```Date: Jan 24, 2013 3:33 AM
Author: William Elliot
Subject: G_delta

On Mon, 21 Jan 2013, Butch Malahide wrote:> On Jan 21, 2:59 am, William Elliot <ma...@panix.com> wrote:You've proven all of this and yet claim you don't know (point set) topology?> > Ok, the other difference is> > a)      regular & every point a G_delta> > and> > b)      every point the countable intersection of closed nhoods of the> > point.> >> > That is for all p, there's some closed Kj, j in N with> > for all j in N, p in int Kj & {p} = /\_j Kj.> >> > Yes, a) -> b) but I doubt the converse.> > In other words, is a) strictly stonger than b).> >> > Now if we assume a Lindelof space, is> > a) still strictly stronger than b?> >> > I think it is.  Are you of the same opinion?> > Let Y be the real line with the topology generated by the usual open> sets and the set Q of all rational> numbers. Offhand, it seems to me that Y is a Lindelof space which has> property b) but is not regular. Let's see.> > Is Y Lindelof? Q is Lindelof because it's countable. Y\Q is Lindelof> because the subspace topology on Y\Q is just the usual topology of the> irrational numbers. Y is Lindelof because it's the union of two> Lindelof subspaces, Q and Y\Q.> > Does Y have property b)? The real line with the usual topology has> property b). If K is a closed neighborhood of p in the real topology,> then it is a closed neighborhood of p in the stronger topology of Y.> Hence Y has property b).> > Is Y regular? No, because Q is a neighborhood of zero which contains> no closed neighborhood of zero.>
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