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I was thinking about this theorem

If E is an infinite subset of a compact set K, then E has a limit point in K.

I know that's already asked here, for example here

But, I think that I get an alternative proof, but I'm not sure if my proof it's all right.

My proof for this theorem:

Let $\{x_n\}_{n \in \mathbb{N}}$ a sequence such that $\{x_n\}_{n \in \mathbb{N}} \subset E \subset K$, by the compactness of $K$ there is a subsequence $\{x_{n_{j}}\}_{j \in \mathbb{N}}$ such that $x_{n_{j}} \rightarrow a \in K$.

Then $a$ is limit point of $E$.

Indeed $\forall \ \varepsilon > 0 \ \ \exists \ j_0 \in \mathbb{N} $ such that $j>j_0 \Rightarrow x_{n_{j}} \in B(a,\varepsilon)$

As $E$ has infinite point we have $B(a,\varepsilon) \cap E \neq \emptyset$, therefore $a$ is a limit point of $E$ [QED].

Thank you.

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  • $\begingroup$ No. There are compact Hausdorff spaces in which the only convergent sequences are eventually constant. $\endgroup$ – DanielWainfleet Mar 23 '18 at 10:35
  • $\begingroup$ I've got it, Thank you! $\endgroup$ – Kutz Mar 23 '18 at 19:01
  • $\begingroup$ Statements about open sets have "dual" statements about closed sets. Are you familiar with the characterization of compactness in terms of the Finite Intersection Property (FIP) involving families of closed sets? $\endgroup$ – DanielWainfleet Mar 23 '18 at 22:35
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This is not correct and it could not be correct because at no moment you used the fact that $E$ is infinite (you mentioned it, but you did not use it). Note that if $(\forall n\in\mathbb{N}):x_n=e$, for some element $e\in E$, all you proved is that some subsequence converges to $e$. In fact, the whole sequence converges to $e$, but that doesn't prove that $e$ is a limit point of $E$.

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  • $\begingroup$ But if $e \notin E $ in this case $ x_n \neq e$ because $\{x_n\} \subset E$ $\endgroup$ – Kutz Mar 23 '18 at 10:44
  • $\begingroup$ @Kutz That doesn't answer my objection. $\endgroup$ – José Carlos Santos Mar 23 '18 at 10:48
  • $\begingroup$ I've got it. Thank you! $\endgroup$ – Kutz Mar 23 '18 at 19:00

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