Re: Compact connected Hausdorff

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.
>
>The above lemma was present some time ago. It seemed quickly
>posted for it's necessary that U be nonnul and proper subset.
>
>I'm of the mind that "of some component" was to be read as
>"of any component" and that Hausdorff is tacid assumption.
>
>Is the surmise correct? Anyway, let C be a component of open U and
>assume the negation, that cl C /\ (X - U) is empty, ie
> (cl C) - U = nulset
>
>>From that comes
> C = cl C proper subset U.
>
>However how to prove the lemma eludes me. Even knowing the equivalence of
>components and quasi-components in continuums, I see no way of using that.
>Have you suggestion?

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.

Since X is connected, the components of U cannot also be closed in X.

Let A be a component of U and let p be an element of cl(A)-A.

Claim p is in X-U.

Suppose not.

p not in X-U ==> p is in U ==> p is in some component of U.

Thus, assume p is in B where B is a component of U. By the choice of
p, B can't be equal to A, hence B is disjoint from A.

Since p is in B, some open neighborhood N of p lies entirely in B,
hence, since B is disjoint from A, N is also disjoint from A. But p is
in cl(A) implies that every neighborhood of p contains a point of A,
contradiction.

Therefore p is in X-U, as claimed.

This proves the lemma.

quasi
.

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