By Samuel Kaplan

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A+ QED Consider a subset A o f E . The i n t e r s e c t i o n o f a l l t h e o r d e r c l o s e d s e t s c o n t a i n i n g A i s an o r d e r c l o s e d s e t , hence t h e smallest one c o n t a i n i n g A. We w i l l c a l l i t t h e o r d e r c l o s u r e o f A. I n g e n e r a l A(1) i s n o t o r d e r c l o s e d , hence i s a p r o p e r s u b s e t o f t h e o r d e r c l o s u r e o f A. A(3) and so on. = Let A ( 2 ) = ( A ( 1 ) )( 1 ) 9 I t may r e q u i r e a t r a n s f i n i t e s e q u e n c e o f t h e s e i t e r a t i o n s t o o b t a i n t h e o r d e r c l o s u r e o f A.

C1 Let A , B be s u b s e t s of E. (i) o r d e r c l o s u r e (AIJ B) IJ (ii) ( o r d e r c l o s u r e A) = (order closure B). O r d e r c l o s u r e ( A n B ) c ( o r d e r c l o s u r e A) n ( o r d e r c l o s u r e B ) , and t h e i n c l u s i o n i s p r o p e r in general. ( i i i ) If, i n ( i i ) , A , B a r e R i e s z i d e a l s , w e h a v e e q u a l i t y . 16. (a) F o r R i e s z i d e a l s 11,12,H o f E : (i) H n (I1 (ii) if E (b) 18. I1 Q 12, t h e n H = (fdnI,) 3 ( H n 1 2 ) .

3) is order c l o s e d , hence i s t h e o r d e r c l o s u r e o f I . Proof. By ( 5 . 1 ) a n d ( 4 . 5 ) , we n e e d o n l y show t h a t ( I ( 1 ) ) + i s c l o s e d under t h e o p e r a t i o n o f a d j o i n i n g suprema o f a r b i t r a r y subsets of itself. idempotent. Now t h i s o p e r a t i o n i s e a s i l y shown t o b e Since (I('))+ i s o b t a i n e d by a p p l y i n g t h e o p e r a - t i o n t o I + , we a r e t h r o u g h . QED We w i l l s e l d o m e n c o u n t e r o r d e r c l o s e d R i e s z s u b s p a c e s i n t h i s work.

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The Bidual of C(X)I by Samuel Kaplan
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