Empty Intersection iff Subset of Complement/Corollary

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Corollary to Empty Intersection iff Subset of Complement

Let $A, B, S$ be sets such that $A, B \subseteq S$.

Then:

$\exists X \in \powerset S: \paren {A \cap X} \cup \paren {B \cap \complement_S \paren X} = \O \iff A \cap B = \O$

where $\overline X$ denotes the relative complement of $X$ in $S$.


Proof 1

Let there exist such a set $X$.

Then:

\(\ds \) \(\) \(\ds \paren {A \cap X} \cup \paren {B \cap \complement_S \paren X} = \O\)
\(\ds \) \(\leadstoandfrom\) \(\ds A \cap X = \O \land B \cap \complement_S \paren X = \O\) Union is Empty iff Sets are Empty
\(\ds \) \(\leadstoandfrom\) \(\ds A \subseteq \complement_S \paren X \land B \subseteq X\) Empty Intersection iff Subset of Complement
\(\ds \) \(\leadstoandfrom\) \(\ds A \cap B = \O\)

$\blacksquare$


Proof 2

We have:

result $\paren {A \cap C} \cup \paren {B \cap \map \complement C} = \O \iff B \subseteq C \subseteq \map \complement A$:

where the universe $\Bbb U$ is posited.


Let $S$ take the position of $\Bbb U$.

Let $C = X$.

Then we have:

$\paren {A \cap X} \cup \paren {B \cap \relcomp S X} = \O \iff B \subseteq X \subseteq \relcomp S A$


Thus we have shown that:

$B \subseteq \relcomp S A$

and it follows from Empty Intersection iff Subset of Complement that:

$A \cap B = \O$

$\blacksquare$


Sources

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