Vector spaces with no Complete norms Can anyone give me an example of a vector space $V$ such that there is no norm which is complete on $V$? 
 A: Space $\mathcal{P}$ of all polynomials in one variable is not complete with respect to any norm as we know that dimension of an infinite-dimensional Banach space is at least $\mathfrak{c}$.
Edit: Proof of the fact that infinite dimensional vector space has dimension at least $\mathfrak{c}$
If not, then dim $X$ $< \mathfrak{c}$.
So, $\exists$ a countable basis of $X$, say {$v_n$: n $\in \mathbb{N}$}.
Let $X_n$ = span{$v_i$ : i=1,2,...,n}.
Clearly, $X$= $\bigcup_{n \in \mathbb{N}}$  $X_n$.
$X_n$ is finite dimensional subspace of $X$, hence closed.
Claim: $\mathring{X_n}$ = $\emptyset$.
If not, then $\exists$ x $\in$ $\mathring{X_n}$.
Thus, $\exists$ r $>$ 0 such that $B(x,r)$ $\subseteq$ $X_n$.
Let z $\in$ $X$ and y = $\frac{r}{2 \Vert z \Vert}$ z.
Then $\Vert y-x \Vert$ =  $\frac{r}{2} < r$.
So, y $\in B(x,r) \subseteq$ $X_n$.
Now, z=$\frac{2 \Vert z \Vert}{r}$ (y-x) $\in$ $X_n$ (since $X_n$ is a subspace).
Hence, $X$ $\subseteq$ $X_n$.
Thus, $X$ = $X_n$ (which is a contradiction since $X$ is infinite dimensional).
Now, $\mathring{\overline{X_n}}$ = $\emptyset$.
Therefore, $X_n$ is nowhere dense.
Thus, $X$ is countable union of nowhere dense sets (which contradicts Baire Category Theorem).
Hence, dim $X$ $\leq \mathfrak{c}$.
A: I assume you want $V$ to be a vector space over $\mathbb{R}$ or $\mathbb{C}$.  Since the latter can be viewed as a special case of the former, let's say $V$ is an $\mathbb{R}$-vector space to fix ideas.
Hints:
1) A finite-dimensional subspace is always closed.
2) A proper closed subspace is nowhere dense.
3) There is a famous theorem about countable unions of nowhere dense subsets of a complete space.
Addendum: I decided that my comment calling for a lack of other answers after having already given an answer myself was something of a conflict of interest, so I have made this answer CW.  
