n o t s t r i c t : i.e.,
and ( 1 , i i ) y l
=
that ( 1 , i ) x,
x,y,
+
x,y,;
x,
y,.
Note t h a t ( i , i i i ) i s , i n t h i s c o n t e x t , t h e s t a t e m e n t t h a t x 1 y 2 =
=
which follows i m m e d i a t e l y f r a n ( 2 ) .
+
x2yI
Thus l e t u s p r o c e e d t o
prove (2).
We know t h a t x 1
=
R
{ x I L I x , 1 , x,
F u r t h e r , assume t h a t x,
=
x,.
=
{xZLI x,
1, a n d y
that
( i ) { x i L ) a n d [ x , ~ ) are m u t u a l l y c o f i n a l , and t h a t
(ii)
R
R
=
{y
L
I
R
y 1.
By ( 4 . 0 2 : 1 6 ) we know t h a t t h e Conway c u t s
L ( x l L , x l R ) and (x, , x Z R ) are e q u i v a l e n t : i.e.,
(26)
R
{ x , 1 a n d {x, 1 a r e m u t u a l l y c o i n i t i a l .
4.07
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
-
Using t h e i n d u c t i o n h y p o t h e s i s we may a p p l y ( 0 )
147
(2) t o e s t a b l i s h
the following.
(27)
If
(i)
L
X,
( i i ) If x1 ( i i i ) If x,
(iv)
If
XI
R
L R
2 x P L then
R
5 x, 2 x,
L
L
x, y
then
X,
then
X,
R
L
R
y + x2y
R
y + x2y
R
5 x Z R t h e n x, y + x l y
Assume t h a t x1 L 5 x, L
PROOF.
L
+ xly
.
L
L L
-
x1
-
x2 Y
-
x2 Y
-
R R
L R
R L
x1 y
S i n c e yL
2
x2
L +
R
5 x, Y
5
x1
L
Y
+
+
R
x2 Y
+
X2Y X1Y
XlY X2Y
L R R
L
-
.
L L
x, y
R R
- x 1 y . L R
-x1 Y
*
R L
- x , y .
< y , and s i n c e t h e i n d u c t i o n
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( x , L , x Z L : y L , y ) h o l d s : i.e.,
(28)
L x, y
+
L L L L x, y 5 x1 Y
+
L x2 y.
Adding x,yL = x 2 y L t o b o t h sides of ( 2 8 ) we a r r i v e a t t h e f o l l o w i n g . L L L L (29) x , y + x , y + x 2
x,
L L +
x2y
L
L
+
x, y.
R e a r r a n g i n g terms i n ( 2 9 ) g i v e s ( 2 7 , i ) . A s s u m e t h a t x , R 5 x, R
.
Since y
<
yR, and s i n c e t h e i n d u c t i o n
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( x , R , x , R :y,yR) holds: i.e., R R R R (30) x , y + x 2 y 5 x,
+
R R
,
Adding x2yR = x , y R t o b o t h s i d e s of (30) we a r r i v e at t h e f o l l o w i n g . R R
(31) x , y
+
x2y
R +
R R R R R x2 Y 5 x1 y + x l y + x i ? Y .
Rearranging terms in (31) g i v e s ( 2 7 , i i ) .
148
Norman L . A l l i n g Assume t h a t x 1 L S x, L
.
Since y
4.07
yR , and s i n c e t h e i n d u c t i o n
<
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( x , L , ~ , ~ : y R, )y holds: i.e.,
(32)
L R x1 Y
+
L L x 2 y 6 x1 y
+ x2
L R
,
Adding x 2 y R = x l y R t o b o t h s i d e s of ( 3 2 ) we a r r i v e a t t h e f o l l o w i n g .
(33)
L R R L L x1 y + x2y + x 2 y S x 1 y
X,Y
+
L R R + x , y .
R e a r r a n g i n g terms i n ( 3 3 ) g i v e s ( 2 7 , i i i ) . Assume t h a t x 1R 6 x, R
.
S i n c e yL
<
y, and s i n c e t h e induction
h y p o t h e s i s h o l d s , we may i n v o k e ( 1 ) a n d c o n c l u d e t h a t P ( ~ , ~ , x , ~ : y ~ , y ) holds: i.e.,
(34)
R R L x1 y + x, y 5 Adding x l y L
R
(35) x, y + xly
L
=
R L +
R
x2 y .
x 2 y L t o b o t h s i d e s o f ( 3 4 ) we a r r i v e a t t h e f o l l o w i n g .
R L
+ X , Y
R L < x , y
L +
X2Y
+
x,
R
y.
R e a r r a n g i n g terms i n (35) g i v e s ( 2 7 , i v ) ; p r o v i n g a l l
Of
(27).
0
Combining ( 2 6 ) a n d ( 2 7 ) we see t h a t we h a v e p r o v e d t h e f o l l o w i n g .
(36)
(i)
L (xl Y
+
L
(x, Y (ii)
L
txl y L
tx, y
+
+
xly X2Y
L L
xlyR
x,yR coinitial. +
-
L L
x1 y
L L
,
x1
R
t
R
-
x2 Y I x 2 Y L R R x1 Y I x , Y
-
x2
L R R Y I x2 Y
xly
R
+
x2yR L x1y
+
X,Y
+
L
-
xlRyR} and
-
x, y I a r e m u t u a l l y c o f i n a l .
-
R L x1 y 1 a n d
R R
xZRyLa ~re m u t u a l l y
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.07
By ( 4 . 0 2 : 1 6 )
149
a n d ( 3 6 ) we s e e t h a t ( 2 ) h o l d s a t t h e n e x t s t a g e i n t h e
induction.
We may now u s e ( 2 ) a t t h e l e v e l ( w h i c h h a s j u s t b e e n p r o v e d ) t o p r o v e (1).
T h u s ( 1 ) i s e s t a b l i s h e d a t t h e n e x t h i g h e r l e v e l of t h e i n d u c t i o n .
-
Thus, by i n d u c t i o n , ( 0 ) (37)
If 0
< x
=
xy.
<
y , i n No, t h e n 0
<
xy.
By ( 1 1 , P ( O , x : O , y ) h o l d s s t r i c t l y ; t h u s 0
PROOF. xy
and 0
( 2 ) have been proved.
=
Oy + x 0
<
0.0
+
0
Conway g a v e t h e f o l l o w i n g i l l u m i n a t i n g v e r s i o n of t h e d e f i n i t i o n of xy C24, p . 191:
(39)
xy
=
(xy - (x xy
4.08
+
- x
L L
(x - x )(Y
L y 1,
-
)(Y R
-
XY
Y ) , xy
+
(xR
-
(xR -
X)(Y X)(Y
R
-
Y)J y
L
11.
THE ASSOCIATIVE LAW FOR MULTIPLICATION I N No
LEMMA.
L e t x , y , a n d z b e i n No; t h e n ( x y ) z
=
x ( y z ) , which w i l l be
d e f i n e d t o b e x y z , as u s u a l . PROOF.
We w i l l p r o c e e d by i n d u c t i o n , on b ( x )
d e f i n i t i o n we s e e t h a t we h a v e t h e f o l l o w i n g .
+
b(y)
+
b z).
BY
Norman L . A l l i n g
150
4.08
Expanding t h e e x p r e s s i o n s f o u n d i n ( 0 ) ( i n ( 1 ) t h r o u g h ( 8 ) ) a n d ( 0 ' ) ( i n ( 9 ) t h r o u g h (1611, a n d i n d u c t i o n g i v e s r i s e t o t h e f o l l o w i n g .
(1)
x Ly z
+
xy Lz
x y zL
-
x Ly L z
-
xLyzL
-
xyLzL + x Ly L 2 L ,
(2)
x Ry z
+
xy R z + x y z L
-
x Ry R z
-
xRyzL
-
xyRzL + x Ry R z L ,
(3)
x Ly z
+
xy R z
+
xyzR
-
xL y R z
-
xLyzR - x y R z R + x L y R z R ,
(4)
x Ry z
+
xy L z
+
xyzR
-
x Ry L z
-
x R y z R - xyLzR
(5)
x Ly z
+
xy L z + x y z R
-
x Ly L z
-
xLyzR
-
(6)
x Ry z
+
xy R z
+
xyzR
-
x Ry R z
-
xRyzR
- xyRzR + x Ry R zR ,
(7)
x Ly z
+
xy R 2
+
xyzL
-
xLy R z
-
xLyzL
-
xyRzL
(8)
x Ry z
+
x y Lz
+
x y zL
-
x Ry L z
-
xRyzL
-
xyLzL + x Ry L z L ,
(9)
x Ly z
+
xy L z + x y z L
- xLy L z -
xLyzL
- xyLzL
(10)
L x yz
+
xy
R2
+
xyzR
- xLyR z -
xLyzR
-
xyRzR + ,LyRzR,
(11)
x Ry z
+
xy L z
+
xyzR
-
xRyLz
-
xRyzR
-
xyLzR + x Ry L zR ,
(12)
x Ry z
+
xy R z
+
xyzL
-
x Ry R z
-
xRyzL
-
xyRzL + x Ry R z L ,
(131 x L YZ
+
xy L z
+
x y zR
- xLyL z - xLyzR - xyLzR
(14)
x LYZ
+
xy R z
+
x y zL
-
x LyR z
- xLyzL -
(15)
x Ry z
+
XY L Z
+
x y zL
-
x Ry L z
-
+
xRyzL
-
+
x Ry L z R ,
xyLzR + x L y L zR ,
+
xLyRzL,
+ x Ly LzL ,
+ x LyL zR ,
xyRzL + x Ly R z L , xyLzL
t
x Ry LzL ,
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.08
(16)
R
x yz
+
R
xy z
+
xyzR
-
R R
x y z
-
R R R
xRyzR - xyRzR + x y z
151
.
The r e a d e r s h o u l d n o t e t h a t each o f t h e s e l a s t 1 6 expressions i s
determined by t h e c h o i c e of a n " L "
o r a n "Rfl o v e r e a c h o f x , y , a n d z.
( T h i s may most e a s i l y be s e e n by l o o k i n g a t t h e s u p e r s c r i p t s on t h e extreme
r i g h t hand term of each e x p r e s s i o n . ) (i,j,k),
(L,L,L),
These e x p r e s s i o n s w i l l t h e n be c a l l e d
where i , j , and k a r e e i t h e r L o r R. (2) is the expression ( R , R , L ) ,
etc.
Thus ( 1 ) i s t h e e x p r e s s i o n From t h i s a n a l y s i s we s e e
t h a t t h e following holds:
On t h e o t h e r hand,
S i n c e t h e s e two e x p r e s s i o n s ( ( 1 7 ) and ( 1 8 ) ) a r e i d e n t i c a l , w e have proved t h e Lemma. Thus we h a v e p r o v e d t h e f o l l o w i n g c o r o l l a r y t o t h e Lemma, and much t h a t precedes i t : THEOREM.
No i s an o r d e r e d commutative r i n g , w i t h 1 # 0.
Note, i n p a s s i n g , t h a t t h e r e i s a s i m p l e r u l e f o r d e s c r i b i n g t h e v a r i o u s o p t i o n s of xyz i n ( 1 7 ) and ( 1 8 ) ; namely t h e f o l l o w i n g . ( i ) A l l 1 6 c h o i c e s of
i,j,lE{L,R}.
(i,j,k)
o c c u r a s o p t i o n s of x y z , where
( i i ) 8 of t h e s e a r e l e f t o p t i o n s and 8 a r e r i g h t o p t i o n s .
( i i i ) ( i , j , k ) i s a l e f t o p t i o n i f f i t has an odd ( r e s p . even) number of L ' s
(resp. R ' s ) i n it.
( i v ) ( i , j , k ) i s a r i g h t o p t i o n i f f i t h a s a:) even
( r e s p . odd) number of L ' s ( r e s p . R ' s ) i n i t . The f o l l o w i n g was noted by Conway C24, p . 191.
(C.f.,
(4.06:41.)
4.08
Norman L . A l l i n g
152
xy
(19)
=
- x
Ixy - ( x L
xy - ( x - x ) ( Y 4.09
-
L
-
L
R
y ) , xy - ( x - x ) ( Y - y R R L y 1, xy - (x - x ) ( Y - y 1 ) . )(y
R
)I
ON NUMBERS G I V E N BY REFINEMENTS OF ( T I M E L Y ) CONWAY C U T S
We have s e e n i n S e c t i o n 4.02 t h a t e a c h XENO e q u a l s (LIR], where ( L , R ) i s a t i m e l y Conway c u t i n No: i . e . ,
( L , R ) i s a Conway c u t i n O b ( x ) .
Let L ' a n d R' be s u b s e t s of No s u c h t h a t ( i ) L ' < ( X I < R ' ;
(ii)for
a l l xL EL t h e r e e x i s t s x L ' E L ' s u c h t h a t xL 6 x L ' ; a n d ( i i i ) f o r a l l R R x E R t h e r e e x i s t s x ~ ' E R ' s u c h t h a t xR' 5 x
.
Then ( L ' , R ' )
w i l l be
c a l l e d a r e f i n e m e n t of ( L , R ) . be a r e f i n e m e n t of ( L , R ) ;
Let (L',R')
PROOF.
Let I
L
< (y} <
a n d I' = ( Y E N O :
R},
=
(LIR}.
L'
<
(y]
<
R'}.
we s e e t h a t X E I 'a n d t h a t I ' i s a s u b i n t e r v a l of t h e
By ( O , ( i ) - ( i i i ) ) i n t e r v a l I.
= (YENO:
then ( L ' I R ' )
Since x i s t h e s i m p l e s t element i n I , i t is c e r t a i n l y t h e
s i m p l e s t element of 1'. be a r e f i n a n e n t of ( L 1 , R 1 ) ,
Let (Ltl,R1')
ments of ( L , R ) ; (L',R')
and l e t ( L ' , R ' )
be refine-
t h e n ( L f f , R t l ) i s a r e f i n e m e n t of ( L , R ) .
i s p o s s i b l y an u n t i m e l y , r e p r e s e n t a t i o n of x ; w h e r e b y a
r e p r e s e n t a t i o n t h a t i s p o s s i b l y u n t i m e l y ( r e s p . i s u n t i m e l y ) we mean a n o t b e s u b s e t s of O b ( x ) Ob(x)).
x , b u t f o r which L ' a n d R ' may ( r e s p . t h e u n i o n of L ' and R' is n o t a s u b s e t of
i n No s u c h t h a t ( L ' I R ' )
Conway c u t ( L ' , R ' )
=
( S e e S e c t i o n 4.02 f o r d e t a i l s . )
We have s e e n t h a t 0
EXAMPLE 0 .
of No f o r which L ' s e n t a t i o n of 0.
