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Chapter 6 (C(X),X)-Duality
CHAPTER 6
(C(X) ,X) -DUALITY
X , Y w i l l d e n o t e compact H a u s d o r f f s p a c e s .
C(X) i s t h e s e t
o f r e a l c o n t i n u o u s f u n c t i o n s on X , a n d n ( X ) w i l l d e n o t e t h e f u n c t i o n o f c o n s t a n t v a l u e 1 on X .
Under t h e u s u a l p o i n t w i s e
d e f i n i t i o n s o f a d d i t i o n , s c a l a r m u l t i p l i c a t i o n , a n d o r d e r , C(X) i s a R i e s z s p a c e w i t h n(X) f o r a n o r d e r u n i t .
The MI-norm
d e t e r m i n e d by n(X) i s p r e c i s e l y t h e s t a n d a r d supremum n o r m , and u n d e r t h i s norm, C(X) i s c o m p l e t e . Thus C(X) i s a n M I - s p a c e .
I n more d e t a i l , f o r f , g E C ( X ) ,
f v g a n d fAg a r e g i v e n b y ( f V g ) ( x ) = m a x ( f ( x ) , g ( x ) ) a n d ( f A g ) ( x ) = m i n ( f ( x ) , g ( x ) ) f o r a l l xEX. f+(x)
=
( f ( x ) ) + , and f - ( x )
in particular, =
Ifl(x)
=
/f(x)I,
( f ( x ) ) - f o r a l l x€X.
Although X is n o t a v e c t o r s p a c e , i t i s u s e f u l (and s u g g e s t i v e ) t o t h i n k o f C(X) a s " d u a l " t o X.
This has been
done by v a r i o u s w r i t e r s , e s p e c i a l l y i n t h e c o n t e x t o f C a t e g o r y Theory ( c f .
[49],
523.2 or [4T).
We w i l l f o r m u l a t e t h e r e l a -
t i o n s b e t w e e n C(X) a n d X f r o m t h i s v i e w p o i n t .
Since X has a
n a t u r a l i m b e d d i n g i n t o t h e d u a l C'(X) o f C(X), C'(X) w i l l t h u s play the r o l e of "bidual"
t o X.
I n p a r t i c u l a r , we w i l l f e e l
free t o denote f ( x ) by ( f , x ) . X w i l l b e c a l l e d t h e K a k u t a n i - S t o n e s p a c e o f C(X).
165
Chapter 6
166
5 3 0 . The t o p o l o g y o f s i m p l e C o n v e r g e n c e o n C ( X )
X i s o f c o u r s e s e p a r a t i n g o n C ( X ) by v e r y d e f i n i t i o n : f # g means t h e r e e x i s t s xEX
such t h a t f ( x )
# g(x).
Dually,
C(X) i s s e p a r a t i n g on X , a n d i n d e e d , i n a v e r y s t r o n g s e n s e : (Urysohn) f o r e v e r y p a i r o f d i s j o i n t , c l o s e d , non-empty s u h s e t s Z1,Z2 o f X , ( i i ) f(x)
=
Remark. -___
t h e r e e x i s t s fEC(X) s a t i s f y i n g ( i ) 0 < f (n(X),
1 f o r a l l xEZ1,
(iii) f(x)
=
0 f o r a l l xEZZ.
I l e n c e f o r t h , we w i l l c a l l an f E C ( X ) w i t h t h e a b o v e
properties a Urysohn f u n c t i o n f o r t h e ( o r d e r e d ) p a i r (Z,,Z,). A s o n e would e x p e c t , we d e f i n e o ( C ( X ) , X )
a s t h e topology
on C ( x ) o f p o i n t w i s e c o n v e r g e n c e on X : a n e t { f } i n CL
C(X) c o n -
v e r g e s t o fEC(X) i n a ( C [ X ) , X ) i f f ( x ) = l i m f ( x ) f o r a l l xEX. CY.N
We w i l l c a l l i t by i t s common name: t h e t o p o l o g y o f s i m p l e convergence.
And we c a n d e f i n e o ( X , C ( X ) ) a s t h e t o p o l o g y o n X
o f p o i n t w i s e c o n v e r g e n c e o n C(X): a n e t { x } i n X c o n v e r g c s t o c1
xEX i n ~ ( y C, ( X ) )
if f ( x )
=
limolf(x,)
for a l l fEC(X).
Since
x
completely regular,o(X,C(X)) i s simply t h e o r i g i n a l topology on X .
We e x a m i n e o(C(X) , X ) . Norm c o n v e r g e n c e ( i n C ( X ) )
and s i m p l e c onvergence.
i m p l i e s b o t h o r d e r convergence
T h e r e i s no o t h e r i m p l i c a t i o n b e t w e e n
the t h r e e convergences: Let X = a m , t h e one-point (Alexandroff) c o m p a c t i f i c a t i o n o f N , and f o r e a c h n , l e t en b e t h e e l e m e n t o f C(X)
h a v i n g v a l u e 1 on n E N a n d v a l u e 0 e l s e w h e r e .
t h e s e q u e n c e {ne,}
Then
c o n v e r g e s t o 0 s i m p l y b u t n e i t h e r normwise n n o r i n t h e o r d e r , a n d t h e s e q u e n c e {c e . ) ( n = 1 , 2 , . * . ) o r d e r 1 1 c o n v e r g e s t o l(X) b u t d o e s n o t c o n v e r g e e i t h e r s i m p l y o r no r m w i s e .
is
16 7
(C(X) ,X)-Duality
For a monotonic n e t , however, w e have one a d d i t i o n a l implication:
( D i n i t h e o r e m ) F o r a m o n o t o n i c n e t { f } i n C(X) cr f E C (X), the following a r e equivalent : (30.1)
1'
{fci. 1 c o n v e r g e s t o f n o r m w i s e ;
2'
{fa} c o n v e r g e s t o f s i m p l y ;
We n e e d o n l y show 2'
Proof. -__
implies lo.
and
For c o n c r e t n e s s ,
assume { f } i s a d e s c e n d i n g n e t w i t h i n f f ( x ) = 0 f o r a l l xEX. a a Consider E > 0 . F o r e a c h xEX, c h o o s e a ( x ) s u c h t h a t (x) <
E.
Since f
cr (XI
i s c o n t i n u o u s , t h e r e e x i s t s an o p e n
n e i g h b o r h o o d W(x) o f x s u c h t h a t f
(XI
(Y) <
E
f o r a l l yiW(x).
Now X i s c o m p a c t , s o t h e r e e x i s t x l , - - ,*x n s u c h t h a t { W ( x l ) , . - . , W ( x n ) } c o v e r s X. Then a > a ( € ) implies fa(y) <
Choose a(€) 2 w ( x , ) , . * . , a ( x n ) . E
f o r a l l yEX.
QED
I n common w i t h norm c o n v e r g e n c e a n d o r d e r c o n v e r g e n c e , simple convergence h a s t h e f o l l o w i n g e a s i l y v e r i f i e d , p r o p e r t y
(30.2)
I f { f } a n d {p } c o n v e r g e s i m p l y t o f a n d g r e s p e c t i v e l y , c1
t h e n {fcrVg,),
a
{fclAgcl},
Ifc1 + g a l , a n d {Afcl}
o f IR) c o n v e r g e s i m p l y t o f v g , f A g , f
+
(A
any element
g , a n d Af r e s p e c t i v e l y .
Chapter 6
168
Otherwise s t a t e d , t h e l a t t i c e o p e r a t i o n s and v e c t o r s p a c e o p e r a t i o n s a r e continuous under simple convergence.
F o r a R i e s z s u b s p a c e F o f C(X) t o b e s i m p l y
(30.3) Corollary.
c l o s e d , i t s u f f i c e s t h a t F+ be s i m p l y c l o s e d .
i s simply c l o s e d .
Note a l s o t h a t C ( X ) +
The f o l l o w i n g i s t h e c r u c i a l p a r t o f M . H .
Stone's proof
of t h e Stone-Weierstrass theorem.
( 3 0 . 4 ) _____ Theorem.
(Stone).
F o r a s u b l a t t i c e A o f C(X), t h e
following are equivalent:
'1
A i s simply closed;
2'
A i s norm c l o s e d .
We n e e d o n l y show t h a t 2'
P roof.
i m p l i e s 1'.
Assume A i s
norm c l o s e d , a n d c o n s i d e r f i n t h e s i m p l e c l o s u r e o f A . E
> 0.
We show t h e r e e x i s t s gEA
such t h a t
i A
( i ) F o r e a c h xoEX, t h e r e e x i s t s g
\I g
- fll
Fix
5 3 ~ .
such t h a t
xO gXO
(XI
gx (x,) 0
2
f ( x ) - 3~
< f(xo)
+
f o r a l l xEX,
E *
In e f f e c t , s i n c e f i s i n t h e simple c l o s u r e of A, then f o r
169
(C(X) , X ) - D u a l i t y e a c h xEX, t h e r e e x i s t s g
€ A such t h a t J g
( g x .(XI
-
0
f(x)
<
-
E ;
(xo)
- f(xo)
xO
xO
< r-, -
and from t h e s e c o n d o f t h e s e i n e q u a i t i e s ,
t h e r e e x i s t s a n o p e n n e i g h b o r h o o d W x) o f x s u c h t h a t ( y ) 2 f ( y ) - 3~ f o r a l l y 6 W ( x ) . X b e i n g compact, t h e r e xox e x i s t { x l , . . - , x n } s u c h t h a t IW(x,), * . , W ( x n ) } c o v e r s X. Set g
Since x
was a r b i t r a r y i n ( i ) , w e now d r o p t h e s u b s c r i p t .
0
We t h u s h a v e { g x l x E X } c A s u c h t h a t f o r e v e r y x , gx 3 ~ l l ( X ) and gx(x) < f(x) +
c o v e r s X.
Set g
=
m
tliTlgx
i 3 ~ l l ( X ;) s o we a r e t h r o u g h .
+
3~ f o r a l l y E V ( x ) .
