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.

  • $\begingroup$ 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 $\endgroup$ Nov 19, 2021 at 21:04
  • $\begingroup$ @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 $\endgroup$
    – erz
    Nov 20, 2021 at 6:53
  • $\begingroup$ The closures are needed because the projective spaces are Hausdorff. $\endgroup$ Nov 20, 2021 at 7:24

1 Answer 1


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.


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