Appendix

656

APPENDIX

k

.

9

81

(2) Moremer i f St = {a;: i < a(t)}we can ae%unce tlrat t E W' * a(t) = aQandforsome U c a * ; i E U , t E W ' * 4 = d a n d 8 = { s ' : i ~ U ) a n d 8:(1, = 8&loe+ 8&4) = 8{(2)for t( I), t ( 2 ) E w',and 8!(1) = 8&a), i # j E w' *

t(l), t(2) E u.

Proof. W.1.o.g. W = A, 8, E A and let K = p when p is regular, and K = p+ otherwise. Clearly K < A and K is regular. For any a < A, cf a = K , clearly S , n a is a bounded subset of a, so let h(a) < a be a bound of it. By 1.3(3) {a < A: of a = K } is stationary, and by 1.3(1) on some stationary W 1c {a < A: cf a = K } h haa constant value 8. As ISl
THEOREM 1.5: (1) I f ASK = A, t h there ie a family 3 ' of 2" functiOne from A into A, such that for any d i e t i n c t f , E B (i < a < K ) and any ordinals y, < A (i < a), (5 < A: for every i < a,f,(5) = y;} # 0. (2) If A<' = A there are subsets S; (i < 2 9 of A, d that for any disjoint nun-empty U,V c A, IUl + I Vl < K , the eet UieUS, - UtevS, has caraimiity A. Proof. (1) Let {(At, (q: 5 < a;), (j!: 5 < a;)):i < A} be an enumeration 5 < a), (j,: 5 < a ) ) such that: of all triples (A, (0,: (i) A is a subset of A, IA1 < K . (ii) 0,is a subset of A and 5( 1) # 5(2) implies Cc,l, # Cc(a,. (iii) a < K and j, < A. (Clearly the number of triples is A, aa A<" = A.) Now for every set B c A we define a function f B : A + A. We define f & ) aa follows: if B n A; = Cisf B ( i ) = j:. Otherwise f B ( i ) = 0. If Batn (5 < a < K ) are distinct and j, < A ( 5 < a ) we can find A c A, IAI < K so that B , n A m distinot; ao for some i, A, = A, a; = a, C: = B,,, n A , j: = j,, so fBaCc,(i) = j,. Hence ( f B : B c A} is a family satisfying our requirements. . (2) Immediate by (l), for if B = (f,: i < 2"}, let St = {a < A:

ft(4 = 01.

THEOREM 1.6: (1) I f S ie an infinite family of subsets of I , then 2oe can jEndt,,EI,A,ES(n < w)suchthat (i) t, E A, iff n = m, or (ii) t , E A , i , n # m,or

f l . 9

8 11

657

FILTERS AND STATIONARY SETS

(iii) t, E A , iff n c m, or (iv) t, E A , iff n 2 m.

For every n < o there ie k < o such that i f IS1 2 k, S a family of then we canjhd t,, A , (i < n) a% above.

(2) 8et8

Proof. (1) We define by induction on n, t,, A,, S, such that: ( a ) So = S , #,+, c S,, S, is infinite. (p) S i = { A E S,: t, E A } and Si = { A E S,: t, $ A } a m non-empty. (y) A n ~ S i o A n $ S ; o S n + # lS i e S n + l= S : . This is easy as we can always find a suitable t,, since S, is infinite, and then it is easy to choose A , and S, + By Ramsey’s theorem for a 2-place function with range of cardinality 4 (see 2.1) the conclusion is easy. (2) Essentially the same proof. 1.4: (1)Ded A is the first cardinal p , such that no tree with DEFINITION A nodes has 2 p branches (a tree is a partially ordered set T such that for every x E T {y E T : y < x} is well-ordered. A branch is a maximal linearly ordered subset). (2) Ded, A is the first regular cardinal which is 2 Ded A.

THEOREM 1.7: (1) Swppo~eS i8 a family of subset8 of I , I injnite, IS1 2 DedJIl. Then there are, for every n, element8 to, . . .,tn-l of I , 8wh that for every w c n there is A, E S such that: ti E A , e i E w for i c n. (2) Suppose S ie a family of subset8 of I , I is jnite,

Then there are element8 to, . . .,t, A,,,ES such thd t, E A, oi E w.

IS1 >

y

Zisn ( ).

