Commutativity Archives - Stumbling Robot

Prove $A \cup B = B \cup A$ and $A \cap B = B \cap A$ .

Proof $(A \cup B = B \cup A).$ If $x$ is any element in $A \cup B$ then, by definition of union, we have $x \in A$ or $x \in B$ . But, if $x$ is in $A$ or $B$ , then it is in $B$ or $A$ , and by definition of union, this means $x \in B \cup A$ . Therefore, $A \cup B \subseteq B \cup A$ .
The other inclusion is identical: if $x$ is any element of $B \cup A$ , then we know $x \in B$ or $x \in A$ . But, $x \in B$ or $x \in A$ implies that $x$ is in $A$ or $B$ ; and hence, $x \in A \cup B$ . Therefore, $B\ \cup A \subseteq A \cup B$ .
Hence, $A \cup B = B \cup A$ .∎

Proof $(A \cap B = B \cap A).$ If $x$ is any element in $A \cap B$ , then we know by definition of intersection that $x \in A$ and $x \in B$ . Hence, $x \in B$ and $x \in A$ , and so, $x \in B \cap A$ . Therefore, $A \cap B \subseteq B \cap A$ .
The reverse inclusion is again identical: if $x$ is any element of $B \cap A$ , then we know $x \in B$ and $x \in A$ . Hence, $x \in A$ and $x \in B$ . This implies $x \in A \cap B$ . Hence, $B \cap A \subseteq A \cap B$ .
So, $A \cap B = B \cap A$ .∎

Stumbling Robot

Tag: Commutativity

Prove the commutative laws of union and intersection