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Topic: G_delta
Replies: 28   Last Post: Jan 26, 2013 3:50 AM

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William Elliot

Posts: 2,637
Registered: 1/8/12
Posted: Jan 24, 2013 3:33 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On Mon, 21 Jan 2013, Butch Malahide wrote:
> On Jan 21, 2:59 am, William Elliot <> wrote:

You've proven all of this and yet claim you don't know (point set)

> > 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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