Lemma 5.9.2. Let X be a Noetherian topological space.
Any subset of X with the induced topology is Noetherian.
The space X has finitely many irreducible components.
Each irreducible component of X contains a nonempty open of X.
Lemma 5.9.2. Let X be a Noetherian topological space.
Any subset of X with the induced topology is Noetherian.
The space X has finitely many irreducible components.
Each irreducible component of X contains a nonempty open of X.
Proof. Let T \subset X be a subset of X. Let T_1 \supset T_2 \supset \ldots be a descending chain of closed subsets of T. Write T_ i = T \cap Z_ i with Z_ i \subset X closed. Consider the descending chain of closed subsets Z_1 \supset Z_1\cap Z_2 \supset Z_1 \cap Z_2 \cap Z_3 \ldots This stabilizes by assumption and hence the original sequence of T_ i stabilizes. Thus T is Noetherian.
Let A be the set of closed subsets of X which do not have finitely many irreducible components. Assume that A is not empty to arrive at a contradiction. The set A is partially ordered by inclusion: \alpha \leq \alpha ' \Leftrightarrow Z_{\alpha } \subset Z_{\alpha '}. By the descending chain condition we may find a smallest element of A, say Z. As Z is not a finite union of irreducible components, it is not irreducible. Hence we can write Z = Z' \cup Z'' and both are strictly smaller closed subsets. By construction Z' = \bigcup Z'_ i and Z'' = \bigcup Z''_ j are finite unions of their irreducible components. Hence Z = \bigcup Z'_ i \cup \bigcup Z''_ j is a finite union of irreducible closed subsets. After removing redundant members of this expression, this will be the decomposition of Z into its irreducible components (Lemma 5.8.4), a contradiction.
Let Z \subset X be an irreducible component of X. Let Z_1, \ldots , Z_ n be the other irreducible components of X. Consider U = Z \setminus (Z_1\cup \ldots \cup Z_ n). This is not empty since otherwise the irreducible space Z would be contained in one of the other Z_ i. Because X = Z \cup Z_1 \cup \ldots Z_ n (see Lemma 5.8.3), also U = X \setminus (Z_1\cup \ldots \cup Z_ n) and hence open in X. Thus Z contains a nonempty open of X. \square
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Comment #9444 by Elías Guisado on
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