Re: Compact connected Hausdorff

On Dec 4, 2005 5:02 AM CT, William Elliot 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?

Since X is compact there exists an open cover K with a
finite subcover; that is X = \/ K_i for some finite
indexing set. Choose K such that U is one of our K_i's
and denote the collection without U as K_j; that is
X = \/ K_i = U \/ (\/ K_j). Now since X is connected
U /\ (\/ K_j) must not be empty, for otherwise it would
be a separation. Thus cl(U) /\ (\/ K_j) is non-empty
and contains an element of X - U.

Maybe? I'm not 100% on this one, but that's what I came
up with.

Regards,
Kyle
.

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