,E I 8wh that for every w c n there is

Procf. (1) Let A = 111, p = Ded, A, and w.1.o.g. I = A, and J c I, IJI < A implies I{A nJ: A ~ L 9 ) 1< p . Let for a s A S: = { ( A n a,a): A EL^). So clearly I S:[ < p for a < A. Let ( A , a) 1 /3 = (A n /3, min{a, 18)). On Uarh St we define the following partial order: ( A l , a l ) 5
658

APPENDIX

1s: - sy

=

I

u-

8ESb)

s :

(tEsg:8

zs

t}


a
EXERCISE 1.1: Show that if p < Ded, A, III = A, then there is a family S of subsets of I such that the conclusion of 1.7(1)fails even for n = 2. EXERCISE 1.2: Show that the bound in 1.7(2)is the best possible. PROBLEM 1.3: Can we replace Ded,lIl by DedlIl in 1.7(1)? EXERCISE 1.4: Prove x < Ded A iff there is an ordered set J, IJ I 2 x, and a dense subset I E J, IIl zs A. (Hint: For * use a lexicographic order on the tree plus its branches. For -= the nodes will be intervals of I.) LEMMA 1.8: (1)Suppose 8, i8 TegUlar, < N, for t E W ,I Wl = NB+, then for 8ome W' G W, l w'l = NB+, and V , IVl < N, and for every distinct s # t E W',S, n S, E V . If W E (6< K,,, :cf S 2 Ka} is stationary, we can have W' stationary too.

AP.,

§ 21

FILTERS AND STATIONARY SETS

659

(2) Imtead "8, is regular" we can demand lStl < cf X, for t E W . If W = gB+ny 8,G HE+,,, we can a 8 8 U W y ES, - V impJim y 2 a f o r Y E W'.

Proof. (1)Similar to 1.4: we can assume for some p lStl = p for every t ; W = h,Sa c h.Thenforsomestationary W1 G {a < &+n:Cfa = p+}, and y < N,, for every a E W1, S, n a c y ; and S, c a for 4' < a. Now we prove by downward induction on 1 5 n that there is V , c y , IVl < HE+,, and I{a E W': S, n y c V,}l = a,,. For Z = n Vn = y ; for 1 = 0 we get our conclusion, and if V,+l is defined, let V , + , = Ui Vl, 'IVlI < HE+,, Vj increasing (i < KB+l);one of the Vj is as required on V l . (2) Similarly.

A.2. Partition theorems

THEOREM 2.1 (Ramsey's Theorem): (1) For any infinite ordered set I , and n-place function f from I , with range of cardinality < No there is an infinite set J c I a h thud if I , Z E "J, 3, 8 increasing, then f ( I ) = f (8). ( 2 ) For any n, k, 1 < w there is m = m,(n, k, 1) such that if I is an ordered set of cardinality r m , and f an n-place function from I with range of cardinality sZ, then for some J c I , IJ I = k; I , Z E nJ, I , Z increasing =- f(3) = f ( t ) .

Proof.(1) For simplicity we assume I = w . We prove the assertion by induction on n. For n = 0 or n = 1 there is nothing to prove, so assume we have proved for n, and we shall prove for n + 1. We now define by induction on k c w , natural numbers Z(k) and infinite sets 8, c w such that (i) Z(k) < Z(k 1 ) s 8, C Sk-1, (ii) for any m E s k , Z(k) < my (iii) for any m,, . . ., m,,-l c k, m ESk

+

f V(m0),. - .,J(mn-l)yW ) )= f(l(mo),- . -

3

4mn-1), m).

We choose Z(0) = 0, So = {i: 0 < i c w}, and if Z(k), SI, are defined, let E , be the equivalence relation on Sk defined by: iE,j o for any mo,. . .,mn-1 s k, f(Z(m0),..., Z(mn-l), i) = f(+,), . . ., J(mn-l),j ) . clearly 8, has only finitely many equivalence classes, so at least one of = them is inhite; choose Z(k + 1) from such a class and let {i: i E S ~Z(k , + 1) < i, iE,$(k + 1)).

660

APPENDIX

[-., § 2

Now define on I' = {Z(k): k c w } a functionf',f'(Z(k,), .. .,t(k,,-l)) = f (Z(ko),...,Z(k,,_,), Z(E)) for all large enough k. Now by the induction hypothesis (applied to I*,f ') we get our conclusion easily. (2) We leave it to the reader.

For any n, k < w we can define an increasing sequence h that the, following h o h : Supposef is an n - p h fandim from 1(:)2 into k, then t7we are h, : I r 2 --t l(:)22, rn c n, and k* c k Buch that (i) hm(q) has tenstla 4J(q))* (ii) q 4 u * hm(7)Q It&). (iii) hmis one-to-one. (iv) I f 9 , ~ ~ 2 ( m c n)aredistinct,thenk* = f(ho(qo), ...,h,,-l(q,,-l)).

THEOREM Z(i) (i <

2.2:

w)w

Proof. Let Z(0) = 0 and if Z(i) is defined, we want Z(i + 1) to be such that:

(*I

If IStl 2 Z(i + 1)for i E I, 111 = n2":), and g is an n-place function from UaerSainto k then there are distinct a:, ui E 8, (aE I)and an n-place function g' from I into k such that g(a;'& . . .,a:(&) = g'(a(l), .. .,a(n)) for any distinct a ( l ) , . . . , a ( n ) ~and I anyj(l),...,j(n)c{O, 1).