, x m } s u c h t h a t {V(x,)
.
f -
For e a c h x , choose a n e i g h b o r -
E.
hood V(x) o f x s u c h t h a t g x ( y ) < f(y) Again t h e r e e x i s t s { x , , . .
2
, . . * ,V(xm) 1
< f + Then g E A a n d f - 3 ~ n ( X )5 g -
QED
(30.5)
Corollary.
(i)
Norm c l o s e d R i e s z i d e a l s o f C ( X )
a r e simply closed.
( i i ) M I - s u b s p a c e s o f C(X) a r e s i m p l y c l o s e d .
531. The d u a l i t y b e t w e e n t h e norm c l o s e d R i e s z
i d e a l s of C(X)
and t h e c l o s e d ( o r open) s u b s e t s o f X
F o r e v e r y s u b s e t Q o f X , we s e t QL
=
{fEC(X)] f(x)
=
0 for all
a n d f o r e v e r y s u b s e t A o f C ( X ) , we s e t
~€41;
Chapter 6
170 Z(A) Z(A)
{x€X I f ( x )
=
=
0 f o r a l l fEA). (In l i n e with our d u a l i t y
i s c a l l e d t h e z e r o - s e t o f A.
a p p r o a c h , w e would p r e f e r t o d e n o t e Z ( A )
b y A&, b u t Z ( R ) i s
the established notation.) We h a v e i m m e d i a t e l y :
( 3 1 . 1 ) For e v e r y A c C ( X )
,
Z(A) i s a c l o s e d s u b s e t o f X .
In the opposite d i r e c t i o n , given
Q
c X,
it i s e a s i l y
v e r i f i e d t h a t QL i s a R i e s z i d e a l a n d s i m p l y c l o s e d , s o a f o r t i o r i norm c l o s e d .
( 3 1 . 2 ) For e v e r y
Thus:
Q c X,
Q'
Now i t i s c l e a r t h a t
i s a norm c l o s e d R i e s z i d e a l .
(0)' =
QL
M o r e o v e r , by t h e Urysohn t h e o r e m , t h i s property.
( 3 1 . 3 ) For e v e r y
Dually ,
0is
We t h u s h a v e :
Q c X, Z(QL)
=
(q
0.
the closure of
Q i n X).
the largest set with
171
(C(X) , X ) - D u a l i t y ( 3 1 . 4 ) For e v e r y s u b s e t A o f C(X),
i s t h e norm c l o s e d
(Z(A))'
Riesz i d e a l H g e n e r a t e d by A.
-__ Proof.
H c (Z(A))'
by ( 3 1 . 2 ) .
clusion consider f € ( Z ( A ) ) I , B
=
{ g E H 10 c: g 5 f } .
i s f i l t e r i n g upward. we show f ( x )
=
f(x) > 0.
a n d we c a n assume f > 0 .
H e n c e , b y t h e D i n i Theorem ( 3 0 . 1 1 , i f
I f f(x)
Then xeZ(A).
=
it w i l l follow t h a t
0 , w e are through.
c a n assume h ( x ) > f ( x ) . =
Suppose
I t f o l l o w s t h e r e e x i s t s hEII, w i t h
h(x) > 0 ( t h e r e e x i s t s kEAwith k(x) # 0 ; s e t h
t h a t g(x)
Set
B i s n o t empty ( i t c o n t a i n s 0 ) and i t
s u p g E B g ( x ) , f o r e v e r y xEX,
Consider x .
fEfl.
For t h e o p p o s i t e con-
Then g
=
=
l k l ) , and w e
hnf i s an element of B such
f(x).
QED In particular,
(31.5) For e v e r y Riesz i d e a l I o f C ( X ) ,
(Z(1))'
i s t h e norm
closure of I.
Summing u p ,
(31.6) There i s a one-one correspondence between t h e f a m i l y o f c l o s e d s u b s e t s 2 o f X a n d t h a t o f norm c l o s e d R i e s z i d e a l s II of C(X):
H <->
Z i f and o n l y i f Z = Z(H)
i f and o n l y i f H
=
ZL.
Chapter 6
172 Given H <-->
Z,
s e t W = X\Z:
Thus t h e r e i s a o n e - o n e
c o r r e s p o n d e n c e b e t w e e n t h e norm c l o s e d R i e s z i d e a l s o f C ( X ) and t h e open s e t s o f X .
T h i s corresp o n d en ce can be g iv en w i t h -
out reference t o the closed s e t s .
First a definition.
A func-
t i o n f on a l o c a l l y compact s p a c e T i s s a i d t o v a n i s h a t i n f i n i t y i f f o r every -~ compact.
E
> 0, the set {tET
1
If(t)
1
> E} i s
The f o l l o w i n g i s e a s i l y v e r i f i e d .
( 3 1 . 7 ) There i s a o n e - o n e c o r r e s p o n den ce between t h e f a m i l y o f
open s u b s e t s W of X and t h a t o f t h e norm c l o s e d R i e s z i d e a l s
H o f C(X): H <-->
fEH}
W i f and o n l y i f W
=
{ x G X \ f ( x ) # 0 f o r some
i f and o n l y i f H i s t h e s e t o f c o n t i n u o u s f u n c t i o n s on
W which v a n i s h a t i n f i n i t y .
H e n c e f o r t h Z w i l l a l w a y s d e n o t e a c l o s e d s e t of
Notation.
X , a n d W an open s e t .
F o r a s u b s e t A o f C ( X ) , W(A) w i l l b e
t h e s e t X\Z(A). The s e t o f c o n t i n u o u s f u n c t i o n s v a n i s h i n g a t i n f i n i t y on a l o c a l l y compact s p a c e T w i l l b e d e n o t e d by C w ( T ) .
Note
t h a t i t i s an M-space u n d e r t h e supremum norm, a n d i s an MEs p a c e i f and o n l y i f T i s c o m p a c t . U s i n g t h e above n o t a t i o n , we c a n s t a t e p a r t o f ( 3 1 . 7 ) a s
follows: I f X
= W IJ Z ,
W a n d Z c o m p l e m e n t a r y , t h e n ZL = Cw(W) .
A n o t h e r o b s e r v a t i o n on C w ( W ) . elements of C(X) of W.
Let Ho c o n s i s t o f t h e
e a c h h a v i n g f o r i t s s u p p o r t a comp.act s u b s e t
Then Ho i s a R i e s z i d e a l , i s c o n t a i n e d i n C w ( W ) ,
norm d e n s e i n t h e l a t t e r .
We n e e d o n l y v e r i f y t h e l a s t
and i s
(C(X) ,X) - D u a l i t y statement.
173
Tn e f f e c t , s i n c e W i s c o m p l e t e l y r e g u l a r , Ho i s s o , by ( 3 0 . 4 )
s i m p l y d e n s e i n C'(W),
( o r t h e Dini Theorem), i s
norm d e n s e i n i t .
We c h a r a c t e r i z e v a r i o u s t y p e s o f R i e s z i d e a l s o f C(X) by p r o p e r t i e s of X . by i n t Q
.
We w i l l d e n o t e t h e i n t e r i o r o f a s e t Q i n X
Recall that a r e g u l a r c l o s e d s e t i s one which i s
t h e c l o s u r e o f an o p e n s e t , a n d a r e g u l a r o p e n s e t i s o n e which i s t h e i n t e r i o r o f a c l o s e d s e t .
We h a v e s e e n ( 3 1 . 2 ) t h a t f o r Q c X , QL i s a norm c l o s e d Riesz i d e a l .
F o r a n o p e n s e t W i n X , WL
(31.8)
___ Proof. WL
F o r an o p e n s e t , w e c a n s a y m o r e :
Let 2
X\W.
=
i s a b a n d o f C(X)
T h e n , as i s e a s i l y v e r i f i e d ,
(ZL)d, hence i s a band
=
QE D
For a Riesz i d e a l H o f C(X),
(31.9) C o r o l l a r y 1.
o r d e r c l o s u r e tI
P roof. simplicity, and H can
=
.'2
=
( i n t Z(H))'.
We c a n a s s u m e H i s norm c l o s e d . denote Z(H)
Also f o r
s i m p l y by Z a n d W(H) b y W.
So W
=
X\Z
S i n c e C(X) i s A r c h i m e d e a n , w h a t we w a n t t o p r o v e
b e s t a t e d : Hdd
=
( i n t Z)'.
A s we remarked i n ( 3 1 . 8 ) ,
2
Chapter 6
174 Hd
=
W"
(x\W)*
But t h e n Hd =
,
(w)" ,
s o , by t h e same r e m a r k , H d d
=
j i n t z)'.
=
QED
For a R i e s z i d e a l f1 o f C(X), t h e
(31.10) Corollary 2.
following are equivalent: lo
t h e o r d e r c l o s u r e o f H i s C(X);
'2
Z(H) h a s empty i n t e r i o r ( s o , b e i n g c l o s e d , i s nowhere
rare,
dense -
i n t h e Bourbaki terminology);
WIN) i s d e n s e i n X .
3'
The f o l l o w i n g a r e a l s o c o r o l l a r i e s , b u t we l i s t them a s theorems.
( 3 1 . 1 1 ) ____ Theorem.
F o r a norm c l o s e d R i e s z i d e a l H o f C ( X ) ,
the
following are equivalent:
1'
H i s a band;
'2
Z(H) i s a r e g u l a r c l o s e d s e t ;
'3
W ( H ) i s a r e g u l a r open s e t .
P roof. Z(H)
=
i f Z(H) H
=
I f H i s a band, t h e n , by (31.9) and (31.3),
m, so
Z(H) i s a r e g u l a r c l o s e d s e t .
Conversely,
i s a r e g u l a r c l o s e d s e t , t h e n , by ( 3 1 . 4 ) and ( 3 1 . 9 ) ,
(Z(H))'
=
order closure H. QED
(C(X) ,X) -Duality
175
I f a s e t i n X i s b o t h open and c l o s e d , w e w i l l c a l l i t clopen.
F o r a norm c l o s e d R i e s z i d e a l lI o f C ( X ) ,
( 3 1 . 1 2 ) Theorem.
the
following are equivalent: 1'
H i s a p r o j e c t i o n band;
2'
Z(II) is clopen;
3'
W(t1)
-__ Proof..
i s clopen.
