Date: Jan 18, 2013 4:23 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?

>

> [My previous reply was typed in a hurry and had a bunch of typos, some

> of which are corrected here.]

>

> 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.

>

> 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 Y}. Let Z = {y in Y: g({y}) is

> bounded}.

Did you intend g(S) = f^-1(S) = { x in X | f(x) in S }?

> 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.

>