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

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: compact
Posted: May 15, 2013 7:30 AM
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On May 15, 2:06 am, William Elliot <ma...@panix.com> wrote:
> On Tue, 14 May 2013, Butch Malahide wrote:
> > On May 14, 11:15 pm, William Elliot <ma...@panix.com> wrote:
> > > A point x is an accumulation point of A
> > > when for all open U nhood x, U /\ A is infinite.

>
> > > If S is compact, then every infinite set A has an accumulation point.
> > > Proof.
> > > If not, then for all x, there's some open U_x nhood x with finite U /\ 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 an accumulation point, is S compact?
>
> > omega_1
>
> 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.



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