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Re: compact
Posted:
May 15, 2013 12:27 AM
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On May 14, 11:15 pm, William Elliot <ma...@panix.com> wrote:
> A point x is an axxumulation 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
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