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Given that:

$S$ is a compact set in the topological space $(X, \mathcal T)$

$T\subset S$ has no accumulation points in S

How do I show that $T$ is finite?

Stefan Hamcke
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jorgea
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  • Why the label about topological vector spaces? – Siminore Jun 19 '13 at 13:51
  • Sorry! I have updated the labels. – jorgea Jun 19 '13 at 13:53
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    Put a neighborhood around every point, and compact gives you a finite covering. Can you get only finite points in each of those neighborhoods? – Ncat Jun 19 '13 at 14:00
  • Check my answer here http://math.stackexchange.com/questions/424473/compactness-and-limit-points/424507?noredirect=1#comment907053_424507, actually I proved the contronominal. – Edoardo Lanari Jun 19 '13 at 14:07
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    @Lano Careful, sequences don't capture all the information in general topological spaces. – Pedro Jun 19 '13 at 14:28
  • Thanks for the primary school precisation, I know it works perfectly only in first countable spaces, just read carefully what I wrote. If it's infinite it has a countable subset, which has no accumulation point and you get the absurd. – Edoardo Lanari Jun 19 '13 at 14:41
  • Do you mean limit point? I think accumulation point only applies to sequences. – Stefan Hamcke Jun 19 '13 at 14:53
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    @PeterTamaroff True; but any sequence with pairwise distinct terms has an accumulation point nonetheless. And from any infinite set you can build a sequence of this kind. What isn't true in general is that the accumulation point is the limit of a subsequence, but this is not relevant for this question. – egreg Jun 19 '13 at 15:08
  • @egreg: exactly what I meant :) – Edoardo Lanari Jun 19 '13 at 15:16
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    @Lano How unfortunate you chose to reply in that way to my comment. I did miss your idea, but I think that remark is uncalled for. – Pedro Jun 19 '13 at 15:49

1 Answers1

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For every $s \in S$ we find a neighbourhood $U_s$ that intersects $T$ in only at most finitely many points, so $U_s \cap T$ is finite (possibly empty). This is because no point in $S$ is an accumulation point of $T$ by assumption.

As the $U_s$ ($ s \in S$) form an open cover of $S$ by compactness we find a finite subcover $U_{s_1},\ldots,U_{s_n}$.

So $$T= T \cap S \subset T \cap (\cup_{i=1}^n U_{s_i}) = \cup_{i=1}^n (T \cap U_{s_i})$$

where the first follows from $T \subset S$, the inclusion from the fact that we cover $S$, and the last equality is the standard distributive property of intersection over unions.

It follows that $T$ is a subset of a union of finitely many finite sets, so itself finite.

Henno Brandsma
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