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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
Re: G_delta
Posted: Jan 21, 2013 3:59 AM
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On Sun, 20 Jan 2013, Butch Malahide wrote:
> On Jan 20, 12:14 am, William Elliot <> wrote:
> >
> > Your proof generaly follows the proof for f in C(omega_1,S),

> > where S is regular Lindelof and ever point is G_delta, that
> > f is eventually constant.  There are some differences in the
> > premises of the two theorems that I'm going to puzzle upon
> > and try to harmonize.

> The puzzle is whether I needed to put that silly adverb "hereditarily"
> in front of "Lindelof". Now that you mention it, just plain Lindelof
> is good enough. I only used "hereditarily Lindelof" to show that Z is
> Lindelof. It's easy to see (as shown in the last step of the argument
> I posted) that Y\Z contains at most one point. It follows from the
> other assumptions that Y\Z is a G_{delta}, i.e., Z is an F_{sigma} in
> Y; and of course an F_{sigma} subspace of a Lindelof space is Lindelof.

Ok, the other difference is
a) regular & every point a G_delta
b) every point the countable intersection of closed nhoods of the

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?

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