S u p p o s e H 3 Hd
=
C(X),
but Z(H)
some xEZ(H) i s i n t h e c l o s u r e o f W ( I I ) . t1 v a n i s h e s on
Conversely, i f X =
T t
v a n i s h e s on x , c o n t r a d i c t i n g ( l l ( X ) ) ( x ) Z1lJ
=
So
Then e v e r y e l e m e n t o f
x a n d e v e r y e l e m e n t o f Hd v a n i s h e s on x .
follows a l l of C(X)
C(X)
i s n o t open.
=
Z2, w i t h Z1,Z2 d i s j o i n t , t h e n c l e a r l y ,
( Z , P 3 (Z$. QED
Finally
,
( 3 1 . 1 3 ) Theorem.
For a R i e s z i d e a l H o f C ( X ) ,
are equivalent: 1' H i s a maximal R i e s z i d e a l ;
2'
Z(f1)
1.
consists of a single point.
the following
176
Chapter 6 -__ Proof.
Assume H i s m a x i m a l , a n d c h o o s e xEZ(f1).
a p r o p e r Riesz i d c a l o f C ( X )
w i t h 11.
{x}'
is
and c o n t a i n s [ I , hencc c o i n c i d e s
I t f o l l o w s from ( 3 1 . 3 ) t h a t Z ( l I )
=
{x}.
T h a t '2
i m p l i e s lo i s c l e a r . Qlill
Consider a c l o s e d s e t Z o f X. Cw(W), W
= X\Z.:
As we h a v e s e e n , ZL =
Tn a d d i t i o n , C(X)/ZL = C(Z) ( c f .
(33.6))
5 3 2 . The d u a l i t y b e t w e e n t h e M l l - s u b s p a c e s o f C ( X )
and t h e u p p e r s e m i c o n t i n u o u s d e c o m p o s i t i o n o f X
P a r a l l e l i n g 531, we c o l l e c t h e r e t h e o r e m s r e l a t i n g t h c M I - s u b s p a c e s o f C(X) t o t h e t o p o l o g y o f X. immediately a r i s e s .
A difficulty
E v e r y MIL-subspace F c o n t a i n s l l ( X ) ,
t h e a n n i h i l a t o r o f F i s always cmpty.
hence
A substitute for this
a n n i h i l a t o r i s t h e dccomposition of X i n t o t h e sets of cons t a n c y o f F.
(The a n n i h i l a t o r __ does e x i s t
-
in C'(X)!
And
i s closely r e l a t e d t o these s e t s of constancy.) By a d e c o m p o s i t i o n o f X , w e w i l l mean a f a m i l y x
=
{Za}
o f m u t u a l l y d i s j o i n t , c l o s e d , n o n - e m p t y s u b s e t s o f X whose
union i s X.
The d e c o m p o s i t i o n s o c c u r r i n g i n o u r s u b j e c t a r e
of a special kind.
A decompositionx
=
{Za} o f X i s c a l l e d
uppersemicontinuous i f f o r e v e r y c l o s e d s e t Z i n X , t h e union of a l l the Za's intersecting 2 i s also closed.
I t i s e a s i l y v e r i f i e d t h a t i f X s > Y i s a continuous mapping ( r e m e m b e r , X , Y a r e a l w a y s compact H a u s d o r f f ) , t h e n t h e
(C(X) , X ) - D u a l i t y decomposition
2
=
{s
-1
(y) / y € s ( X ) }
177
of X i s uppersemicontinuous.
Conversely, every uppersemicontinuous decomposition
X i s d e t e r m i n e d b y a c o n t i n u o u s mapping o f X :
2
=
{Za} o f
In e f f e c t , l e t
X A> L b e t h e mapping w h i c h a s s i g n s t o e a c h xEX t h e Z t a i n i n g i t , and d e n o t e h y J f i n e d b y t h i s mapping q .
(Z,J)
cona t h e i n d u c t i v e t o p o l o g y on 5 d e -
The r e s u l t i n g t o p o l o g i c a l s p a c e
is called the q u o t i e n t s p a c e o f X d e t e r m i n e d by
c a l l e d t h e q u o t i e n t map, a n d ? very d e f i n i t i o n o f ? ,
the quotient topology.
is
By t h e
q i s continuous.
( 3 2 . 1 ) The q u o t i e n t s p a c e
Proof.
t ,q
(x,y) i s
compact H a u s d o r f f .
N o t e f i r s t t h a t an e q u i v a l e n t f o r m u l a t i o n o f t h e
u p p e r s e m i c o n t i n u i t y o f 5 i s t h a t f o r e v e r y open s e t W o f X , union o f a l l t h e 2 ' s c o n t a i n e d i n W i s a l s o open. c1
a s u b s e t Q o f X whole i f it i s a union o f Z ' s . whole i f Q
=
q
-1( q ( Q ) ) .
c1
the
Let u s c a l l
Q is then
T h u s , by t h e d e f i n i t i o n o f t h e
i n d u c t i v e t o p o l o g y , i f a n o p e n s e t W i n X i s w h o l e , t h e n q(W)
i s open i n
(2,y).
I t f o l l o w s t h a t t o show
(2,y)
is Hausdorff,
i t s u f f i c e s t o show t h a t d i s t i n c t Z ' s h a v e d i s t i n c t w h o l e (Y
neighborhoods i n X.
So c o n s i d e r Z
j o i n t open neighborhoods Wl,W2
"1
#
and c h o o s e d i s -
Za2,
o f Zal,Za r e s p e c t i v e 1y . 2
V1 b e t h e u n i o n o f a l l t h e Z ' s c o n t a i n e d i n
a
union of a l l those contained i n W2.
Since
2
Le t
W1, a n d V2 t h e i s uppersemicon-
t i n u o u s , V1 a n d V2 a r e o p e n , s o w e a r e t h r o u g h .
That
(t,7)
i s compact f o l l o w s from t h e c o n t i n u i t y o f q . QED
178
Chapter 6 mapping X
Returning to a continuous
__ >
resulting uppersemicontinuous decomposition
of X, form the quotient space ( X , y j . ical mapping ( 2 , J j
-
Y and the
2 = I S -1 ( y ) 1 y E s (X)}
Then s induces the canon-
2: Y such that the following diagram
commutes.
Then
(32.2)
2 is a homomorphism of
(X,r)onto Y.
The verification is simple.
We can now proceed with o u r examination of the MIl-subspaces of C(X).
Given a subset A of C ( X ) ,
a set Z in X will
be called set of constancy of A if (i) e v e r y f E A i s constant on 2 , and (ii) 2 is maximal with respect to this property.
If
A consists of a single element f, then t h e s e t s of constancy
are the sets { f - l ( X ) I X E I R l .
(32.3)
Given a subset A of C ( X ) ,
the set of sets of constancy
of A is an uppersemicontinuous decomposition of X.
(C(X) ,X)-Duality
Proof. -__
179
The family of continuous mappings { f l f E A 1 of X
into R defines a single continuous mapping X -> S IR
A
so
has the product topology). {
Then s(X)
IRA (as always,
is compact Hausdorff,
s - l(y) I y E s ( X ) 1 s an uppersemicontinuous decomposition of X.
But the sets s-l[y) are precisely the sets of constancy of A. QED
For any decomposition continuous), let F ( 2 )
=
x
of X (not necessarily uppersemi-
{fCC(X)lf
is constant on every
ZCX} .
Then, a s can b e verified b y straightforward computation, F(X) is an MIL-space.
(32.4) Given a subset A of C ( X ) ,
sets of constancy.
Then F(x)
let
b e the collection of its
is precisely the Ma-subspace I:
of C ( X ) generated by A.
____ Proof.
A s we have noted above, F(2)
and therefore contains F.
is an MIL-subspace
Conversely, consider fEF(x);
show f is in the norm closure of F, hence lies in F. [30.4),
we
By
it is enough t o show that f is in the simple closure
of F.
Lemma.
For every Z l , . . + , Z n € X ,a l l distinct, F contains
_ _ I
a Urysohn function for (Zl, lJnZ.) ( c f . the beginning of 5 3 0 ) . 2 1
Chapter 6
180
T h e r e e x i s t s R E F s u c h t h a t g(Z,) g(Z,)
tains A).
#
g ( Z z ) ( s i n c e I: c o n -
and g ( Z 2 ) a r e e a c h a s i n g l e r e a l n u m b e r , s o
by a d d i n g a n a p p r o p r i a t e m u l t i p l e o f I ( X ) t o g , i f n e c e s s a r y ,
w e c a n assume g ( Z 2 )
=
0.
Then,, h y m u l t i p l y i n g g b y an a p p r o -
p r i a t e s c a l a r , w e c a n assume g ( Z 1)
=
1.
The e l e m e n t (gVo)AI(X)
o f F i s now a Urysohn f u n c t i o n f o r ( Z l , Z 2 ) ; d e n o t e i t b y E ~ . S i m i l a r l y , F c o n t a i n s a Urysohn f u n c t i o n g i f o r e a c h p a i r (ZIJZi)
(i
=
3,...,n).
A;gi
i s t h e n a Urysohn f u n c t i o n f o r
( Z l , lJ;Zi). I t f o l l o w s f r o m t h e Lemma t h a t f o r e v e r y f i n i t e s u b s e t { x l , * - - , x n }o f X , t h e r e e x i s t s g E F s u c h t h a t g ( x i ) i
=
l,...,n.
=
f(xi)
€or
Thus f i s i n t h e s i m p l e c l o s u r e o f F .
OED i t s sets of constancy a r e s i n g l e
I f A i s s e p a r a t i n g on X , points.
(32.5)
(32.4) then reduces t o
( W e i e r s t r a s s - S t o n e Theorem - l a t t i c e v e r s i o n ) .
A c C(X) i s s e p a r a t i n g on X ,
If
t h e n t h e M I - s u b s p a c e o f C(X)
g e n e r a t e d b y A i s C(X) i t s e l f .
