Date: Nov 18, 2012 10:34 PM
Author: William Elliot
Subject: Re: definition of closure in topological space question

On Sun, 18 Nov 2012, David Hartley wrote:
> <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).
>

A set K is a closed (compact) nhood of A
when K closed (compact) and A int K.

> 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).


Excellent. If only all spaces were regular, then Cn A could be
the definition of closure of A.

BTY, your proof doesn't require the space to be Hausdorff, T1 or T0.