Is there an easy example of a vector space which can not be endowed with the structure of a Banach space Let $V$ be a real vector space. 
Is there always a norm on $V$ such that $V$ is complete with respect  to this norm?
If not, is there an easy counterexample?
 A: $\mathbb{R}^\mathbb{N}$ in the product topology is an example. It's a completely metrisable topological vector space, but there can be no compatible norm because all neighbourhoods of 0 are unbounded.
A: Let $\kappa$ be an infinite cardinal.
If $B$ is a Banach space such that the least size of a dense subset of $B$ is $\kappa$, then $B$ is of size $\kappa^{\aleph_0}$.
So the only possible sizes of Banach spaces are powers with the exponent $\aleph_0$.
As was pointed out above, every infinite dimensional Banach space has (Hamel) dimension at least $2^{\aleph_0}$.
If a Banach space has size $>2^{\aleph_0}$, then its size is actually equal to the dimension.  It follows that an infinite cardinal can only be the dimension of a Banach space if it is of the form $\kappa^{\aleph_0}$.
But there are many cardinals that are not of that form and for every cardinal $\kappa$ there is a vector space of dimension $\kappa$.  
The first infinite cardinal not of the form $\kappa^{\aleph_0}$ is $\aleph_0$, 
as was pointed out above.  The next cardinal is $\aleph_1$, which is of the form 
$\kappa^{\aleph_0}$ iff the continuum hypothesis holds.
The only cardinals for which we can say for sure that they are not of the form $\kappa^{\aleph_0}$ are suprema of increasing chains of cardinals of countable length, like $\aleph_\omega$, the sup of the $\aleph_n$.
On the other hand, any two vector spaces are isomorphic iff they have the same dimension.  Also, there are vectorspaces of all dimensions.
So, the question whether a vector space has a norm that turns it into a Banach space really only asks which cardinals are dimensions of Banach spaces.
For every cardinal $\kappa$, $\ell^2(\kappa)$ is a Banach space (even Hilbert!) of density $\kappa$ and dimension $\kappa^{\aleph_0}$.
It follows that an infinite dimensional vector space is isomorphic to a Banach space if and only if its dimension is of the form $\kappa^{\aleph_0}$ for some $\kappa$.  (Actually, $(\kappa^{\aleph_0})^{\aleph_0}=\kappa^{\aleph_0}$, so $\lambda$ is of the form $\kappa^{\aleph_0}$ iff $\lambda^{\aleph_0}=\lambda$.)
A: No. Suppose $V$ is a normed space of (Hamel) dimension $\aleph_0$, with basis $\{v_1, v_2, v_3, \ldots \}$ say. Then $V$ is the union of a countable family of finite-dimensional subspaces, namely $\langle v_1 \rangle, \langle v_1, v_2 \rangle, \langle v_1, v_2, v_3 \rangle, \ldots$. Proper closed subspaces of a normed space are nowhere dense, so $V$ is a countable union of nowhere dense sets, and so is incomplete by the Baire category theorem.
