Union and intersection of a set with the empty set

Prove that $A \cup \varnothing = A$ and that $A \cap \varnothing = \varnothing$ .

Proof. If $x \in A \cup \varnothing$ , then $x \in A$ or $x \in \varnothing$ by definition of union. Since $x \notin \varnothing$ (by definition of $\varnothing$ ), we must have $x \in A$ . Hence, $A \cup \varnothing \subseteq A$ .
On the other hand, if $x \in A$ , then $x \in A \cup \varnothing$ by definition of union, so $A \subseteq A \cup \varnothing$ .
Therefore, $A \cup \varnothing = A$ .∎

Proof. If $x \in A \cap \varnothing$ , then by definition of intersection $x \in A$ and $x \in \varnothing$ . But, by definition of $\varnothing$ , there is no $x$ such that $x \in \varnothing$ . Hence, $A \cap \varnothing$ must be empty. By uniqueness of the empty set then, we have $A \cap \varnothing = \varnothing$ .∎

Stumbling Robot

Union and intersection of a set with the empty set

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