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?

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    $\begingroup$ The Zariski topology does not make $\mathbb{R}^n$ a topological vector space. Addition is not continuous. $\endgroup$ – Eric Wofsey Oct 6 '20 at 3:57
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    $\begingroup$ In fact, every non-Hausdorff topological vector space has the form of the example in your answer: it is a product of a Hausdorff topological vector space with an indiscrete vector space. $\endgroup$ – Eric Wofsey Oct 6 '20 at 4:00
  • $\begingroup$ @EricWofsey: (#1) Oops. I should have known better --- this sort of thing is why we have schemes. My algebraic geometry is getting rusty, I guess. Thanks. $\endgroup$ – Jacob Manaker Oct 7 '20 at 5:33
  • $\begingroup$ @EricWofsey: (#2) That sounds like it would make an interesting answer.... $\endgroup$ – Jacob Manaker Oct 7 '20 at 5:34

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.)


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