We can do this, for let Zr(0) = k, Z:(a + 1) be 2 to the power 2 E;"(n29. Now for any suitable I,Xu, g n[lr(a)n21(i)]n, and 21(1+1)-1(') let I = n2":) for simplicity, and now define by induction on a c n2'") sets W , and elements ,:a a; such that: (i) W, E L. a s j < n2l(:)} and W,+lE W,, (ii) I n s,i = . ..: * -2lW - a) for j i ,-:, (iii) a: # af E S , n W,, (iv) if b o y .. .,b,,-a E {uj: y c 2, p c a} u U {S,:a c j c n2"{)}and m 5 n - 1, then

w,

I".:

g ( ~ o , . . . Y ~ r n - l , u ~ , b ~ , . . . ~ b ,=, - g(boy...,bm-l,af,bm,.. a) .,bn-d.

If we have defined for fi c a, we first choose Wa to satisfy (i) and (ii) and then we can eaaily satisfy (iii)and (iv).Clearly a!, ui me as required in (*). Now we prove the theorem by induction on i. For i = 0 there is nothing to prove. Suppose we have proved for i and we shall prove for

A P . 9

5 21

66 1

PARTITION THEOREMS

i + 1. We apply (*) for I = l(O2 x n,f (or more exactly, some suitable extension of it) and B<,,,> = {v: v E l({+l)2, r) Q Y ) , and get'distinct p&,> E B<,,,), a = 0, 1 and an n-place function f' from into k, so a(n-1) thatf @%b).o>, &&,1>, * * 9 P = f'(do), ~ ( 1 ) s- - * q(n - 1))Now we use the induction hypothesis on i for f ' and get suitable h; (m < n), k*. Now define h, so that for r) E i22, hm(r))= h&(r))and for

-

r ) E '2,

9

hn(r)Ya))= P7v.m>.

THEOREM 2.3: Suppose n, k < w and f is an n-placefunction from O>2 into k. Then we can jind functions h:, hz from 2 into 2 (for rn < n) and k* < k and natural numbers lo(i),P(i) such that (i) lo(i),P(i)increase with i, (ii) h:(r)) 7m Zength lo(l(q)),hi(r))h.9 length ll(l(r)));h: and h i are one-to-one, (iii) r ) Q v iflhi(r))Q h%4, (iv) h i m -4 %(r)), E 9 , distinct, (v) for any i and q0, . . .,

k*

-

= f(h:(r)o), *

- 9

hLl(7S-l)).

Proof of 2.3.Define l ( i )as in 2.2; and apply 2.2 to get for each i < w , < n) and kr < k. For everyj the number of possible (h,,* '+2: m < n ) is finite, hence by Konig's lemma there is an increasing sequence i ( a ) < w (a < w ) and k* so that h,,{(") t j 2 2 = hm,f(j) t j r 2 for a 2 j and k&) = k*. Now for r) ~ ' let 2 r)' E *('I2 be defined by r)'[a] = r)[a], a < j, r)'[a]= 0, j I; a < i ( j ) and let %(.I) = h,,dq'), h h ) = h,,~o,(r))and clearly they prove 2.3. m < n, functions hm,{(m

THEOREM 2.4: Suppose f , is an n(m)-placefunction from O > 2 into a jinite set. Then there are functions ho,h1 : ">2 +-O>2 such that: (i) Z(r)) = l(v) implies l(ho(r)))= l(ho(v)),l(hl(r)))= l(hl(v));ho and hl are one-to-one. (ii) 7 Q Y iifs hl(v)Q hl(v), (iii) hl(r))4 ho(r)), (iv) i f m I i I min{a, ,!I}, ql E "2, vz E 82 (for l < n(m)),vl t i = r i, ~ ( 1 )z ~ ( 2 ) i z vlca, i , , t h n fm(ho(70),*

-

r

2

r

ho(7n(m)-1)) = fm(ho(vo),*

Proof. Immediate by repeated uae of 2.3.

*

9

ho(vn(m)-l))-

662

[M., §2

APPENDIX

DEFINITION2.1 : A + (K); if for every n-place function f from h into p there are 8 5 A, 181 = K and a < p such that for any increming sequence a E n 8 , f (a) = a.

THEOREM 2.6: an@)+ -.(A+):+'. Proof. By induction on n. For n = 0 it is easy, so suppose we have proved it for n, and we shall prove it for n + 1. Let x = p = an@); and define an increasing sequence pi (i < p') of ordinals < x + such that (*) if 8 c pi, lLSl 5 p, then for every y < x+ there is anordinaly' < ~ i + l s u c h t h a t f o r a n y a.o. ,. , s n - , ~ 8 f ( s 0,..., 8n-1,y) = f(sO,...,an-l,y') a n d y ' E B e y E 8 . Thisiseasilydone byinducpi. Now define by tion, by cardinality considerations; let p* = Uiep+ induction on i < &+, yi < /3f+l, such that for any so,. ., an-1 E { y j : j < i},

.

f

(80,

---

9

sn-19

ri) = f

(80,

---

9

an-l,p*h

j < i * rj Z:

Yi.

