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

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William Elliot

Posts: 1,600
Registered: 1/8/12
Re: compact
Posted: May 15, 2013 3:06 AM
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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?



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