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Everything is in the context of metric spaces.

My definition of compactness is: every sequence has a convergent subsequence.

Let $B_1, B_2 ...$ be a countable open cover of a compact set $A$. Suppose there is no finite subcover. Then, $\exists$ $a_1 \in A \setminus B_1$,

$\exists$ $a_2 \in A \setminus (B_1 \cup B_2)$,

$\exists$ $a_3 \in A \setminus (B_1 \cup B_2 \cup B_3)$,

$\exists$ $a_n \in A \setminus (B_1 \cup \cdots \cup B_n)$,

But $A$ is compact, so $\exists$ a subsequence $a_{n_k}$ such that $\lim \limits_{k \to \infty} a_{n_k} = a \in A$.

there exists $n_0$ such that $a \in B_{n_0}$ so all but finitely many elements in the subsequence $a_{n_k}$ are inside $B_{n_0}$.

Please help me complete the argument to arrive at a contradiction. I did not understand it in lecture.

I know there are infinitely many $a_{n_k}$'s inside $B_{n_0}$ and only finitely many $a_{n_k}$'s outside $B_{n_0}$.

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    $\begingroup$ What's your definition of compactness? It's usually defined to mean that every open cover has a finite subcover. Then the contradiction is already in your second sentence. $\endgroup$ – joriki Feb 6 at 1:16
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    $\begingroup$ What you write as your definition of compactness is the definition of sequential compactness. The two are not equivalent. They're equivalent for metric spaces. Are you only considering metric spaces? I think your question needs a bit more context. $\endgroup$ – joriki Feb 6 at 1:20
  • $\begingroup$ done............. $\endgroup$ – ironX Feb 6 at 1:21
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Note that $a_i \not \in B_j$ if $i > j$. Since $(a_{n_k})_k$ is a subsequence, there exists $k_0$ such that $n_{k_0} > n_0$ and in particular, we have that $n_l > n_0$ for all $l > k_0$. This is absurd since it shows that infinitely many terms of $(a_{n_k})_k$ lie outside of $B_{n_0}$.

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  • $\begingroup$ You haven't used the metric. You can't conclude compactness from sequential compactness without using the metric because the implication only holds in metric spaces, not generally in topological spaces. $\endgroup$ – joriki Feb 6 at 1:29
  • $\begingroup$ The assumption is that our metric space is compact, no? $\endgroup$ – qualcuno Feb 6 at 1:30
  • $\begingroup$ No, it only seems so because of the OP's terminological unorthodoxy. You purportedly proved from "every sequence has a convergent subsequence" that "every open cover has a finite subcover". In standard terminology, that's proving compactness from sequential compactness, and that's not possible in general, only in metric spaces. $\endgroup$ – joriki Feb 6 at 1:32
  • $\begingroup$ Agreed, but I'm not seeing the mistake. Maybe we're using that $A$ is metric in an intermediate step? $\endgroup$ – qualcuno Feb 6 at 1:33
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    $\begingroup$ The proof should essentially be this one, right? As you point out, we haven't used the metric whatsoever. Anyway, thanks for spotting that, I was confused as well :P $\endgroup$ – qualcuno Feb 6 at 1:42

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