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