Define an n-place function f ' on p+ :

f ' ( i ( o ) , .**ni(n -

1)) =f(3/i(o),...,ri(n-l),B*).

By the induction hypothesis there is 8' c p + , 181 ' = A+ so that for any i(0) i ( n - 1 ) ~ 8 ' f'(i(O),...,i(n , - 1)) = ao, for some fixed ao. Now S = {yi: i E 8')proves the theorem.

THEOREM 2.6: For every n, m < w there is k = k ( n , m ) < w (and k(n, 1) = 0 ) such that whenever h = 5&)+ the follolving ho&: I f f is an m-place function from n 2 X into x (or any set of cardinality 5 x), then there is I s such that: (i), ( ) E I and i f q E I n n'A, then l{a: a < A, f ( a ) E I } ! = x+, (ii), if qo, . . .,qm-1, yo, . . .,vmW1E I and ( q o - -) (vo, . . .) (see V I I , Definitions 2.3 and 3.1), then f (qo, . . .) = f (yo, . . ).

-

-

.

Remark. We do not try to get the best k(n, m). Proof. We fist prove for m = 1 that we can choose k(n, 1) = 0, and then prove by induction on n. Clearly we can assume n, m > 0. Case 1.m = 1. Let k = 0, so A = x+. We define, by induction on j 5 n, for any q E n - f h a set I, E {v: V E " ~ A , q Q v or q = v } and ordinals an(q),. . ., an- j ( q ) < x such that: (i) q E I, and if v E I, n "'A, then l{a c A: v-(a) E I,,}] = x+,

AP.9

5 21

663

PARTITION THEOREMS

(ii) foranyvEI,,n'A,n - j 5 i 4 n,f(v) = q(r)). For j = 0, r ) €"A, I,,= {r)}. Suppose we have defined I , for any 1 7 ~ " - j + l h , and let r ) ~ ~ - f A For . any P < A, the sequence (a,,(r)^(j?)), . ,t~,,-,+~(r)~@))) is well defined, and there are only 5 2 = x such sequences, hence there is a sequence (a,,,. ..,a,,-,+,) which we get for A = x+ pa. Let a,,(r)) = a,,,. .,t~,,-~+,(r)) = a,,-,+l and Qn-,(r)) = f (4and

..

.

"u

I , = (7)

{In^: (%(r)-(P)),

*

-

*

9

%-j+l(f

)>

= (a,,,* *

.,%-,+l)}*

Clearly this satisfies our demands. Case 2. n = 1. This follows by 2.6, taking k(n, a)= m

- 1.

+

Case 3. Suppose we have proved for n, m, and we shall prove for n 1,m. Let p = &,a+,,,+,(x), k(n l , m ) = k(n,m) m2 m 2. Define an m-place function g on n2A:

+

+

g(r),,

.. .,r),,,-,)

= {(h,

Po, . . .,P,,-,

a):

+ +

h a function, w -c myh: w +my

Pi < p (1 < m ) and a = f (v,, . . .,v,,- 1) where 1 E w * Vl = r)h(l)-(Pi), 1 $ w =+ Vl = q3.

Clearly the range of g haa cardinality I2'. By the hypothesis of Cam (3) there is a set I , c IZ A which satisfies: (i)' ( ) E I, and r ] E I, n " > A implies I{a < A: f ( a ) E I,}! = (2")+, (ii)' if +j,i j E I,, q i j , then g(ij) = g(ij). We cas find I, c I, such that implies I{a < A: f ( a ) E I , ) ~= x+ (i)" ( ) € I , , and r ) €1,n (define I, n 'A by induction on i). In view of (ii)' it suffices to find for each r ) E "A n I, a set I,,c {r)^(a): a < p}, iI,l = x+ such that:

-

(*I

If ij,q are similar sequences of length m from I*= U{I,,:~EI,n~A}uI,andij[I] I n = $1 r n t h f ( 7 j ) =

f (6).

For this it suffices to prove Theorem 2.7 below. for i < x ( X i well ordered THEOREM 2.7: 8wppme IX,l = amnc,,,+l,(x)+ by 5 ) and f k an mplace fundion f r m Uiex X i into a set of cardinality

664

LAP., $ 2

APPENDIX

I

S X . Then there are sets Y , E Xi, YiI = X + 8w:h that: if s(Z),t(1) E Ti(,, (1 < m) a d i(ll) = i(Za) [8(11) < a($,) o t(ll) < t(l,)] then