<
(0) < R ' ;
-
[
then ( L ' , R ' )
I f t h e u n t o n of L ' and R'
u n t i m e l y r e p r e s e n t a t i o n of 0. n o t an i s o l a t e d o c c u r r e n c e .
I}.
Let L ' and R ' be any s u b s e t s
is possibly an untimely r e p r e is non-empty,
then (L',R')
is a n
The r e a d e r c a n e a s i l y v e r i t y t h a t t h i s i n
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.09
(2)
153
Let y
=
L R L R { y Iy ]&NO, a n d l e t IyL',yR') b e a r e f i n e m e n t of f y , y 1.
x
=
(x
+
y
L' +
Y, x
+
x
YL'I
R'
+ y, x
+ yR'].
By c o n d i t i o n ( 0 1 , a n d t h e f a c t t h a t No i s a n o r d e r e d g r o u p
PROOF.
( 4 . 0 8 ) , we know t h a t f o r each xL a n d x R t h e r e e x i s t s x L ' a n d xR' s u c h t h a t
<
xL + y 5 xL' + y
x
y
+
<
xR'
+
y S xR
+
y ; a n d we know t h a t f o r each y L
and yR t h e r e e x i s t s y L ' a n d y R ' s u c h t h a t x + yL 5 x x + y R ' I x + yR .
(31
+
yL'
< x
+
y
<
0
L' L' L' L' R' xy=(x y + x y - x y , x y + x y - x L' R' L' R' R' L' R' L' x Y + X Y - x y , x y + x y - x y }. PROOF of ( 3 ) .
U s i n g ( 0 ) a n d t h e f a c t t h a t No i s a n o r d e r e d r i n g
( 4 . 0 8 1 , we know t h a t t h e f o l l o w i n g h o l d .
(4)
F o r each xL a n d yL t h e r e e x i s t x L ' a n d y L ' f o r w h i c h
(i)
L
x y
+
xyL
-
xLyL 6 xL'y
+
xy
L' - x L' y L' < x y .
( i i ) F o r each xR a n d y R t h e r e e x i s t x R ' a n d y R ' f o r w h i c h
R x y
+
xyR
-
xRyR 5 x R ' y + xy
R'
( i i i ) For each xL a n d y R t h e r e exist
L
X Y
(iv)
+
R
PROOF of
L'
+
+
x y R'
- xL' y R' >
xy.
xyL
-
xRyL 2 xR'y + xyL'
-
xR'yL'
> xy.
4).
b y
consequence, 0 t h u s xL y
xLyR h xL'y
xL' a n d y R ' f o r w h i c h
f o r e a c h xL, t h e r e e x i s t s xL' 2 x L ; a n d f o r each y L t h e r e
By ( O ) ,
exists y
-
xy.
F o r each xR a n d yL t h e r e e x i s t x R ' a n d y L ' f o r w h i c h X Y +
(i)
xyR
- x R' y R ' <
x yL
L
< -
.
Thus, 0
xy - x
L'
< (x - xL')(y - y
y - xy
L'
+
xLyL 5 xL'y + x y L '
xL'
-
Y
L'
xL'yL'
L'
L (x - x )(y - yL).
AS a L L L L 6 X Y - X Y - x y + x y , a n d
<
) 5
xy.
Norman L. A l l i n g
154
4.09
x R , there e x i s t s x R ' 5 xR : a n d f o r each y R t h e r e R' R R Thus 0 < ( x - x R ' ) ( y - y ) 2 ( x - x ) ( y - y 1. As a R' R' R' R' R R R R
( i i ) By ( 0 1 , f o r e a c h
exists y
R'
4 yR.
<
consequence, 0 R
t h u s x y + xyR
-
xy
-
x
y - xy
xRyR 5 x
R'
+ x
-
y + xy R'
y
xR'yR'
( i i i ) By ( 0 1 , f o r e a c h xL, t h e r e e x i s t s x
R'
L'
2
x
L
L R xL'yR' 5 x y + xy
-
L R x y
By ( 0 ) . f o r each x R , t h e r e e x i s t s x
R'
4 x
Thus 0
- xL ' y
consequence, 0 > xy t h u s xy
< xL ' y
+
xyR'
-
- xy
R'
xL')(y +
+ x y ,and
; and f o r each yR t h e r e
L R x ) ( y - y 1. A s a x L ' y R' t x y - x Ly - x y R + x Ly R, a n d R'
y
5 yR.
x y - x y
< xy.
-
exists y
(iv)
> (x -
-
2 xy
) 2 (x
-
.
R
; a n d f o r e a c h yL t h e r e
L
.
consequence, 0
>
R' x y
+
R ' L' R xy - x y - x y L ' + x y L xy x y - xyL + x R y L , a n d L' R ' L' R L R L xy x y 5 x y + xy - x y i p r o v i n g ( 4 ) .
exists Y
t h u s xy
L'
<
2 Y
Thus 0
> (x - x
R'
)(y
-
y
L'
1 L (x
-
R
L
x ) ( y - y 1.
AS
a
-
R'
-
To c o m p l e t e t h e proof of ( 3 ) , we c a n r e a s o n as we d i d i n t h e proof of (1).
a
PROPERTIES OF DIVISION I N No
4.10
THEOREM. t h a t xy = 1 .
-
Were x then -x(z)
Let x b e i n No, w i t h x n o t 0.
T h e r e e x i s t s y i n No s u c h
Thus, No i s a n o r d e r e d f i e l d .
<
0 , a n d were -x t o h a v e a m u l t i p l i c a t i v e i n v e r s e z i n No,
1 = x(-z),
showing t h a t x h a s a m u l t i p l i c a t i v e i n v e r s e i n No.
Thus t o p r o v e t h e Theoren i t s u f f i c e s t o p r o v e t h e f o l l o w i n g . (0)
>
Assume t h a t x
0 in
No.
T h e r e e x i s t s a YENO s u c h t h a t xy = 1 .
I n t h i s s e c t i o n , t h e f o l l o w i n g r e s u l t w i l l a l s o b e of u s e . (1)
T h e r e e x i s t s u b s e t s L and R of t h e s e t of p o s i t i v e elements of No such t h a t L
<
t i m e l y (4.02).
R , w i t h x = { O , L I R ) , f o r which t h i s r e p r e s e n t a t i o n i s
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.10
PROOF of ( 1 ) . timely (4.02).
By d e f i n i t i o n x
>
Since x
t h a t 0 5 xL ( 4 . 0 2 : 1 4 ) .
=
[x
L
0 , and s i n c e 0
Let L
=
L
[ x : xL
R x 1 , the r e p r e s e n t a t i o n being
I
[I},
=
155
t h e r e must e x i s t a n xL s u c h
> 01, and
let R
=
[x
R
1.
Then x
=
[O,LIR}; e s t a b l i s h i n g ( 1 ) .
( 1 ) We may write x a s [ O , x L l x R } , w i t h e a c h
Conway d e f i n e d t h e m u l t i p l i c a t i v e i n v e r s e y of x as f o l l o w s
being timely. [24,
xL > 0 , t h e r e p r e s e n t a t i o n
p . 211.
(1
+
(xL
L L x)y ) / x , ( 1
-
+
(xR
-
R R x)y ) / x I.
Conway writes t h e f o l l o w i n g a f t e r t h i s d e f i n i t i o n .
"Note t h a t ex-
p r e s s i o n s i n v o l v i n g yL and yR a p p e a r i n t h e d e f i n i t i o n o f y . t h a t r e q u i r e s u s t o ttexplaintt t h e d e f i n i t i o n .
It is this
The e x p l a n a t i o n i s t h a t we
r e g a r d t h e s e p a r t s of t h e d e f i n i t i o n as d e f i n i n g new o p t i o n s f o r y i n terms So even t h e d e f i n i t i o n of t h i s y is an i n d u c t i v e one. t [ T h i s
of o l d o n e s .
i s i n a d d i t i o n t o t h e t t o t h e r t l i n d u c t i o n s by which we s u p p o s e t h a t i n v e r s e s L
for the x
and x
R
have a l r e a d y been f 0 u n d . 1 ~ ~C24, p . 21.1
In a footnote t
[24, p . 211 Conway g i v e s t h e f o l l o w i n g v e r y i n s t r u c t i v e example. Example 0. as [0,21}.
Let x R
=
3.
Thus x c a n be w r i t t e n , as i t is above i n ( l ) ,
.
Hence [ x ) i s empty, and xL
=
2.
By (21, t h e g e n e r a l f o r m u l a
for y is
Let yoL
=
0.
Note t h a t yoL
BY ( 3 ) , we s e e t h a t y I R
=
Let us make no c h o i c e of a y o
S i n c e t h e r e i s no y o R , t h e r e i s n o y 1
1/2.
U s i n g ( 3 ) , we s e e t h a t y Z L = 1 / 4 . y S R = 11/32, y G L = 21/64, y,R
< 1/3.
=
T h i s t h e n y i e l d s y,R = 318, y,L
43/128,
...
=
R
L
. .
5/16,
Writing t h i s i n t h e customary
Norman L. A l l i n g
156
way g i v e s y = ( 0 , 1/4, 5/16, 21/64,
... I
4.10
1/2,
3/8,
11/32,
The decimal v e r s i o n of t h i s i s even more i n s t r u c t i v e . {O,
0.25,
0.3125,
0,328125,
... I
0.5,
0.375,
0.34375,
431128,
...
I.
It is t h e following: 0.3359375,
... ) .
I n t r y i n g t o u n d e r s t a n d ( 2 ) more f u l l y , n o t e t h a t e x c e p t f o r 0 t h e o p t i o n s g i v e n i n (2) a r e d e t e r m i n e d i n p a r t by t h e c h o i c e of w h i c h o p t i o n o n e c h o o s e s t o c o n s i d e r , r i g h t or l e f t .
Let u s make t h e f o l l o w i n g
def i ni t i o n s .
(4)
(i)
R
-
x)y ) / x ,
be d e f i n e d t o be ( 1 + ( x L
-
x)y ) / x ,
Let [R:Ll(y) b e d e f i n e d t o be ( 1 + ( x
( i i ) l e t [L:R](y)
L
R
R
L
( i i i ) l e t [ L : L l ( y ) b e d e f i n e d t o be ( 1
+
( x L - x ) y L ) / x L , and
l e t [R:Rl(y) b e d e f i n e d t o be ( 1
+
(xR
(iv)
-
R R x)y )/x
.
Then, a c c o r d i n g t o (21,
PROOF of ( 0 ) .
Note t h a t (0) i s t r u e i f x
<
are no s i m p l e r p o s i t i v e e l e m e n t s ; t h u s f o r b ( x ) f o r which ( 0 ) i s t r u e f o r a l l X E N O , w i t h 0
< x
-
1.
Note a l s o t h a t t h e r e
2, (0)i s t r u e .
and b ( x )
<
Let a e o n
a.
We w i l l s t a r t o f f t h e f i n i t e i n d u c t i o n , as i n t h e e x a m p l e a b o v e , by d e f i n i n g y o t o b e 1.
(6)
Recall t h a t 1 = {Ol}: t h u s
yoL = 0, and there are no Y, R . Consider t h e f o l l o w i n g i n e q u a l i t y :
(7,n) ( i ) x y k <
1 , f o r a l l y,
L : and ( i l l 1
N o t e t h a t (7,O) i s t r u e , y I R be d e f i n e d t o be [L:L](y,); hypothesis,
<
xyn R , a l l yn R
.
Let u s d e f i n e y I L t o b e [R:L](y,),
and l e t
then, using our ( t r a n s f i n i t e ) induction
I n t r o d u c t i o n t o t h e s u r r e a l number F i e l d No
4.10
S i n c e No i s a n o r d e r e d r i n g , (7,l) h o l d s .
Let ncN such t h a t y n L a n d
Let y n + l L and y n+ 1
ynR a r e d e f i n e d , For which ( 7 . n ) holds.
= (1 +
(xR
-
x)yk)/x R , o r
(1
+
(xL
-
x)ynR)/xL; and
( i i i ) y n + l R = C L : L I ( ~ ~=) ( 1
+
(xL
-
x)y:)/xL,
+
(xR
-
x)yn )/x
=
Y,+~
(i)
(ii) Y
(iv)
and t h e ex-
be d e f i n e d a s f o l l o w s :
p r e s s i o n [*:.](.)
(9)
157
[R:L](y,)
~ += [L:R](yn) ~
=
Yn+l
=
[R:R](yn)
Now l e t t h e e x p r e s s i o n
(1
=
or
R
R
.
be d e f i n e d as f 011ow9 :
{ a : - ) ( . )
( i i i ) xyn+lR
-
1 = ((x
-
x )(I
L
-
xy,
(iv)
XY,+~
-
1 = ((x
-
R x )(1
-
'Yn
Using
(lo),
L R
we see t h a t t o prove t h a t (7,n+l) h o l d s , i t s u f f i c e s t o
prove t h e f o l l o w i n g :
PROOF of (11). that x
-
x
R
<
0,
Since
1
-
x < x
xynL
>
R
,
a n d s i n c e xynL
0, a n d h e n c e ((x
-
<
1 ( 7 , n , i ) , we s e e
R x )(1
-
xynL))/xR
-
158
Norman L. A l l i n g
{R:L](yn)
<
0; proving ( i ) .
>
s e e t h a t x - xL
<
0; proving ( i i ) .
x
-
x
R
<
0,
L
Since
-
1
xy
<
xynR
Since
x - xL > 0 , 1 - xy n proving ( i i i ) .
-
0, 1
Since
xL
<
xL
a n d since 1
((x - x
0, a n d
L
)(1
x, and since x y t
>
0, and ( ( x
x
< xR ,
<
0, and ( ( x
n
< x,
4.10
-
L x )(1
and s i n c e 1
-
<
xy
R x )(1
<
-
<
xynR ( 7 , n , i i ) , we
xynR))/xL
=
{L:R}(yn)
1 , ( 7 , n , i ) , w e see t h a t
x y t ) ) / x L = {L:Ll(yn) n
-
>
0;
( 7 , n , i i ) , we see t h a t xynR))/x
R
=
(R:R}(yn)
>
0;
p r o v i n g ( i v ) , a n d h e n c e a l l of ( 1 1 ) . T h u s we s e e t h a t ( 7 , n + l ) h o l d s .
Hence, by i n d u c t i o n , (7,111 holds f o r
all ncN.
(12)
F o r a l l m , ncN, y,
PROOF of ( 1 2 ) .
Since x
>
L
<
yn
R
. By (7.m) a n d (7,111. x y m L
L e t m , ncN.
0 , we see t h a t (12) f o l l o w s .
o
L R Let y b e d e f i n e d t o b e ( y n I y n 1 .