( 3 2 . 4 ) s t a t e s t h a t i f we s t a r t w i t h an ME-subspace F o f C(X), t a k e t h e s e t 2 o f i t s s e t s o f c o n s t a n c y , t h e n t a k e F ( , ) , we a r r i v e b a c k a t F .
Dually,
(32.6) Every uppersemicontinuous decomposition
2
of X i s the
(C(X) ,X) - D u a l i t y
181
s e t o f s e t s o f c o n s t a n c y o f t h e MIL-subspace F ( X ) w h i c h i t
determines.
Proof.
We n e e d o n l y show t h e m a x i m a l i t y o f e a c h Z E
Specifically, consider
Zo€X
and x e Z o ; w e h a v e t o show t h e r e
e x i s t s fEF(z) such t h a t f ( x ) # f(Zo). space
(x,J)
determined b y x .
Denote b y Y t h e q u o t i e n t
Y i s compact H a u s d o r f f ( 3 2 . 1 ) ,
h e n c e t h e r e e x i s t s hEC(Y) s u c h t h a t h ( q x ) # h ( q ( Z o ) ) . f
=
hoq.
5.
Set
Then f i s a c o n t i n u o u s f u n c t i o n on X w h i c h i s c o n -
s t a n t on e v e r y Z € z , h e n c e i s i n
F(2)
- and s a t i s f i e s
f(x) # f(Zo).
QED Summing u p ,
( 3 2 . 7 ) T h e r e i s a o n e - o n e c o r r e s p o n d e n c e b e t w e e n t h e MIL-subs p a c e s F o f C(X) a n d t h e u p p e r s e m i c o n t i n u o u s d e c o m p o s i t i o n s of X.
F <-->;!
i f and o n l y i f
2
2
i s t h e s e t of sets of con-
s t a n c y o f F i f a n d o n l y i f F c o n s i s t s o f t h e e l e m e n t s o f C(X) c o n s t a n t on e v e r y Z E
2.
033. The MIL-homomorphisms o f C ( X )
For t h e following d i s c u s s i o n , it i s convenient t o use t h e notation
( f , x ) i n place of f(x)
(cf. the beginning of the
Chapter 6
182 Chapter).
Each xEX
d e f i n e s a f u n c t i o n on C(X) by f +>
(f,x),
and s i n c e C(X) i s s e p a r a t i n g on X , d i s t i n c t x ' s d e f i n e d i s t i n c t functions.
We c a n t h e r e f o r e i d e n t i f y e a c h x w i t h t h e f u n c t i o n
i t d e f i n e s , and h e n c e f o r t h w e w i l l do t h i s : x w i l l be c o n s i d e r e d a f u n c t i o n on C(X).
M o r e o v e r , by t h e v e r y d e f i n i t i o n
o f a d d i t i o n a n d s c a l a r m u l t i p l i c a t i o n i n C(X), x i s a c t u a l l y a linear functional.
Even m o r e , f r o m t h e d i s c u s s i o n o f t h e
b e g i n n i n g o f t h e C h a p t e r on t h e l a t t i c e o p e r a t i o n s i n C ( X ) , e v e r y xEX i s , i n f a c t , a n MI-homomorphism o f C(X) ( i n t o IR). And t h e y a r e t h e o n l y o n e s :
( 3 3 . 1 ) F o r a f u n c t i o n $ o n C(X), t h e f o l l o w i n g a r e e q u i v a l e n t
'1
$ i s a n MI-homomorphism o f C(X) i n t o R ;
2O
$EX.
P roof. -
I t r e m a i n s t o show t h a t ' 1
implies 2
0
.
We n o t e
f i r s t t h a t two MI-homomorphisms $1 , $ 2 o f C(X) i n t o 1R a r e i d e n t i c a l i f and o n l y i f (+1)-1(0) h o l d s f o r 4.
$-'lo) i s
=
Now assume 1'
($2)-1(0).
t h e n a R i e s z i d e a l H o f C(X), a n d
$ b e i n g a s i n g l e l i n e a r f u n c t i o n a l - i s a maximal o n e .
Z(H)
consists of a single point x (31.13).
Hence
But t h e n x -1( 0 )
=
H,
s o w e have $ = x . QED
C o n s i d e r a c o n t i n u o u s mapping X <-
S
Y.
For each fEC(X),
f o s i s a c o n t i n u o u s f u n c t i o n on Y, s o an e l e m e n t o f C(Y). t h u s h a v e a mapping f
I--->
f o s o f C(X) i n t o C ( Y ) .
We
We w i l l c a l l
(C(X) ,X) - D u a l i t y
183
t h i s mapping t h e t r a n s p o s e o f s , a n d d e n o t e i t b y s t . f o r a l l f E C ( X ) and yEY, ( s t f , y )
=
(f,sy).
I t i s t r i v i a l t o v e r i f y t h a t C(X)
S
t
> C(Y)
Thus,
Mn-
i s an
homomorphism.
We show t h a t , c o n v e r s e l y , f o r e a c h Mn-homomorph-
T i s m C(X) ->
C(Y), t h e r e e x i s t s a unique c o n t i n u o u s
X
( 3 3 . 2 ) Theorem.
X,Y,
=
(Stone).
mapping
T:
Given two c o m p a c t H a u s d o r f f s p a c e s
t h e r e i s a one-one correspondence between t h e c o n t i n u o u s
m a p p i n g s s o f Y i n t o X a n d t h e Ma-homomorphisms T o f C(X) i n t o C(Y).
T <->
s i f a n d o n l y i f (f,sy)
= (Tf,y)
for a l l ffC(X)
a n d yEY.
Proof.
-~
I t r e m a i n s t o show t h e e x i s t e n c e o f s , g i v e n T .
F o r e a c h yEY, yoT i s an M ~ - h o m o m o r p h i s m o f C(X) i n t o R , h e n c e , by (32.1),
an element o f X.
o f Y i n t o X; d e n o t e i t b y s .
f E C ( X ) a n d yEY.
We t h u s h a v e a m a p p i n g y k--> So ( f , s y )
=
(Tf,y) for all
i t i s immediate from t h i s i d e n t i t y t h a t f o r
e v e r y n e t { y a } i n Y a n d yoEY, i f ( g , y o ) = l i m a ( g , y a ) gEC(Y), t h e n ( f , s y o ) = l i m a ( f , s y a ) continuous.
That T
=
f o r a l l fEC(X).
i s a n MIL-subspace o f C(Y) ( 1 8 . 2 ) , a n d T - ’ ( O )
R i e s z i d e a l o f C(X).
for all Thus s i s
s t f o l l o w s f r o m t h e same i d e n t i t y .
As we know, f o r a n MIL-homomorphism C(X)
s(Y)
yoT
T
--A
C(Y), T(C(X))
i s a norm c l o s e d
And f o r a c o n t i n u o u s m a p p i n g X <---1 i s a c l o s e d s u b s e t o f X and { s (x) ( x € s ( Y ) } i s an
S
Y,
184
Chapter 6
uppersemicontinuous decomposition o f Y.
Y b e a c o n t i n u o u s mapping a n d T
( 3 3 . 3 ) L e t X
T-'(O) -1 {s
=
(s(Y))'and
(x) ( x E s ( Y ) l
s(Y)
=
=
st .
Then
Z(T-'(O));
i s t h e s e t of s e t s of constancy of
T ( C ( X ) ) , a n d T(C(X)) i s t h e s e t o f e l e m e n t s o f C(Y) c o n s t a n t -1 on e v e r y s (x),
Otherwise s t a t e d , T - l ( O ) and s(Y) c o r r e s p o n d t o e a c h o t h e r u n d e r ( 3 1 . 6 ) , and T(C(X)) and { s each o t h e r under (32.7).
-1
(x) I x € s ( Y ) }
correspond t o
The p r o o f o f ( 3 3 . 3 ) i s s t r a i g h t -
forward.
(33.4) Corollary. (i)
s i s a s u r j e c t i o n i f a n d o n l y i f T i s an i n j e c t i o n .
( i i ) s i s an i n j e c t i o n i f and o n l y i f T i s a s u r j e c t i o n .
We a p p l y t h e a b o v e t o t h e m a t e r i a l o f 5 § 3 1 , 3 2 .
Let F b e
an MIL-subspace o f C ( X ) a n d 2 i t s s e t o f s e t s o f c o n s t a n c y , and d e n o t e t h e q u o t i e n t s p a c e
( Z J ) by
Y.
Let Y
q u o t i e n t map.
Then an a d a p t a t i o n o f t h e a r g u m e n t f o r ( 3 2 . 6 ) , t p l u s ( 3 3 . 4 ) a b o v e , g i v e s u s t h a t C(Y) %> C ( X ) i s an MILi s o m o r p h i s m o n t o F.
do t h i s :
Thus F c a n b e i d e n t i f i e d w i t h C(Y).
We
(C(X) , X ) - D u a l i t y Let F b e an M I - s u b s p a c e o f C ( X ) ,
(33.5)
constancy o f F, Y q u o t i e n t map.
=
(2,y) t h e
185
2
t h e s e t of s e t s of
q u o t i e n t s p a c e , and Y
Then u n d e r t h e d e f i n i t i o n ( f , q y )
=
(f,y) for a l l
f E F and qyEY, w e h a v e F = C ( Y ) .
Now l e t Z b e a c l o s e d s u b s e t o f X a n d H Let X <-
S
Z b e t h e i n j e c t i o n map.
MI-homomorphism o n t o . ~
Then C ( X )
~
>c(z)
=
Z(H)).
i s an
can be i d e n t i f i e d w i t h C ( Z ) .
(33.6)
For e v e r y c l o s e d s u b s e t Z of X ,
C(X)/ZL
( f , z ) for a l l qf€C(X)/Z'
Remark. The mapping C ( X ) s t r i c t i o n map f k--> f l Z .
T i e t z e E x t e n s i o n Theorem.
L
onto C(Z).
Thus
We do t h i s :
C(X)/H
=
t
Factoring it:
t " we o b t a i n a n M a - i s o m o r p h i s m (s ) o f C ( X ) / H
definition (qf,z)
ZL ( s o Z
=
__ > C ( Z )
=
C(Z)
under the
and z E Z .
is c l e a r l y t h e r e -
(33.6) then a l s o follows from t h e
(C(X) ,X) -DUALITY
X , Y w i l l d e n o t e compact H a u s d o r f f s p a c e s .