Proof. We define for n 4 m , i < x sets XT such that ( 1 ) X! = x xn+l i c - xn i,IXFl=~m(rn-n+l)(~)+, (2) if 8(1), t(1)EX;,), 2 < m , I{i(l):i(1)r i(lo)}l 4 n, and 1.3

then f (8(0),. . ., s(m - 1 ) ) = f (t(O),. . ., t ( m - 1)). We define Xr by induction on n. For n = 0, Xp = X i and clearly (1) and ( 2 ) hold. Suppose we have defined for n, and we shall define XT+l by induction on i. Suppose we have defined X;+l for j < i; ohoose @ c X:, IS:[ = m for i < a < X . By 2.6 we ccm easily find XP+l c X:, IX;L+lI = arn(rn-n)(X)+ such that ( 2 ) holds for n + 1, i (lo) = i when i (1) 4 i * s(l), t(1)E Xzlf and i ( l ) > i =. s(l), t(1)E&). By the induction hypothesis on n clearly ( 2 ) holds. Now let Y , = Xf end clearly the Y,'; rn as required. THEOREM 2.8: Suppme IS1 > A, S is the h d n o f f , and for x € 8 , If (x)l < A ( f ( x ) is a set). Then there ie S* € 8 , IS*! = 181 8w:h thut s # t E S* implie8 e $ f (t).

Proof. Cwe 1. IS1 = p is regular. W.1.o.g. S = p; x E S =.f ( x ) s p and define Si by induction on i s A, so that Si is a maximal subset of p - a f [where 4 = sup Uj,, 4, a? = sup U { f (B) u B < a:)] which satisfies 8 # t E st s-8 $ f (t). If for some i, ISigrl = p we finish. Otherwise, i 4 A * a? < p, and notice that j < i, s E p - a? implies f ( 8 ) n S , # 0. Now clearly the 8,'s &re pairwise disjoint and for any 8 E p - a!; f ( 8 ) n Si # 0 for i < A, hence If (a)] 2 A, contradiction.

v}:

Cme 2. IS1 = p is singular. Let p = Z,,,p(i), K = ofp, p ( i ) , i < K , is a strictly increasing sequence, A, K < p(0). W.1.o.g. we can assume that S = p, end x € 8 e-f ( x ) G 8; end a < /3 =. /3 $ f ( a ) (as we can replace 8 by any S' c S , IS'I = p) and /3 # y ~ ( p ( i ) , ~ ( i )=+ ) {a: p ( i ) < a < p ( i ) + ) implies B $ f ( y ) (by Case 1). Now we define by

f i . 9

8 21

665

PARTITION THEOREMS

induction on t; < x = max{K+, A + } sets 8:c (p(i),p(i)+)(for i < K) such that (i) a €8:impliesf(a) is disjoint to U,<,8{, (ii) a E Sl,/3 E Ueec 8: implies /3 < a , (iii) 1S:I = p(i) or It$! < p(i) and St is a maximal subset of (p(i),p ( i ) + ) satisfying (i) and (ii). Clearly for any t; < x, ct # /3 E =- a$f(/3) hence if l U f < x S ~=l p we finish. Otherwiae, for < p(j(t;)).As x is every t; < x there isj([) < K such that IUf K there is j0 < K such that IUl = x where U = {t; < x:j(t;) 5 jO};so for all t ; U, ~ l8f'l < p ( j o ) .There is a such that p ( j o ) , sup Uc6uS{o < a < p ( j o ) + so by (iii) for all t; E U,f ( a ) n U,<,o#{ p 0 hence lf(a)I 2 x > A, contradiction.

u,<%t$

EXERCISE 2.1: Suppose I c 0 2 A , and v 4 7 , 7 E I =- v E I and for r ) E = > A n I define A,, = l{i < A: r)^(i) E I)I. (1) I f a = n < w,f:I+XandcfA,, > Xforeveryr]En>AnIthen there are J c I and al, I 5 n such that ( ) E J and E J n " > A implies h, = I{i < A: r)^(i) E J}I and 7 E J impliesf(7) = aI(,,). ( 2 ) If a = w and for every 7 E I , cf X, > Xn, then there are J E I and aIIM above.

EXERCISE 2.2: Reprove 2.7 as follows: Msume IX,l = A rn > 1, X , = h x {i} (for i < p ) and f 5 x+ and define

=

D,,,(x+)+,

and use 2.6; and get suitable S E A, 1 8 1 = x+ w.1.o.g. 8 = x+ and let Y,= (xi, ~ (+i1)) (or xi, x(i + 1)-ordinalproducts).

EXERCISE 2.3: Reprove 2.6 similarly, assuming h = p = IRangefl, 2' 5

g(ao,

--

*9

x+ using

amn - 1 ) =

= {(c, (i:,

*

-

- - (ic-l, i mI ( -m1- 1 ) ) ) : . .), . ..,(a,;-1,. ..)), Z(k) < n,i:

iP,o)),

c = f((q8,.

*

* * * 9

< mn}

get 8 s A, IS( = x+ from 2.6, w.1.o.g. S = x+ and let

I = { ( a o , .. . , a I ) :1

5

n, and for each k

5

1

+ A and 0 < a. < A}.

