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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.

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    $\begingroup$ Maybe do a simple case first. Prove that: The union of a chain of subgroups is a subgroup.NOTE: not arbitrary union of subgroups; just union of a chain of subgroups. $\endgroup$ – GEdgar Mar 21 '20 at 20:50
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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.

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