Metric Space Compact iff Complete in All Equivalent Metrics
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Theorem
Let $M_1 = \left({A, d_1}\right)$ be a metric space.
Then $M_1$ is compact iff $M_2 = \left({A, d_2}\right)$ is a complete metric space whenever $d_2$ is equivalent to $d_1$.
Proof
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Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$: Complete Metric Spaces
