# Empty Intersection iff Subset of Complement

## Corollary to Intersection with Complement is Empty iff Subset

$S \cap T = \O \iff S \subseteq \relcomp {} T$

where:

$S \cap T$ denotes the intersection of $S$ and $T$
$\O$ denotes the empty set
$\complement$ denotes set complement
$\subseteq$ denotes subset.

### Corollary

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

 $\ds S \cap T$ $=$ $\ds \O$ $\ds \leadstoandfrom \ \$ $\ds S \cap \relcomp {} {\relcomp {} T}$ $=$ $\ds \O$ Complement of Complement $\ds \leadstoandfrom \ \$ $\ds S$ $\subseteq$ $\ds \relcomp {} T$ Intersection with Complement is Empty iff Subset

$\blacksquare$

## Proof 2

$S \subseteq T \iff S \cap \relcomp {} T = \O$

Then we have:

 $\ds$  $\ds S \nsubseteq \relcomp {} T$ $\ds$ $\leadstoandfrom$ $\ds \neg \paren {\forall x \in S: x \in \relcomp {} T}$ Definition of Subset $\ds$ $\leadstoandfrom$ $\ds \exists x \in S: x \notin \relcomp {} T$ Denial of Universality $\ds$ $\leadstoandfrom$ $\ds \exists x \in S: x \in T$ Definition of Set Complement $\ds$ $\leadstoandfrom$ $\ds x \in S \cap T$ Definition of Set Intersection $\ds$ $\leadstoandfrom$ $\ds S \cap T \ne \O$ Definition of Disjoint Sets

Thus:

 $\ds$  $\ds S \cap T = \O$ $\ds$ $\leadstoandfrom$ $\ds \forall x \in S: x \in \relcomp {} T$ $\ds$ $\leadstoandfrom$ $\ds S \subseteq \relcomp {} T$

$\blacksquare$