C(X) i s t h e s e t
o f r e a l c o n t i n u o u s f u n c t i o n s on X , a n d n ( X ) w i l l d e n o t e t h e f u n c t i o n o f c o n s t a n t v a l u e 1 on X .
Under t h e u s u a l p o i n t w i s e
d e f i n i t i o n s o f a d d i t i o n , s c a l a r m u l t i p l i c a t i o n , a n d o r d e r , C(X) i s a R i e s z s p a c e w i t h n(X) f o r a n o r d e r u n i t .
The MI-norm
d e t e r m i n e d by n(X) i s p r e c i s e l y t h e s t a n d a r d supremum n o r m , and u n d e r t h i s norm, C(X) i s c o m p l e t e . Thus C(X) i s a n M I - s p a c e .
I n more d e t a i l , f o r f , g E C ( X ) ,
f v g a n d fAg a r e g i v e n b y ( f V g ) ( x ) = m a x ( f ( x ) , g ( x ) ) a n d ( f A g ) ( x ) = m i n ( f ( x ) , g ( x ) ) f o r a l l xEX. f+(x)
=
( f ( x ) ) + , and f - ( x )
in particular, =
Ifl(x)
=
/f(x)I,
( f ( x ) ) - f o r a l l x€X.
Although X is n o t a v e c t o r s p a c e , i t i s u s e f u l (and s u g g e s t i v e ) t o t h i n k o f C(X) a s " d u a l " t o X.
This has been
done by v a r i o u s w r i t e r s , e s p e c i a l l y i n t h e c o n t e x t o f C a t e g o r y Theory ( c f .
[49],
523.2 or [4T).
We w i l l f o r m u l a t e t h e r e l a -
t i o n s b e t w e e n C(X) a n d X f r o m t h i s v i e w p o i n t .
Since X has a
n a t u r a l i m b e d d i n g i n t o t h e d u a l C'(X) o f C(X), C'(X) w i l l t h u s play the r o l e of "bidual"
t o X.
I n p a r t i c u l a r , we w i l l f e e l
free t o denote f ( x ) by ( f , x ) . X w i l l b e c a l l e d t h e K a k u t a n i - S t o n e s p a c e o f C(X).
165
Chapter 6
166
5 3 0 . The t o p o l o g y o f s i m p l e C o n v e r g e n c e o n C ( X )
X i s o f c o u r s e s e p a r a t i n g o n C ( X ) by v e r y d e f i n i t i o n : f # g means t h e r e e x i s t s xEX
such t h a t f ( x )
# g(x).
Dually,
C(X) i s s e p a r a t i n g on X , a n d i n d e e d , i n a v e r y s t r o n g s e n s e : (Urysohn) f o r e v e r y p a i r o f d i s j o i n t , c l o s e d , non-empty s u h s e t s Z1,Z2 o f X , ( i i ) f(x)
=
Remark. -___
t h e r e e x i s t s fEC(X) s a t i s f y i n g ( i ) 0 < f (n(X),
1 f o r a l l xEZ1,
(iii) f(x)
=
0 f o r a l l xEZZ.
I l e n c e f o r t h , we w i l l c a l l an f E C ( X ) w i t h t h e a b o v e
properties a Urysohn f u n c t i o n f o r t h e ( o r d e r e d ) p a i r (Z,,Z,). A s o n e would e x p e c t , we d e f i n e o ( C ( X ) , X )
a s t h e topology
on C ( x ) o f p o i n t w i s e c o n v e r g e n c e on X : a n e t { f } i n CL
C(X) c o n -
v e r g e s t o fEC(X) i n a ( C [ X ) , X ) i f f ( x ) = l i m f ( x ) f o r a l l xEX. CY.N
We w i l l c a l l i t by i t s common name: t h e t o p o l o g y o f s i m p l e convergence.
And we c a n d e f i n e o ( X , C ( X ) ) a s t h e t o p o l o g y o n X
o f p o i n t w i s e c o n v e r g e n c e o n C(X): a n e t { x } i n X c o n v e r g c s t o c1
xEX i n ~ ( y C, ( X ) )
if f ( x )
=
limolf(x,)
for a l l fEC(X).
Since
x
completely regular,o(X,C(X)) i s simply t h e o r i g i n a l topology on X .
We e x a m i n e o(C(X) , X ) . Norm c o n v e r g e n c e ( i n C ( X ) )
and s i m p l e c onvergence.
i m p l i e s b o t h o r d e r convergence
T h e r e i s no o t h e r i m p l i c a t i o n b e t w e e n
the t h r e e convergences: Let X = a m , t h e one-point (Alexandroff) c o m p a c t i f i c a t i o n o f N , and f o r e a c h n , l e t en b e t h e e l e m e n t o f C(X)
h a v i n g v a l u e 1 on n E N a n d v a l u e 0 e l s e w h e r e .
t h e s e q u e n c e {ne,}
Then
c o n v e r g e s t o 0 s i m p l y b u t n e i t h e r normwise n n o r i n t h e o r d e r , a n d t h e s e q u e n c e {c e . ) ( n = 1 , 2 , . * . ) o r d e r 1 1 c o n v e r g e s t o l(X) b u t d o e s n o t c o n v e r g e e i t h e r s i m p l y o r no r m w i s e .
is
16 7
(C(X) ,X)-Duality
For a monotonic n e t , however, w e have one a d d i t i o n a l implication:
( D i n i t h e o r e m ) F o r a m o n o t o n i c n e t { f } i n C(X) cr f E C (X), the following a r e equivalent : (30.1)
1'
{fci. 1 c o n v e r g e s t o f n o r m w i s e ;
2'
{fa} c o n v e r g e s t o f s i m p l y ;
We n e e d o n l y show 2'
Proof. -__
implies lo.
and
For c o n c r e t n e s s ,
assume { f } i s a d e s c e n d i n g n e t w i t h i n f f ( x ) = 0 f o r a l l xEX. a a Consider E > 0 . F o r e a c h xEX, c h o o s e a ( x ) s u c h t h a t (x) <
E.
Since f
cr (XI
i s c o n t i n u o u s , t h e r e e x i s t s an o p e n
n e i g h b o r h o o d W(x) o f x s u c h t h a t f
(XI
(Y) <
E
f o r a l l yiW(x).
Now X i s c o m p a c t , s o t h e r e e x i s t x l , - - ,*x n s u c h t h a t { W ( x l ) , . - . , W ( x n ) } c o v e r s X. Then a > a ( € ) implies fa(y) <
Choose a(€) 2 w ( x , ) , . * . , a ( x n ) . E
f o r a l l yEX.
QED
I n common w i t h norm c o n v e r g e n c e a n d o r d e r c o n v e r g e n c e , simple convergence h a s t h e f o l l o w i n g e a s i l y v e r i f i e d , p r o p e r t y
(30.2)
I f { f } a n d {p } c o n v e r g e s i m p l y t o f a n d g r e s p e c t i v e l y , c1
t h e n {fcrVg,),
a
{fclAgcl},
Ifc1 + g a l , a n d {Afcl}
o f IR) c o n v e r g e s i m p l y t o f v g , f A g , f
+
(A
any element
g , a n d Af r e s p e c t i v e l y .
Chapter 6
168
Otherwise s t a t e d , t h e l a t t i c e o p e r a t i o n s and v e c t o r s p a c e o p e r a t i o n s a r e continuous under simple convergence.
F o r a R i e s z s u b s p a c e F o f C(X) t o b e s i m p l y
(30.3) Corollary.
c l o s e d , i t s u f f i c e s t h a t F+ be s i m p l y c l o s e d .
i s simply c l o s e d .
Note a l s o t h a t C ( X ) +
The f o l l o w i n g i s t h e c r u c i a l p a r t o f M . H .
Stone's proof
of t h e Stone-Weierstrass theorem.
( 3 0 . 4 ) _____ Theorem.
(Stone).
F o r a s u b l a t t i c e A o f C(X), t h e
following are equivalent:
'1
A i s simply closed;
2'
A i s norm c l o s e d .
We n e e d o n l y show t h a t 2'
P roof.
i m p l i e s 1'.
Assume A i s
norm c l o s e d , a n d c o n s i d e r f i n t h e s i m p l e c l o s u r e o f A . E
> 0.
We show t h e r e e x i s t s gEA
such t h a t
i A
( i ) F o r e a c h xoEX, t h e r e e x i s t s g
\I g
- fll
Fix
5 3 ~ .
such t h a t
xO gXO
(XI
gx (x,) 0
2
f ( x ) - 3~
< f(xo)
+
f o r a l l xEX,
E *
In e f f e c t , s i n c e f i s i n t h e simple c l o s u r e of A, then f o r
169
(C(X) , X ) - D u a l i t y e a c h xEX, t h e r e e x i s t s g
€ A such t h a t J g
( g x .(XI
-
0
f(x)
<
-
E ;
(xo)
- f(xo)
xO
xO
< r-, -
and from t h e s e c o n d o f t h e s e i n e q u a i t i e s ,
t h e r e e x i s t s a n o p e n n e i g h b o r h o o d W x) o f x s u c h t h a t ( y ) 2 f ( y ) - 3~ f o r a l l y 6 W ( x ) . X b e i n g compact, t h e r e xox e x i s t { x l , . . - , x n } s u c h t h a t IW(x,), * . , W ( x n ) } c o v e r s X. Set g
Since x
was a r b i t r a r y i n ( i ) , w e now d r o p t h e s u b s c r i p t .
0
We t h u s h a v e { g x l x E X } c A s u c h t h a t f o r e v e r y x , gx 3 ~ l l ( X ) and gx(x) < f(x) +
c o v e r s X.
Set g
=
m
tliTlgx
i 3 ~ l l ( X ;) s o we a r e t h r o u g h .
+
3~ f o r a l l y E V ( x ) .
, x m } s u c h t h a t {V(x,)
.
f -
For e a c h x , choose a n e i g h b o r -
E.
hood V(x) o f x s u c h t h a t g x ( y ) < f(y) Again t h e r e e x i s t s { x , , . .
