# Is the projectivization of a topological vector space Tychonoff?

Let $$E$$ be a locally convex topological vector space over $$\mathbb{R}$$. The projectivization $$PE$$ is the quotient of $$E\backslash\{0_{E}\}$$ with respect to the equivalence relation $$e\sim f$$ if $$e=\lambda f$$.

Is $$PE$$ a Tychonoff (i.e. completely regular Hausdorff) space?

As far as I can tell, the theorems about the quotient uniform spaces do not apply. On the other hand, it is plausible to expect that this is a known fact.

I can show that $$PE$$ is completely Hausdorff, i.e. any two points can be separated by a real-valued continuous function. Indeed, if $$e\not\sim f$$, take $$\mu,\nu\in E^{*}$$ such that $$\left<\mu,e\right>=1, \left<\nu,e\right>=0, \left<\mu,f\right>=0, \left<\nu,f\right>=1$$, and consider the map $$\mu\oplus \nu:E\to \mathbb{R}^2$$. By the definition of the quotient, this map induces a map $$\varphi: PE\to P\mathbb{R}^2=S^1$$. Since the latter is Tychonoff, we can separate the images of the classes of $$e$$ and $$f$$ by a continuous function.

• I am afraid that your proof of the complete Hausdorffness can contain a gap because for the locally convex space $\mathbb R^\omega$ the projective space is not Urysohn: for any sequence of non-empty open sets $U_1,\dots,U_n$ in $P\mathbb R^\omega$ the intersection $\overline U_1\cap\dots\cap\overline U_n$ is not empty. The fact that the projective space of $\mathbb R^\omega$ is not Urysohn (and hence not completely Hausdorff) has been first noticed by Gelfand and Fuks, the reference to their paper can be found in: doi.org/10.1016/j.topol.2021.107909 Nov 19, 2021 at 21:04
• @TarasBanakh thank you, you are right, and here is the gap: the map $\mu\oplus\nu$ does not work because it has a non-trivial kernel. Perhaps you could post your comment as an answer? Also, are you sure that the closures are needed? take the pre-images of $U_i$'s, they contain "shifted" subspaces of finite co-dimension, which necessarily intersect
– erz
Nov 20, 2021 at 6:53
• The closures are needed because the projective spaces are Hausdorff. Nov 20, 2021 at 7:24

The projective space $$PE$$ of a topological vector space $$E$$ is Hausdorff but in general is not Tychonoff, not functionally Hausdorff and even not Urysohn (let us recall that a topological space is Urysohn if any distinct points have disjoint closed neighborhoods).
As a suitable counterexample, consider the countable product of lines $$E=\mathbb R^\omega$$. The projective space $$P\mathbb R^\omega$$ is superconnected in the sense that for any non-empty open sets $$U_1,\dots,U_n$$ in $$P\mathbb R^\omega$$ the intersection of their closures $$\overline U_1\cap\dots\cap\overline U_n$$ is not empty. This pathological property of the projective space $$P\mathbb R^\omega$$ was first noticed by Gelfand and Fuks in 1967.
A countable counterpart of the projective space $$P\mathbb R^\omega$$ is the projective space $$\mathbb QP^\infty$$, whose topology has been characterized in this paper.