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!


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.

  • $\begingroup$ That's true. If a subspace is dense $G_\delta$ then so is any translate, but then taking an intersection contradicts Baire's theorem. $\endgroup$ – Sean Eberhard Mar 24 '13 at 15:39
  • $\begingroup$ What intersection do we take? $\endgroup$ – Henno Brandsma Mar 24 '13 at 15:42
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Mar 24 '13 at 15:45
  • 1
    $\begingroup$ 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. $\endgroup$ – Sean Eberhard Mar 24 '13 at 15:46
  • $\begingroup$ 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. $\endgroup$ – 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!

  • 2
    $\begingroup$ $(0,1)$ is a vector space? $\endgroup$ – Robert Israel May 13 '16 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.