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It is well known that completeness is not a property maintained by homeomorphism ($\cong$) between metric spaces.

Let $\mathcal{C}$ denote the class of all complete metric spaces satisfying that for all $X\in\mathcal{C}$ and $Y$ metric space, if $X \cong Y$ then $Y$ is also complete. Class $\mathcal{C}$ is the largest class of complete metric spaces that is closed under homeomorphism.

Let $\mathcal{C_0}$ denote the class of all compact metric spaces. We know that $\mathcal{C_0}\subseteq \mathcal{C}$.

Is it the case that $\mathcal{C_0}=\mathcal{C}$?

Anguepa
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1 Answers1

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Yes, as any non-compact metric space has an equivalent (same topology yielding) metric that is not complete. E.g cf. this post for ideas.

Henno Brandsma
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