(13)
( i ) (xyIL
-
1
<
0, and ( i i ) 0
<
The f o l l o w i n g h o l d s :
(xyIR
-
1.
I n order t o p r o v e ( 1 3 1 , l e t u s e s t a b l i s h t h e f o l l o w i n g .
(14)
R
1
+
xR (y - [R:L](yn)),
- xLynR = 1
+
L x (y
-
[L:R](yn)),
L
-
xLyk
1
+
L x (y
-
[L:L](yn)),
xynR
-
xRyt = 1
+
xR ( y
-
[R:R](yn)).
(i)
X
Y
+
xy
(ii)
L x y
+
xy n
L (iii) x y
+
xy
xR y
+
(iv)
-
n
xRyk
=
=
and
<
1
<
xyn
R
.
I n t r o d u c t i o n t o t h e s ur r ea l number f i e l d No
4.10
R x y
PROOF of ( 1 4 ) . =
1
f
x R (y
xLy + x y R
n
1
+
-
(1
L
L 1 + x (y R
X Y f X Y
( x R - x ) y n L ) / xR )
f
(xL - x)ynR)/xL)
f
L
- x Ly,
-
(1
+
R
- x
R
L
1
=
Yn
R
1
=
-
L
XYn
+
L
- x
1
=
f
xL(y
( x Ly - 1 - x Ly, L
f
(xL - x ) y t ) / x R (x y
f
L
= 1 +
-
- x
1
x ) y n R ) / xR )
-
-
1
- x Ry nL
xynL))
f
[R:L](yn)), proving ( i ) .
R
+
[L:RI(y,)).
proving ( i i ) .
xynL)) =
L
x (y - C L : L I ( ~ , ) ) , proving ( i i i ) .
yn
R
+ xynR)) =
-
1 + xR (y
=
-
[R:RI(y,)),
proving ( i v ) .
o
Note t h a t by ( 1 4 , i i i ) a n d ( g , i i i ) , we h a v e
PROOF of ( 1 3 ) .
x Y
1 + xR ( y
=
R (x y
f
n
- ( 1 + (xR
1 +X(Y
1
=
R = l + ( xL y - 1 - x Ly R + x y n R ) ) =
-"n
x L (y - (1
X Y f X Y
xynL - xRynL
f
159
-
Y,
-
1 = xL ( y
[L:L](yn))
xL ( y
=
-
yn+lR)
<
0.
By
( 1 4 , i v ) , ( 9 , i v ) , a n d t h e d e f i n i t i o n o f y , we h a v e
R x Y
- x
XY,
f
-
-
1 = xR ( y
C R : R 1 ( y n ) ) = xR ( y
-
+ xynR
-
xLynR
-
-
1 = xL ( y
[L:R](yn))
=
xL ( y - y n + l L )
R ( g , i ) , a n d t h e d e f i n i t i o n of y , we h a v e x y + x y
R
C R : L I ( ~ , ) ) = x (y
PROOF of
R
t h e r e are no 1
-
yn+,L)
(15).
.
>
0 ; proving ( 1 3 , i i ) .
-
Let xy
By (131, z
z.
L < 1 < zR
n
<
0 ; proving
Were z
<
=
1 S z
L
Since 1 5
f o r sane z
z S. 1 , z
L
-,
0.
By ( 1 4 , i ) ,
L - l = x R( y -
(Ol}; t h u s l L
=
1 t h e n by ( 4 . 0 2 : 1 4 ) ,
R
(4.02:14),
>
o
Recall t h a t 1
.
-'Yn
0 o r zR S 1 f o r some z ; b o t h o f w h i c h a r e a b s u r d .
absurd.
yntlR)
By ( 1 4 , i i ) , ( 9 , i i ) , a n d t h e d e f i n i t i o n of y , we h a v e
(13,i).
x Ly
*
Y,
Were 1
< z
0 and z 5
t h e n by
o r l R 5 z f o r some l R ; b o t h o f w h i c h a r e
1.
By ( t r a n s f i n i t e ) i n d u c t i o n we have p r o v e d (0).
4.10
Norman L. A l l i n g
160
S i n c e ( 0 ) i m p l i e s t h e Theorem, t h e Theorem i s p r o v e d .
Having made a l l t h e s e c a l c u l a t i o n s , we can now s e e more
CONCLUSION.
2 ) comes a b o u t .
c l e a r l y how t h e e x p r e s s i o n define y ( =
o
I Y L I Y R 1 ) and prove t h a t xy
must know t h a t ( x y )
L
<
1
<
=
I n o r d e r t o prove ( 0 ) we m u s t 1.
I n o r d e r f o r xy t o be 1 , we
(13) must hold.
xyIR: i . e . ,
I n checking t o s e e
t h a t (13) does i n d e e d h o l d , e x p r e s s i o n s of t h e t y p e t h a t o c c u r on t h e l e f t hand s i d e of ( 1 4 ) m u s t be c o n s i d e r e d .
I n order t o study these expressions,
q u a n t i t i e s of t h e t y p e t h a t occur o n t h e r i g h t - h a n d s i d e of ( 9 ) a r i s e . These elements of No engender ( 2 ) ' and p r o v i d e t h e f o r m u l a e t h a t move t h e f i n i t e i n d u c t i o n frcm s t a g e n t o s t a g e n
1.
+
Even though t h i s l i n e o f r e a s o n i n g p r o v i d e s a m o t i v a t i o n f o r ( 2 1 , d o e s n o t r e d u c e t h e a u t h o r ' s admiration of Conway's i n s i g h t .
see t h a t t h e r i n g No i s a f i e l d seems r e m a r k a b l e i n d e e d .
it
Indeed, t o
To p r o v e i t i n
t h e way t h a t Conway d i d seems t o t h e a u t h o r l i t t l e s h o r t of i n s p i r e d . 4.20
DISTINGUISHED SUBCLASSES OF No
(1 ]
We h a v e shown t h a t element 1 i n No,
Let
l e t (n
n.1
+
11.1
=
nEN
i s t h e e l e m e n t 0 i n No, and t h a t
f o r which no1
+ 1 = {0,1,2,
{0,1,2,
=
...,n -
...
l,nl].
{Ol
is t h e
, n - 1 1 ) i n No; t h e n
Thus, by f i n i t e induc-
t i o n , t h e f o l l o w i n g i s proved:
(0)
For a l l n i n N , n.1
=
(O,l,2,
... , n
-
11 )EN,
I t i s convenient t o i d e n t i f y ncN w i t h n.lENo, s u b s e m i - g r o u p of No.
-1
(n)).
and t h u s r e g a r d N as a
We can a l s o c o n s i d e r t h e element I N [ ] , and c a l l i t w
as Conway d o e s [24, p . 121. (1)
(= b
Clearly n
<
w f o r a l l nEN, t h u s
No is a non-Archimedean f i e l d . The class On of von Neumann o r d i n a l s w a s d e s c r i b e d i n s e c t i o n 1 . 0 2 .
R e c a l l t h a t 0 i s t h e empty s e t , t h a t 1
=
{O}, etc.
Recall a l s o t h a t i f a
i s i n On, t h e n i t s s u c c e s s o r i s a u n i o n {a]. I t i s n a t u r a l t o a s s o c i a t e
4.20
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
OEOn w i t h f ( 0 )
=
0
t o let f(1)
[]]ENo,
=
f ( 6 ) = { ( f ( u ) ) a < B 11.
=
1
=
{O]]eNo,
Let BEOn and l e t
Then, one can e a s i l y see t h a t
f is a n o r d e r - p r e s e r v i n g map of On i n t o No.
b(f(8))
161
F u r t h e r , f o r a l l $€On,
B.
=
Fran t h i s we s e e , among o t h e r t h i n g s , t h a t
No i s a p r o p e r class. On o c c a s i o n , we w i l l i d e n t i f y BEOn w i t h f(B)cNo, even though t h e y a r e (To s e e t h i s t a k e No f o r example t o
q u i t e d i f f e r e n t objects i n s e t theory.
be g i v e n by t h e C u e s t a D u t a r i c o n s t r u c t i o n , g i v e n i n S e c t i o n 4.02.) ELEMENTS OF No H A V I N G F I N I T E BIRTHDAY
4.21
Let nEN, and c o n s i d e r t h e f o l l o w i n g a s s e r t i o n s .
(0,n) lNnl
= 2
n
,
10nl
=
2
n
-
1 , and 1M
n
I
n+ 1
=
2
...
,
PROOF.
C l e a r l y (0,O)
For each x
holds.
one p o i n t i n M
,
, we
(LIRIeN n+
=
( = 0 n+l
1.
(O,l),
hold.
that (Mn[
=
2n+’
-
F u r t h e r , i f b o t h L and R a r e non-empty t h e n x
Clearly Nn+l
1.
t h a t has a s u c c e s s o r i n M
l e a s t e l e m e n t of M n ) “+I
I
=
2”+l- 1
-
1
+
Let neN f o r which ( 0 , n )
can t a k e L and R t o c o n s i s t of a t m o s t
can be t a k e n t o be t h e immediate s u c c e s s o r of x
n
1.
For a l l neN, ( 0 , n ) h o l d s
(1)
M
-
*
n’
L
in M
n‘
By ( 0 , n ) we know
has one p o i n t i n i t f o r e a c h e l e m e n t of
p l u s t w o more p o i n t s ,
{ ] u ) ( u being t h e
and { v l ] ( v b e i n g t h e g r e a t e s t element i n M , ) . 2
=
2
n+l
R
., e s t a b l i s h i n g
(O,n+l).
Thus
162
Norman L. A l l i n g
4.21
S i n c e No i s an o r d e r e d f i e l d ( 4 . 1 0 ) , i t i s a f i e l d of c h a r a c t e r i s t i c 0, thus
(2)
t h e prime f i e l d of No i s t h e f i e l d Q of r a t i o n a l numbers. Any r a t i o n a l number c a n , of c o u r s e , be w r i t t e n as a / b , f o r acZ and
bsN.
a / b w i l l be s a i d t o b e i n r e d u c e d f o r m i f a a n d b a r e r e l a t i v e l y C l e a r l y e v e r y r a t i o n a l number x may be w r i t t e n i n r e d u c e d form.
prime.
F u r t h e r , r e d u c e d forms a r e u n i q u e .
Let a / b b e t h e r e d u c e d form f o r x .
x w i l l be c a l l e d d y a d i c i f t h e r e e x i s t s neZ+ for which b = 2 b e t h e s e t of a l l d y a d i c numbers.
.
Let D
C l e a r l y D i s a s u b r i n g of Q t h a t con-
Further,
t a i n s 1/2.
(3)
n
D i s t h e s m a l l e s t s u b r i n g of Q t h a t c o n t a i n s 1 / 2 .
(1) If b = 2n,
Let XED be w r i t t e n as a / b i n r e d u c e d form.
LEMMA 1 .
f o r sane ncZ+, t h e n ( i i ) x = { x
-
2-"1x
2-")
+
i n No.
Note t h i s r e p r e -
s e n t a t i o n may n o t be t i m e l y .
Let n
PROOF.
t h e n x is i n Z , x i s i n N
= 0;
1x1 '
d e f i n e t o b e m ) i s of t h e form [ m - 1 1 } (4.20:O). s i m p l e s t e l e m e n t of No b e t w e e n m - 1 a n d m + 1 .
-
s i m p l e s t e l e m e n t b e t w e e n -m - 1 a n d -m ( i i ) , provided n
Note t h a t I m
0.
-
>
is.
(4)
D e f i n e z t o be [x
(5)
22 = z + z =
(6)
z
+
x
-
2-"
<
22
F u r t h e r , -m
+
1 ) ( r e s p . [-m
whereas {m
-
-
+
-
2-"Ix
x
-
< z
2-"12
Then
+ 2-"].
+
+ x + 2-".
x
+
2-"];
thus,
Fran ( 4 ) we see t h a t
is the
x h a s t h e form 1 ) -m
1 ) ) is
+
11) ( r e s p .
0 and assume t h a t ( i ) and ( i i ) h o l d s f o r n
+ 1))
IZ
Clearly m is then t h e
+ 1 , showing t h a t
11 m
n o t a t i m e l y r e p r e s e n t a t i o n of m ( r e s p . -m), Now l e t n
a n d 1x1 ( w h i c h we
-
1.
[I
-m
I n t r o d u c t i o n t o t h e surreal number f i e l d No
x - 2
-n
<
163
z < x+2-".
On combining ( 7 ) a n d (61, w e f i n d t h a t 2x - 2 -(!l-l)
< x +
ZX - 2-(n-1)
<
- 2 -n <
z
< zx
+
22
< x +
z
+ 2-" < 2x + P - 1 )
; hence,
2-(n-1)*
A p p l y i n g t h e i n d u c t i o n h y p o t h e s i s t o 2 x , we know t h a t 2x i s t h e s i m p l e s t element i n No s u c h t h a t t h e f o l l o w i n g h o l d s :
-
2x
(10)
<
2- ( I P 1 )
2x
<
2x + 2
-(n-l)
F r a n ( 9 ) we know t h a t 22 i s i n t h e f o l l o w i n g i n t e r v a l :
(ZX
-
2 - ( n - l ) ,2x + 2 - ( n - 1 ) ) ,
(11)
t h e same o p e n i n t e r v a l i n No t h a t c o n t a i n s 2x
F r a n ( 7 ) we know t h a t
(10).
2
-
<
2-n
x
<
z + 2-".
Adding x t o b o t h s i d e s of ( 1 1 ) g i v e s u s
(12)
- 2 -n <
x + 2
< x
2x
+ 2
+
2-".
U s i n g ( 8 ) a n d ( 1 0 ) we s e e t h a t 2x i s t h e s i m p l e s t element i n t h e interval I
=
(x
+
z
-
x + z
2-",
+
2-").
U s i n g ( 5 ) we know t h a t 22 i s t h e
s i m p l e s t e l e m e n t i n I , a n d t h u s we s e e t h a t 2x dered f i e l d , x
=
z , p r o v i n g Lemma 1.
=
22.
Since No i s an or-
o
A s u b c l a s s S o f No w i l l be c a l l e d s y m m e t r i c i f
XES implies - x d .
U s i n g i n d u c t i o n , one e a s i l y sees t h a t t h e f o l l o w i n g i s t r u e : (13)
For a l l acOn, O a ,
M
a
and N
are s y m m e t r i c .
4.21
Norman L. A l l i n g
164
I f any of t h e s e sets h a s a g r e a t e s t element x , t h e n i t w i l l be c a l l e d t h e r a d i u s of t h e s e t i n q u e s t i o n .
