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. $\square$

Comment #9443 by on

In the second paragraph, it seems that “take one of the $X_i$ and expand it to an irreducible component $Y$” needs Lemma 5.8.3, point (3), which requires Zorn's lemma. It can be done without the axiom of choice: the second paragraph may be replaced by

Suppose $X_i\subset Y$, where $Y$ is some irreducible subset of $X$. By the argument in the first paragraph, $Y\subset X_j$ for some $j$. Since the original union does not have redundant members, $X_i=Y=X_j$. Therefore $X_i$ is maximal in the set of irreducible subsets of $X$, that is, $X_i$ is an irreducible component of $X$.

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