Theorem N14.1. If is compact ( sequentially compact, for metric spaces), then it is complete, i.e. all Cauchy sequences converge.
Proof. Let be a Cauchy sequence. Then there is a convergent subsequence ; we want to show . Let . Cauchy implies there is such that for all ,. Since the subsequence converges, , so there is such that . Then for all ,∎
Corollary N14.2. with the usual distance metric is a complete metric space.
Proof. Every Cauchy sequence in is bounded, so it is contained in a cube for large, which is compact. Hence it converges. ∎
Theorem N14.3. (real-valued functions) with the uniform metric is also complete.
Proof. Given a Cauchy sequence ,: for all there is such that . So for each , is a Cauchy sequence in , which has a limit , and pointwise.
Claim: uniformly. Given , let be such that . Then for all and all ,
since . So for all ,.∎
If is complete and is closed, then is complete: a Cauchy sequence in converges in , and the limit is in . The continuous functions form a closed subset of :