TITLE: Compactness of a metric space
QUESTION [26 upvotes]: If a metric space $(X,d)$ is compact then for every equivalent metric $\sigma$, $(X,\sigma)$ is complete. This is because, for any Cauchy sequence in $(X,\sigma)$ has a convergent subsequence due to fact $(X,\sigma)$ is a compact metric space, hence original sequence is convergent. My question is, does the converse also hold ?

In other words, let $(X,d)$ be a metric space such that for every equivalent
metric $\sigma$ on $X$ is complete. Does this imply $(X,d)$ is
compact?

REPLY [2 votes]: Here is yet another proof.
Suppose $(X,d)$ is not compact. Since it is a metric space, it follows that there is a sequence $\bar x = (x_n)_{n\in \mathbf N}$ with no convergent subsequence. We will find an equivalent metric $d'$ which makes a subsequence of $\bar x$ a Cauchy sequence, so $d'$ is necessarily incomplete.
We may assume without loss of generality that $\bar x = (x_n)_n$ contains no $d$-Cauchy subsequence (otherwise, $d=d'$ is incomplete). It follows that there is some $\varepsilon>0$ such that for all $n$, there is some $m>n$ such that $d(x_n,x_m)>\varepsilon$. We may assume without loss of generality that $d(x_n,x_m)>\varepsilon$ for all $n,m$ (passing to a subsequence if necessary).
Now, for each $n$, let $B_n$ be the ball around $x_n$ of radius $\varepsilon/2$, and write $X'$ for $X\setminus \bigcup_n B_n$. Now, define a metric $d'$ by the following formula:
$$
d'(x,y) = \begin{cases}
d(x,y) & x,y\in X' \\
d(x_n,y)+\varepsilon/n & x\in B_n, y\in X'\\
d(x,x_n)+\varepsilon/n & x\in X', y\in B_n\\
\varepsilon/\min(m,n) & x\in B_n, y\in B_m, n\neq m \\
d(x,y)/n & x,y\in B_n
\end{cases}
$$
It remains to check that:

$d'$ is a metric (the only nontrivial axiom is the triangle inequality, but it is easily checked by considering all possible cases),
$d'$ is equivalent to $d$ (this is very easy),
$(x_n)_n$ is Cauchy in $d'$ (this is trivial).