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Topic: Onto [0,1]
Replies: 40   Last Post: Apr 29, 2013 10:16 PM

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David C. Ullrich

Posts: 2,752
Registered: 12/13/04
Re: Onto [0,1]
Posted: Apr 22, 2013 10:56 AM
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On Sun, 21 Apr 2013 18:55:29 -0700, William Elliot <marsh@panix.com>
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.





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