Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
William Elliot

Posts: 1,664
Registered: 1/8/12
Re: G_delta
Posted: Jan 19, 2013 4:13 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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. In fact,
that every countable set has an upper bound is equivalent to uncountable
cofinality.

> > > 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.

> > Is Z closed?
>
> It will be shown that Y\Z contains a single point c, the eventual
> value of the function f, and so Z = Y\{c}. To ask whether Z is closed
> is to ask whether c is an isolated point of Y. It could, but it
> doesn't have to be.
>

> > > 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.

>



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.