Thus
f o r a l l a e o n , t h e r a d i u s of N a , M a ,
(14)
PROOF.
is a.
Ow i s a s u b s e t of D .
LEMMA 2.
0 , = 1-2,
and O a t ,
W e know t h a t 0 , is empty, t h a t 0 , = {O], t h a t 0 ,
-1,
-1/2,
1 , 21, e t c .
1/2,
0,
[-l,O,l],
=
Let n d be s u c h t h a t ( i ) O n i s a
s u b s e t of D , and ( i i ) t h e d i s t a n c e b e t w e e n s u c c e s s i v e e l e m e n t s i n 0 n i s 2
-k
,
f o r some k s Z + . S i n c e Oa and D are b o t h symmetric ( 1 3 ) . i t s u f f i c e s t o show t h a t O n +
i s a s u b s e t of t h e s e t of D t , f o r a l l nrN. g r e a t e s t element of 0 i n D.
n'
t h e n u i s i n N , and { u l } = u
v
induction hypothesis C
,
-(k+l)
.
t
.
If u is t h e
1 (4.20:0),
n
+
. As
-
2-k
-
which i s
n o t e d above
Let v be t h e immediate s u c c e s s o r of u i n On.
v ) / 2 = x , t h e n x i s i n D.
(15)
+
Assume t h a t u is n o t t h e g r e a t e s t element i n 0
( l ) , On i s f i n i t e .
x + 2
Let u be i n On
Let ( u
+
u , which we w i l l c a l l c, i s i n D and by t h e + Thus u = x - 2 - ( k t l ) , and v = f o r sane kcz
.
By Lemma 1 , ( u l v ) = x .
Summarizing what we have shown t h a t
if v is t h e immediate s u c c e s s o r of u i n O n t ,
then {ulv)
=
(u
+
v)/2,
and t h a t i t i s i n N n .
From t h i s we see t h a t On+l s a t i s f i e s ( i ) and ( i i ) a b o v e , a n d hence we have proved Lemma 2.
LEMMA 3 .
PROOF.
Ow is a s u b r i n g of
Recall t h a t 2
R
make no c h o i c e at a l l f o r 2
.
-
No t h a t c o n t a i n s 112.
I),
t h u s we may take 2L t o be 1 and may R 2h Let h = (01 l } ; t h u s hL 0 , and h = 1 (1
-
.
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.21
2hL - 2 L h L , 2Rh
=
{2Lh
+
=
{h
2hL - hLI h
h
<
+
1
<
h
2hR
-
h
R
-
2RhR
=
{h
I
. w
2Lh + 2hR
h
11.
+
-
Let x a n d y b e i n 0
und er s u b t r a c t i o n .
w
c o n s t r u c t i o n of No ( 4 . 0 2 1 ,
.
+
2hL
We know t h a t 0
<
h
=
1 ; showing
L
a n d Iy 1, a n d ( y
R
T h u s e a c h o p t i o n of x
o p t i o n s of
x
+
+
1 a r e f i n i t e s u b s e t s of 0
y a n d of x y i s i n 0
y a n d xy a r e f i n i t e i n number, x
THEOREM.
Ow =
+
w
.
Assume
b(y)'
w h e r e b(x)
+
b(y)
S i n c e t h e s e t of
y a n d xy a r e i n 0
w
.
0
D.
By Lemma 2, Ow i s a s u b s e t of D .
PROOF.
is closed
L R As s u c h , {x 1 , (x ) a r e f i n i t e
w'
k.
1 ; thus
we may r e g a r d x a s a n element of
t h a t t h e sum an d p r o d u c t of e a c h e l e m e n t o n Ok i s i n 0 =
2h
<
2 Rh L]
T h i n k i n g of No a s t h e C u e s t a
CD(Ob(x)) a n d y as a n e l e m e n t of CD(Ob(y)). s u b s e t s of O b ( x l ,
-
2 L h R , 2Rh
F i r s t n o t e t h a t , by ( 1 4 1 , 0
t h a t 1 / 2 cN2, a n d hence 1 1 2 ~ 0
Dutari
I
Since 1 is t h e simplest p o s i t i v e element,
1.
+
2hR
+
+
165
of Q t h a t c o n t a i n s 112.
By Lemma 3, Ow is a s u b r i n g
By (31, D is t h e smallest s u b r i n g of Q t h a t con-
t a i n s 1 / 2 ; t h u s D = Ow.
COMMENT.
The number
3 is i n Ow, b u t 1 / 3 is n o t ; t h u s 0 is n o t a w
field.
Mw
4.30
(0)
A number x i n
-n < x x
+
< n,
1, x
+
No w i l l be c a l l e d
and x = 112,
x
(x - 1 , x
114,
+ 114, x + 1 / 8 ,
THEOREM 0 ( C 2 4 , p p . 2 4 - 2 5 ] ) .
x
-
i f t h e r e e x i s t s ncN f o r w h i c h
x
-
... , x
... , x - 1/2", ... I 1/2", ... 1, [24, p . 241,
1/8, +
(1) Each deD i s a real number in No.
i n No, t h e n so are - x , x + y, a n d x y . ( i i i ) For e a c h real number x in No, l e t L = {qsQ: q < x ) a n d l e t R = {qcQ: q > XI. Then x = {L I R ) . (ii) If
a n d y a r e r e a l numbers
Norman L . A l l i n g
166
Given any g a p ( L , R ) i n Q , t h e n {LIR) i s a real number i n No.
(iv)
( i ) By Lemma 1 of S e c t i o n 4.21,
PROOF.
-
{x
2-"1
x + 2-n].
={-x - 2
-
-n
2-"
I
-x
+
y, x + y
-
-
-
(x - x ) ( y
x
2-")
+
2-m) x
2-"
+
+
- ( x - x R )(Y -
XY
L
y 1, xy
- (x -
(x
-
2-n))(y
-
(y
-
-
(x
-
2-"))(y
-
( y + 2 - 7 , xy
2 - 9 , xy
{xy - (=
I X Y - ,-(n+m)
I
xy
+
-
2-ml y x
+
+
2-m).
y
=
-x
-
Y
R
R y 11
- (x
2 T , xy
)I
-
= +
2 - 9 1 (y
(x + 2-"))(y
( 2 - n ) ( 2 - m ) l xy
-
(-
I
-
(y
+
2-9
-
(y
-
2-9
2 - n ) ( 2 - m ) , xy
-
=
(2-")(2-91
2 - ( n + m ) 1 ; s h o w i n g t h a t xy i s a r e a l number i n No.
< x ) and l e t R = { q E Q : q > X I . S i n c e by d e f i n i t i o n t h e r e e x i s t s nEN s u c h t h a t -n < x < n , L and R a r e non-empty. Clearly
( i i i ) Let L = { q E Q : (0)
-
-
R x )(Y
-
(x
and y = { y
y , x + y + 2 - m ) ; showing t h a t x + y i s
{xy - ( x {xy
( i i ) Let x a n d y b e r e a l
Using (4.08:19) we know t h a t xy =
L L x ) ( y - y 1,
-
L
-
xy
(x
2-"1
may be w r i t t e n as
a n d t h u s - x i s a r e a l number i n No.
a r e a l number i n No. {xy
-
(x
=
2-"),
+
dED
By ( 4 . 0 9 : 1 ) , x i s r e a l .
n u m b e r s i n No, w i t h x
(X
4.30
L and {x
coinitial. (4.02:16),
-
2-"]
q
a r e m u t u a l l y c o f i n a l a n d R and { x + 2-"]
By ( 4 . 0 2 : 1 6 ) , x { L I R ) is real.
As we have
=
(LIR].
are mutually
( i v ) Let ( L , R ) b e a g a p i n Q .
o
see i n S e c t i o n 4.21, 0
w
i s t h e r i n g D of d y a d i c n u m b e r s .
- D
S i n c e D i s d e n s e i n t h e f i e l d of real numbers R , a number r i n R associated w i t h s u b s e t s L
Clearly L < R.
=
{acD: a
< r)
a n d R = {bED: b
>
i s a t i m e l y r e p r e s e n t a t i o n of x.
Let x
=
= w.
c a n be
r ] of O w .
Let x = { L I R ] , a n d n o t e t h a t x i s n o t i n 0
{No,<,bl i s a f u l l class of s u r r e a l numbers ( 4 . 0 3 : 2 ) , b ( x ) on D.
By
w
.
Since
Thus ( L , R )
f ( r ) . Let f be t h e i d e n t i t y map
Clearly f is a n o r d e r - p r e s e r v i n g map of R i n t o M
w'
The d e f i n i t i o n s
o f a d d i t i o n a n d m u l t i p l i c a t i o n i n R , o n t h e o n e h a n d , a n d i n No o n t h e o t h e r a r e s u c h t h a t f i s a n i n j e c t i v e homomorphism of R i n t o M U . e.g.,
(See
S e c t i o n 1 . 1 6 , w h e r e much of t h i s i s c a r r i e d o u t i n t h e c a t e g o r y of
ordered g r o u p s . )
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.30 (1)
167
C l e a r l y f is a n isomorphism of R o n t o (XENO: x is a r e a l number). Let u s i d e n t i f y R w i t h {XENO: x r e a l ) , by means of f . M
u
t o be ( N ( } ( 4 . 2 0 ) .
C l e a r l y w i s t h e greatest e l e m e n t i n M
is t h e least element i n M
(2)
<
C l e a r l y , d-
d-,
d + E Mu
<
d
- R,
w
.
I
d e f i n e d + t o be {d d e f i n e d - t o b e {d
(3)
For example, r e c a l l t h a t w was d e f i n e d
contains other elements.
-
and
-0 =
{IN]
Given d E D we c a n
d + 1 , d + 1/2,
-
1, d
1/2,
... , d
... , d -
d + , f o r a l l d i n D.
and d
w'
-
<
2-"
d-
<
d
+
l/n,
... ]
... [
l/n,
and
d].
Note t h a t
< d+ <
d
+
2-" f o r a l l n d .
F i n a l l y , we c a n g i v e t h e f o l l o w i n g d e s c r i p t i o n of Mu.
(4)
M
w
is t h e u n i o n o f t h e f o l l o w i n g d i s j o i n t s u b s e t s : ( i ) R, ( i i ) { + w ) ,
a n d ( i i i ) id*: dsD). PROOF.
We h a v e s e e n t h a t M
w
contains all set d e f i n e d i n ( i ) - ( i i i ) .
I t i s a l s o clear t h a t t h e s e t s described i n ( i ) - ( i i i ) are d i s j o i n t . X E M ~ ;then b(x) 5 w.
-
t h a t b ( x ) = w, l e t L
If b ( x ) < w, t h e n X E O ~ , a n d h e n c e i s i n R . (dEOw: d
< XI, and l e t
i s a C u e s t a D u t a r i c u t i n Ow, a n d x i s empty t h e n x
= -w.
=
{L[R).
R = {dcOw: d
Assume t h a t L and R a r e non-empty;
If (L,R) is a gap i n D, t h e n XER
Dedekind-cut i n D.
( L , R ) is n o t a gap i n D; t h e n i t h a s a c u t p o i n t dED.
If doR, t h e n x = d
-
.
0
> XI; then
If R Is e m p t y , x
- D.
-
Let
Assume
w.
(L,R)
If L
thus (L,R) is a
Assume now t h a t
If dcL, t h e n x = d'.
168
Norman L. A l l i n g Conway [ 2 4 , p . 2 4 1 d e f i n e s x t o b e
B I B L I O G R A P H I C NOTE.
number i f and o n l y if - n (X
-
- 1, x
1/2, x
-
<
1/3,
4.30
x
< n
... I
f o r sane i n t e g e r n , a n d x
x
+
1 , x + 1/2, x + 113,
We h a v e c h o s e n t o u s e s i m p l e r numbers, 1/2",
Ira
=
...1".
e a c h of which i s i n D ,
r a t h e r t h a t t h e numbers l / n , as Conway d o e s [ 2 4 , p . 2 4 1 , s i n c e some of
.
t h e s e numbers a r e i n a r e i n N
Note t h a t t h u s t h e r e p r e s e n t a t i o n ( x
1/3, 1 / 5 , e t c . a r e i n Nu.
-
is a r e f i n e m e n t of ( x
1/2n)
For example 1/2 a n d 114 a r e in D , w h i l e
l/n, x
+
l/n).
-
By ( 4 . 0 9 : l )
1/2",
X
+
they are
e q u i v a l e n t ; t h u s t h e q u e s t i o n of which r e p r e s e n t a t i o n o n e uses s e e m s n o t t o be of major i m p o r t a n c e .
THE MAP XENO
4.40
+
w
x
ENO
+
F o l l o w i n g t h e usual p r a c t i c e s i n an o r d e r e d f i e l d K , f o r XEK l e t 1x1 =
max.(x,
-XI.
For x a n d y i n K , write x
1x1 5 n l y l a n d ( y I 4 n l x l . s u r a t e iff x (0)
a
a
y.
a
y i f t h e r e e x i s t s ncN s u c h t h a t
F u r t h e r , we w i l l s a y t h a t x a n d y a r e commen-
Clearly
i s an e q u i v a l e n c e r e l a t i o n o n K .
Let y / ( a ) d e n o t e t h e e q u i v a l e n c e c l a s s e s mod Clearly
O/(a)
i s {Ol.
a
t o which y belongs.
C l e a r l y K is Archimedean i f and o n l y i f K h a s two
e q u i v a l e n c e classes: ( 0 ) a n d K* ( = {xcK: x # 0 1 ) .
Let y b e i n No*.
<
x
01).
<
nlyl}.
Let J ( y )
=
{xcNo: t h e r e e x i s t s ncN s u c h t h a t l y l / n
C l e a r l y J ( y ) i s a n o n - e m p t y i n t e r v a l i n No'
Using (4.02:7),
( = (XENO:
x >
we know t h a t t h e r e e x i s t s a s i m p l e s t e l e m e n t i n J(y),
which Conway c a l l s t h e l e a d e r of y [24, p . 311. T h u s t h e l e a d e r of YENO*
is t h e s i m p l e s t p o s i t i v e element i n y / ( p ) .
The s i m p l e s t l e a d e r of a l l i s 1 , t h e o n l y p o s i t i v e n u m b e r b o r n o n d a y 1 .
Conway d e f i n e s wo t o be 1 .
He l e t s w 1 be t h e s i m p l e s t l e a d e r t o t h e r i g h t
4.40
I n t r o d u c t i o n t o t h e surreal number f i e l d No
of 1 i n No, namely w, which i s i n N
w
(4.30).
