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Re: Perfectly Uncountable
Posted:
Jun 26, 2012 11:10 AM
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On Tue, 26 Jun 2012 01:52:08 -0700, William Elliot <marsh@panix.com>
wrote:
>On Mon, 25 Jun 2012, David C. Ullrich wrote:
>> On Mon, 25 Jun 2012 09:28:24 -0500, David C. Ullrich
>> <ullrich@math.okstate.edu> wrote:
>
>> >Clarify something first: Do you have some reason to think that
>> >this is actually true?
>
>> Forget I asked that. This notion of "scattered" is new to me.
>> If S is compact Hausdorff and not scattered then S has a
>> perfect subset. QED, by the result above and Tietze.
>>
>To prove that a not scattered compact Haudorff space S
>can be continuously mapped to [0,1].
>
>S has a not empty perfect kernel K.
>Case K has a multi-point component.
> There's a Urysohn function f that maps K onto [0,l].
> f can be extended to all of S by Tietze's extension theorem.
>
>Case K has no multi-point component.
> K is totally disconnected. K is uncountable.
> K can be continuously mapped onto the Cantor set C.
>As easy as you claim that to be, I don't see it.
Write K = K_0 union K_1, where each K_j is nonempty, closed,
and K_0 and K_1 are disjoint.
Since K_j is open and K has no isolated points, K_j must be
infinite.
Since K_j is infinite, we can repeat the argument. Write
K_0 = K_00 union K_01 and K_1 = K_10 union K_11,
such that [etc.].
And now repeat again, splitting each K_jk into two pieces.
Now, for each x in K there is an infinite string b_0 b_1 ...
of 0's and 1's such that x lies in K_[b_0 b_1 ... b_n] for
every n. Define f(x) to be the element of the Cantor
set corresponding to the string b_0 b_1 ... .
The fact that each K_alpha is open for every
finite sequence alpha of 0's and 1's shows that f is continuous.
>Finally, does C continuously map onto [0,1]?
Come on now. Read the wikipedia article on the Cantor
set. I haven't looked, but this must be in there, it's
a basic property of the Cantor set. (If it's not there
let us know and it will be there soon.)
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