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Topic: compact
Replies: 9   Last Post: May 18, 2013 9:56 PM

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William Elliot

Posts: 1,706
Registered: 1/8/12
compact
Posted: May 18, 2013 9:56 PM
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On Wed, 15 May 2013, Butch Malahide wrote:
> On May 15, 2:06 am, William Elliot <ma...@panix.com> wrote:

> > A point x is a saturation point of A
> > when for all open U nhood x, |U /\ A| = |A|
> >
> > If S is compact, then every infinite set A has saturation point.
> > Proof.
> > If not, then for all x, some open U_x nhood x with |U_x /\ A| < |A|.
> > Since C = { U_x | x in S } covers S,
> > . . there's a finite subcover { U_x1,.. U_xj }
> > Thus A = A /\ (U_x1 \/..\/ U_xj } = (U_x1 /\ A) \/..\/ (U_xj /\ A),
> > is finite which of course it isn't.
> >
> > If every infinite A subset S has a saturation point, is S compact?

>
> Yes. Suppose S is not compact. If kappa is the minimum cardinality of
> an open cover of S which has no finite subcover, then there exists A
> subset S such that |A| = kappa and A has no saturation point.
>

Let C be a cover of S with minimal cardinality.
Let eta = |C|, well order C = { U_xi | xi < eta }.
and for all nu < eta, let C_nu + { U_xi | xi <= nu }.

By minimality, for all nu < eta, there's some a_nu in S - \/C_nu.
A = { a_nu | nu < eta } is infinite. Otherwise, there'd be some
nu < eta with A subset \/C_nu, so a_nu not in A.

Let x be a saturation point of A.
Some nu < eta with x in U_nu.

x in V = \/{ U_xi | xi <= nu }
V /\ A = { a_xi | xi <= nu }

Thus, provided |A| = |C| can be proved,
the contradiction, |V /\ A| < |A|.



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