Suppose you have a Noetherian topological space $X$, and its finitely many irreducible components are $X_1,\dots,X_n$. If $U\subseteq X$ is open, why are the irreducible components of $U$ precisely the intersections $U\cap X_i$ which aren't empty?
The thing that's tripping me up is I don't see how to go from components in $U$ to those in $X$. For suppose $U\cap X_i=(U\cap Z_1)\cup(U\cap Z_2)$, where $Z_1,Z_2$ are closed in $X$. To show $U\cap X_i$ is irreducible, it has to be either $U\cap Z_1$ or $U\cap Z_2$. If not, there exist points $x,y\in U$ where $x\in X_i\setminus Z_1$ and $y\in X_i\setminus Z_2$. My hunch says there should be a way to go up and find a proper decomposition of $X_i$ in $X$ as a union of closed sets to get a contradiction.
I run into the same issue trying to show $U\cap X_i$ is maximal among irreducible sets in $U$.
What's the right way to see this property?
Best Answer
Let $X$ be any topological space. The irreducible components of $X$ meeting $U$ correspond to the irreducible components of any open subset $U$ of $X$, via intersecting with $U$.
In this paragraph, we will give a bijection between irreducible closed subsets of $X$ meeting $U$ and the irreducible closed subsets of $U$.
In this paragraph, we will show that the bijection above maps components to components in both directions.