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.



Is a normed space which is homeomorphic to a Banach space complete?
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.


