# Algebraic closed set systems are closed under unions of chains

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.

• 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. – GEdgar Mar 21 '20 at 20:50

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.