Closure Operator from Closed Sets

Theorem

Let $S$ be a set.

Let $\CC$ be a set of subsets of $S$.

Let $\CC$ be closed under arbitrary intersections:

$\forall \KK \in \powerset \CC: \bigcap \KK \in \CC$

where $\bigcap \O$ is taken to be $S$.

Define $\cl: \powerset S \to \CC$ by letting:

$\map \cl T = \bigcap \set {C \in \CC: T \subseteq C}$

Then $\cl$ is a closure operator whose closed sets are the elements of $\CC$.

Proof

First we will show that $\cl$ is a closure operator.

Inflationary

Let $T \subseteq S$.

By Set Intersection Preserves Subsets/General Result/Corollary, $T \subseteq \map \cl T$.

Since this holds for all such $T$, $\cl$ is inflationary.

$\Box$

Increasing

Let $T \subseteq U \subseteq S$.

Let $\TT$ and $\UU$ be the sets of elements of $\CC$ containing $T$ and $U$, respectively.

Since Subset Relation is Transitive, every set containing $U$ contains $T$, so $\UU \subseteq \TT$.

By Intersection is Decreasing, $\bigcap \TT \subseteq \bigcap \UU$.

Thus $\map \cl T \subseteq \map \cl U$.

$\Box$

Idempotent

Let $T \subseteq S$.

By the premise, the intersection of a subset of $\CC$ is in $\CC$.

Thus in particular $\map \cl T \in \CC$.

Therefore:

$\map \cl {\map \cl T} \subseteq \map \cl T$

Since $\cl$ is inflationary:

$\map \cl T \subseteq \map \cl {\map \cl T}$

By definition of set equality:

$\map \cl {\map \cl T} = \map \cl T$

Since this holds for all $T \subseteq S$, $\cl$ is idempotent.

$\Box$

Finally, we need to show that the elements of $\CC$ are the closed sets with respect to $\cl$.

If $C \in \CC$, then since $\cl$ is inflationary:

$C \subseteq \map \cl C$

But since $C \subseteq C$, $\map \cl C \subseteq C$.

Thus by definition of set equality:

$\map \cl C = C$

so $C$ is closed with respect to $\cl$.

Suppose instead that $C$ is closed with respect to $\cl$.

Then $\map \cl C = C$.

Since $\CC$ is closed under intersections, $C \in \CC$.

$\blacksquare$