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Regular Closed Subsets and Subspaces
Posted:
Mar 26, 2005 11:00 PM
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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)
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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
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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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