On Sun, 21 Apr 2013 18:55:29 -0700, William Elliot <firstname.lastname@example.org> wrote:
>On Sun, 21 Apr 2013, Butch Malahide wrote: > >> On Apr 21, 2:56 am, William Elliot <ma...@panix.com> wrote: >> > Can an uncountable compact Hausdorff be continuously mapped onto [0,1]? >> >> Let X be the ordinal omega_1 + 1 with its order topology. X is an >> uncountable compact Hausdorff space. X can not be continuously mapped >> onto [0,1]. (Hint: X is scattered.) Whether X can be discontinuously >> mapped onto [0,1] is independent of ZFC. > >Whoops, wrong question. >Can a perfect compact Hausdorff space be continuously mapped onto [0,1]?
As has been pointed out, _given_ that you're so obnoxious about other people's English, you should be more careful with yours! The answer to the question as stated is obviously yes, since for example [0,1] is a perfect compact Hausorff space.
The answer to the question you meant to ask is also yes, it seems to me. Say K is a perfect compact Hausdorff space. Either K has a connected subset containing more than one point or not.
If C is a subset of K containing p and q, p <> q, then it follows from Tietze that there is a continuous f : K -> [0,1] with f(p) = 0 and f(q) = 1; now f(C) must be connected, qed.
On the other hand, if K has no connected subsets with more than one point: It's easy to construct a continuous map from K onto the Cantor set, qed. (Say K is the disjoint union of the closed set A_0 and A_1. Map A_0 to the left half of the Cantor set and A_1 to the right half. Now A_0 is the disjoint union of the closed sets A_00 and A_0,1... repeat countably many times and you've defined a continuous map onto the Cantor set.)
> >BTW, countable, (locally) compact Hausdorff spaces are imperfect.