Re: Compact connected Hausdorff


quasi wrote:
> 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
Connected components of a topological space Y must be closed in Y (the
closure of a connected set is connected) but not necessarily open
unless the space is assumed locally connected.For example Y= the
rationals in R (reals) with the relative topology ;the components are
Points which are not open .Regards,Stuart

.

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