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Let $(X,d)$ be a metric space and $A \subset X$ a subset.

Definition 1: We call $A$ coverage compact, if each open cover of $A$ has a finite subcover.

Definition 2: We call $A$ sequentially compact, if every sequence in $A$ has a convergent subsequence.

Bolzano-Weierstrass-Theorem: $A$ is coverage compact iff $A$ is sequentially compact.

My Proof of "$\implies$": Let $A$ be coverage compact and $(U_i)_{i \in I} \subset X$ a open cover of $A$ and $(x_k)_{k \in \mathbb{N}} \subset A$ a sequence. Then there exists a finite subset $K \subset I$ so that $A \subset \bigcup_{i \in K} U_i := \hat{U}$.

If there would exist a subsequence of $x_k$ converging to $x^* \in A$, in each of the finitely many $U_i$ for $i \in K$ there could only be finitely many members of $x_k$ since otherwise there would be a converging subsequence. But this is a contradiction to the fact that $x_k$ has infinitely many members.

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No, it is not correct:

  1. You use the symbol $k$ for two different things.
  2. Why is it that $U_k$ cannot have infinitely many $x_k$'s?
  3. What are the $U_k$'s?
  4. The set $\{x_k\,|\,k\in\mathbb N\}$ may well be finite, in spite of the fact that there are infinitely many $k$'s.
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  • $\begingroup$ For 2: If a subsequence would converge to an $x^* \in U_{\ell}$ for an $\ell \in K$, then in every neighbourhood of $x^*$, there have to be infinitely many members of the subsequence, right? $\endgroup$ – Viktor Glombik Mar 12 '19 at 14:37
  • $\begingroup$ Right, in the sense that thare are infinitely many $k$'s such that $x_k\in U_l$. But you can have $k\neq k'$ and $x_k=x_{k'}$. $\endgroup$ – José Carlos Santos Mar 12 '19 at 15:02
  • $\begingroup$ Unfortunately, I still don't see why this would ruin my proof: Sure, there can be members of the sequence, which are identical, for example, you could have every even member be one and each odd by 0. But then there would exist two convergent subsequences. Even if the the sequence would iterate over the first $n$ numbers for some finite $n$, the would still be a convergent subsequence. Even if the sequence would, in some order, be equal to every point in the set, you could still find a point whose every neighbourhood contains infinitely many members of the sequence, so a convergent subsequence. $\endgroup$ – Viktor Glombik Mar 13 '19 at 0:21
  • $\begingroup$ You are right, I accidentally put $\implies$ instead of $\impliedby$ and edited accordingly. $\endgroup$ – Viktor Glombik Mar 13 '19 at 13:45

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