Regular Closed Subsets and Subspaces

From: William Elliot (marsh_at_agora.rdrop.com)
Date: 03/27/05 

Date: Sun, 27 Mar 2005 04:00:06 +0000 (UTC)

If U is a regular open subset of a regular open subspace A,
then U is regular open subset of the overspace S.

U is regular open when U = int cl U = interior of closure of U
K regular closed when K = cl int K

A.cl U = closure of U within A = A /\ cl U
A.int U = interior of U within A = A /\ int(S\A \/ U) 

--
U regular open within regular open A ==> U regular open
Proof with the tacit assumptions U subset A subset S
U = A.(int cl) U = A.int A.cl U = A /\ int(S\A \/ A.cl U)
 	= A /\ int(S\A \/ (A /\ cl U)) = A /\ int(S\A \/ cl U)
 	= int A /\ int(S\A \/ cl U) = int(A /\ (S\A \/ cl U))
 	= int(A /\ cl U) = int A /\ int cl U = A /\ int cl U
int cl U = int cl A/\U subset int cl A /\ int cl U
 	= A /\ int cl U subset int cl U
int cl U = A /\ int cl U = U
-- 
Now the dual theorem
    If K is a regular closed subset of a regular closed
    subspace A, then K is regular closed subset of the overspace S.
would be expected, in fact seems likely (ie, no obvious
counterexample).  However it defies a dual version of the
previous proof.  Here's some details
K regular closed within regular closed A ==> K regular closed ???
K = A.(cl int) K = A.cl A.int K = A /\ cl(A /\ int(S\A \/ K))
 	= cl A /\ cl(A /\ int(S\A \/ K)) = cl(A /\ int(S\A \/ K))
Thus quickly
 	cl K = K;  cl int K subset cl K = K
Leaving to prove
 	K = cl K subset cl int K
Any suggestions, comments?
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