Set of Subsets is Cover iff Set of Complements is Free

Theorem

Let $S$ be a set.

Let $\CC$ be a set of sets.

Then $\CC$ is a cover for $S$ if and only if $\set {\relcomp S X: X \in \CC}$ is free.

Proof

Let $S$ be a set.

Let us recall the definition of free:

Let $\SS$ be a set of sets.

Then $\SS$ is a free set of sets if and only if the intersection of $\SS$ is empty.

That is:

$\bigcap \SS = \O$

Necessary Condition

Let $\CC$ be a cover for $S$:

$S \subseteq \bigcup \CC$

Suppose:

$\set {\relcomp S X: X \in \CC}$

is not free.

Then there exists an $x \in S$ such that:

$\forall X \in \CC: x \in \relcomp S X$

But by the definition of (relative) complement:

$\forall X \in \CC: x \notin X$

That is:

$x \notin \bigcup \CC$

which contradicts:

$S \subseteq \bigcup \CC$

$\Box$

Sufficient Condition

Let:

$\set {\relcomp S X: X \in \CC}$

be free.

Suppose:

$\bigcup \CC$

is not a cover for $S$.

Then there exists some $x \in S$ such that:

$\forall X \in \CC: x \notin X$

Hence:

$\forall \relcomp S X \in \CC: x \in X$

which contradicts the supposition that:

$\set {\relcomp S X: X \in \CC}$

is free.

$\blacksquare$