Date: Nov 18, 2012 10:44 AM
Author: David Hartley
Subject: Re: definition of closure in topological space question

In message <Pine.NEB.4.64.1211172248550.8793@panix1.panix.com>, William 
Elliot <marsh@panix.com> writes
>> Make it *closed* neighbourhoods of A in 2 and then it's equivalent to
>> usual closure in T1 normal spaces, even regular spaces. (Probably it's
>> equivalent if and only if the space is regular.)

>
>More than T1 is needed for by 2, within the cofinite reals, cl {0} = R.
>
>Can you show the equivalence for normal T1 spaces?



Let Cn(A) be the intersection of all closed neighbourhoods of A, (where
a closed neighbourhood is a closed set C such that there is an open set
U with A c= U c= C).

Claim.
A space X is regular iff Cn(A) = Cl(A) for every subset A of X.

If X is not regular, then there exists x e X and a (closed) subset A of
X such that x is not in A but every nbhd. of x meets every nbhd. of A.
But then x is in every closed nbhd. of A and so is in Cn(A). Hence Cn(A)
=/= Cl(A).

Conversely, if Cn(A) =/= Cl(A) for some A, then regularity fails at any
x in Cn(A) \ Cl(A).
--
David Hartley