Set of Closed Elements wrt Closure Operator under Subset Operation is Complete Lattice
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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 $14.25$