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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