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Let $(X,d)$ be a metric space and consider on $X$ the topology induced by the metric $d$.

Let $\{x_n\}_{n\in \mathbb{N}}$ be e sequence of points in $X$. Is it true that if there exists $x\in X$ such that $d(x,x_n)\rightarrow \infty$ then $\{x_n\}_{n\in \mathbb{N}}$ can not be contained in any compact subset of $X$ (for the topology induced by $d$)?

The answer is yes if compact sets in $(X,d)$ must be bounded, but I don't know if it's true.

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If $d(x,x_n)$ is unbounded, then $\{B(x, n): n \in \mathbb{N}\}$ is an open cover of $(X,d)$ without a finite subcover (A finite subcover would reduce to the largest radius ball and this misses many $x_n \in X$)

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Yes, compact subsets of a metric space must be bounded. In fact, a metric space is compact if and only if it is complete and totally bounded, which is stronger than being bounded.

To see that boundedness is necessary, note that compactness and sequential compactness are the same for metric spaces. In an unbounded space you can construct a sequence $(x_n)$ such that $d(x_i, x_j) > 1$ for all $i \ne j$. Such a sequence cannot have a convergent subsequence.

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  • $\begingroup$ An unbounded sequence can have a convergent subsequence. Take the sequence $(1, 1, 1, 2, 1, 3, 1, 4, 1, 5, \ldots)$, then take the subsequence of odd indexed terms. It's constantly one. But the even indexed terms clearly are unbounded. $\endgroup$ – David Bowman Mar 16 '17 at 20:16
  • $\begingroup$ You're right, I misspoke. Now fixed. $\endgroup$ – Nathaniel Mayer Mar 16 '17 at 20:18

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