169
I n g e n e r a l , wx w a s d e f i n e d by
+
Conway as f o l l o w s , where r and s r a n g e o v e r R :
(1)
wx
=
(0,rw
xL
Isw
xR
1 C24, p . 311.
I t w i l l be c o n v e n i e n t , o n o c c a s i o n , t o r e s t r i c t r a n d s i n ( 1 ) t o be
S i n c e Qt and R t are m u t u a l l y c o f i n a l a n d m u t u a l l y c o i n i t i a l , t h e
i n .'Q
v a l u e of w x g i v e n by t h e r e s u l t i n g d e f i n i t i o n (which we w i l l c a l l (1,Q')) i s t h e same as t h a t g i v e n by ( 1 ) .
L e t u s m a k e s u r e t h a t t h e d e f i n i t i o n , as g i v e n i n ( I ) , agrees w i t h those g i v e n i n t h e p a r a g r a p h p r e c e d i n g ( 1 ) . 0 =
[Ol], UJ
-1
which i s 1 .
[Ol],
1 =
.
is, a c c o r d i n g t o ( 1 ) .
thus w -1
=
1
=
[lo],
{I], =
0
=
Let u s d e t e r m i n e what
(O,rll = w. thus w-l
t h u s by ( 1 1 , w
[Olrl, a n e l e m e n t we
d e f i n e d i n (4.30) t o be 0'.
For x , y , a n d z i n K (2)
<<
x
y i f nx
Thus x
(3)
0
<
wu
, we
w i l l write
y f o r a l l nEN.
<
y iff x
y a n d x a n d y a r e incommensurate.
Clearly x
<<
LEMMA 0.
Let u a n d v be i n No.
y and y
<<
z implies x
<<
(1) u
z.
<
v i f and o n l y if ( i i )
<< wv. PROOF.
2b(v)). (1).
<<
<
+
Let u s p r o c e e d by i n d u c t i o n o n m a x . ( b ( u )
Assume t h a t ( i ) h o l d s .
<
u
R
6 v,
or
b(v), 2b(u),
Using i n d u c t i o n a n d t h e d e f i n i t i o n of w
i t i s e v i d e n t t h a t a l l wu a r e p o s i t i v e .
e x i s t s a uR w i t h u
+
By ( 4 . 0 2 : 1 4 ) ,
( b ) there e x i s t s v
L
U
( a ) there
w i t h u S vL
<
v.
If
Norman L. A l l i n g
170
( a ) h o l d s then wu << w' induction, w
U
R
<<
R
.
I f uR
-
v then w
w v , and u s i n g (31, w
U
approximately a s a b o v e , we s e e t h a t w ( i i ) . Now assume t h a t ( i i ) h o l d s .
v t h e n of c o u r s e mu
=
U
<<
4.40
<<
U
R
< v t h e n , by
V
<<
w ; proving t h a t ( i ) implies
Assume f o r a moment t h a t u L v .
>
If u
w v , which i s a b s u r d .
If u =
v t h e n we may u s e t h e
> > w V , which a g a i n i s
W'
Thus we have shown t h a t ( i i ) I m p l i e s ( i ) . o
LEMMA 1 . uy.
If u
If ( b ) h o l d s , t h e n , a r g u i n g
wV.
f a c t t h a t ( i ) implies ( i i ) t o conclude t h a t absurd.
wV.
( i ) For a l l xcNo+ t h e r e e x i s t s a unique ycNo s u c h t h a t x
a
F u r t h e r , ( i i ) b(wy) 5 b ( x ) . L R L x may be w r i t t e n i n t h e form {O,x ( x 1 w i t h x
By ( 4 . 1 0 : l ) .
PROOF'.
P r o c e e d i n g by i n d u c t i o n o n
and xR p o s i t i v e , and chosen t o b e i n Ob(x).
R b ( x ) , we see t h a t f o r e a c h xL ( r e s p . x ) t h e r e e x i s t s a unique yL ( r e s p . y
R
such t h a t xL = uy
L
(resp. xR
R s i m p l e a s xL ( r e s p . x 1.
X I , then ( i ) is proved.
R a
wy
1, w i t h wy
L
R ( r e s p . wy ) a t l e a s t a s
I f x i s commensurate w i t h o n e of i t s o p t i o n s , s a y Note a l s o t h a t b ( u y ' ) 2 b ( x ) , s h o w i n g t h a t ( i i )
Assume t h a t x i s c o m m e n s u r a t e w i t h none of i t s o p t i o n s ; then by
holds.
Lemma 0 , r w y {yLlyR].
L
<< x <<
R SW'
Then y i s i n No.
,
f o r a l l r , SER'.
Let y b e d e f i n e d t o b e
Let I be d e f i n e d t o be {ZENO: xL
L
l e t J be d e f i n e d t o be (zcNo: rwy
<<
z
<<
<
z
<
R swY , f o r a l l r , s E R + } .
i s commensurate w i t h n o n e of i t s o p t i o n s , I
=
J.
x
R
1 , and
Since x
Since x i s , by d e f i n i -
t i o n , t h e s i m p l e s t element i n I and since wy is d e f i n e d t o be t h e s i m p l e s t e l e m e n t i n J , we s e e t h a t x
= w'.
I n t h i s c a s e , of c o u r s e , b ( x )
A s t o t h e uniqueness of y , t h a t f o l l o w s frcm Lemma 0 .
THEOREM.
For a l l x and y i n No, wX+'
= wXwy.
=
b(uy).
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.40
Proceeding by i n d u c t i o n on b ( x )
PROOF. =
Then by Lemma 0, X > 0 and Y > 0 .
uy.
(4)
X
=
{O,aw
X
L bw
X
R
+
b(y), let X
171
=
w
X
and l e t Y
By d e f i n i t i o n ,
L
R
] and Y = (0,cwy IdaY ) , where a , b , c , d a r e i n R t .
Then XY =
L
(5)
10. awx
(6)
L (0, a w x ty
+'
+
+
cw '+Y
cw '+Y
L
L
-
acw
-
acw
L
L
ty
ty
L ~
L
bw
, bw
R
R
ty
+'
t
dw"Y
t
dw"y
R
-
bdwX ty
R
R
I
R
-
R R bdwX ty
I
S i n c e ( 5 ) and ( 6 ) a r e i d e n t i c a l , t h e Theorem is proved. Applying t h e Theorem, we o b t a i n t h e f o l l o w i n g c o r o l l a r y :
(7)
For a l l XENO, w
-X
=
l/w
X
.
F u r t h e r , t h e map t h a t takes XENO t o w x i s
a n o r d e r - p r e s e r v i n g homomorphism o n t o t h e class A of a l l l e a d e r s i n
No; t h u s A i s an o r d e r e d group, under m u l t i p l i c a t i o n , t h a t is o r d e r l s u n o r p h i c t o t h e a d d i t i v e group (No,+) of No.
4.41
FINITE LINEAR COMBINATIONS OF w - x ( 1) ,
... , w - x ( n )
OVER R
We w i l l be concerned i n t h i s s e c t i o n w i t h f i n i t e l i n e a r combinations
Norman L . A l l i n g
172
of elements of t h e form w Y o v e r R.
4.41
Let aeR, and l e t Y E N O .
The f o l l o w i n g
a r e obvious.
(0)
(i)b 0.~')
= 0,
0
f o r all YENO, and ( i i ) b ( a w 1 = b ( a ) , f o r all aeR.
Since b ( - x ) = b(x), and having c o n s i d e r e d ( O , ( i ) ) ,
L
R
Let a = { a la J E R .
f i e l d of Mu.
Since b ( a ) 4
W,
w e know ( b y c o n v e n t i o n
R L t h a t ( a L , a ) i s a t i m e l y r e p r e s e n t a t i o n f o r a: t h u s each a
(4.02:15)),
and each a
F u r t h e r , w e saw i n (4.30) t h a t R i s a sub-
Ow.
=
0.
A s we
As b e f o r e , l e t D denote t h e r i n g of a l l dyadic numbers ( 4 . 2 1 ) .
saw i n S e c t i o n 4 . 2 1 , D
>
assume t h a t a
R . is i n D.
Since a { aL wY
LEMMA.
awy
PROOF.
Let {a
=
L'
I
>
0 , we may a l s o assume t h a t each aL
0.
aR wY I . {dcD: 0
=
>
<
d
<
a) and l e t [ a
R'
) = (dED: d
>
a].
S i n c e D i s dense i n R we know t h a t L R ( a L ' , a R ' ) i s a refinement of ( a ,a ) (4.09:O).
- D.)
( C l e a r l y ( a L 1 , a R ' ) i s t i m e l y i f and o n l y i f a d a
= (3''
I
By ( 4 . 0 9 : 1 ) ,
aR').
By d e f i n i t i o n ,
aJ
{aLwY
=
aLwy
R
+
awy
+
-
L awy
R
aLJ
,
L
aLw Y , a Rwy + a wY aRwY
+
au'
L
-
R
-
R
aRwY
I
L aRwY I .
There a r e f o u r e x p r e s s i o n s f o l l o w i n g t h e e q u a l i t y s i g n i n ( 3 1 , w h i c h we w i l l c o n s i d e r i n o r d e r .
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.41 L
aLwY + y
L
-
L
<
y,
wy
aLwY
<<
L
-
aLwY + ( a
=
aL)WY
L
Since aL < a ,
L
o <
a - a
S i n c e t h e r e e x i s t s aL’
( 4 . 4 0 , Lemma 1 ) .
my
.
173
>
.
Since
a L , we s e e
t h a t t h e following holds.
R
Consider now t h e second e x p r e s s i o n i n ( 3 ) : aRwY
(a
+
-
aR)wY
(4.40, Lemma 1 ) . (5)
a R J+ a J
R
.
< aR, a
Since a
-
aR
<
(a
-
- a w
<
YR --
<<
y R , wY
wy
R
Thus R
R
-
L YR a )w
Lemma 1 ) .
(4.40,
Since y
0.
aRJ
< o <
aL’J
<
am’.
R A aLwY + a u y - aL wY
Consider now t h e t h i r d e x p r e s s i o n i n ( 3 ) : aLwY
R
aRwY+ aw
.
o <
Since aL < a ,
a
- aL .
Since y
<
y R , ,Y
L
-
aRwY
=
<<
J
R
Thus R
(6)
auy
< a R ’ u Y < aLwY + aWy - aL w YR .
Consider now t h e l a s t e x p r e s s i o n i n ( 3 ) : aRwY
(a
+
-
aR)wY
( 4 . 4 0 , Lemma 1 ) .
(7)
L
.
Since a
+
COROLLARY 0. =
<
Since y
0.
L
<
y , Y.
L
<<
UY
L aWY - aR W yL .
that t h e following h o l d s : awy
C
aR
=
There e x i s t s aR’ < a ; t h u s ,
a J < aR’mY < aRwY
+
-
L
R
Hence, b y ( 4 ) - ( 7 ) , auy
Then A
< aR, a
aRwY+ a0y
{aLw-‘
{aL’WyI a R ’ u Y ) . By ( 4 . 0 9 : 1 )
= =
iaLwyI a R u Y ) ; proving t h e Lemma.
Let a , ccR, x +
cw-y,aw-x
, we c o n c l u d e
+
<
y be i n No, A
=
aw- X and l e t C
cL~-YlaRw-X +cW-y,aW-x + cRu-Y].
=
c w- Y
.
Norman L. A l l i n g
174
By d e f i n i t i o n ,
PROOF.
(8)
A
C = {AL
t
4.41
+
C, A
CLI AR
+
c
C, A +
+
R
}.
Using Lemma 0, t h e r i g h t hand s i d e of ( 8 ) i s s e e n t o equal
T h i s may a l s o be g e n e r a l i z e d as f o l l o w s .
COROLLARY 1 .
L e t nEN l e t , a i and ci be i n R, f o r i
<...<
-x(l) + Let A = alw
l e t x(1)
c = c w- x ( l )
t
1
(10)
“alL (a, (a,
+
+
R +
x ( n ) be i n No.
...
t
c
-x(n).
n
c , ) w-x(l)
+
clL).w-x(l)
+
cl)-w - x ( l )
+
( a , + c1R ~ * w - ~ ( ’ )t
PROOF.
Let
x ( n ) be i n No.
... ... ...
+
(a
t
(an
+
( a nR
t
(an
+
n
m i s not equal t o 0.
a
4 . 4 0 h o l d s , w- x ( ’ )
>>
>
< ,
cn)*w- x ( n ) , cnL)-w-x(n)~
+
+
+
cn)*w- x ( n )
#
c:)-w-~(n)].
alw-’(’)
0 or A
... , <
... >> w - x ( n ) ,
BIBLIOGRAPHIC NOTE.
and l e t
anw- x ( n ) ,
+
= 1,.
<
o
..,n, and l e t x(l)
... + anw- x ( n ) .
Let m be t h e l e a s t element i n [ l ,
Then A
Since x(1)
=
1 ,...,n, and
=
l e t aicR, f o r i
nEN,
Let A
a l l of t h e a i l s a r e 0.
PROOF.
...
+
Apply t h e d e f i n i t i o n of a d d i t i o n and Lemma 0.
PROPOSITION.
... , <
Then A + C
...
=
0 a c c o r d i n g as am
< ,
Assume t h a t not
...*n) >
such t h a t
0 o r am < 0.
x ( n ) , a n d since Lemma 0
Of
Section
Sane of t h e r e s u l t s i n t h i s s e c t i o n may be new.
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.50 4.50
THE SIGN-EXPANSION
F o l l o w i n g Conway [24, p p . 30-311, l e t a
Let X E N O , w i t h b ( x ) = B .
B, l e t
175
L ( x ) = [yEOa: y
a
x is not i n 0
a'
< X I , and l e t Ra(x)
= ( ~ € 0 y~ >:
XI.
t h u s ( L ( x ) , R ( x ) ) is a Cuesta Dutari c u t i n 0 a
.
[L,(x)] R a ( x ) ] , and n o t e t h a t x EN a a a 1 B, and let x
>
.
Let x
a, a
=
Now a s s u m e t h a t
a , x f xa.
ath a p p r o x i m a t i o n
t o x [24, p .
( x ) ) i s a s u b s e t of L
(x) ( r e s p .
Conway c a l l s x
x.
=
Since B
>
S i n c e f3
<
the
291. (0)
If a. ,< a , I 6, t h e n ( x a
PROOF. R
a1
Since L
( x ) ) , x and x
a1
a0
a0
1
(x) (resp. R
b o t h f i l l (4.02)
we s e e t h a t L a o ( x a l )
=
[Lao(xa,)I R
{L ( X I / R
a0
(xa
=
= 1
La
(X) 0
a0
x
a0
a0
t h e Conway c u t ( L
a n d Ra a0
.