Ak+lak < ak+l < Ak+lak

666

[AP., § 3

APPENDIX

A.3. Vsrious results LEMMA3.1:IfK < A 5 x,xBinguZarandforeveryp < xAD(~,A,A,K) holde, then AD(x, A, A, K) hide (see Definition VII, 1.11). Proof. Clearly if AD(p, A, A, K ) holds, 8 is a set of cardinality A, there is a (A, K)-family of p subsets of 8. Let x = p,, p, < x ; and let 8' be a (A, K)-family of subsets of A, 8' = {A,: i < po}. As lA,l = A, there is a (A, K)-fadY, 8,,of p1subsets of A, for i < po. Clearly (J,
z,<,,o

EXERCI8E 3.1: Prove A D ( A , A, A, 1) and natural implications.

LEMMA 3.2: Suppo8e AD(x,A, p, K ) holds. (1) Then An 2 x; 80 x = 2A, 2 K implies An = 2". (2) If h > 2", N, = min{A,: A: 2 A} and A s Nu, then for 8ome /3, A 2 N, = /3 > N, > 2%. Proof. (1) Let 8 = {A,: i < x} be a (p, +family of subsets of A. Choose B, c A,, lB,l = K ; so i # j+ B, # B,, hence.AK= I{B c A: IBI = K}I 2

l{B,: i <

x}l

=

x.

K , p correspond to N,, p, x). On K-contradictory orders and K-skeleton like sequences see Definitions VIII, 3.1 and 3.2 (VIII, Section 3 is the only place they are used).

(2) Immediate by VII, 1.9 (A,

THEOREM 3.3: For every A = A

+

K+, K

K-cOntradkt("yOrder8 Of Ca?'di?U&ty A.

regular, there are 2Apai&e

We prove 3.3 by a series of claims. DEFINITION 3.1: The orders I,J will be called strongly K-contradictory if they have cofinalities 2 K and there are no orders I,,J , with cofinality 2 K such that there is a model M with an anti-symmetric relation < and K-SkeletOn like sequences (a8:8 E I , + I*)and (b,: 8 E J , + J*) such that: (1) For every t E J* and 8' E I , there is 8 E I , , 8, < 8 , M C a, < b,. ( 2 ) For every ~ E I and * 8 l ~ Jthere , is ~ E J ,8,, < 8 , M Cb, < a,. (I*is I with order the inverse of the order of I.)

Remark. By the definition of K-skeleton like sequences (DefinitionVIII, E J * there are 8 O E I , , 8, E I* such that

3.1) (1) implies that for every t

dp.,

I!31

VARIOUS RESULTS

667

I , + I* k 80 I; 8 s 81 implies M C a, < b,. Similarly for (2). We shall use this many times. CLAIM 3.4: Any t

~

8 K - C O~d ~ a d i ~c t 0 p . yOrder8 ~ ~are K-COd?'&i&Wy. y

Proof.Immediate. QUELZTION 3.2: Is the converse true?

CLAIM 3.6: If I,J have distinct cojklitim > K , then I , J are strongly K-drdid0p.y.

Proof. W.1.o.g. c f J = p > A = c f I , A 2 K . Suppose I,J are not strongly K-contradictory, and we shall get a contradiction. By Definition 3.1, there are M ,I,, J,, (as: 8 E I , + I * ) , (b,: t E J , + J * ) satisfying the conditions mentioned there. Let (44: a < A) be an increasing unbounded sequence in I, and for each a < h choose t ( a ) E J and t'(a) E J , such that J , + J * C t'(a) I; t I; t(a) implies b, < a,(,). As J has cofinality > A, there is a bound t ( * ) E J to {t(a):a < A}. So for every a < A, J 1 + J * C t'(a) I; t ( * ) I; t(a),hence b,(.) < a,(,). This contradicts (1) from Definition 3.1 (by the remark to it).

CLAIM 3.6: If A i8 a regular cardinal > K , K regular, I = Zaeh I:, J = I,<,, J,* and {a < A: cf a 2 K and I,, J , are 8trongly K-contradictory} # 0 mod D(A) ( A = 0 mod D(A)ifSA - A E D(A),A G A), then I , J are ~trongly~-COdrdictory. Proof. Suppose there are a model N,orders I , , J 1 and sequences (a,: 8 E I1 + I*), (b,: t E J 1 + J *) satisfying the conditions from Definition 3.1, and we shall get a contradiction. As I,, J , are non-empty, we can choose 8, = 8(a)E I,, t, = t(a)E J,. Now we define ordinals a, = a ( i )for i < h such that: (i) j < i < A implies a, < a, < A, (ii) for a limit ordinal 6, ad = sup{a,: i < a}, (iii) if 8 E I*,'I + I* C 8 I; 8 ~ , + 1 then ) N C a, < btca(,),, (iv) if t E J * , J 1 + J * C t I; tact+,)then M C b, < a,,,,,,. Let us define by induotion: Case 1. i = 0; then a. = 0. Case 2. i regular).

= S is

limit; then

a,

=

Ufi, a,

(it exists as i < A and h is

668

APPENDIX

bP.9

63

Case 3. a, is defined and we s h d define a,+1.

