Set of Closed Elements wrt Closure Operator under Subset Operation is Complete Lattice

Theorem

Let $S$ be a set.

Let $\cl$ be a closure operator on the power set $\powerset S$ of $S$.

Let $\mathscr C$ be the set of all subsets $T$ of $S$ such that:

$\map \cl T = T$

Then the algebraic structure $\struct {\mathscr C, \subseteq}$ forms a complete lattice.

Proof

Recall the closure axioms:

\((\text {cl} 1)\)	$:$		$\cl$ is inflationary:	\(\ds \forall X \subseteq S:\)		\(\ds X \)	\(\ds \subseteq \)	\(\ds \map \cl X \)
\((\text {cl} 2)\)	$:$		$\cl$ is increasing:	\(\ds \forall X, Y \subseteq S:\)		\(\ds X \subseteq Y \)	\(\ds \implies \)	\(\ds \map \cl X \subseteq \map \cl Y \)
\((\text {cl} 3)\)	$:$		$\cl$ is idempotent:	\(\ds \forall X \subseteq S:\)		\(\ds \map \cl {\map \cl X} \)	\(\ds = \)	\(\ds \map \cl X \)

First we note that from Closure Axiom $\text {cl} 1$: Inflationary we have that:

$\map \cl S = S$

and so:

$S \in \mathscr C$

Let $\AA \subseteq \mathscr C$.

Thus $\AA$ is a set of subsets $T$ of $S$ for all of which $\map \cl T = T$.

From Intersection of Closed Sets is Closed:

$\ds \cap \AA \in \mathscr C$

The result follows from Set of Subsets which contains Set and Intersection of Subsets is Complete Lattice,

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.25$