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

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William Elliot

Posts: 1,443
Registered: 1/8/12
Re: G_delta
Posted: Jan 20, 2013 1:14 AM
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On Sat, 19 Jan 2013, Butch Malahide wrote:
> On Jan 19, 3:13 am, William Elliot <ma...@panix.com> wrote:
> >
> > > > > > Does this generalize to every uncountable limit ordinal eta,
> > > > > > that f in C(eta,R) is eventually constant and thusly the Cech
> > > > > > Stone compactification of of eta is eta + 1?  Does eta need to
> > > > > > have an uncountable cofinality for this generalization?

> >
> > > > > Yes, the same argument that works for omega_1 also works for any
> > > > > ordinal of uncountable cofinality.

> >
> > > > > Let X be a linearly ordered topological space (i.e., a linearly
> > > > > ordered set with its order topology) in which every increasing
> > > > > sequence converges. [Examples: any ordinal of uncountable
> > > > > cofinality; the long line; any countably compact LOTS.] Call a
> > > > > subset of X "bounded" if it has an *upper* bound in X,
> > > > > "unbounded" otherwise. Observe that (1) the union of countably
> > > > > many bounded sets is bounded, and (2) the intersection of
> > > > > countably many unbounded closed sets is unbounded.

> >
> > > > Does not (1) hold because every increasing sequence converges?
> > > > Wouldn't every countable set has an upper bound suffice?

> >
> > > Yes, "every countable set has an upper bound" would suffice for (1),
> > > but not for (2).

> >
> > Why not?  I've reviewed a proof for that theorem and see than instead
> > of supremum or limits, that upper bounds would suffice.

>
> I doubt it.
>

You're correct. Dedekind completeness is require or
at least every increasing sequence has a limit.

If X has a top, then from the top on, a function is constant.
Thus wlog, X is unbounded above.

Your proof generaly follows the proof for f in C(omega_0,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.

Nice proof.

> > In fact, that every countable set has an upper bound is equivalent to uncountable cofinality.
>
> True (aside from the trivial case where X has a greatest element) but
> irrelevant. For totally ordered (but not necessarily well-ordered)
> sets, having uncountable cofinality is not enough to force the
> intersection of two closed cofinal subsets to be nonempty. Let zeta =
> omega^* + omega, the order type of the integers. Let X be an ordered
> set of type zeta times omega_1. Every countable subset of X has an
> upper bound. The order topology of X is the discrete topology. Every
> uncountable subset of X is a closed cofinal subset. Partition X into
> two disjoint uncountable subsets A and B. Then A and B are closed
> cofinal subsets of X whose intersection is empty. Moreover, since X is
> discrete, C(X,R) is the set of all functions from X into R; it is easy
> to see that not all of them are eventually constant.
>

> > > > > Let Y be a topological space which is hereditarily Lindelof and such
> > > > > that, for each point y in Y, the set {y} is the intersection of
> > > > > countably many closed neighborhoods of y. [Example: any separable
> > > > > metric space.]

> >
> > > > > THEOREM. If X and Y are as stated above, then every function f in
> > > > > C(X,Y) is eventually constant.

> >
> > > > > PROOF. We may assume that X has no greatest element. For S a subset of
> > > > > Y, let g(S) = {x in X: f(x) is in S}. Let Z = {y in Y: g({y}) is
> > > > > bounded}.

> >
> > > > Is assuming X has no greatest element, an additional premise?
> >
> > > No. The assumption that X has no greatest element is made without loss
> > > of generality, because the contrary case is trivial: if X has a
> > > greatest element, then every function with domain X is eventually
> > > constant.

> >
> > Uncountable cofinality implies X has no max.

>
> Sure. So what?
>




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