Re: Compact connected Hausdorff

On Mon, 05 Dec 2005 08:55:44 +0100, Jannick Asmus
<jannick.news@xxxxxx> wrote:

>On 05.12.2005 00:17, quasi wrote:
>> On Sun, 04 Dec 2005 23:59:04 +0100, Jannick Asmus
>> <jannick.news@xxxxxx> wrote:
>>
>>
>>>On 04.12.2005 20:11, quasi wrote:
>>>
>
>>>>
>>>>The components of U must be open in U, hence open in X.
>>>
>>>Why that? Is that really true?
>>>
>>>J.
>>
>>
>> A component of U must be open in U (a component of any space is both
>> open and closed in that space).
>
>First, let me say that in this context 'component' means 'connected
>component' to me. Just to avoid confusion.
>
>Could you prove that a component is open *without* any additional
>assumption on U (e.g., U has only fintely many components or U is
>locally connected)?

No, I can't -- your question is right on target, making me aware my
misconception. Thanks.

>>
>> But since U was specified as open in X, it follows that the components
>> of U are also open in X (a relatively open subset of an open set is
>> open).
>>
>> What am I missing?
>>

I see that I was missing something basic -- components are always
closed, but not always open.

quasi
.

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