Proposition: Let $(X,d)$ be a metric space and $\lbrace x_n \rbrace_{n=1}^\infty \subset X$ be a convergent sequence with $x_n \rightarrow x_0, n \rightarrow \infty$. Show that $K = \lbrace x_n \mid n \in \mathbb{N} \cup \lbrace 0 \rbrace \rbrace$ is a compact set.
Question: It is easy to see how to do this with open covers via the standard definition of compactness. What I am wondering is how one would prove this theorem using sequential compactness (since it is equivalent to regular compactness for metric spaces), if it is even possible to do so.
Best Answer
Suppose $\{z_n\}$ is a sequence in $K$.
If the sequence visits infinitely often a point, then we get a convergent subsequence. Thus we can assume the sequence only visits each point a finite number of times (in particular, $K$ is infinite).
Choose $k(1)$ so that $z_{k(1)}$ is in $K'=\{x_n:n>0\}$ and let $p(1)$ be the least integer such that $z_{k(1)}=x_{p(1)}$.
Suppose we have selected $k(n)$ and $p(n)$. Choose $k(n+1)$ to be the least integer $m$ such that $m>k(r)$, $z_{m}\in K'$ and $z_m=x_r$ for some $r>p(n)$; $p(n+1)$ will be the least such $r$.
Note that $z_{k(n)}=x_{p(n)}$, and that $\{x_{p(n)}\}$ is a subsequence of the convergent sequence we started with.