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It is well-known that up to topological isomorphism there is exactly one Hausdorff topological vector space (say, over $\mathbb{C}$) of a given dimension $n$, namely $\mathbb{C}^n$ with the euclidean topology. What happens if we drop the Hausdorff condition, can we still classify all topological vector spaces? For example, there is one $\mathbb{C}^n_{triv}$ with the trivial (indiscrete) topology, and more generally we have $\mathbb{C}^k_{triv} \times \mathbb{C}^{n-k}$ for $0 \leq k \leq n$. Are these all?

Here is how one might start. If $V$ is a topological vector space of dimension $n$, then we can consider its subspace $K=\overline{\{0\}}$, which has some dimension $k$. The quotient $V/K$ is a topological vector space of dimension $n-k$. But it is also Hausdorff, so that it is isomorphic to $\mathbb{C}^{n-k}$. One checks that $K$ has the trivial topology (if $C \subseteq K$ is closed and $p \in C$, then $C-p \subseteq K$ is closed and contains $0$, so that $C-p=K$ and hence $C=K+p=K$.) But it is not clear a priori if this quotient splits.

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The quotient actually splits, without any assumption on dimensions of the topological vector space $X$.

Let $Y$ be any algebraic complement subspace of $K:=\overline{\{0\}}$, that is the restriction to $Y\times K$ of the addition map is a bijective (linear) map $s:Y\times K\to X$. I claim it is a homeomorphism, hence a TVS isomorphism.

As you said, $K$ has the trivial (aka indiscrete) topology, that is, $K$ is the only nbd of $0$ in $K$. Also, recall that a TVS $X$ ($T_0$ or not) always possesses a basis of closed nbd's of $0$. But if $V$ is a closed nbd of $0$ in $X$, $V=V+K$, since the closure of any $x$ is $x+K$. Then also $s((V\cap Y) \times K)=(V\cap Y) + K=V+K=V$, which means that $s$ is a homeomorphism.

In conclusion, $X$ splits in the TVS direct sum of the (Hausdorff, dense in $X$) subspace $Y$ and the (indiscrete, closed in $X$) subspace $K$. As a consequence, the restriction of the quotient map $\pi:X\to X/K$ to $Y$ is also an isomorphism $\pi_{|Y}:Y\to X/K$.

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    $\begingroup$ Thank you! For the record, here is a proof of the fact you mention that there is a basis of closed nbd's of $0$: Let $U$ be a nbd of $0$. Choose a nbd $V$ of $0$ with $V-V \subseteq U$. Then $\overline{V} \subseteq U$, because for $x \in \overline{V}$ we have $(x + V) \cap V \neq \emptyset$, and hence $x \in V-V \subseteq U$. This works in every topological group. $\endgroup$ – HeinrichD Apr 3 '17 at 18:43

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