(XI]
0
(x
=
a1
xa
0
= R
.
a0
(x), R
a1
a0
( X I ) i n No,
Thus
(XI. a0
(Xa 1
1
=
a0
Let u s c a l l ycNo a p r e d e c e s s o r of x, a n d write y < t x i f t h e r e e x i s t s a
<
B such that y
a s u c c e s s o r of y.
=
x
.
I f y i s a n p r e d e c e s s o r of x we w i l l refer t o x as
x i s t h e immediate s u c c e s s o r of y i f y
i s no z ~ N of o r which y
<
z
<
x.
<
x, and t h e r e
C l e a r l y 0 i s t h e o n l y e l e m e n t i n No t h a t
has no p r e d e c e s s o r . (1)
Under < t , No i s a p a r t i a l o r d e r c l a s s .
PROOF.
Assume t h a t x o c t x 1 a n d x 1
ct
y ; t h e n by ( 0 1 , x o
p r o v i n g t h a t No i s a p a r t i a l l y - o r d e r e d c l a s s u n d e r
Let t h e s e t of p r e d e c e s s o r s of x, p r N o ( x )
(2)
=
ct XI.
p r ( x ) i s a w e l l - o r d e r e d s e t u n d e r < t . Under < t , No i s a tree.
c t y;
176
Norman L . A l l i n g PROOF.
C l e a r l y acB
We w i l l r e f e r t o
4.50
xacprNo ( x ) i s an o r d e r - p r e s e r v i n g b i j e c t i o n .
+
a s t h e t r e e o r d e r on No.
See e . g . ,
c57, pp. 292-
3201 o r [55, pp. 315-3263 f o r r e f e r e n c e s t o t h e t h e o r y o f t r e e s .
Since 0
i s t h e element of No t h a t h a s no p r e d e c e s s o r , 0 i s t h e r o o t of No. Let G be a n o r d e r e d group G .
(3)
For ycG l e t
s i g n ( y ) be + i f y
>
0,
( i i ) s i g n ( y ) be 0 i f y
=
0 , and l e t
<
0.
(i)
( i i i ) s i g n ( y ) be
-
if y
Returning our a t t e n t i o n a g a i n t o No,
(4)
l e t o ( x ) ( a ) be d e f i n e d t o be s i g n ( x
-
xa), for a l l a
B U ( X ) E [ + } w i l l be c a l l e d t h e Sign e x p a n s i o n of x. and
as follows: l e t
[ * I Bis a n o r d e r e d s e t .
-<
0
<
+.
bl(Z Y
= Y,
Y
b(x).
Let u s o r d e r
6
Next l e t C be t h e u n i o n of ( { k } )Beon,
is an o r d e r e d s e t .
and b ' ( E ) =On.
-,
0,
Then, under t h e l e x i c o g r a p h i c o r d e r i n g ,
C i n t o O n by t a k i n g e a c h p o i n t of [ + I B t o 6.
( I + ) B ) B E Y ; then E
<
L e t b' map
Let L y be t h e u n i o n o f
Clearly E be t h e u n i o n of ( 2 y ) y e 0 n ,
F u r t h e r t h e f o l l o w i n g h o l d , f o r a l l BcOn.
B I B L I O G R A P H I C NOTE.
L y i s d e f i n e d [55, p p . 3 1 6 1 t o b e t h e f u l l
b i n a r y t r e e of h e i g h t Y .
E h r l i c h [ 2 8 1 c a l l s L t h e f u l l b i n a r y t r e e of
h e i g h t On.
The h e i g h t f u n c t i o n of t r e e t h e o r y [55, p p . 3151 i s t h e f u n c -
tion bl.
(5)
(i)
u maps No i n t o
(ii)
b'.o
z,
= b,
( i i i ) o ( N ) i s c o n t a i n e d i n (*}',
B
(iv)
d o B ) i s c o n t a i n e d i n E 8'
and
177
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.50
One c a n d e f i n e t h e r e l e v a n t o r d e r s o n Z Y and o n I d i r e c t l y .
However,
many fewer s p e c i a l cases must be c o n s i d e r e d i f we p r o c e e d a s f o l l o w s .
L e t Z y A = { t ~ ( f , O ] ~ :t h e r e e x i s t s 6
< bY - ( s ) ,
t(a)
Clearly b
zY1
= f,
maps 1 Y
Y
-
b Y ^ ( t )< Y s u c h t h a t f o r a l l a
a n d f o r a l l a f o r which b - ( s ) 2 a Y o n t o Y.
<
Y , t h e n t ( a ) = 01.
Let Eye b e l e x i c o g r a p h i c a l l y o r d e r e d .
Then
is a n o r d e r e d s e t .
For seZy, w i t h b ( s )
< 6, a n d (6)
s o cZy
B , l e t s A c Z y As u c h t h a t s * ( u ) =
=
s - ( a ) = 0 , f o r a l l a f o r which 6 6 a
Let s o ~ E ya,n d l e t B o
1
, and
that so
+
-
Now l e t
byA(soA) BO.
sl.
-
=
=
b y A ( s A ) ,f o r a l l s c Z y .
By d e f i n i t i o n , B o
b(so).
S1€Ey,
for a l l a
S(u),
< Y.
seZy + s n e Z Y A i s a b i j e c t i o n s u c h t h a t b ( s )
PROOF. -
=
<
and l e t B1 = b ( s l ) .
Y.
Then Assume
I f B o = B = B 1 , t h e n i t i s clear t h a t s o A f s l A . Assume
< E l , t h e n s o ^ ( B o ) 0 a n d s l A ( B 0b) f ; t h u s s o A C s l A . Assume t h a t b 0 > B l r t h e n s l n ( B 1 ) 0 a n d S O A ( B 1 ) t f; t h u s s o A f s l A . Hence t h e mapping s + s A is i n j e c t i v e . Let t E E y n , a n d l e t b y A ( t ) = B ; t h e n 8 < Y . t h a t B,
Let s =
<
6.
tls:
-
i . e . , l e t s be t h e r e s t r i c t i o n of t t o t h e s e t of a l l o r d i n a l s
Then c l e a r l y seZy a n d s- = t .
Let Y 6 6 6
c , a n d l e t t Y E EY
o
6
A
b e e x t e n d e d t o i y ( t y )=
d e f i n i n g t ( a ) t o b e 0, f o r a l l a f o r which Y 5 a
Y
f ol 1owing
(7)
<
6.
Y i s a n o r d e r - p r e s e r v i n g map of Z Y
i6
(ii)
i5 6 - i 6y = i y 5,
and
( i i i ) i Yy is t h e i d e n t i t y mapping of EyA.
into
by
C l e a r l y we h a v e t h e
.
(i)
t6EEgA,
Z6*,
4.50
Norman L. A l l i n g
178
Using ( 7 ) we s e e t h e f o l l o w i n g .
(8)
The d e f i n i t i o n of o r d e r on C i s independent of t h e c h o i c e o f Y, a n d i t r e n d e r s Z ( r e s p . C ) i n t o an ordered c l a s s ( r e s p . an o r d e r e d s e t ) .
Y
< {Yj.
Given any s u b s e t S of C , t h e r e e x i s t Y E O n s u c h t h a t b ' ( S )
a well-defined order-preserving i n j e c t i o n , for
mapping SES
+ s " E i~s ~t h e n
a l l such Y .
Let S" be t h e image of S I n C Y under t h i s mapping.
(9)
The
I n t h e f o l l o w i n g pages, " w i l l always have t h i s meaning. Following Conway C24, p.301, we have t h e f o l l o w i n g . LEMMA.
a is an o r d e r - p r e s e r v i n g map of No i n t o E .
PROOF.
Let x and y be i n No, w i t h x
t i o n on b(x)
+
f o r which x 5 yL
Let u s p r o c e e d by i n d u c -
By ( 4 . 0 2 : 1 4 ) w e know t h a t ( i ) there exists sane y
b(y).
<
< y.
xR f o r which
y , o r ( i i ) t h e r e e x i s t s some
x
<
x
R
L
5 y.
If ( i ) holds then we may use t h e i n d u c t i o n h y p o t h e s i s t o conclude t h a t a ( x ) S
L
< a ( y ) ; which shows t h a t
o ( y L ) and t h a t a ( y )
<
u(x)
<
w e may use t h e i n d u c t i o n h y p o t h e s i s t o conclude t h a t u ( x ) 5 a ( y ) ; which shows t h a t o ( x )
<
I f ( i i ) holds
a(y).
a(x
R
R
and a(X )
o(y).
THE STRUCTURE OF E AND THE SIGN-EXPANSION
4.51
The o r d e r e d c l a s s E w a s d e f i n e d and s t u d i e d i n S e c t i o n 4.50.
operator 0.
then
a s follows: l e t - ( + I
o p e r a t e on
For each S E E , l e t
b'(s);
(0)
-
-9
i s i n Z.
_ _
( s) =
-3
9,
=
-, - ( - I
= +,
Let t h e
and -(O) =
be d e f i n e d a s f o l l o w s : - s ( a ) = - s ( a ) , f o r a l l a Clearly
f o r a l l SEE, and f o r a l l XENO, a ( - x )
=
a(X).
<
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.51
Henceforth we w i l l write -s a s
179
Let 0 b e t h e e l e m e n t i n { + l o :
-9.
i . e . , 0 i s t h e empty mapping; t h a t i s i t i s j u s t t h e empty s e t .
Conway d e f i n e d x
a
124, p . 291, c a l l i n g i t t h e a% approximation o f x.
We d e a l t w i t h t h i s n o t i o n , u s i n g t h e same n o t a t i o n and terminology, i n F u r t h e r , u s i n g i t we i n t r o d u c e d t h e idea of t h e o r d e r
S e c t i o n 4.50.
No, under which No is a t r e e .
(1)
o f s t o t h e set a.
If a, 6 a 1 5 8 , t h e n (sa
PROOF.
(sa )
1 010
-
On
Let us now c o n s i d e r t h e same i d e a s f o r E.
Let S E E , and l e t B = b ’ ( s ) . the restriction
1 a0
I f a 5 B , l e t sa Clearly s
=
s
a0
= s.
= s(a:
i . e . , l e t su be
Further,
.
(slal)lao,which i s j u s t s l a o .
L e t us c a l l teE a p r e d e c e s s o r of s, and write t < t s, i f there e x i s t s
a < 8 such t h a t t
=
s
a
(cf. (4.50)).
If t i s an p r e d e c e s s o r of s we w i l l
refer t o s as a s u c c e s s o r of t ( c f . ( 4 . 5 0 ) ) .
s i s t h e immediate s u c c e s s o r
of t i f t < t s, and i f there is no ueL f o r which t < t u < t s. C l e a r l y 0 i s t h e o n l y element i n L t h a t has no p r e d e c e s s o r . (2)
Under < t , E is a p a r t i a l order class. PROOF.
Assume t h a t so
8,
and
9,
t; then by (11,
9,
C t t.
o
Let t h e set of p r e d e c e s s o r s of s, p r E ( s ) , be {teE: t < t s}.
(3)
p r ( s ) i s a well-ordered set under < t . Under < t , 1 is a tree.
PROOF.
C l e a r l y aEB
+
s E p r z ( s ) i s an o r d e r - p r e s e r v i n g b i j e c t i o n . a
Let < t be c a l l e d tree o r d e r on L.
Note: 0 i s t h e root of t h e t r e e E.
F o l l ~ i n gConway c24, Theorem 181, we have t h e f o l l o w i n g .
Norman L. A l l i n g
180
THEOREM.
is an order-preserving map of No onto
Q
PROOF.
Q
4.51
r,
w i t h b'
s u f f i c e s t o prove t h a t Let
Y
S E ( ~ }
b.
was d e f i n e d i n (4.501, where i t w a s shown t o be an order-
preserving map i n t o Z (Lemma 4.50) f o r which b t * o = b ( 4 . 5 0 : 5 ) .
Cy.
-0 =
.
(I
is s u r j e c t i v e .
For each B
< Y,
Thus, i t
Let YEOn and assume t h a t o ( O y )
t h e r e e x i s t s a unique y E N
5
=
such t h a t
0
o(yg) = SB.
Let x = { ( y 8 : s
(4)
8
< s and
B
-
<
Yll ( y B : s8
>
s and B
< Yl).
Since (No,<,bl is a f u l l c l a s s of surreal numbers (4.03:2), b ( x ) 6 Y.
< Y.
Let s
>
s
B
Note t h a t a(x)(B)
(resp. s
<
6 Y , b t ( o ( x ) ) 6 Y.
COROLLARY.
4.52
+
-
s ) i f f s(B) B
Hence
Q(X) =
(resp. - ) i f f x
> y B (resp.
thus a ( x ) l Y
+ (resp. -);
=
x
<
s.
iff 8 Since b ( x ) y
s , and b ( x ) = Y .
( r , < , b ' } is a f u l l c l a s s of s u r r e a l numbers.
THE NEAREST COMElON PREDECESSOR OF A SUBCLASS OF C
Let S be a non-empty s u b c l a s s of C . predecessor of S i f a S t s , f o r each SES.
aEE w i l l be c a l l e d a common
A cmmon predecessor c of S w i l l
be c a l l e d t h e n e a r e s t common predecessor of S, i n symbols c = ncp(S), i f given any common predecessor a of S t h e n a S t c .
(0)
S has a nearest common predecessor. PROOF.
Let m i n . ( b t ( S ) )
of t h e c h o i c e of seS,
=
Y , and l e t
for a l l a 6 61.
Further, i t is unique.
r
=
{a <
Clearly
Y:
r
S(a)
is independent
i s a lower-saturated
I,) I ) . subset of Y (1.021, which may b e e m p t y ( f o r example i f S = { ( +( Let 8 be t h e l e a s t ordinal t h a t i s g r e a t e r than a l l of t h e elements of Let S ~ E S ,and l e t c ( a ) a l l S E S and a l l a
<
=
s , ( a ) , f o r a l l ae[O,B).
B, c(a)
=
c is a common predecessor of S.
r.
By t h e c h o i c e of B , f o r
s ( a ) ; t h u s f o r a l l SES,
sB
-
c ; showing t h a t
Now l e t a be a common p r e d e c e s s o r of S ,
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.52 and l e t B '
=
c
.
=
ncp(S)
Since a
b'(a).
sBl,
=
181
f o r a l l SES, B ' 5 6, a n d a
=
c
Thus
B"
Now l e t d b e a n e a r e s t common p r e d e c e s s o r of S; t h e n d S t c A s a consequence c = d .
and c St d .
EXAMPLES.
( i ) ncp([s,l)
=
so.
( i i ) I f so is a common p r e d e c e s s o r of
S o , a n d S is t h e u n i o n S o and [ s o ) , t h e n s o = n c p ( S ) .
Let c = n c p ( S ) .
