# proof that a metric space is compact if and only if it is complete and totally bounded

A metric space is compact if and only if it is complete and totally bounded.

Proof.  Let $X$ be a metric space with metric $d$. If $X$ is compact, then it is sequentially compact and thus complete. Since $X$ is compact, the covering of $X$ by all $\epsilon$-balls must have a finite subcover, so that $X$ is totally bounded.

Now assume that $X$ is complete and totally bounded. For metric spaces, compact and sequentially compact are equivalent; we prove that $X$ is sequentially compact. Choose a sequence $p_{n}\in X$; we will find a Cauchy subsequence (and hence a convergent subsequence, since $X$ is complete).

Cover $X$ by finitely many balls of radius $1$ (since $X$ is totally bounded). At least one of those balls must contain an infinite number of the $p_{i}$. Call that ball $B_{1}$, and let $S_{1}$ be the set of integers $i$ for which $p_{i}\in B_{1}$.

Proceeding inductively, it is clear that we can define, for each positive integer $k>1$, a ball $B_{k}$ of radius $1/k$ containing an infinite number of the $p_{i}$ for which $i\in S_{k-1}$; define $S_{k}$ to be the set of such $i$.

Each of the $S_{k}$ is infinite, so we can choose a sequence $n_{k}\in S_{k}$ with $n_{k} for all $k$. Since the $S_{k}$ are nested, we have that whenever $i,j\geq k$, then $n_{i},n_{j}\in S_{k}$. Thus for all $i,j\geq k$, $p_{n_{i}}$ and $p_{n_{j}}$ are both contained in a ball of radius $1/k$. Hence the sequence $p_{n_{k}}$ is Cauchy.

## References

• 1 J. Munkres, Topology , Prentice Hall, 1975.
Title proof that a metric space is compact if and only if it is complete and totally bounded ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded 2013-03-22 18:01:06 2013-03-22 18:01:06 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 40A05 msc 54D30