Absorption Laws (Set Theory)/Union with Intersection

Theorem

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

Proof 1

\(\ds \)

\(\)

\(\ds \paren {S \cap T} \subseteq S\)

Intersection is Subset

\(\ds \)

\(\leadsto\)

\(\ds S \cup \paren {S \cap T} = S\)

Union with Superset is Superset‎

$\blacksquare$

Proof 2

\(\ds x\)

\(\in\)

\(\ds S \cup \paren {S \cap T}\)

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(\in\)

\(\ds S \lor \paren {x \in S \land x \in T}\)

Definition of Set Intersection and Definition of Set Union

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(\in\)

\(\ds S\)

Disjunction Absorbs Conjunction

$\blacksquare$

Also see

• Intersection Absorbs Union, where it is proved that $S \cap \paren {S \cup T} = S$

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

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $1$. Sets: Exercise $3$
• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $3$