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.

>