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. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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