Subset Intersection Set Difference is Empty Iff Subset of Second Set

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Theorem

Let $S$ and $T$ be sets.

Let $A \subseteq S$.

Then:

$A \cap S \setminus T = \O$ if and only if $A \subseteq T$

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:

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

\(=\)

\(\ds \O\)

\(\ds \leadstoandfrom \ \ \)

\(\ds A \setminus T\)

\(=\)

\(\ds \O\)

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

\(\ds \leadstoandfrom \ \ \)

\(\ds A\)

\(\subseteq\)

\(\ds T\)

Set Difference with Superset is Empty Set

$\blacksquare$