TITLE: Finite dimensional topological vector space non complete
QUESTION [5 upvotes]: I know that all finite-dimension Hausdorff topological vector spaces are complete.
I would like to know if there is a example of a topological vector space (over the real or complex numbers) that is not Hausdorff and is also not complete.
I tried to build a counter example, but I couldn't. First, I tried to put the topology generated by the basis $\mathscr{B}=\{(-a,a)\times \mathbb{R}: a>0\}$ in $\mathbb{R}^2$, but this space is complete. Then I tried to take a known example of non-complete space, so I chose the product topology with infinite copies of $\mathbb{R}$, but this space has no finite dimension and is Hausdorff.

REPLY [2 votes]: Denote by $\mathscr{F}_E(y)$ be the filter of neighborhoods of a point $y$ in the TVS $E$.
Let $V$ be a finite dimensional TVS and denote with $N$ the closure of $\{0\}$, this is again a vector sub-space as it is the closure of a vector sub-space. Further, every open neighbourhood of $0$ contains all of $N$.
Now $V/N$ is a quotient of a topological group by a closed sub-group, hence it is Hausdorff space. In other words $V/N$ is a Hausdorff TVS, since it is also finite dimensional it is complete.
So let $\mathscr{F}$ be a Cauchy filter in $V$. By continuity of the projection map we have that $\phi(\mathscr{F})$ is a Cauchy filter in $V/N$, hence admits a limit $\overline{x}$ by completeness. We will now see that any lift $x$ of $\overline{x}$ is a limit of $\mathscr{F}$ in $V$.
Let $Z \in \mathscr{F}_{V/N}(\overline{x})$, since $\mathscr{F}_{V/N}(\overline{x}) \subset \phi(\mathscr{F})$, there exists $F \in \mathscr{F}$ such that $\phi(F) \subset Z$. Therefore, $F \subset \phi^{-1}(\phi(F)) \subset \phi^{-1}(Z)$, which implies that $\phi^{-1}(Z) \in \mathscr{F}$.
We claim that $\mathscr{F}_V(x) \subset \mathscr{F}$. Let $U+x \in \mathscr{F}_V(x)$ and consider $W \in \mathscr{F}_V(0)$ such that $W+W \subset U$. Define $Z=\phi(W+x)=\phi(W)+\overline{x} \in \mathscr{F}_{V/N}(\overline{x})$. As we have seen  $\phi^{-1}(Z) \in \mathscr{F}$, that is, $W+x+N=\phi^{-1}(\phi(W+x)) \in \mathscr{F}$. Observing that $W+x+N \subset W+x+W \subset U+x$, we obtain $U+x \in \mathscr{F}$ as desired.