I'

By the remark to Definition 3.1, there is 8l E I* such that 8 E I*, + I* C 8 s 8l implies M C a, < be,,,,,. Similarly, there is t1 E J*

such that t E J * , J l + J* C t 5 tl implies bt < aNMt,,,,. Let a,+1be the firetordinal > ~ s u c h t h a t I 1 + I*C8cr(t+1) < d , J 1 + J * CtCr(i+l)< tl. Clearly {a,: i < A, i limit} is a closed unbounded subset of A, so it ED(A).Hence by the hypothesis there is a limit 8 < A such that O f f a d 2 K (80 Cledy Cf 6 = Cf a d 2 K ) and lcr(d),Jcr(d) &l'0 Stroll& tc-contradictory. h t US define I + = In(d),J + = JMb), It = I {8at): i < a}, JZ = J {tN0:i < 8). We define the model N such that IN1 = 1611, x N = {(a, b): (b, a ) E <") and N = (INI,
r

+

Proof. By 1.3(3) A = {a: a < A, cf a = K } is a stationary subset of A, and by 1.3(2) there are pairwise disjoint, stationary A, c A (i< A). For any set W c h and a < A let I,,, be K if a E U B e w A sand K + otherwise; and let I, = C,
{a: of a 2

K;

I,,,, I,,, are strongly K - c o n t r a d i d r y }

includes A,, hence by claim 3.6, I,, I, are strongly K-contradictory. Our conclusion follows immediately.

CLAIM 3.8: If A i8 eingular and > K , K regular, then tirere ia a family of 2Apairwi8e 8 t r q l y ~ - c o n t ~ a c E i c t ~d r yr 8 of cardinality A.

zQcy

Proof. Let A = A, where p = cf A, K < A,, h, i n c r e d g , A, regular and choose a regular x, p + K + I x < A, and let A, (i < p ) be disjoint stationary subsets of {a:a < x, cf a = K}. Let K, be a family of 2% pairwise strongly K-contradictory orders of cardinality A,. For any f E K , (i.e., a function with domain p, f (a)E K,) and fl < x

n,<,,

~ ~ ,

D.,

o 31

669

VABIOUS RESULTS

let I,,,, be f ( a ) when /3 E A,, and K when /3 # UaeP A,. Let I , = ZBexIF,8;m in 3.7 we o m prove (by 3.5, 3.6) that I,, I , me strongly tc-contradiatory when f # g E 9,. As II,l = h and In,<,9 , ,l = 2% = 2 A we Gnish.

n,<,,

nu<,,

Proof of T h e m 3.3. Immediate by Claims 3.7, 3.8, and 3.4.

DEFINITION 3.2: ( 1 ) An m-place relation R over a set 8 is wnneoted [antisymmetric] iff for every distinct .so,. E S there is a permutation a of m suoh that R(8,(0), ., & ( , , , - I ) ) holds [does not hold]. (2) If R is an m-plaoe relation over 8, u a permutation over 8, then

..

' R

= ((80,

- ..

Y

-

h - 1 ) : <~o(o)Y * * Y %(m-i))

E R}.

LEMMA 3.9: ( 1 ) 8-e R ie an mplace, connected and antiqpnmetric relation over w, and w b a A,-m-indiecernible sequence, A , = {.Nxo, * * %I}. Then tkre i.9 a relation R*(xo,. .,xgm- which is a Boolean combination of imhnaa of R,such t h d when i c k # 1

.

9

R*(mi,mi + 1 ,...,mi +m

- l,mk,mk + 1,...,mk + m - 1 , m2,mi + 1, . . . , m i + m - 1 )

holds iff k c 1. (2) 8uppo8e R ie an mplace, connected and antieyntmetriC reldtion Over m = ( 0 , ., .,m - 1). Thenfor 8ome n < m - 1 and permWion a of m R'(O,. . ,n - 1, n, n 1 , n 2 , . . ., m 1) i ,R'(O,. ..,n - 1 , n + 1 , n, n + 2,. ..,m - 1).

.

+

-

+

Proof. (1) Let R* be

A {B(Xh(o),. . .,X h ( m - l ) ) t : h

&

fUM3tiOIl from m hlto %,

h one-to-one,and

t = 0 o ' t # 1 * R(h(O),.. .,W(m - 1)) holds), and let

P ( i , k , l ) = R*(im,im + l , . . . , i m

(eo P ia a three-placerelation over w).

+m - l,kmy..-sh--m)

670

P.,8 3

APPENDIX

By the R-m-indiscernibility of w , 0 < k < 1 implies P(0, k, I ) holds. Suppose it holds for 0 < 1 < k, and we shall get a contradiction. By the R-m-indiscernibility of w we can assume 1 = 1, k = 2. Let u be a permutation of m, n < m - 1, then

P ( 0 , 1,. . ., m - 1) ifF R'(O,l ,...,n , m + n + 1 , 2 m + n + 2 , 2m+m-1) (as w ie R-m-indiscernible) iff R'(O ,...,n,2m + n + l , m + n + 2, ..., m + m - 1) (as P(0, 2, 1) holds) ifF Ro(0,..., n - l , m + n , 2 m + n + 1, m n 2, ...,m + m 1) (as w is R-m-indiscernible) iff P ( 0,...,n - 1 , 2 m + n , m + n + 1, 2m n 2, ..., 2m + m 1) (as P(0,2, 1) holds) iff R'(O ,..., n - l , n + 1,n,n + 2,..., m - 1) (as w is R-m-indiscernible)

...,

+ +

-

+ +

-

.

