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Re: G_delta
Posted:
Jan 21, 2013 5:00 AM
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On Jan 21, 2:59 am, William Elliot <ma...@panix.com> wrote:
> 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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