TITLE: A $T_0$ topological vector space is Hausdorff
QUESTION [7 upvotes]: I know a $T_0$ topological group is $T_1$, and if we have a topological vector space, there should be a way of using scalar multiplication to get disjoint neighborhoods. Right?

REPLY [4 votes]: As suggested in the comments, a topological group which is $T_1$ is already Hausdorff. Here is a proof of this fact.
Remember a space is Hausdorff if and only if the diagonal is closed in the product space. Let $X$ be our space, with operation $\cdot$.
Define $f \colon X \times X \to X$ by $f(x,y)=x\cdot y^{-1}$. Such map is continuous and since the space is $T_1$, $\{\operatorname{id}\}$ is a closed set. The preimage under $f$ of the set $\{\operatorname{id}\}$ is the diagonal, which is closed by continuity, hence $X$ is Hausdorff.