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.
Best Answer
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.