# How to show this set is irreducible

Proposition. Let $$X$$ be a topological space and $$U\subseteq X$$ an open set of $$X$$. If $$Z$$ is a closed irreducible set of $$X$$ that meets $$U$$ then $$Z\cap U$$ is a closed irreducible set of $$U$$.

Note. Irreducibility is an intrinsic property that does not depend on ambient space (irreducible in $$U$$ is the same as irreducible in $$X$$)

I'm having trouble writing a proof of this: $$Z\cap U$$ is clearly a closed set of $$U$$ but none of the equivalent definitions of irreucibility that i know of seem to work, for example if $$Z\cap U=F\cup G$$ with $$F,G$$ closed in $$Z\cap U$$ then $$Z\cap U=(Z\cap U)\cap (F^*\cup G^*)$$ for some $$F^*,G^*$$ closed in $$Z$$ and thus in $$X$$. I don't know how to continue from here, no rearranging of the last expression works. Why does $$U$$ need to be open?

Thanks

• To see why $U$ needs to be open, let $Z$ be the parabola $y=x^2$, which is obviously irreducible, and intersecting $Z$ with a line parallel to $x$-axis (but not the $x$-axis itself) gives two points. – user10354138 Jun 10 at 15:19
• @user10354138 i dont see how that helps, im just starting with algebraic geometry. Can you explain it please? – Pedro Jun 10 at 15:30

If $$Z\cap U=(F_1\cap U)\cup(F_2\cap U)$$ with $$F_1,F_2$$ closed sets in $$X$$, then $$Z\subseteq F_1\cup F_2\cup (Z\cap U^c)$$. Since $$Z\cap U\neq\varnothing$$, the closed set $$Z\cap U^c$$ is not the whole of $$Z$$. Thus irreducibility of $$Z$$ implies either $$F_1\supseteq Z$$ or $$F_2\supseteq Z$$, hence either $$F_1\cap U$$ or $$F_2\cap U$$ must equal $$Z\cap U$$.