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

Theorem: 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 $ϵ$-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 for all $k$. Since the $S_k$ are nested, we have that whenever $i,j\ge k$, then $n_i,n_j\in S_k$. Thus for all $i,j\ge k$, $p_{n_i}$ and $p_{n_j}$ are both contained in a ball of radius $1/k$. Hence the sequence $p_{n_k}$ is Cauchy.

Title	proof that a metric space is compact if and only if it is complete and totally bounded
Canonical name	ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded
Date of creation	2013-03-22 18:01:06
Last modified on	2013-03-22 18:01:06
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	5
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 40A05
Classification	msc 54D30