Re: Compact connected Hausdorff

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)?

>
> 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?
>
> quasi

J.
.

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