From "A course in Universal Algebra" of Burris and Sankappanavar, exercise 6 page 24.
Given a set $A$ and a family $K$ of subsets of $A$, $K$ is said to be closed under unions of chains if whenever $C ⊆ K$ and $C$ is a chain (under $⊆$) then $\bigcup C \in K$; and $K$ is said to be closed under unions of upward directed families of sets if whenever $D ⊆ K$ is such that $A_1, A_2 \in D$ implies $A_1 ∪ A_2 ⊆ A_3$ for some $A_3 \in D$, then $\bigcup D \in K$. A result of set theory says that $K$ is closed under unions of chains iff $K$ is closed under unions of upward directed families of sets
(Schmidt) A closed set system $K$ for a set $A$ is called an algebraic closed set system for $A$ if there is an algebraic closure operator on $A$ such that the closed subsets of $A$ are precisely the members of $K$. If $K ⊆ Su(A)$, show that $K$ is an algebraic closed set system iff $K$ is closed under (i) arbitrary intersections and (ii) unions of chains.
I am aware that a closed set system is closed under arbitrary intersections. I can't figure out how to show the logical equivalence between the closure under unions of chains and the algebraic closure operator property.