Re: Compact connected Hausdorff

On Dec 4, 2005 7:03 AM CT, Narcoleptic Insomniac wrote:

> 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.

A friend of mine just pointed out that the lemma in
question asks about the closure of *some component* of
U.

I completely neglected that fact, and to be quite
honest, I'm not sure what is meant by "component".

What is meant by "component" and does the approach still
seem logical?

Regards,
Kyle
.

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