```Date: Jan 19, 2013 2:05 AM
Author: William Elliot
Subject: Re: G_delta

On Tue, 15 Jan 2013, Butch Malahide wrote:> On Jan 14, 11:10 pm, 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?> 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?Is Z closed? > CLAIM. Each point z in Z has a neighborhood V_z such that g(V_z) is> bounded.> > PROOF OF CLAIM. Let {U_n: n in N} be a countable collection of closed> neighborhoods of z whose intersection is {z}. Assume for a> contradiction that each set g(U_n) is unbounded. Since f is> continuous, each g(U_n) is an unbounded closed set. By property (2)> above, g({z}) = /\{g(U_n): n in N} is unbounded, contradicting the> assumption that z is in Z.> > Thus the set Z is covered by open sets V_z such that g(V_z) is> bounded. Since Y is hereditarily Lindelof, it follows that Z is> covered by countably many open sets V such that g(V) is bounded. In> view of property (1)  above, it follows that g(Z) is bounded. Since> g(Y) = X is unbounded, Y\Z is nonempty. All we have left to show is> that Y\Z consists of a single point. Assume for a contradiction that Y> \Z contains two distinct points c and d. Thus g({c}) and g({d}) are> unbounded closed subsets of X. By property (2) above, the intersection> of g({c}) and g({d}) is unbounded; in particular, it is nonempty. Let> x be a point in the intersection of g({c}) and g({d}). Then c = f(x) => d, contradicting the assumption that c and d are two distinct points.>
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