Let $\mathbb{P}^n$ be the projective space of dimension $n$ with the Zariski topology. Then the diagonal of $\mathbb{P}^n \times \mathbb{P}^n$ is clearly a closed subset. Since a topological space $X$ is Hausdorff if and only if the diagonal of $X \times X$ is closed, we conclude that $\mathbb{P}^n$ is Hausdorff. In fact, the above argument extends to any quasi-projective variety of $\mathbb{P}^n$.

Question: I find this conclusion strange, since it is known that in general the Zariski topology does not give rise to Hausdorff spaces. Am i missing something here?


The theorem you quote is only valid if $X\times X$ is endowed with the product topology.
But $\mathbb P^n\times \mathbb P^n$ is endowed with another, stronger (=more closed subsets) topology.
(That stronger topology is determined by the requirement that the zero sets of bihomogeneous polynomials in both sets of variables be closed.)

  • $\begingroup$ You are welcome, Manos. $\endgroup$ – Georges Elencwajg Feb 24 '15 at 17:33

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