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.
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.