Hence it suffices to prove (2). (2) Let nu = min{i 5 m: u(i) # i or i = m}, assume there are no such u and n; and we shall prove by downward induction on nu that u EC = (0': P'(o,. ..,m - 1) = R ( 0 , .. .,m - 1)). Clearly our assumption means u €2, n < m - 1 implies u(n, n + 1)E C((n, n + 1) is the permutation interchanging n and n + 1).If nu = m this is trivial, nu = m - 1 is impossible, and nu = m - 2 follows by the assumption. Suppose we have proved for n + 1, and nu = n, and let u(i) = n (so clearly n < i < m) and let uo = u(i, i - l)(i - 1, i - 2). (n + 1,n); clearly j < nu implies u o ( j )= u ( j ) =j, and uo(n) = u ( i ) = n, hence no, > nu, hence uo €27. But clearly u = uo(n,n + 1) (n + l , n + 2 ) . . . ( i - 1,i)soclemlyuEZ.

-

EXERCISE 3.3: Suppose 111 2 2m, R an m-place connected and antisymmetric relation over a, < l , < orders on I,and (I,< l), (I,< I) are R-m-indiscerniblesequences. Then (1)or (2) holds: (1) I = I , u I,, (Il, I , disjoint); and < l , < a are identical on each 1,(1 = 1, 2) and 81 E I , implies 81 < 82, 8I < a 81. (2) I = I , u I , U I , (the II'spairwise disjoint); < l , < a are identical on I,, 11, U 131 S m - 2 and 81 E I , implies 81 <' 8 , <' 83, 81 < a 82 < 83.

,

AP.,

3 31

VARIOUS RESULTS

671

DEFINITION 3.3: A dependency relation on a set W , is a relation R between members of Wand subsets of W (xdepends on w ) satisfying

the following conditions (where w is called independent if, for no

x E w, does x depend on w - {x}): (0) x depends on {x}. ( 1 ) Exchange principle : if {x$:i < a} is not independent, then for i < p}. some p < a,xp depends on {xi: (2) Finite character: x depends on w iff x depends on some finite u E w (so we have monotonicity: if x depends on u,uE w , then x

depends on w ) . (3) (Weak) transitivity : if w, u are independent, x depends on u, and every y ~ depends u on w , then x depends on w.

DEFINITION 3.4 : A nice dependency relation on W is defined similarly, strengthening (3) to (3)’ Full transitivity: if x depends on u,and every y ~ depends u on w (where u,vE W ) , then x depends on w. LEMMA 3.10: For a dependency relation on W, any w E W , any two maximal independent subsets of w have the same power. Also, every independent subset of w can be extended to a maximal independent subset of w. Proof. The second sentence follows from the finite character (2).For the first, suppose uoc w is a maximal independent subset of w,and u c w is an independent subset of w. It suffices to prove luol2 I u I so assume this fails. If JuI is infinite, for every X E U there is a finite w, E uoon which x depends. By cardinality consideration for some > luJand clearly finite uh G uo,u’= {x:w, = ui}is infinite, hence Iu’J uh,u’satisfy the assumptions every x E u‘ depends on ui,So uh U u’, and uh is finite. So w.1.o.g. uo is finite. We choose a on w,uo,u counterexample with minimal Iuo-uI.Let uo= {xi:i= 1 , n} with uon u = {xi:i= l , m } . Choose x,~u-u,(exists as luol< lul). So {xi:iQ m} is independent (being G u),but {xt:i< n} is not (as X ~ E U depends on uo= {xi:i = 1, n} and x,(iQ n ) are distinct as xo4 uo).So let I , (0, ...,m } c I E (0, ..., n} be maximal s.t. {x(:i~I}is independent and w.1.o.g. I = (0, ..., k} so m < k < n. def Clearly, ui = {xi:icI)is independent, and every member xiof uo depends on it (if ~ E by I (0) and monotonicity, if i$I, by the exchange principle for (xo,xl,...,xk,xi}x, depends on ui).Now for

672

APPENDIX

LAP.,

83

every x E u, x depends on u,, by assumption and every y E u,, depends on u;. As u,,,u; are independent, x depends on u;. So u;,u satisfy the assumptions, but lui - uI = I{i :m < i < k}l < I{i:m < i B n}l = Iu,,-uI as k < n and u;,u form a counterexample (as IuJ < luJ < lu,,l). This is a contradiction to the choice of uo,u.