Total Boundedness is not Preserved under Homeomorphism

Theorem

Let $M = \struct {A, d}$ be a totally bounded metric space.

Let $M' = \struct {A', d'}$ be a metric space.

Let $M$ be homeomorphic to $M'$.


Then it is not necessarily the case that $M'$ is totally bounded.


Proof


This theorem requires a proof.
In particular: According to S&S (citation below), this is proved somehow using the metric $\delta = \dfrac d {1 + d}$, given some metric space $M = \struct {A, d}$, but the derivation of this is obscure.
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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
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