Re: what sets admit connected Hausdorff topological spaces?

From: W. Dale Hall (mailtowd-hall_at_pacbell.net)
Date: 11/02/04 

Date: Tue, 02 Nov 2004 17:04:05 GMT

David C. Ullrich wrote:
> On Tue, 02 Nov 2004 09:42:05 -0600, David C. Ullrich
> <ullrich@math.okstate.edu> wrote:
>
>
>>On Tue, 02 Nov 2004 08:52:19 -0500, David Bernier
>><david250@videotron.ca> wrote:
>>
>>
>>>{} and {1}, obviously.
>>>
>>>What about N, the non-negative integers?
>>
>>Isn't it well-known that such a space must
>>have cardinality at least c?
>
>
> Never mind. I don't know how it happened that
> I saw the word "compact" in the subject line.
>
>
>>Hmm, in any case it's true, and follows from
>>things that are certainly well-known: If K is
>>a compact Hausdorff space then the continuous
>>real-valued functions on K separate points.
>>In particular if K has more than one point
>>then there exists a non-constant real-valued
>>continuous function, and if K is connected
>>its range must be a non-trivial interval.
>>
>>Surely any set of cardinality > c admits
>>such a topology. (For example "surely" you
>>can partition such a set into subsets of
>>cardinality c and then define some compactification
>>of a sort of star-shaped thingie...)
>>
>>
>>>David Bernier
>>
>>
>>************************
>>
>>David C. Ullrich
>
>
>
> ************************
>
> David C. Ullrich

Now I'm pretty confused. After I saw your note, I saw Zundark's citation
of the Bing paper, and wonder whether Bing's countable space isn't
already compact. If not, surely the one-point compactification is
countable iff the original space is, and "should" also be connected and
Hausdorff in the same case.

That is, it "should" yield a compact connected countable Hausdorff
space.

The error I'm making must lie in that "should", but I don't see it.

Dale

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