
Re: definition of closure in topological space question
Posted:
Nov 18, 2012 10:44 AM


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

