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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.∎