
Re: definition of closure in topological space question
Posted:
Nov 18, 2012 10:34 PM


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.

