TITLE: Finite union of Noetherian topological spaces is Noetherian
QUESTION [6 upvotes]: Suppose $X_1,...,X_n\subset X$, and each $X_i$ is a Noetherian topological space (with the induced topology). Thats is, each of them satisfies the descending chain condition for closed subsets. I'd like to understand why $\bigcup_{i=1}^nX_i$ is also Noetherian (with the induced topology). I saw this result here but without the proof included, and I want to make sure I have the right idea.
Is this just because, if we have a chain of sets $Z_i$ closed in $\bigcup_{i=1}^nX_i$, then each $Z_i$ is closed in each $X_i$ and so we use the descending chain condition for the $X_i$s?

REPLY [6 votes]: You're exactly right -- if you want to show that the descending chain condition holds for $\bigcup_i X_i,$ start with your chain
$$
Y = \bigcup_i X_i\supseteq Z_1\supseteq Z_2\supseteq\dots\supseteq Z_j\supseteq\cdots,
$$
with each $Z_j$ closed in $Y.$ Then for any $i,$ we obtain a chain of closed subsets by intersecting with $X_i$:
$$
X_i \supseteq Z_1\cap X_i\supseteq Z_2\cap X_i\supseteq\dots\supseteq Z_j\cap X_i\supseteq\cdots.
$$
Write $Z_j\cap X_i = Z_{i,j}.$ For each $i,$ the chain $Z_{i,1}\supseteq Z_{i,2}\supseteq\cdots$ terminates: say $Z_{i,r} = Z_{i,r_i}$ for each $r\geq r_i.$
Now, you need to show that this implies that the original chain terminates. Let $m = \max_i\{r_i\}.$ I claim that $Z_m = Z_r$ for any $r\geq m.$ Suppose that $z\in Z_m$ and $r\geq m.$ Then for any $i$ such that $z\in X_i$ (some such $i$ must exist since $Y = \bigcup_i X_i$), we have $Z_{i,m} = Z_{i,r},$ so $z\in Z_r\cap X_i.$ In particular, $z\in Z_r,$ so that $Z_m\subseteq Z_r,$ and we are finished.