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

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William Elliot

Posts: 1,490
Registered: 1/8/12
Re: compact
Posted: May 15, 2013 10:34 PM
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On Wed, 15 May 2013, dullrich@sprynet.com wrote:
> On Wed, 15 May 2013 00:06:25 -0700, William Elliot <marsh@panix.com>
> >
> >> > 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,

>
> "has cardinality less than |A|", not "is finite".
>

Wrong place for that comment. See ** below.

> > 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),


** which since A is infinite, is < |A|, which of course it isn't.



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