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Re: Perfectly Uncountable
Posted:
Jun 25, 2012 10:42 AM
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On Mon, 25 Jun 2012 09:28:24 -0500, David C. Ullrich
<ullrich@math.okstate.edu> wrote:
>On Sun, 24 Jun 2012 20:28:33 -0700, William Elliot <marsh@panix.com>
>wrote:
>
>>On Sun, 24 Jun 2012, David C. Ullrich wrote:
>>> On Sun, 24 Jun 2012 02:43:20 -0700, William Elliot <marsh@panix.com>
>>>
>>> >Let S be an uncountable perfect (without isolated points) compact
>>> >Hausdroff space. S can be continuously mapped onto [0,1].
>>> >
>>> >Is this reasoning correct?
>>> No.
>>>
>>> >Let C be a connected component of S.
>>> >By hypothesis and the fact that S is locally compact,
>>> >C is multi-pointed and normal. Thus it can be continuously
>>> >mapped onto [0,1] by some Urysohn function f.
>>>
>>> I don't know exactly where your error is, since I don't follow
>>> the reasoning. But let S be the Cantor set. Then C contains
>>> just one point.
>>>
>>> >Since C is closed and S normal, f can be extended to all of S by Tietze's
>>> >extension theorem.
>>
>>Let S be a perfect, locally connected normal T1 space.
>>Then S continuously maps onto [0,1].
>>Proof.
>>Let C be a connected component of S.
>>C is clopen, thus normal T0, multi-pointed and connected.
>>It follows that there's some points a,b and a Urysohn function
>>f:C -> [0,1] with f(a) = 0, f(b) = 1. Since C is connected,
>>f is a surjection and since C is closed, f can be continuously
>>extended to all of S.
>>
>>How's that for correctness and pointing out my mistake?
>
>Somewhat lame, adding an un-needed extra hypothesis.
>Case 1: There exists a connected subset C containing
>more than one point. Do what you said.
>Case 2: S is totally disconnected. It's not hard to
>construct a continuous function from S onto the
>Cantor set.
>
>>Alas, it's no help to show that a not scattered, compact Hausdorff
>>space continuously maps onto [0,1].
>
>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.
>>BTW, isn't the Cantor set scattered?
>
>What??? Wikipedia says "scattered" means that every nonempty
>subset has an isolated point. So a perfect space cannot be scattered.
>
>
>
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