2
, . . * ,V(xm) 1
< f + Then g E A a n d f - 3 ~ n ( X )5 g -
QED
(30.5)
Corollary.
(i)
Norm c l o s e d R i e s z i d e a l s o f C ( X )
a r e simply closed.
( i i ) M I - s u b s p a c e s o f C(X) a r e s i m p l y c l o s e d .
531. The d u a l i t y b e t w e e n t h e norm c l o s e d R i e s z
i d e a l s of C(X)
and t h e c l o s e d ( o r open) s u b s e t s o f X
F o r e v e r y s u b s e t Q o f X , we s e t QL
=
{fEC(X)] f(x)
=
0 for all
a n d f o r e v e r y s u b s e t A o f C ( X ) , we s e t
~€41;
Chapter 6
170 Z(A) Z(A)
{x€X I f ( x )
=
=
0 f o r a l l fEA). (In l i n e with our d u a l i t y
i s c a l l e d t h e z e r o - s e t o f A.
a p p r o a c h , w e would p r e f e r t o d e n o t e Z ( A )
b y A&, b u t Z ( R ) i s
the established notation.) We h a v e i m m e d i a t e l y :
( 3 1 . 1 ) For e v e r y A c C ( X )
,
Z(A) i s a c l o s e d s u b s e t o f X .
In the opposite d i r e c t i o n , given
Q
c X,
it i s e a s i l y
v e r i f i e d t h a t QL i s a R i e s z i d e a l a n d s i m p l y c l o s e d , s o a f o r t i o r i norm c l o s e d .
( 3 1 . 2 ) For e v e r y
Thus:
Q c X,
Q'
Now i t i s c l e a r t h a t
i s a norm c l o s e d R i e s z i d e a l .
(0)' =
QL
M o r e o v e r , by t h e Urysohn t h e o r e m , t h i s property.
( 3 1 . 3 ) For e v e r y
Dually ,
0is
We t h u s h a v e :
Q c X, Z(QL)
=
(q
0.
the closure of
Q i n X).
the largest set with
171
(C(X) , X ) - D u a l i t y ( 3 1 . 4 ) For e v e r y s u b s e t A o f C(X),
i s t h e norm c l o s e d
(Z(A))'
Riesz i d e a l H g e n e r a t e d by A.
-__ Proof.
H c (Z(A))'
by ( 3 1 . 2 ) .
clusion consider f € ( Z ( A ) ) I , B
=
{ g E H 10 c: g 5 f } .
i s f i l t e r i n g upward. we show f ( x )
=
f(x) > 0.
a n d we c a n assume f > 0 .
H e n c e , b y t h e D i n i Theorem ( 3 0 . 1 1 , i f
I f f(x)
Then xeZ(A).
=
it w i l l follow t h a t
0 , w e are through.
c a n assume h ( x ) > f ( x ) . =
Suppose
I t f o l l o w s t h e r e e x i s t s hEII, w i t h
h(x) > 0 ( t h e r e e x i s t s kEAwith k(x) # 0 ; s e t h
t h a t g(x)
Set
B i s n o t empty ( i t c o n t a i n s 0 ) and i t
s u p g E B g ( x ) , f o r e v e r y xEX,
Consider x .
fEfl.
For t h e o p p o s i t e con-
Then g
=
=
l k l ) , and w e
hnf i s an element of B such
f(x).
QED In particular,
(31.5) For e v e r y Riesz i d e a l I o f C ( X ) ,
(Z(1))'
i s t h e norm
closure of I.
Summing u p ,
(31.6) There i s a one-one correspondence between t h e f a m i l y o f c l o s e d s u b s e t s 2 o f X a n d t h a t o f norm c l o s e d R i e s z i d e a l s II of C(X):
H <->
Z i f and o n l y i f Z = Z(H)
i f and o n l y i f H
=
ZL.
Chapter 6
172 Given H <-->
Z,
s e t W = X\Z:
Thus t h e r e i s a o n e - o n e
c o r r e s p o n d e n c e b e t w e e n t h e norm c l o s e d R i e s z i d e a l s o f C ( X ) and t h e open s e t s o f X .
T h i s corresp o n d en ce can be g iv en w i t h -
out reference t o the closed s e t s .
First a definition.
A func-
t i o n f on a l o c a l l y compact s p a c e T i s s a i d t o v a n i s h a t i n f i n i t y i f f o r every -~ compact.
E
> 0, the set {tET
1
If(t)
1
> E} i s
The f o l l o w i n g i s e a s i l y v e r i f i e d .
( 3 1 . 7 ) There i s a o n e - o n e c o r r e s p o n den ce between t h e f a m i l y o f
open s u b s e t s W of X and t h a t o f t h e norm c l o s e d R i e s z i d e a l s
H o f C(X): H <-->
fEH}
W i f and o n l y i f W
=
{ x G X \ f ( x ) # 0 f o r some
i f and o n l y i f H i s t h e s e t o f c o n t i n u o u s f u n c t i o n s on
W which v a n i s h a t i n f i n i t y .
H e n c e f o r t h Z w i l l a l w a y s d e n o t e a c l o s e d s e t of
Notation.
X , a n d W an open s e t .
F o r a s u b s e t A o f C ( X ) , W(A) w i l l b e
t h e s e t X\Z(A). The s e t o f c o n t i n u o u s f u n c t i o n s v a n i s h i n g a t i n f i n i t y on a l o c a l l y compact s p a c e T w i l l b e d e n o t e d by C w ( T ) .
Note
t h a t i t i s an M-space u n d e r t h e supremum norm, a n d i s an MEs p a c e i f and o n l y i f T i s c o m p a c t . U s i n g t h e above n o t a t i o n , we c a n s t a t e p a r t o f ( 3 1 . 7 ) a s
follows: I f X
= W IJ Z ,
W a n d Z c o m p l e m e n t a r y , t h e n ZL = Cw(W) .
A n o t h e r o b s e r v a t i o n on C w ( W ) . elements of C(X) of W.
Let Ho c o n s i s t o f t h e
e a c h h a v i n g f o r i t s s u p p o r t a comp.act s u b s e t
Then Ho i s a R i e s z i d e a l , i s c o n t a i n e d i n C w ( W ) ,
norm d e n s e i n t h e l a t t e r .
We n e e d o n l y v e r i f y t h e l a s t
and i s
(C(X) ,X) - D u a l i t y statement.
173
Tn e f f e c t , s i n c e W i s c o m p l e t e l y r e g u l a r , Ho i s s o , by ( 3 0 . 4 )
s i m p l y d e n s e i n C'(W),
( o r t h e Dini Theorem), i s
norm d e n s e i n i t .
We c h a r a c t e r i z e v a r i o u s t y p e s o f R i e s z i d e a l s o f C(X) by p r o p e r t i e s of X . by i n t Q
.
We w i l l d e n o t e t h e i n t e r i o r o f a s e t Q i n X
Recall that a r e g u l a r c l o s e d s e t i s one which i s
t h e c l o s u r e o f an o p e n s e t , a n d a r e g u l a r o p e n s e t i s o n e which i s t h e i n t e r i o r o f a c l o s e d s e t .
We h a v e s e e n ( 3 1 . 2 ) t h a t f o r Q c X , QL i s a norm c l o s e d Riesz i d e a l .
F o r a n o p e n s e t W i n X , WL
(31.8)
___ Proof. WL
F o r an o p e n s e t , w e c a n s a y m o r e :
Let 2
X\W.
=
i s a b a n d o f C(X)
T h e n , as i s e a s i l y v e r i f i e d ,
(ZL)d, hence i s a band
=
QE D
For a Riesz i d e a l H o f C(X),
(31.9) C o r o l l a r y 1.
o r d e r c l o s u r e tI
P roof. simplicity, and H can
=
.'2
=
( i n t Z(H))'.
We c a n a s s u m e H i s norm c l o s e d . denote Z(H)
Also f o r
s i m p l y by Z a n d W(H) b y W.
So W
=
X\Z
S i n c e C(X) i s A r c h i m e d e a n , w h a t we w a n t t o p r o v e
b e s t a t e d : Hdd
=
( i n t Z)'.
A s we remarked i n ( 3 1 . 8 ) ,
2
Chapter 6
174 Hd
=
W"
(x\W)*
But t h e n Hd =
,
(w)" ,
s o , by t h e same r e m a r k , H d d
=
j i n t z)'.
=
QED
For a R i e s z i d e a l f1 o f C(X), t h e
(31.10) Corollary 2.
following are equivalent: lo
t h e o r d e r c l o s u r e o f H i s C(X);
'2
Z(H) h a s empty i n t e r i o r ( s o , b e i n g c l o s e d , i s nowhere
rare,
dense -
i n t h e Bourbaki terminology);
WIN) i s d e n s e i n X .
3'
The f o l l o w i n g a r e a l s o c o r o l l a r i e s , b u t we l i s t them a s theorems.
( 3 1 . 1 1 ) ____ Theorem.
F o r a norm c l o s e d R i e s z i d e a l H o f C ( X ) ,
the
following are equivalent:
1'
H i s a band;
'2
Z(H) i s a r e g u l a r c l o s e d s e t ;
'3
W ( H ) i s a r e g u l a r open s e t .
P roof. Z(H)
=
i f Z(H) H
=
I f H i s a band, t h e n , by (31.9) and (31.3),
m, so
Z(H) i s a r e g u l a r c l o s e d s e t .
Conversely,
i s a r e g u l a r c l o s e d s e t , t h e n , by ( 3 1 . 4 ) and ( 3 1 . 9 ) ,
(Z(H))'
=
order closure H. QED
(C(X) ,X) -Duality
175
I f a s e t i n X i s b o t h open and c l o s e d , w e w i l l c a l l i t clopen.
F o r a norm c l o s e d R i e s z i d e a l lI o f C ( X ) ,
( 3 1 . 1 2 ) Theorem.
the
following are equivalent: 1'
H i s a p r o j e c t i o n band;
2'
Z(II) is clopen;
3'
W(t1)
-__ Proof..
i s clopen.
