Lemma 5.9.4. Let $X$ be a topological space. Let $X_ i \subset X$, $i = 1, \ldots , n$ be a finite collection of subsets. If each $X_ i$ is Noetherian (with the induced topology), then $\bigcup _{i = 1, \ldots , n} X_ i$ is Noetherian (with the induced topology).

Proof. Let $\{ F_ m\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X' = \bigcup _{i = 1, \ldots , n} X_ i$ with the induced topology. Then we can find a decreasing sequence $\{ G_ m\} _{m \in \mathbf{N}}$ of closed subsets of $X$ verifying $F_ m = G_ m \cap X'$ for all $m$ (small detail omitted). As $X_ i$ is noetherian and $\{ G_ m \cap X_ i\} _{m \in \mathbf{N}}$ a decreasing sequence of closed subsets of $X_ i$, there exists $m_ i \in \mathbf{N}$ such that for all $m \geq m_ i$ we have $G_ m \cap X_ i = G_{m_ i} \cap X_ i$. Let $m_0 = \max _{i = 1, \ldots , n} m_ i$. Then clearly

$F_ m = G_ m \cap X' = G_ m \cap (X_1 \cup \ldots \cup X_ n) = (G_ m \cap X_1) \cup \ldots (G_ m \cap X_ n)$

stabilizes for $m \geq m_0$ and the proof is complete. $\square$

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