Sequentially Compact Metric Space is Complete
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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