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Let $X$ be a non-empty set. Suppose that $d_1$ and $d_2$ are two possibly different metrics on $X$. Let $\tau_i$ denote the topology generated by the metric $d_i$ ($i\in\{1,2\}$).

The following are known:

  • $\tau_1=\tau_2\equiv\tau$;
  • $(X,d_1)$ is a totally bounded metric space;
  • $(X,d_2)$ is a complete metric space.

Question: Is the topological space $(X,\tau)$ compact?

This is an extra twist added to the classical result that any totally bounded and complete metric space is compact. The twist is that one is given two different metric spaces, one being totally bounded and the other complete, even though they generate the same topology. I wonder whether this twist preserves the compactness of the common topology.


In other words:

  • Any totally bounded and complete metric space is compact, but...
  • ...is any totally bounded and complete$\textit{ly metrizable}$ metric space compact?
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It's not correct. Take $X_1 = (0,1)$ and $X_2 = \mathbb R$ (identified with $X_1$ via some homeomorphism). Then with the standard metric on both $X_i$, $X_1$ is totally bounded and $X_2$ is complete. Both are not compact.

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  • $\begingroup$ I'm a little bit concerned about identification via homeomorphism, as metric properties are usually not topological. (Or is complete metriz$\textit{ability}$ a topological property? I'm not sure.) Can one find explicitly (i) either a totally bounded metric on $\mathbb R$ that generates the usual topology on $\mathbb R$; (ii) or a complete metric on $(0,1)$ that generates the usual topology? Does the homeomorphism between $(0,1)$ and $\mathbb R$ assure this? $\endgroup$ – triple_sec Nov 18 '14 at 13:17
  • $\begingroup$ @triple_sec: I think our example are basically the same... You can identify $(1, \infty)$ with $(0,1)$ with the map $x\mapsto 1/x$. Then your $d_2$ is the "pullback" of the usual metric on $(0,1)$. $\endgroup$ – user99914 Nov 18 '14 at 22:09
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Update: NOT true.

Counterexample: Consider $\mathbb N$ with the discrete topology, which is clearly not compact. Let $d_1$ be the discrete metric, which is complete, and $d_2(n,m)\equiv|1/n-1/m|$. Now, $d_2$ can be shown to be a totally bounded metric. It is also not difficult to check that both $d_1$ and $d_2$ generate the discrete topology on $\mathbb N$.

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