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

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William Elliot

Posts: 1,609
Registered: 1/8/12
Re: G_delta
Posted: Jan 14, 2013 3:52 AM
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On Mon, 14 Jan 2013, Butch Malahide wrote:
> On Jan 14, 12:04 am, William Elliot <ma...@panix.com> wrote:
> >
> > In 1st countable spaces, every point p is G_delta.
> > In fact, if every point of a compact Hausdorff space S,
> > is G_delta, then S is 1st countable.

>
> Oh, right, I'd forgotten that.
>

> > Define g:omega_1 -> omega_1 + 1, eta -> eta.
> > Assume S is compact Hausdorff and f in C(omega_1, S).
> >
> > Is there some h in C(omega_1 + 1, S) with f = hg?

>
> Are you just asking whether a continuous function from omega_1 to a
> compact Hausdorff space S can always be extended to a continuous
> function on omega_1 + 1? That sounds like it should be true. I'm not a
> topologist, but I took a course in topology back in 1957-58. I'm not
> sure I remember enough topology to answer a technical question like
> that, but I'll take a stab at it.


> First suppose S = [0, 1]. (Gotta walk before you can run.) I believe
> that a continuous real-valued function on omega_1 must be eventually
> constant. (I may even remember how to prove that, but it's late and I
> don't want to do any hard thinking now.) In that case, just map the
> point omega_1 to the same constant, and everything is fine.


Yes, that's correct. In fact it's true for regular Lindeloff spaces
for which every point is a G_delta (whence the G_delta question).

If S is comapct Hausdorff, then every point is G_delta iff S is 1st
countable. If S is countable, then every point is G_delta.
Thus the problem is to find h for uncountable, not 1st countable S.

> Now, can't the result for an arbitrary compact Hausdorff space S be
> derived from the [0, 1] case? Say, by embedding S in a cube, or
> something like that?


Yes,
e:S -> [0,1]^F, x -> prod{ f(x) | f in F }
where
F = C(S,[0,1])
is an embedding for any Tychonov T0 space S.

How can that be used?



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