Absorption Laws (Set Theory)/Intersection with Union

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Theorem

$S \cap \paren {S \cup T} = S$


Proof 1

\(\displaystyle \) \(\) \(\displaystyle S \subseteq \paren {S \cup T}\) Set is Subset of Union
\(\displaystyle \) \(\leadsto\) \(\displaystyle S \cap \paren {S \cup T} = S\) Intersection with Subset is Subset‎

$\blacksquare$


Proof 2

\(\displaystyle x\) \(\in\) \(\displaystyle S \cap \paren {S \cup T}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle S \land \paren {x \in S \lor x \in T}\) Definition of Set Intersection and Definition of Set Union
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle S\) Conjunction Absorbs Disjunction

$\blacksquare$


Also see

These two results together are known as the Absorption Laws, corresponding to the equivalent results in logic.


Sources