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.
My question
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.
Following the tips of GEdgar i've realized that the answer was pretty simple, sometimes we just need some encouragement ^_^
Algebraic closure implies closure under unions of chains
Let $X$ be a chain with $X_1 \subseteq X_2 \subseteq ...$ (using countable index for simplicity).
$C(\bigcup X)=\bigcup\{C(Y)\ |\ \ (Y \subseteq \bigcup X) \land (|Y| \in \mathbb{N})\}$ (for algebraicity)
For every $Y$ we can find an $X_n$ such that $Y \subseteq X_n$, this prove $C(Y) \subseteq C(X_n)=X_n$. Taking the unions from both sides we get: $C(\bigcup X) \subseteq \bigcup X$
The other verse of inclusion is trivial, hence equality is established and $K$ is closed under unions of chains.
Closure under unions of chains implies algebraic closure
$C(X)=C(\bigcup\{Y\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\})=C(\bigcup\{C(Y)\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\})$
$\{C(Y)\ |\ \ (Y \subseteq X) \land (|Y| \in \mathbb{N})\}$ is upward directed so its union is closed and the algebraic proprerty is proved.