Proposition 5

Let $X$ be a topological space. Every irreducible subset of $X$ is contained in an irreducible component of $X$ and $X$ is the union of its irreducible components.

To prove the first assertion, it is sufficient, by virtue of Zorn's Lemma, to prove that the set 3 of irreducible subsets of $X$ is inductive. Let $\mathscr{G}$ be a subset of $\mathfrak{S}$ totally ordered by inclusion; we show that the union $E$ of the sets $F \in \mathscr{G}$ is irreducible. Let $U, V$ be two open sets of $X$ which meet $E$; as $\mathscr{G}$ is totally ordered, there exists a set $F \in \mathscr{G}$ meeting $U$ and $V$; as $F$ is irreducible, $U \cap V$ meets $F$ and hence also $E$, which proves that $E$ is irreducible and hence that 3 is inductive. The second assertion follows from the first, for every subset of $X$ consisting of a single point is irreducible.