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

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David C. Ullrich

Posts: 3,085
Registered: 12/13/04
Re: compact
Posted: May 16, 2013 10:42 AM
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On Wed, 15 May 2013 16:26:54 -0500, quasi <quasi@null.set> wrote:

>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?


A typo, sort of. Later in the same post he said
"is finite", where he should have said "has
cardinality less than |A|". So I decided to post
a comment, and then I inserted the comment in
the wrong place, this paragraph looking so much
like that paragraph...

>
>>> which of course it isn't.
>
>quasi





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