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

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William Elliot

Posts: 1,601
Registered: 1/8/12
compact
Posted: May 15, 2013 12:15 AM
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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?





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