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?


  • $\begingroup$ 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. $\endgroup$ – user10354138 Jun 10 at 15:19
  • $\begingroup$ @user10354138 i dont see how that helps, im just starting with algebraic geometry. Can you explain it please? $\endgroup$ – 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 $Z\cap U$.


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