Does there exists a non-Hausdorff topological vector space with a proper convex open subset?

Question

Wikipedia's page on the Hahn-Banach theorem mentions a technique that converts a non-Hausdorff topological vector space $X$ into a (Hausdorff) locally convex one: apply the weak topology (induced by $X^*$). However, the argument assumes that $X$ contains a proper, convex, open set.

The standard example of a non-Hausdorff topological vector space is $\mathrm{Spec}{(\mathbb{R}[\vec{x}])}$ with the Zariski topology. This does not contain a proper open convex subset.

What is an example of a non-Hausdorff topological vector space to which this technique applies?

Answer 1

Yes, there is! Let $\mathcal{T}$ be the indiscrete topology on $\mathbb{R}$ and consider $X=\mathbb{R}\times(\mathbb{R},\mathcal{T})$. This is a topological vector space (exercise); and $(-1,1)\times\mathbb{R}$ is an example of a proper convex open subset.

In fact one can show more:

1. $X$ has a topology induced by the seminorm $\|(x,y)\|=|x|$.
2. $X$ is locally convex (even without the "weak topology trick").
3. $X^*\cong\mathbb{R}^*\oplus0$; all continuous functionals on $X$ are constant on the second direct factor.

I leave these claims, as well as the construction of an infinite-dimensional example, to the interested reader.

Finally, a general remark: the Zariski topology is much more pathological than one needs to obtain non-Hausdorffness. You should maybe consider re-reading Stein and Seebach.

(This answer inspired by Answer #330413: Is there an example of a, non-Hausdorff, topological vector space which has bounded subspaces.)