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$(E,d)$ be a metric space and $(x_n)_{n\in\mathbb N}\subseteq E$ be relatively compact.

Assume that there is a $x\in E$ such that every convergent subsequence converges to $x$. Does it already follow that $(x_n)_{n\in\mathbb N}$ is convergent and converges to $x$?

The convergence of $(x_n)_{n\in\mathbb N}$ to $x$ is equivalent to the assertion that every aribtrary subsequence is convergent and converges to $x$; but we only know that every subsequence, for which we a priori assume that it is convergent, converges to $x$.

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Let the original sequence be $\sigma$, and let $K=\operatorname{cl}\{x_n:n\in\Bbb N\}$; by hypothesis $K$ is compact. If $\sigma$ does not converge to $x$, there are an $\epsilon>0$ and a subsequence $\sigma_1=\langle x_{n_k}:k\in\Bbb N\rangle$ such that $d(x_{n_k},x)\ge\epsilon$ for all $k\in\Bbb N$. Clearly $\sigma_1$ is a sequence in the compact set $K$, so it has a convergent subsequence $\sigma_2=\langle x_{n_{k_i}}:i\in\Bbb N\rangle$. But $\sigma_2$ is also a subsequence of $\sigma$, so by hypothesis it converges to $x$, contradicting the fact that $d(x_{n_{k_i}},x)\ge \epsilon$ for each $i\in\Bbb N$. Thus, $\sigma$ must converge to $x$.

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  • $\begingroup$ I was thinking of counterexamples in $l^2(\Bbb R)$ $\endgroup$ – weierstrash Dec 27 '20 at 6:13
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What about the following example in the interval (0,2): $1,\frac{1}{2},1,\frac{1}{3},1,\frac{1}{4},1,\frac{1}{5},1,\frac{1}{6},1,\frac{1}{7},\dots$

Every convergent subsequence converges to 1, but the entire sequence does not converge

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  • $\begingroup$ the sequence set is not compact. $\endgroup$ – weierstrash Dec 27 '20 at 5:59
  • $\begingroup$ right, fixed (if I understand the correctly the definition of relatively compact, then (0,2) is relatively compact in [0,2] ) $\endgroup$ – Igor Shinkar Dec 27 '20 at 6:06
  • $\begingroup$ it needs to be only relatively compact, i.e. its closure needs to be compact. en.wikipedia.org/wiki/Relatively_compact_subspace $\endgroup$ – Igor Shinkar Dec 27 '20 at 6:08
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    $\begingroup$ the subsequence ${\frac{1}{n}}$ converges to $0$ $\endgroup$ – weierstrash Dec 27 '20 at 6:08
  • $\begingroup$ but 0 is not in the (0,2) $\endgroup$ – Igor Shinkar Dec 27 '20 at 6:08

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