S u p p o s e H 3 Hd
=
C(X),
but Z(H)
some xEZ(H) i s i n t h e c l o s u r e o f W ( I I ) . t1 v a n i s h e s on
Conversely, i f X =
T t
v a n i s h e s on x , c o n t r a d i c t i n g ( l l ( X ) ) ( x ) Z1lJ
=
So
Then e v e r y e l e m e n t o f
x a n d e v e r y e l e m e n t o f Hd v a n i s h e s on x .
follows a l l of C(X)
C(X)
i s n o t open.
=
Z2, w i t h Z1,Z2 d i s j o i n t , t h e n c l e a r l y ,
( Z , P 3 (Z$. QED
Finally
,
( 3 1 . 1 3 ) Theorem.
For a R i e s z i d e a l H o f C ( X ) ,
are equivalent: 1' H i s a maximal R i e s z i d e a l ;
2'
Z(f1)
1.
consists of a single point.
the following
176
Chapter 6 -__ Proof.
Assume H i s m a x i m a l , a n d c h o o s e xEZ(f1).
a p r o p e r Riesz i d c a l o f C ( X )
w i t h 11.
{x}'
is
and c o n t a i n s [ I , hencc c o i n c i d e s
I t f o l l o w s from ( 3 1 . 3 ) t h a t Z ( l I )
=
{x}.
T h a t '2
i m p l i e s lo i s c l e a r . Qlill
Consider a c l o s e d s e t Z o f X. Cw(W), W
= X\Z.:
As we h a v e s e e n , ZL =
Tn a d d i t i o n , C(X)/ZL = C(Z) ( c f .
(33.6))
5 3 2 . The d u a l i t y b e t w e e n t h e M l l - s u b s p a c e s o f C ( X )
and t h e u p p e r s e m i c o n t i n u o u s d e c o m p o s i t i o n o f X
P a r a l l e l i n g 531, we c o l l e c t h e r e t h e o r e m s r e l a t i n g t h c M I - s u b s p a c e s o f C(X) t o t h e t o p o l o g y o f X. immediately a r i s e s .
A difficulty
E v e r y MIL-subspace F c o n t a i n s l l ( X ) ,
t h e a n n i h i l a t o r o f F i s always cmpty.
hence
A substitute for this
a n n i h i l a t o r i s t h e dccomposition of X i n t o t h e sets of cons t a n c y o f F.
(The a n n i h i l a t o r __ does e x i s t
-
in C'(X)!
And
i s closely r e l a t e d t o these s e t s of constancy.) By a d e c o m p o s i t i o n o f X , w e w i l l mean a f a m i l y x
=
{Za}
o f m u t u a l l y d i s j o i n t , c l o s e d , n o n - e m p t y s u b s e t s o f X whose
union i s X.
The d e c o m p o s i t i o n s o c c u r r i n g i n o u r s u b j e c t a r e
of a special kind.
A decompositionx
=
{Za} o f X i s c a l l e d
uppersemicontinuous i f f o r e v e r y c l o s e d s e t Z i n X , t h e union of a l l the Za's intersecting 2 i s also closed.
I t i s e a s i l y v e r i f i e d t h a t i f X s > Y i s a continuous mapping ( r e m e m b e r , X , Y a r e a l w a y s compact H a u s d o r f f ) , t h e n t h e
(C(X) , X ) - D u a l i t y decomposition
2
=
{s
-1
(y) / y € s ( X ) }
177
of X i s uppersemicontinuous.
Conversely, every uppersemicontinuous decomposition
X i s d e t e r m i n e d b y a c o n t i n u o u s mapping o f X :
2
=
{Za} o f
In e f f e c t , l e t
X A> L b e t h e mapping w h i c h a s s i g n s t o e a c h xEX t h e Z t a i n i n g i t , and d e n o t e h y J f i n e d b y t h i s mapping q .
(Z,J)
cona t h e i n d u c t i v e t o p o l o g y on 5 d e -
The r e s u l t i n g t o p o l o g i c a l s p a c e
is called the q u o t i e n t s p a c e o f X d e t e r m i n e d by
c a l l e d t h e q u o t i e n t map, a n d ? very d e f i n i t i o n o f ? ,
the quotient topology.
is
By t h e
q i s continuous.
( 3 2 . 1 ) The q u o t i e n t s p a c e
Proof.
t ,q
(x,y) i s
compact H a u s d o r f f .
N o t e f i r s t t h a t an e q u i v a l e n t f o r m u l a t i o n o f t h e
u p p e r s e m i c o n t i n u i t y o f 5 i s t h a t f o r e v e r y open s e t W o f X , union o f a l l t h e 2 ' s c o n t a i n e d i n W i s a l s o open. c1
a s u b s e t Q o f X whole i f it i s a union o f Z ' s . whole i f Q
=
q
-1( q ( Q ) ) .
c1
the
Let u s c a l l
Q is then
T h u s , by t h e d e f i n i t i o n o f t h e
i n d u c t i v e t o p o l o g y , i f a n o p e n s e t W i n X i s w h o l e , t h e n q(W)
i s open i n
(2,y).
I t f o l l o w s t h a t t o show
(2,y)
is Hausdorff,
i t s u f f i c e s t o show t h a t d i s t i n c t Z ' s h a v e d i s t i n c t w h o l e (Y
neighborhoods i n X.
So c o n s i d e r Z
j o i n t open neighborhoods Wl,W2
"1
#
and c h o o s e d i s -
Za2,
o f Zal,Za r e s p e c t i v e 1y . 2
V1 b e t h e u n i o n o f a l l t h e Z ' s c o n t a i n e d i n
a
union of a l l those contained i n W2.
Since
2
Le t
W1, a n d V2 t h e i s uppersemicon-
t i n u o u s , V1 a n d V2 a r e o p e n , s o w e a r e t h r o u g h .
That
(t,7)
i s compact f o l l o w s from t h e c o n t i n u i t y o f q . QED
178
Chapter 6 mapping X
Returning to a continuous
__ >
resulting uppersemicontinuous decomposition
of X, form the quotient space ( X , y j . ical mapping ( 2 , J j
-
Y and the
2 = I S -1 ( y ) 1 y E s (X)}
Then s induces the canon-
2: Y such that the following diagram
commutes.
Then
(32.2)
2 is a homomorphism of
(X,r)onto Y.
The verification is simple.
We can now proceed with o u r examination of the MIl-subspaces of C(X).
Given a subset A of C ( X ) ,
a set Z in X will
be called set of constancy of A if (i) e v e r y f E A i s constant on 2 , and (ii) 2 is maximal with respect to this property.
If
A consists of a single element f, then t h e s e t s of constancy
are the sets { f - l ( X ) I X E I R l .
(32.3)
Given a subset A of C ( X ) ,
the set of sets of constancy
of A is an uppersemicontinuous decomposition of X.
(C(X) ,X)-Duality
Proof. -__
179
The family of continuous mappings { f l f E A 1 of X
into R defines a single continuous mapping X -> S IR
A
so
has the product topology). {
Then s(X)
IRA (as always,
is compact Hausdorff,
s - l(y) I y E s ( X ) 1 s an uppersemicontinuous decomposition of X.
But the sets s-l[y) are precisely the sets of constancy of A. QED
For any decomposition continuous), let F ( 2 )
=
x
of X (not necessarily uppersemi-
{fCC(X)lf
is constant on every
ZCX} .
Then, a s can b e verified b y straightforward computation, F(X) is an MIL-space.
(32.4) Given a subset A of C ( X ) ,
sets of constancy.
Then F(x)
let
b e the collection of its
is precisely the Ma-subspace I:
of C ( X ) generated by A.
____ Proof.
A s we have noted above, F(2)
and therefore contains F.
is an MIL-subspace
Conversely, consider fEF(x);
show f is in the norm closure of F, hence lies in F. [30.4),
we
By
it is enough t o show that f is in the simple closure
of F.
Lemma.
For every Z l , . . + , Z n € X ,a l l distinct, F contains
_ _ I
a Urysohn function for (Zl, lJnZ.) ( c f . the beginning of 5 3 0 ) . 2 1
Chapter 6
180
T h e r e e x i s t s R E F s u c h t h a t g(Z,) g(Z,)
tains A).
#
g ( Z z ) ( s i n c e I: c o n -
and g ( Z 2 ) a r e e a c h a s i n g l e r e a l n u m b e r , s o
by a d d i n g a n a p p r o p r i a t e m u l t i p l e o f I ( X ) t o g , i f n e c e s s a r y ,
w e c a n assume g ( Z 2 )
=
0.
Then,, h y m u l t i p l y i n g g b y an a p p r o -
p r i a t e s c a l a r , w e c a n assume g ( Z 1)
=
1.
The e l e m e n t (gVo)AI(X)
o f F i s now a Urysohn f u n c t i o n f o r ( Z l , Z 2 ) ; d e n o t e i t b y E ~ . S i m i l a r l y , F c o n t a i n s a Urysohn f u n c t i o n g i f o r e a c h p a i r (ZIJZi)
(i
=
3,...,n).
A;gi
i s t h e n a Urysohn f u n c t i o n f o r
( Z l , lJ;Zi). I t f o l l o w s f r o m t h e Lemma t h a t f o r e v e r y f i n i t e s u b s e t { x l , * - - , x n }o f X , t h e r e e x i s t s g E F s u c h t h a t g ( x i ) i
=
l,...,n.
=
f(xi)
€or
Thus f i s i n t h e s i m p l e c l o s u r e o f F .
OED i t s sets of constancy a r e s i n g l e
I f A i s s e p a r a t i n g on X , points.
(32.5)
(32.4) then reduces t o
( W e i e r s t r a s s - S t o n e Theorem - l a t t i c e v e r s i o n ) .
A c C(X) i s s e p a r a t i n g on X ,
If
t h e n t h e M I - s u b s p a c e o f C(X)
g e n e r a t e d b y A i s C(X) i t s e l f .
( 3 2 . 4 ) s t a t e s t h a t i f we s t a r t w i t h an ME-subspace F o f C(X), t a k e t h e s e t 2 o f i t s s e t s o f c o n s t a n c y , t h e n t a k e F ( , ) , we a r r i v e b a c k a t F .
