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Let $X$ be a metric space, and $A\subset X$. Show that $\bar A$ is compact iff for every sequence in $A$ there exists a subsequence that converges to a point in $X$. I showed the "forward" direction, but am stuck showing the reverse.

Let $(x_n)\subset \bar A$ be a sequence. if $A$ is closed, then $A=\bar A$ and if a sequence in a closed set converges, it converges to an element of the set. But by assumption every sequence in $A$ has a convergent subsequence (to an element in $A$), and the result follows.

Thing is, what happens if $A$ is not closed? Choose a sequence in $\bar A$ such that every element of the sequence is not in $A$, i.e. $(x_n)\subset\partial A \setminus A$. What guarantees us that it has a convergent subsequence? Thanks!

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    $\begingroup$ Notice that the clause after the "iff" says "for every sequence in $A$...". It does not say "for every sequence in $\overline A$..." as you seem to have read it. $\endgroup$ – Lee Mosher Dec 10 '16 at 14:29
  • $\begingroup$ I'm not sure what you refer your comment to, but this is indeed the problem. I said that if A is closed then the result follows immediately, but if not I don't understand how the fact that every sequence in A converges to $x \in X$ helps, since I can choose a sequence in $\bar A$, which I want to show is compact, that is entirely not in $A$. $\endgroup$ – Yoni Dec 10 '16 at 14:40
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To prove the reverse direction, assume we know that every sequence in $A$ has a convergent subsequence. Let's say your method for proving compactness of $\overline A$ is to start with a sequence $(x_n)$ in $\overline A$ and show that it has a convergent subsequence.

The strategy is to replace the sequence $(x_n)$ with a closely related sequence $(y_n)$ in $A$. Using that $x_n \in \overline A$, pick $y_n \in A \cap B(x_n,2^{-n})$. From the assumption, some subsequence of $(y_n)$ converges to a point $p$. Since $(y_n)$ is a sequence in $A$, it follows that $p \in \overline A$. For any $\epsilon>0$, the ball $B(p,\epsilon/2)$ contains infinitely many terms of the sequence $y_n$. In particular there exists arbitrarily large values of $n$ such that $y_n \in B(p,\epsilon/2)$ and $2^{-n} < \epsilon/2$, and so $x_n \in B(p,\epsilon)$. This shows that some subsequence of $(x_n)$ converges to $p$.

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  • $\begingroup$ What is r that you mentioned? $\endgroup$ – Gitika Oct 19 at 15:09
  • $\begingroup$ Oops, that was a typo, which I fixed. Thanks. $\endgroup$ – Lee Mosher Oct 19 at 15:41
  • $\begingroup$ U r welcome. Also I wanted to know that if $p \in \bar A$ then every open ball centered at p will contain infinite elements of $A$. Why does it contain infinite elements of $(y_n)$ $\endgroup$ – Gitika Oct 19 at 15:44
  • $\begingroup$ Because, as stated, some subsequence of $(y_n)$ converges to the point $p$. $\endgroup$ – Lee Mosher Oct 19 at 15:52
  • $\begingroup$ Oh okay okay..Could you please explain why $x_n$ is in B(p,$\epsilon$)? $\endgroup$ – Gitika Oct 19 at 16:02
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Hint: if $(x_n)\subset \overline A$, we can find a sequence $(y_n) \subset A$ such that for all $n$, $d(x_n,y_n)\le 1/n$.

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