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

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David C. Ullrich

Posts: 3,238
Registered: 12/13/04
Re: compact
Posted: May 15, 2013 11:24 AM
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On Wed, 15 May 2013 00:06:25 -0700, William Elliot <marsh@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,


"has cardinality less than |A|", not "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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