1

I would like to prove that given $X \subset \mathbb{P}^n$ quasi-projective variety, i.e. a locally closed subset, every open and closed subsets of X are quasi-projective varieties.

Let $U\subset X$ be an open subset, then $U\subset X\subset \overline{X}$, so $U$ is a open set within a closed set, so it is locally closed.

Let $A\subset X$ be a closed subset, then $A\subset X$ and I think that if I can prove that I can always find an open set $V$ such that $A\subset V\subset X$ then $A=A \cap \overline{V}$ is a locally closed set. But I am not able to determine if this is right and how to prove it.

Have you any suggestions?

Phi_24
  • 1,114

1 Answers1

0

This is probably not the easiest way to show what you want but here is what I thought about:

For any $x$ in $A$, $x$ is also in $X$ so there is an open $U \in \mathbb{P}^n$ such that $X\cap U$ is closed in $U$ (definition of locally closed subset, eg the equivalency of two definitions of locally closed sets)

Now I claim that $A\cap U$ is closed in $U$ as it is closed in $X\cap U$. Indeed $(X \cap U) \setminus (A\cap U) = (X\setminus A) \cap U$ is open in $X\cap U$ as it is open in $X$ by definition of subset topology.

Dominique Mattei
  • 890