Subset of Set Difference iff Disjoint Set

Theorem

Let $S, T$ be sets.

Let $A \subseteq S$

Then:

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

where:

$A \cap T$ denotes set intersection

$\O$ denotes the empty set

$S \setminus T$ denotes set difference.

Proof

We have:

\(\ds A \cap \paren {S \setminus T}\)

\(=\)

\(\ds \paren {A \cap S} \setminus T\)

Intersection with Set Difference is Set Difference with Intersection

\(\ds \)

\(=\)

\(\ds A \setminus T\)

Intersection with Subset is Subset

Then:

\(\ds A\)

\(\subseteq\)

\(\ds S \setminus T\)

\(\ds \leadstoandfrom \ \ \)

\(\ds A\)

\(=\)

\(\ds A \cap \paren {S \setminus T}\)

Intersection with Subset is Subset

\(\ds \leadstoandfrom \ \ \)

\(\ds A\)

\(=\)

\(\ds A \setminus T\)

As $A \cap \paren {S \setminus T} = A \setminus T$

\(\ds \leadstoandfrom \ \ \)

\(\ds A \cap T\)

\(=\)

\(\ds \O\)

Set Difference with Disjoint Set

$\blacksquare$