Simple Proof Methods

3. Theorem. For any sets $A$ and $B$, the following holds:$$A \cup B = B \cup A.$$ Proof. Let $A$ and $B$ be any sets. To show that $A \cup B = B \cup A$, we need to show that $A \cup B \subseteq B \cup A$ and that $B \cup A \subseteq A \cup B$:
   • [Showing that $A \cup B \subseteq B \cup A$] Let $x$ be any element in the set $A \cup B$. This means that $x$ is in the union of $A$ and $B$. This means that $x \in A$ or that $x \in B$. In other words, it means that $x \in B$ or that $x \in A$. This means that $x$ is in the union of $B$ and $A$, so $x \in B \cup A$. This implies $A \cup B \subseteq B \cup A$.
   • [Showing that $B \cup A \subseteq A \cup B$] Let $x$ be any element in the set $B \cup A$. This means that $x$ is in the union of $B$ and $A$. This means that $x \in B$ or that $x \in A$. In other words, it means that $x \in A$ or that $x \in B$. This means that $x$ is in the union of $A$ and $B$, so $x \in A \cup B$. This implies $B \cup A \subseteq A \cup B$.
   Therefore, we conclude that $A \cup B = B \cup A$ by the definition of set equality.◼