Set Difference with Disjoint Set

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Theorem

Let $S, T$ be sets.

Then:

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

where:

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


Proof

\(\displaystyle S \cap T\) \(=\) \(\displaystyle \varnothing\)
\(\displaystyle \iff \ \ \) \(\displaystyle S\) \(\subseteq\) \(\displaystyle \complement \left({T}\right)\) Intersection with Complement is Empty iff Subset
\(\displaystyle \iff \ \ \) \(\displaystyle S \cap \complement \left({T}\right)\) \(=\) \(\displaystyle S\) Intersection with Subset is Subset‎‎
\(\displaystyle \iff \ \ \) \(\displaystyle S \setminus T\) \(=\) \(\displaystyle S\) Set Difference as Intersection with Complement

$\blacksquare$


Sources