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Re: G_delta
Posted:
Jan 19, 2013 5:22 AM
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On Jan 19, 3:13 am, William Elliot <ma...@panix.com> wrote:
> On Fri, 18 Jan 2013, Butch Malahide wrote:
> > On Jan 19, 1:05 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.
> 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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