Re: Compact connected Hausdorff

On 04.12.2005 20:11, quasi wrote:
> On Sun, 4 Dec 2005 03:02:50 -0800, William Elliot
> <marsh@xxxxxxxxxxxxxxxxxx> wrote:
>
>
>>>Lemma: If space X is compact and connected. For any open set
>>>U of X, the closure of some component of U intersect X - U.
>>


>
> If U is empty or U = X, the lemma is false, so assume the statement of
> the lemma is changed to exclude those cases.
>
> With that correction, the lemma is true, as is your point that "some
> component" can be strengthened to "every component".
>
> However, there's no need to assume either compact or Hausdorff, so
> those requirements can be dropped.
>
> Here's a sketch of the proof:
>
> The components of U must be open in U, hence open in X.

Why that? Is that really true?

J.
.

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