# Subset of Set Difference iff Disjoint Set

## Theorem

Let $S, T$ be sets.

Let $A \subseteq S$

Then:

$A \cap T = \varnothing \iff A \subseteq S \setminus T$

where:

$A \cap T$ denotes set intersection
$\varnothing$ denotes the empty set
$S \setminus T$ denotes set difference.

## Proof

We have:

 $\displaystyle A \cap \paren {S \setminus T}$ $=$ $\displaystyle \paren{A \cap S} \setminus T$ Intersection with Set Difference is Set Difference with Intersection $\displaystyle$ $=$ $\displaystyle A \setminus T$ Intersection with Subset is Subset

Then:

 $\displaystyle$  $\displaystyle A \subseteq \paren{ S \setminus T}$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle$  $\displaystyle A = A \cap \paren{ S \setminus T}$ Intersection with Subset is Subset $\displaystyle \leadstoandfrom \ \$ $\displaystyle$  $\displaystyle A = A \setminus T$ As $A \cap \paren {S \setminus T} = A \setminus T$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle$  $\displaystyle A \cap T = \O$ Set Difference with Disjoint Set

$\blacksquare$