Re: Compact connected Hausdorff

On Sun, 04 Dec 2005 23:59:04 +0100, Jannick Asmus
<jannick.news@xxxxxx> wrote:

>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.

A component of U must be open in U (a component of any space is both
open and closed in that space).

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
.

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