Lemma 5.11.5. Let $X$ be a topological space. The following are equivalent:

1. $X$ is catenary,

2. $X$ has an open covering by catenary spaces.

Moreover, in this case any locally closed subspace of $X$ is catenary.

Proof. Suppose that $X$ is catenary and that $U \subset X$ is an open subset. The rule $T \mapsto \overline{T}$ defines a bijective inclusion preserving map between the closed irreducible subsets of $U$ and the closed irreducible subsets of $X$ which meet $U$. Using this the lemma easily follows. Details omitted. $\square$

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