(1)
S 1 ~ Ss u c h
t h a t so 5 c 5
Assume f o r a moment t h a t e i t h e r ( I ) s
PROOF. (ii) s
< c
holds.
Let 6
=
or 0, f o r a l l SCS.
c ; which i s a b s u r d .
f o r a l l S E S , or
S i n c e c A ( B ) = 0 , a n d since c is a common p r e d e c e s -
bl(c). = +
> c
9,.
W i t h o u t l o s s of g e n e r a l i t y we may assume t h a t ( i )
f o r a l l SES.
sor of S, s^(B) =
There e x i s t s o ,
B)
Thus s
=
+,
Were t h e r e
8,hS
f o r a l l SES.
w i t h s,^(B) = 0 , s,
Hence
BEr,
which i s
absurd.
PROOF. so S c 6
bt(c)
<
8,.
Let
8,
<
8,
be i n {+IY, and l e t c
Since b'(s,)
= Y =
Y ; thus so f c f sl.
LEMMA.
b ' ( s l ) , and s i n c e
Hence s o
< c<
3,
<
By ( 2 1 , j ,
j,
and
3,
are unequal,
9,.
There exists a unique
- [Jol). A s s u m e , f o r a moment
Let Y b e t h e l e a s t e l e m e n t i n b ' ( J ) .
that there exists j,
ncp((jo,jl)).
n c p ( ~ s , , s , ) ) . By ( 1 )
Let J be a non-empty i n t e r v a l i n Z.
j o c J such that b f ( j o ) < b ' ( J PROOF.
=
i n J , for which b * ( j o ) = Y = b t ( j l ) .
< c <
j , , and b ' ( c )
<
Y.
Let c =
S i n c e J is a n i n t e r v a l ,
c i s i n J ; which i s a b s u r d .
THEOREM. joEJ
such t h a t b F ( j o )< b ' ( J PROOF.
(1)
Let J be a non-empty i n t e r v a l i n 2 .
-
The u n i q u e e l e m e n t
{ j , ) )is ncp(J).
Let j , be as d e f i n e d i n t h e Lemma, a n d l e t c
a n d t h e f a c t t h a t J is a n i n t e r v a l , we see t h a t
CEJ.
=
ncp(J).
Using
By c o n s t r u c t i o n
Norman L. A l l i n g
182 bl(j,) 5 bt(c).
that bt(c)
=
t h a t c = j,.
Since c
bt(j,,).
=
n c p f J ) , c St j,; t h u s b l ( c ) 5 b t ( j o ) ; and we see
Since
i s unique having t h e s e p r o p e r t i e s , we s e e
j,
o
B I B L I O G R A P H I C NOTE.
i n Ey has been d e f i n e d . 4.53
4.52
The n e a r e s t common p r e d e c e s s o r of two e l e m e n t s
See, e.g.,
[55, pp. 316-3171.
THE TREE STRUCTURE OF A FULL CLASS OF SURREAL NUMBERS
Let { F , < , b , Y ) b e a c l a s s of s u r r e a l numbers of h e i g h t Y ( 4 . 0 3 ) .
-
P r o c e e d i n g v e r y much a s we d i d f o r N o i n S e c t i o n 4.50, l e t XEF, and l e t {YEF: y < x a n d b ( y ) < a ] , b ( x ) = B ; t h e n B < Y. Let a < 8, l e t La(x) a n d l e t R,(x)
= {YEF:
y
F ( < , a ) , t h u s ( L a ( x ) , R,(x))
> x and b(y) <
a].
S i n c e B > a , x is not i n
is a Cuesta Dutari c u t i n F ( < , a ) .
{La(x)l Ra(x)}, and n o t e that x a c F ( = , a ) .
Let xu
S i n c e B > a , x f x a'
=
Recall
(4.50) t h a t Conway C24, p.291 c a l l s xa t h e u t h a p p r o x i m a t i o n t o x.
Let us c a l l yeF a p r e d e c e s s o r ( c f . ( 4 . 5 0 ) ) of x , a n d write y
( c f . ( 4 . 5 1 ) ) i f there e x i s t s a
<
s u c h t h a t y = xa
.
I f y is a n predeces-
sor of x we w i l l r e f e r t o x as a s u c c e s s o r ( c f . (4.50)) o f y .
x is t h e
immediate s u c c e s s o r of y i f y c t x , and i f t h e r e is no zcF f o r which y
(1)
C l e a r l y 0 is t h e o n l y element i n F t h a t has no p r e d e c e s s o r . Under < t , F is a p a r t i a l o r d e r c l a s s . Let t h e s e t of p r e d e c e s s o r s of x, pr,(x)
(2)
= (YEF: y < t x ) .
p r ( x ) i s a w e l l - o r d e r e d set under < t . Under < t , F is a tree. We w i l l refer t o
< t as t h e tree order on ( F , < , b , Y l ( c f . 4.53). --
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.53
I n S e c t i o n 4.51 we a l s o d e f i n e d t h e same terms o n Z ,
183 h o w e v e r i t was
To prove t h a t t h e two i d e a s are t h e same,
based o n a d i f f e r e n t d e f i n i t i o n .
i t s u f f i c e s t o prove t h e f o l l o w i n g . LEMMA.
Let X E N O , b ( x ) = Y , and l e t u ( x )
PROOF.
Let u s proceed by i n d u c t i o n o n Y .
b ' ( a ( x 1) %
x s
% %'
8
=
Let a
b'(sg).
<
6.
x ) i f f s > s (resp. s a % a %
<
If %
<
Y , a(x
%
= S
% a
+
(resp. -1 i f f x
sa) i f f s ( a ) =
%
%'
A s we have s e e n ( 4 . 5 0 : 5 ) ,
Using t h e f a c t t h a t ( x )
we see t h e f o l l o w i n g : a ( x B ) ( a ) =
(4.50:0),
<
=
= 3.
+
%
( r e s p . -1;
>
x
a
= x
a
(resp.
thus u(x ) %
=
o
u i s an i s a n o r p h i s m of { N o , < t ) o n t o {E,<,}.
COROLLARY.
THEOREM.
Let { F , < , b , Y ) and { F f , b q , < ' , Y }be f u l l c l a s s e s of s u r r e a l
n u m b e r s of h e i g h t 7 .
Each h a s a t r e e o r d e r , C t and
respectively.
The
unique s u r r e a l monomorphism f of F o n t o F ' ( T h e o r a 4.03) i s a n isomorphism of { F , < t I o n t o { F q , < t l ) .
THE PREDECESSOR CUT REPRESENTATION OF A SURREAL NUMBER
4.54
Let { F , < , b , Y } be a class of s u r r e a l numbers of h e i g h t Y ( 4 . 0 3 ) . L e t X E F ( = , B ) ;then %
B. (0)
<
Y.
I n S e c t i o n 4.54 we d e f i n e d x a , f o r a l l a
<
Thus we have t h e f o l l o w i n g .
x
=
{{x : x
PROOF.
a
a
<
x and a
<
{xu: x
>
x and a
< %I}.
S i n c e { F , < , b , Y } and {Z , < , b ' , Y ) a r e i s o m o r p h i c , i t s u f f i c e s
Y
t o e s t a b l i s h (0) f o r { Z y , < , b T , Y ) . To prove t h i s we need only prove ( 0 ) f o r ( r y , < , b f ) . Let y be t h e number d e f i n e d o n t h e r i g h t hand s i d e of ( 0 ) . S i n c e { Z , < , b ' } i s a f u l l c l a s s of surreal numbers ( 4 . 0 3 : 2 ) , b ' ( y ) S. b ' ( x ) =
184 Let a
B.
>
< 8.
Note t h a t x ( a ) =
+
(resp. -1 i f f x
(resp. y
<
x a ) i f f y ( a ) = + ( r e s p . -1;
B, we see t h a t x
=
y.
y
4.54
Norman L. A l l i n g
xa
>
xu ( r e s p . x
thus y[B = x.
The f o l l o w i n g w i l l be c a l l e d t h e p r e d e c e s s o r
< x and a <
>
x and a
(2)
Let (L,R) be t h e p r e d e c e s s o r c u t r e p r e s e n t a t i o n of x.
a
iff
Since b ' ( y ) 5
of x:
< 61).
((xu: xa
(xa: x
x
cut r e p r e s e n t a t i o n
(1)
B),
<
Then I L 1
+
IS(
= Ib(X)I.
4.60
ALTERNATIVE AXIOMS FOR A FULL CLASS OF SURREAL NUMBERS
W e w i l l now g i v e a n a l t e r n a t i v e s e t of a x i o m s f o r a f u l l c l a s s of
surreal numbers of h e i g h t B, t h e f i r s t set of axioms b e i n g g i v e n i n S e c t i o n
4.03.
If B = On, l e t [O,B)
d e n o t e On.
F i r s t we have t h e f o l l o w i n g O R D E R
AXIOM: (0)
Assume t h a t S is a n o r d e r e d class. We w i l l c a l l t h e f o l l o w i n g t h e BIRTH-ORDER AXIOM:
(B)
Assume t h a t t h e r e e x i s t s a map b of S o n t o [ o , ~ ) . S , < , b , B ) s a t i s f i e s (0) a n d (9). For a
-
b-'([O,aI),
<
8, l e t S ( < , a ) =
and l e t S ( - , a ) = b-' ( a ) .
S i n c e Conway o f t e n c a l l s XENO f q s i m p l e r f tt h a n YENO, i f b ( x )
<
b ( y ) , we
w i l l call t h e n e x t axiom a b o u t [ S , <, b ) t h e SIMPLE DENSITY AXIOM:
<
(SD) F o r a l l a t h a t x,
<
y
B , a n d a l l x,
<
x,.
<
x I E S ( = , a ) , t h e r e exists y c S ( < , a ) s u c h
I n t r o d u c t i o n t o t h e surreal number f i e l d No
4.60
105
S i n c e t h e f o l l o w i n g axiom resembles t h e TI -set c o n d i t i o n , a n d i t h a s
5
t h e f u l l n e s s c o n d i t i o n of
i n i t , we w i l l c a l l i t t h e FULL ETA
(4.03:2)
AXIOM:
(FE)
< 6 , and s u b s e t s
Given a L
< (y) <
<
L
R of S ( < , a ) , t h e r e e x i s t s YESs u c h t h a t
R , f o r which b ( y ) S u .
L e t [ S , < , b , B ) s a t i s f y (SD) a n d (FE) ( i n a d d i t i o n of ( 0 )a n d (B)), a n d l e t i t b e c a l l e d a new f u l l class of s u r r e a l numbers of h e i g h t Let J b e a non-empty i n t e r v a l i n S.
LEMMA 0.
ment x o ~ Js u c h t h a t ( b ( x , ) l
<
b(J
-
C l e a r l y t h e r e e x i s t s X,EJ f o r w h i c h b ( x , )
which b ( x , )
T h e r e e x i s t s a n ele-
C l e a r l y x, is u n i q u e .
{x,,]).
-
i t has a least
S i n c e b ( J ) i s a n o n - e m p t y s u b s e t of [ O , B )
PROOF. element a .
=
u.
< x,.
<
Let X , E J f o r
With o u t l o s s of
By t h e T h e S i m p l e D e n s i t y Axiom
(SD), t h e r e e x i s t s y E S ( < , a ) s u c h t h a t x, YEJ. S i n c e b ( y )
u.
Assume, f o r a m o m e n t , t h a t x, b x l .
g e n e r a l i t y , we may assume t h a t x,
B.
<
< x,.
y
S i n c e J is a n i n t e r v a l ,
a , we see t h a t u i s n o t t h e l e a s t e l e m e n t of b ( J ) ; which
i s absurd.
CONWAY'S SIMPLICITY THEOREM. S ( < , a ) be
given.
<
Let a
8,
and let subsets L
There e x i s t s a u n i q u e XES s u c h t h a t L
< (XI <
<
R of
R , for
which b ( x ) i s m i n i m a l .
PROOF. non-empty.
Let J
=
(YES: L
<
[y]
<
R).
By t h e F u l l Eta Axiom ( F E ) , J is
By Lemma 0, J has a u n i q u e e l e m e n t x h a v i n g m i n i m a l b i r t h d a y .
0
Note t h a t s i n c e C o n w a y ' s S i m p l i c i t y Theorem h o l d s , x i s u n i q u e l y
d e t e r m i n e d by L a n d R . d e n o t e x.
F o l l o w i n g Conway, l e t t h e symbol ( L I R I b e u s e d t o
C o n t i n u i n g t o u s e Conway's n o t a t i o n a n d c o n v e n t i o n s C24, p . 41, L
R
L
we w i l l w r i t e x a s ( x ( x 1 , w h e r e [ x 1 L R w i t h y = Iy Iy 1 .
=
R L and [x 1
=
R.
Let y b e i n S,
186
(4)
4.60
Norman L. A l l i n g
x 5 y i f and o n l y i f
(i)
(ii) x
<
Assume t h a t ( i ) h o l d s .
PROOF.
< x
S i m i l a r l y xL
Then x 4 y
5 y , and hence xL
T h u s we s e e t h e f o l l o w i n g : ( * ) yL
L
y for a l l x
<
L
<
. yR, and hence x
y; t h u s ( i i ) h o l d s .
<
y
R
.
Assume now t h a t
Also assume f o r a moment, t h a t (1) i s f a l s e : i . e . , t h a t y < x.
( i i ) holds.
for all x
<
yR f o r a l l y R , and xL
, x R , y L , and y R .
we s e e t h a t b ( y )
<
see t h a t b ( x )
<
b(x).
<
y
< x <
y
R
,
<
a n d (**I xL
y
< x < xR ,
Using t h e Conway S i m p l i c i t y Theorem a n d ( * ) ,
Using t h e Conway S i m p l i c i t y Theorem and ( * * ) , we
b ( y ) ; which i s a b s u r d .
The c o n t r a p o s i t i v e of ( 4 ) i s t h e f o l l o w i n g :
(5)
(i)
x
<
y i f and o n l y i f
( i i ) x 4 yL, for sane y L , or
5 y, f o r sane x
S i n c e x i s t h e s i m p l e s t element
< [ y ) < R f o r which in S f o r which L < [ x < R , b ( x )
o
THEOREM 0.
Every new f u l l c l a s s o f s u r r e a l n u m b e r s of h e i g h t 6,
[ S , < , b , B ] i s a f u l l class of surreal numbers of h e i g h t PROOF.
.
By ( F E ) , t h e r e e x i s t s YESs u c h t h a t L
PROOF.
4 b(y) 5 a .
R
For a l l Conway c u t s (L.R) i n S ( < , a ) , x = [LIRIcS 5 , a ) .