Dually,
(32.6) Every uppersemicontinuous decomposition
2
of X i s the
(C(X) ,X) - D u a l i t y
181
s e t o f s e t s o f c o n s t a n c y o f t h e MIL-subspace F ( X ) w h i c h i t
determines.
Proof.
We n e e d o n l y show t h e m a x i m a l i t y o f e a c h Z E
Specifically, consider
Zo€X
and x e Z o ; w e h a v e t o show t h e r e
e x i s t s fEF(z) such t h a t f ( x ) # f(Zo). space
(x,J)
determined b y x .
Denote b y Y t h e q u o t i e n t
Y i s compact H a u s d o r f f ( 3 2 . 1 ) ,
h e n c e t h e r e e x i s t s hEC(Y) s u c h t h a t h ( q x ) # h ( q ( Z o ) ) . f
=
hoq.
5.
Set
Then f i s a c o n t i n u o u s f u n c t i o n on X w h i c h i s c o n -
s t a n t on e v e r y Z € z , h e n c e i s i n
F(2)
- and s a t i s f i e s
f(x) # f(Zo).
QED Summing u p ,
( 3 2 . 7 ) T h e r e i s a o n e - o n e c o r r e s p o n d e n c e b e t w e e n t h e MIL-subs p a c e s F o f C(X) a n d t h e u p p e r s e m i c o n t i n u o u s d e c o m p o s i t i o n s of X.
F <-->;!
i f and o n l y i f
2
2
i s t h e s e t of sets of con-
s t a n c y o f F i f a n d o n l y i f F c o n s i s t s o f t h e e l e m e n t s o f C(X) c o n s t a n t on e v e r y Z E
2.
033. The MIL-homomorphisms o f C ( X )
For t h e following d i s c u s s i o n , it i s convenient t o use t h e notation
( f , x ) i n place of f(x)
(cf. the beginning of the
Chapter 6
182 Chapter).
Each xEX
d e f i n e s a f u n c t i o n on C(X) by f +>
(f,x),
and s i n c e C(X) i s s e p a r a t i n g on X , d i s t i n c t x ' s d e f i n e d i s t i n c t functions.
We c a n t h e r e f o r e i d e n t i f y e a c h x w i t h t h e f u n c t i o n
i t d e f i n e s , and h e n c e f o r t h w e w i l l do t h i s : x w i l l be c o n s i d e r e d a f u n c t i o n on C(X).
M o r e o v e r , by t h e v e r y d e f i n i t i o n
o f a d d i t i o n a n d s c a l a r m u l t i p l i c a t i o n i n C(X), x i s a c t u a l l y a linear functional.
Even m o r e , f r o m t h e d i s c u s s i o n o f t h e
b e g i n n i n g o f t h e C h a p t e r on t h e l a t t i c e o p e r a t i o n s i n C ( X ) , e v e r y xEX i s , i n f a c t , a n MI-homomorphism o f C(X) ( i n t o IR). And t h e y a r e t h e o n l y o n e s :
( 3 3 . 1 ) F o r a f u n c t i o n $ o n C(X), t h e f o l l o w i n g a r e e q u i v a l e n t
'1
$ i s a n MI-homomorphism o f C(X) i n t o R ;
2O
$EX.
P roof. -
I t r e m a i n s t o show t h a t ' 1
implies 2
0
.
We n o t e
f i r s t t h a t two MI-homomorphisms $1 , $ 2 o f C(X) i n t o 1R a r e i d e n t i c a l i f and o n l y i f (+1)-1(0) h o l d s f o r 4.
$-'lo) i s
=
Now assume 1'
($2)-1(0).
t h e n a R i e s z i d e a l H o f C(X), a n d
$ b e i n g a s i n g l e l i n e a r f u n c t i o n a l - i s a maximal o n e .
Z(H)
consists of a single point x (31.13).
Hence
But t h e n x -1( 0 )
=
H,
s o w e have $ = x . QED
C o n s i d e r a c o n t i n u o u s mapping X <-
S
Y.
For each fEC(X),
f o s i s a c o n t i n u o u s f u n c t i o n on Y, s o an e l e m e n t o f C(Y). t h u s h a v e a mapping f
I--->
f o s o f C(X) i n t o C ( Y ) .
We
We w i l l c a l l
(C(X) ,X) - D u a l i t y
183
t h i s mapping t h e t r a n s p o s e o f s , a n d d e n o t e i t b y s t . f o r a l l f E C ( X ) and yEY, ( s t f , y )
=
(f,sy).
I t i s t r i v i a l t o v e r i f y t h a t C(X)
S
t
> C(Y)
Thus,
Mn-
i s an
homomorphism.
We show t h a t , c o n v e r s e l y , f o r e a c h Mn-homomorph-
T i s m C(X) ->
C(Y), t h e r e e x i s t s a unique c o n t i n u o u s
X
( 3 3 . 2 ) Theorem.
X,Y,
=
(Stone).
mapping
T:
Given two c o m p a c t H a u s d o r f f s p a c e s
t h e r e i s a one-one correspondence between t h e c o n t i n u o u s
m a p p i n g s s o f Y i n t o X a n d t h e Ma-homomorphisms T o f C(X) i n t o C(Y).
T <->
s i f a n d o n l y i f (f,sy)
= (Tf,y)
for a l l ffC(X)
a n d yEY.
Proof.
-~
I t r e m a i n s t o show t h e e x i s t e n c e o f s , g i v e n T .
F o r e a c h yEY, yoT i s an M ~ - h o m o m o r p h i s m o f C(X) i n t o R , h e n c e , by (32.1),
an element o f X.
o f Y i n t o X; d e n o t e i t b y s .
f E C ( X ) a n d yEY.
We t h u s h a v e a m a p p i n g y k--> So ( f , s y )
=
(Tf,y) for all
i t i s immediate from t h i s i d e n t i t y t h a t f o r
e v e r y n e t { y a } i n Y a n d yoEY, i f ( g , y o ) = l i m a ( g , y a ) gEC(Y), t h e n ( f , s y o ) = l i m a ( f , s y a ) continuous.
That T
=
f o r a l l fEC(X).
i s a n MIL-subspace o f C(Y) ( 1 8 . 2 ) , a n d T - ’ ( O )
R i e s z i d e a l o f C(X).
for all Thus s i s
s t f o l l o w s f r o m t h e same i d e n t i t y .
As we know, f o r a n MIL-homomorphism C(X)
s(Y)
yoT
T
--A
C(Y), T(C(X))
i s a norm c l o s e d
And f o r a c o n t i n u o u s m a p p i n g X <---1 i s a c l o s e d s u b s e t o f X and { s (x) ( x € s ( Y ) } i s an
S
Y,
184
Chapter 6
uppersemicontinuous decomposition o f Y.
Y b e a c o n t i n u o u s mapping a n d T
( 3 3 . 3 ) L e t X
T-'(O) -1 {s
=
(s(Y))'and
(x) ( x E s ( Y ) l
s(Y)
=
=
st .
Then
Z(T-'(O));
i s t h e s e t of s e t s of constancy of
T ( C ( X ) ) , a n d T(C(X)) i s t h e s e t o f e l e m e n t s o f C(Y) c o n s t a n t -1 on e v e r y s (x),
Otherwise s t a t e d , T - l ( O ) and s(Y) c o r r e s p o n d t o e a c h o t h e r u n d e r ( 3 1 . 6 ) , and T(C(X)) and { s each o t h e r under (32.7).
-1
(x) I x € s ( Y ) }
correspond t o
The p r o o f o f ( 3 3 . 3 ) i s s t r a i g h t -
forward.
(33.4) Corollary. (i)
s i s a s u r j e c t i o n i f a n d o n l y i f T i s an i n j e c t i o n .
( i i ) s i s an i n j e c t i o n i f and o n l y i f T i s a s u r j e c t i o n .
We a p p l y t h e a b o v e t o t h e m a t e r i a l o f 5 § 3 1 , 3 2 .
Let F b e
an MIL-subspace o f C ( X ) a n d 2 i t s s e t o f s e t s o f c o n s t a n c y , and d e n o t e t h e q u o t i e n t s p a c e
( Z J ) by
Y.
Let Y
q u o t i e n t map.
Then an a d a p t a t i o n o f t h e a r g u m e n t f o r ( 3 2 . 6 ) , t p l u s ( 3 3 . 4 ) a b o v e , g i v e s u s t h a t C(Y) %> C ( X ) i s an MILi s o m o r p h i s m o n t o F.
do t h i s :
Thus F c a n b e i d e n t i f i e d w i t h C(Y).
We
(C(X) , X ) - D u a l i t y Let F b e an M I - s u b s p a c e o f C ( X ) ,
(33.5)
constancy o f F, Y q u o t i e n t map.
=
(2,y) t h e
185
2
t h e s e t of s e t s of
q u o t i e n t s p a c e , and Y
Then u n d e r t h e d e f i n i t i o n ( f , q y )
=
(f,y) for a l l
f E F and qyEY, w e h a v e F = C ( Y ) .
Now l e t Z b e a c l o s e d s u b s e t o f X a n d H Let X <-
S
Z b e t h e i n j e c t i o n map.
MI-homomorphism o n t o . ~
Then C ( X )
~
>c(z)
=
Z(H)).
i s an
can be i d e n t i f i e d w i t h C ( Z ) .
(33.6)
For e v e r y c l o s e d s u b s e t Z of X ,
C(X)/ZL
( f , z ) for a l l qf€C(X)/Z'
Remark. The mapping C ( X ) s t r i c t i o n map f k--> f l Z .
T i e t z e E x t e n s i o n Theorem.
L
onto C(Z).
Thus
We do t h i s :
C(X)/H
=
t
Factoring it:
t " we o b t a i n a n M a - i s o m o r p h i s m (s ) o f C ( X ) / H
definition (qf,z)
ZL ( s o Z
=
__ > C ( Z )
=
C(Z)
under the
and z E Z .
is c l e a r l y t h e r e -
(33.6) then a l s o follows from t h e