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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.

José Carlos Santos
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Kutz
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1 Answers1

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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 $E$.

José Carlos Santos
  • 440,053