LEMMA 1 .
b(y) 5 a.
xR
B (4.03).
Apply t h e Conway S i m p l i c i t y Theorem and Lemma 1 .
THEOREM 1 .
E v e r y f u l l c l a s s o f s u r r e a l n u m b e r s of h e i g h t
B,
[ S , < , b , B ] , i s a new f u l l class of surreal numbers of h e i g h t 6. PROOF.
-
C l e a r l y (0) a n d (8) h o l d .
of S ( < , a ) , and l e t y
(LIR].
As t o ( F E ) , l e t L
<
R be s u b s e t s
Since [ S , < , b , e ] i s a f u l l Class of surreal
n u m b e r s , y c S ( S , a ) ; t h u s (FE) h o l d s .
Let x
<
y be i n S ( - , a ) .
L R Let ( x ,x 1
and ( y L , y R ) b e ( t i m e l y ) Conway c u t r e p r e s e n t a t i o n s f o r x a n d y ( 4 . 0 2 ) .
BY
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
4.60
( 5 ) , ( a ) t h e r e e x i s t s yL s u c h t h a t x S y
x
R
5 y.
L b(y )
<
L
, o r ( b ) t h e r e e x i s t s xR s u c h t h a t
Assume t h a t ( a ) h o l d s , t h e n x 5 y
b(y)
holds, then x
<
L
<
b(x), x f yL; and hence x
= a =
L
xR f y; and hence x
<
x
L
R
< y.
< y
L
<
R
Thus (SD) h o l d s .
COMMENTARY ON THE NEW SET OF A X I O M S . l i k e ( 0 ) and ( B ) seem f o r c e d .
R
S i n c e ( y Iy ) i s t i m e l y ,
y.
Assume t h a t ( b
y. R
<
S i n c e ( x Ix ) i s t i m e l y , b ( x )
xR S y.
187
b(x) = a
=
b(y)
o
C l e a r l y s a n e t h i n g v e r y much
The F u l l E t a Axiom ( F E ) , i s a n a p p a r e n t l y
s l i g h t l y weaker v e r s i o n of the o l d f u l l n e s s axiom (4.03:2).
If a n y t h i n g i s
novel about t h i s new s e t of axioms i t would seem t o b e t h e S i m p l e D e n s i t y Axiom, ( S D ) .
I n s e v e r a l of t h e e x p l i c i t c o n s t r u c t i o n s , t h i s c o n d i t i o n
arises very n a t u r a l l y .
tion (4.02:3,ii),
For example, i t does i n t h e C u e s t a D u t a r i c o n s t r u c -
and i n t h e sign-expansion construction (4.52:2).
One c o u l d make a s l i g h t g e n e r a l i z a t i o n of ( B ) by a l l o w i n g b ( S ) t o b e a n y s u b c l a s s C of On.
T h i s c o u l d be c o n v e r t e d t o (81, b y l e t t i n g g be t h e
o r d e r - p r e s e r v i n g map of C o n t o [O, f3)
, and
by d e f i n i n g a new b i r t h - o r d e r map
b f t o be g - b .
S i n c e t h e F u l l Eta Axiom ( F E ) , d e a l s w i t h t w o t h i n g s , i t c o u l d b e b r o k e n u p i n t o t h e f o l l o w i n g two t h e f o l l o w i n g two axioms: the ETA AXIOM ( E ) t h a t asserts t h a t s u c h a n element y e x i s t s ; and t h e FULLNESS A X I O M ( F )
(E) i s a d e n s i t y ( F ) t e l l s u s some-
t h a t asserts t h a t s u c h a y e x i s t s f o r which b ( y ) 5 a.
c o n d i t i o n and is a n a l o g u e s w i t h a n rl -set c o n d i t i o n . 5 t h i n g a b o u t e l e m e n t s of h e i g h t 2 a i n t h e t r e e s t r u c t u r e of [ S , < , b , B } . a l s o c o n t r o l s how l a r g e b ( S(<,a).
{ - I- 1 )
<
On t h e o t h e r hand, (SD) d e a l s w i t h t h e p o i n t s of h e i g h t
between x, and x , .
C l e a r l y (FE)
=
=
1.
L e t u s s e e i f i t s a t i s f i e s (FE).
t h e empty s e t , L
=
0
=
R.
Let b map
Clearly {S,<,b,f3} satisfies ( 0 ) a n d ( B ) . Let a
Let YES; t h e n L
s e e t h a t {S,<,b,B] s a t i s f i e s (FE).
a in S
(F) + ( E ) .
Let S be any o r d e r e d c l a s s , s u c h t h a t IS1 2 2.
EXAMPLE 0.
a l l of S t o OEOn, and l e t 6
It
is, when a p p l i e d t o a Conway c u t i n
<
f3;
<
then a
{y]
<
R.
=
0.
S i n c e S(<,O)
Since S
By a s s u m p t i o n , I S 1 2 2.
=
S(S,O),
Let x,
<
is
we
x 1 be
188
Norman L. A l l i n g
i n S(=,O). (B)
t
Since t h e r e i s no p o i n t y i n S ( < , O ) , (SD) f a i l s .
PROPOSITION 0 .
If { S , < , b , B } s a t i s f i e s
IS(=,O)l Let S
EXAMPLE 1 .
B = 2.
+
-
(B),
(O),
a n d (SD); a n d i f S
= 1.
[ c , d , e, f , g ) , ordered l e x i c o g r a p h i c a l l y .
Let
Let b ( e ) = 0 , and l e t t h e b i r t h d a y map b map c , d , f , and g t o 1 .
Clearly { S , < , b , 2 ] s a t i s f i e s ( 0 ) and (B).
<
(y]
Note (FE) t a l k s a b o u t e x i s t e n c e
L e t a ( i n axiom ( F E ) ) be 0; t h e n L and R are empty,
not uniqueness.
<
Hence ( 0 )
(FE) h o l d and (SD) f a i l s i n t h i s Example.
is non-empty, t h e n
L
4.60
R f o r a l l YES, (FE) h o l d s a t t h i s l e v e l .
Now l e t a
=
1.
Since Having
d e a l t w i t h t h e Conway c u t ( 0 , 0 ) , we h a v e o n l y t w o more Conway c u t s of S(<,1)
t o investigate:
( i ) ( 0 , i e ) ) and ( i i ) ( [ e l , 0 ) .
p r e s e n c e of c and d s u f f i c e t o s a t i s f y (FE). f and g s u f f i c e t o s a t i s f y (FE); t h u s {S,<,b,2)
I n case ( i i
:n c a s e ( I ) , t h e
, the
satisfies (0)
S i n c e t h e r e i s no element i n S between c and d c S ( - , l ) ,
+
p r e s e n c e of
fB)
+
(FE).
S , < , b , B ] does not
s a t i s f y (SD). Thus from Examples 0 and 1 we see t h a t we have proved t h e f o l l o w i n g :
(SD) i s independent of (0) t (B) + ( F E ) .
THEOREM 2.
Let S be any o r d e r e d c l a s s .
EXAMPLE 2 .
C l e a r l y ( S , < , b , B ] s a t i s f i e s (0) + (B).
o n t o s m o C0,B). Since
Let b be any i n j e c t i o n of S
IS(=,a)l
1 , for all a
=
< 6;
(SD) i s v a c u o u s ,
t h u s ( 0 ) + (B) + (SD) h o l d , f o r t h i s
example. EXAMPLE 3 .
on S.
Let
B
2
3, let S
-
[O,B),
<
B.
and l e t b be t h e i d e n t i t y map
-
A s we saw above, [ S , < , b , B J s a t i s f i e s (0) + (B) + (SD). S i n c e B 2 3,
there exist a
< B
such t h a t a
+
2
Let L
(FE) f a i l s f o r t h i s Conway c u t i n S(<,a + 2 ) . THEOREM 3.
{ a ] and l e t R = { a + 11.
Thus we have t h e f o l l o w i n g :
(FE) i s independent of (0) + (B) + (SD).
BIBLIOGRAPHIC NOTE.
The m o t i v a t i o n f o r t h i s S e c t i o n is d e s c r i b e d a t
t h e e n d of S e c t i o n 4.03.
The Simple Density Axiom grew out of s u c h t h i n g s
as the proof of (5) i n C6, p. 2441.
The i d e a o f u s i n g i t as a n axiom is
4.60
189
I n t r o d u c t i o n t o t h e s u r r e a l number f i e l d No
d u e t o Timothy A . S w a r t z , who U n i v e r s i t y of R o c h e s t e r .
- at
- was
t h e time
Using (0) + (B)
an u n d e r g r a d u a t e a t t h e
(SD), S w a r t z w a s a b l e t o p r o v e
+
an i n t e r e s t i n g embedding theorem f o r s u c h s y s t e m s ( d u r i n g t h e f a l l of
1985).
When h e a d d e d axioms ( 0 ) a n d ( 2 ) of [71 t o h i s s y s t e m s , i t became
e q u i v a l e n t t o (0) + ( 8 ) 4.61
+
(SD)
+
(FE).
CONWAY CUTS, ORDERED BY EXTENSION, AND CUESTA DUTARI CUTS
L e t X be a n o r d e r e d s e t .
Recall (4.02) t h a t C D ( X ) d e n o t e s t h e s e t of Recall a l s o t h a t we p u t a n o r d e r o n
a l l Cuesta Dutari c u t s i n X (1.20).
C D ( X ) i n S e c t i o n 4.02, u n d e r which i t i s a n o r d e r e d s e t ( 4 . 0 2 : 2 ) .
L e t CC(X) b e t h e set of a l l Conway c u t s i n X (1.20). be i n C C ( X ) .
(L,,R,)
L e t (L,,R,)
( L 1 , R I ) w i l l be c a l l e d a n e x t e n s i o n o f ( L , , R , ) ,
and if Lo
is a s u b s e t of L , and R,
i s a s u b s e t of R , .
o r d e r s e t under e x t e n s i o n .
C l e a r l y t h e u n i o n of any non-empty s u b s e t T o f
Clearly
C C ( X ) is a p a r t i a l
C C ( X ) , w h i c h i s o r d e r e d by e x t e n s i o n , is a Conway c u t i n X t h a t i s a n
By Z o r n ' s Lemma ( s e e e . g . ,
e x t e n s i o n o f a l l t h e Conway c u t s i n T .
we have p r o v e d p a r t ( i ) of t h e f o l l o w i n g :
p.33]), (0)
C50,
(1)
Every (L,RlsCC(X) h a s a maximal e x t e n s i o n (L*,R*) i n CC(X).
( i i ) C D ( X ) i s t h e set of a l l maximal e l e m e n t (L*,R*) of C C ( X ) .
PROOF.
I n o r d e r t o p r o v e ( i i ) , assume f o r a moment, t h a t ( L * , R * )
not a Cuesta Dutari cut i n X. u n i o n of L* and R*.
x
>
y , t h e n l e t R**
(L**,R**)
-
If ( i ) t h e r e exists YEL s u c h t h a t x
b e t h e u n i o n of L* and {x} a n d l e t R** e x t e n s i o n of ( L * , R * ) ,
which i s a b s u r d .
R*.
<
y, t h e n l e t L**
C l e a r l y (L**,R**)
is a proper extension of (L*,R*),
t h a t ( i ) a n d ( i i ) d o n o t h o l d ; t h e n L*
-
i s a proper
If ( i i ) t h e r e e x i s t s Y E R s u c h t h a t
b e t h e u n i o n of R* a n d { X I a n d l e t L**
-
is
Then t h e r e e x i s t s x i n X which i s n o t i n t h e
<
{x}
which i s a b s u r d .
< R*.
L*.
Clearly
Assume now
Let L** be t h e u n i o n of
L* and { x } a n d l e t R** R*. Then (L**,R**) i s a p r o p e r e x t e n s i o n of (L*,R*), which i s a b s u r d . C o n v e r s e l y , c l e a r l y e v e r y C u e s t a D u t a r i c u t i n X i s a maximal Conway c u t in X .
Norman L. A l l i n g
190
Let (L,R)cCC(X). a n d l e t R-
letL
(1)
+
=
-
X
Let L- = {xEX: t h e r e exists Y E L s u c h t h a t x 6 y } ,
Let R +
L-.
4.61
(xcX:
=
t h e r e e x i s t s ycR s u c h t h a t x L y } , a n d
+
= X - R .
- -
(i) (L ,R
+
a n d (L , R
( i i ) Let ( L * , R * )
+
a r e maximal extensions of (L,R) i n C C ( X ) .
b e a maximal e x t e n s i o n of (L,R) i n C C ( X ) .
t h e l i n e a l o r d e r i n g o n CD(X), (L-,R-)
Then, i n
<, ( L t , R + ) .
I (L*,R*)
PROOF.
C l e a r l y L- is t h e smallest l o w e r - s a t u r a t e d s u b s e t of X t h a t
contains L.
S i n c e L* is a l o w e r - s a t u r a t e d s u b s e t of X t h a t c o n t a i n s L , L-
is a s u b s e t of L*; hence (L-,R-) smallest u p p e r - s a t u r a t e d
5 (L*,R*) ( 4 . 0 2 : 1 , i i i ) .
C l e a r l y R + is t h e
s u b s e t of X t h a t c o n t a i n s R .
u p p e r - s a t u r a t e d s u b s e t of X t h a t contains R , (L*,R*) S (L+,R+) ( 4 . 0 2 : 1 , i i I ) .
Rt
S i n c e R* i s a n
i s a s u b s e t of R * ;
hence
o
Let ( S , < , b , B ) s a t i s f y ( 0 ) + ( B )
+
( F E ) (4.60), l e t a
<
6 and l e t
(L*,R*) b e I n CD(S(< , a ) ) . (2)
<
If yES(S,a) s u c h t h a t L*
PROOF.
R* t h e n b ( y )
a.
o
For a l l ( L , R ) E C C ( S ( < , a ) t h e r e e x i s t s y s S ( - , a )
PROOF.
=
S i n c e (L*,R*) i s a Cuesta Dutari c u t i n S ( < , a ) , t h e u n i o n of
L* and R* is S ( < , u ) ; t h u s b ( y ) L a .
(3)
<
{y]
such t h a t L
By ( 0 1 , ( L , R ) h a s a maximal e x t e n s i o n ( L * , R * )
w h i c h i s a n e l e m e n t of C D ( X ) . which t h e f o l l o w i n g h o l d s : L*
i s a s u b s e t of R*,
<
<
{y}
<
R.
i n CC(S(<,a)),
By (FE) a n d ( 2 ) , t h e r e e x i s t s y ~ S ( = , a )f o r {y]
<
R*.
we c o n c l u d e t h a t L < { y )
S i n c e L is a s u b s e t of L * a n d R
<
R.