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I have a normed space $(E,||\cdot||)$ which is homeomorphic (as a topological space) to a Banach space $F$.

Does this imply that $(E,||\cdot||)$ is also a Banach space?

I think I read something like this to be true if $E$ (and therefore also $F$) is separable, but I am not totally sure. So, also this special case would be interesting.

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  • $\begingroup$ I was about to make the following answer: in a topological group, there is a natural uniform structure (Bourbaki, TG, III, §3), so in particular, in a topological vector space, topology determines a uniform structure; now completeness is a uniform property (viz., all Cauchy ultrafilters are convergent), so in topological groups it must be preserved by homeomorphisms. But now I've completely confused myself as to whether this answer is correct or not! $\endgroup$
    – Gro-Tsen
    Dec 7, 2016 at 23:13
  • $\begingroup$ @Gro-Tsen: No, I dont think this argument works since my homeomorphism need not be a group isomorphism, so it need not be a morphism of the uniform spaces, in particular the complete topological vector space $\mathbb R$ is homeomorphic to the noncomplete uniform space $ \left] 0 , 1 \right[$ which is not complete ... $\endgroup$
    – Neslihan
    Dec 8, 2016 at 13:06

2 Answers 2

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Let $\bar{E}$ be the norm completion of $E$, which is a Banach space. Then we can consider $E$ as a dense linear subspace of $\bar{E}$, where the subspace topology and the norm topology on $E$ coincide. In particular, since this topology is homeomorphic to $F$, it is completely metrizable, so $E$ is a $G_\delta$ in $\bar{E}$ (Kechris, Classical Descriptive Set Theory, Theorem 3.11). As a dense $G_\delta$, in particular $E$ is comeager in $\bar{E}$. If $x \in \bar{E} \setminus E$, then $E, E+x$ are disjoint comeager subsets of $\bar{E}$, which is absurd by the Baire category theorem. So $E = \bar{E}$ and thus the norm on $E$ is complete.

I think I did not use separability anywhere.

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    $\begingroup$ You can skip the last three lines and say: if $\exists x\in\overline{E}\setminus E$, then $E\cap(E+x)=\emptyset$, contradicting Baire's theorem (as $E$ and $E+x$ are dense $G_\delta$ subsets of $\overline{E}$). $\endgroup$
    – Mizar
    Dec 7, 2016 at 15:03
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    $\begingroup$ So the conclusion is also true if $F$ is a complete metric space (not necessarily a Banach space), isn't it? $\endgroup$ Dec 7, 2016 at 15:35
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    $\begingroup$ @PietroMajer: Yes, indeed. That makes sense because the homeomorphism from $E$ to $F$ doesn't necessarily preserve the linear structure, so we shouldn't expect the linear structure on $F$ to be important. $\endgroup$ Dec 7, 2016 at 16:56
  • $\begingroup$ Thanks a lot! This should work for metrizable abelian groups as well, with the same proof, right? $\endgroup$
    – Neslihan
    Dec 8, 2016 at 9:53
  • $\begingroup$ @Neslihan: I would think so, but you should check the details - I'm less familiar with that case. In particular, you need that the metric completion of $E$ is again a group, so perhaps you need the metric to be translation invariant. $\endgroup$ Dec 8, 2016 at 15:29
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It is an old result of Victor Klee (answering a question of Banach) that a metrizable topological vector space (i.e., there is a translation invariant metric giving the topology) is a complete topological vector space (i.e., w.r.t. the uniformity induced by the $0$-neighborhoods) if there is some complete metric (not necessarily translation invariant) giving the same topology.

The reference is: Victor Klee, Invariant metrics in groups (solution of a problem of Banach), Proc. Amer. Math. Soc. 3, 484 - 487 (1952).

The proof (which is quite similar to the arguments in Nate's answer) can also be seen in Koethe's book Topological Vector Spaces I §15.11.

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