Timeline for Is a normed space which is homeomorphic to a Banach space complete?

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Dec 8, 2016 at 15:29	comment	added	Nate Eldredge		@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.
Dec 8, 2016 at 9:53	comment	added	Neslihan		Thanks a lot! This should work for metrizable abelian groups as well, with the same proof, right?
Dec 7, 2016 at 16:56	comment	added	Nate Eldredge		@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.
Dec 7, 2016 at 15:35	comment	added	Pietro Majer		So the conclusion is also true if $F$ is a complete metric space (not necessarily a Banach space), isn't it?
Dec 7, 2016 at 15:12	history	edited	Nate Eldredge	CC BY-SA 3.0	deleted 78 characters in body
Dec 7, 2016 at 15:03	comment	added	Mizar		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 $\overline{E}$).
Dec 7, 2016 at 14:58	vote	accept	Neslihan
Dec 7, 2016 at 14:53	history	edited	Nate Eldredge	CC BY-SA 3.0	deleted 56 characters in body
Dec 7, 2016 at 14:46	history	answered	Nate Eldredge	CC BY-SA 3.0