```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 bethe definition of closure of A.BTY, your proof doesn't require the space to be Hausdorff, T1 or T0.
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