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I have a question about subsequences of a sequence in a metric space. let $X$ be a metric space and $(x_n)$ be a diverging sequence in $X$. Suppose $(x_n)$ has a subsequence with a limit $a \in X.$ Does it hold true that there exists a subsequence of $(x_n)$ such that it has a limit $b \in X$, $b \neq a$?

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    $\begingroup$ No, unless $X$ is compact. For example, the sequence $0,1,0,2,0,3,0,4,\dots$ has only one limit point $0\in\mathbb{R}$ (but of course it has the other limit point $+\infty\in\bar{\mathbb{R}}$ that is not in $\mathbb{R}$). $\endgroup$ – user10354138 Apr 3 at 10:51
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That assertion holds if and only if $X$ is compact. In fact:

  • If $X$ is compact, then every sequence has a convergent subsequence. So, every divergent sequence has a convergent subsequence.
  • If $X$ is not compact, then it has a sequence $(x_n)_{n\in\Bbb N}$ with no convergent sequence. So, if $x\in X$, the sequence$$x,x_1,x,x_2,x,x_3,\ldots$$has a convergent subsequence, but it has no subsequence that converges to a different limit.
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