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Re: G_delta
Posted:
Jan 18, 2013 4:23 AM
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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.
>
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