# Question on compact sets in metric spaces

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.

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 $x_n \in X$)
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.
• 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. – David Bowman Mar 16 '17 at 20:16