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I'm reading from "ALGEBRAS, LATTICES, VARIETIES" by Ralph N. McKenzie, George F. McNulty and Walter F. Taylor and had a question about closure systems introduced in Chapter 2 on Lattices.

Let $A$ be a set. We say countable chain here to mean a sequence $(A_i)_{i\in\mathbb{N}}$ of subsets of $A$ such that $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots$. We say chain here to mean a family $K$ of subsets of $A$ such that for all $A_1, A_2 \in K$ we have $A_1 \subseteq A_2$ or $A_2 \subseteq A_1$ (i.e. $K$ is totally ordered by $\subseteq$).

Is this statement true: Let $A$ be a set and $\mathscr{C}$ be a closed set system of $A$. $\mathscr{C}$ is closed under the union of countable chains of closed subsets if and only if it is closed under the union of chains of closed subsets.

Asaf Karagila
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Josstopher
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The statement is not true. Let $A$ be the ordinal $\omega_2$. Let ${\mathcal C}_0$ be the set of intervals $[0,\alpha)$ where $\alpha<\omega_2$ has countable cofinality. Let ${\mathcal C}={\mathcal C}_0\cup \{A\}$. ${\mathcal C}$ is the set of closed subsets of a closed set system on $A$ which has the property that any union of a countable chain of closed sets is closed. For this closed set system, $\omega_1\subseteq A$ is a union of a chain of closed sets $\left(\omega_1 = \bigcup_{B\in {\mathcal C}\\ B\subseteq \omega_1} B\right)$ but not a union of a countable chain of closed sets because $\omega_1$ does not have countable cofinality.

Keith Kearnes
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