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

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Posts: 12,067
Registered: 7/15/05
Re: compact
Posted: May 15, 2013 4:48 PM
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dullrich wrote:
>William Elliot 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".

Looks like a finite union of finite sets, no?

>> which of course it isn't.


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