Subset of Set Difference iff Disjoint Set

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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

$\blacksquare$