Lemma 5.8.4. Let X be a topological space and suppose X = \bigcup _{i = 1, \ldots , n} X_ i where each X_ i is an irreducible closed subset of X and no X_ i is contained in the union of the other members. Then each X_ i is an irreducible component of X and each irreducible component of X is one of the X_ i.
Proof. Let Y be an irreducible component of X. Write Y = \bigcup _{i = 1, \ldots , n} (Y \cap X_ i) and note that each Y \cap X_ i is closed in Y since X_ i is closed in X. By irreducibility of Y we see that Y is equal to one of the Y \cap X_ i, i.e., Y \subset X_ i. By maximality of irreducible components we get Y = X_ i.
Conversely, take one of the X_ i and expand it to an irreducible component Y, which we have already shown is one of the X_ j. So X_ i \subset X_ j and since the original union does not have redundant members, X_ i = X_ j, which is an irreducible component. \square
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Comment #9443 by Elías Guisado on
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