Set Difference Relative Complement

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Theorem

Let $A, B \subseteq S$.


Then the set difference between $A$ and $B$ can be expressed as the intersection with the relative complement with respect to $S$:

$A \setminus B = A \cap \complement_S \left({B}\right)$


Proof

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle A \setminus B\) \(=\) \(\displaystyle \left\{ {x \in A: x \notin B}\right\}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Definition of set difference          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \left\{ {x \in A: x \in \complement_S \left({B}\right)}\right\}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Definition of relative complement          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle A \cap \complement_S \left({B}\right)\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Definition of set intersection          

$\blacksquare$


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