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Topic: G_delta
Replies: 28   Last Post: Jan 26, 2013 3:50 AM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: G_delta
Posted: Jan 15, 2013 8:35 PM
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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 S, "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}.

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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