Sequentially Compact Metric Space is Complete

Theorem

Let $M = \struct {A, d}$ be a metric space which is sequentially compact.

Then $M$ is complete.

Proof

Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $A$.

As $M$ is sequentially compact, $\sequence {x_n}$ has a convergent subsequence.

By Convergent Subsequence of Cauchy Sequence in Metric Space, this implies that the entire sequence $\sequence {x_n}$ is convergent.

Hence, $M$ is complete.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces