# Convergent Sequence + Limit is Compact using Sequential Compactness

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.

• If a sequence in $K$ only visits a finite number of points, it certainly has a convergent (constant) subsequence. If it visits an infinite number of points, build a convergent subsequence. – egreg Nov 11 '17 at 23:41
• @egreg it was building that convergent subsequence specifically and showing that it converged that I was having trouble with when I attemped this earlier :) – Foobanana Nov 12 '17 at 1:52
• @egreg mostly the building part :P – Foobanana Nov 12 '17 at 1:53

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.