
Re: G_delta
Posted:
Jan 20, 2013 1:14 AM


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 wellordered) > 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? >

