# why must a normed space homeomorphic to a complete metric space be complete?

Why must a normed space X homeomorphic to a complete metric space Y be complete?

I've solved this in the case where Y is a normed space (considering open balls gives Lipschitz equivalence), but am not sure how to proceed in general. Can we use an adapted open-balls argument even when Y isn't normed?

Maybe there's an equivalent condition for completeness that will come in useful? In a normed space we can use (absolutely convergent)$\implies$(convergent) for series, but this won't work in a metric space.

I've tried using an explicit homeomorphism $f:X\rightarrow Y$ and taking pullbacks and pushforwards along $f$ of a Cauchy sequence in $X$, but without success.

Many thanks for any help with this!

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I don't know if this works, but some thoughts:

It is well known that a metric space $(X,d_X)$ can be homeomorphic to a complete metric space $(Y,d_Y)$ and the metric $d_X$ on $X$ is not complete. E.g. $X = (0,1)$ in the standard metric, compared with $Y = \mathbb{R}$ in the standard (complete) metric. However, for such spaces $X$ it is the case that they are a $G_\delta$ in their completion (they are what is known as topologically complete).

Every normed space $(X,\|\cdot\|)$ has a normed completion $(\hat{X},\|\cdot\|)$ in which $X$ embeds isometrically. So we know from the above that $X$ is then a $G_\delta$ in $\hat{X}$. We want to show that these spaces are actually equal.

So maybe it is possible to show that a dense $G_\delta$ linear subspace of a Banach space is the whole space?

Added (after comments by Sean Eberhard) This is indeed the case: $X$ is a dense $G_\delta$ in $\hat{X}$, so if $y \in \hat{X}\setminus X$, $y + X$ and $X$ are both dense $G_\delta$'s and their intersection is empty (otherwise $y$ would a difference of members of $X$ hence in $X$). This contradicts Baire's theorem, so such a $y$ cannot exist and $X = \hat{X}$, and thus is complete already.

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That's true. If a subspace is dense $G_\delta$ then so is any translate, but then taking an intersection contradicts Baire's theorem. – Sean Eberhard Mar 24 '13 at 15:39
What intersection do we take? – Henno Brandsma Mar 24 '13 at 15:42
I just found that the Banach-Kuratowski-Pettis theorem applies: this says that if $X$ is a non-meagre topological group and $A$ is a subgroup that is a Baire space, $A$ is either meagre in $X$ or clopen in $X$. But this seems overkill, so anything simpler would be welcome. – Henno Brandsma Mar 24 '13 at 15:45
If $X$ is a proper dense $G_\delta$ subspace then so is $x+X$ for any $x\notin X$. But $X\cap(x+X)=\emptyset$, a contradiction to Baire's theorem. – Sean Eberhard Mar 24 '13 at 15:46
Nice answer! And @SeanEberhard: Nice observation! I had also got as far as realizing that $X$ was a dense $G_\delta$ in its completion, but couldn't see how to proceed. – Nate Eldredge Mar 24 '13 at 17:30

This question is not true: $\mathbb{R}$ is homeomorphic to $(0,1)$. $\mathbb{R}$ is a complete metric space whit Euclidean norm and $(0,1)$ is a normed vector space which isn't complete!

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$(0,1)$ is a vector space? – Robert Israel May 13 at 19:31