
Re: G_delta
Posted:
Jan 19, 2013 5:22 AM


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 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?

