Simple Proof Methods

2. Theorem. For any set $A$, the following holds:$$A \cap \emptyset = \emptyset.$$ Proof. Let $A$ be any set. To show that $A \cap \emptyset = \emptyset$, we need to show that $A \cap \emptyset \subseteq \emptyset$ and that $\emptyset \subseteq A \cap \emptyset$:
   • [Showing that $A \cap \emptyset \subseteq \emptyset$] Let $x$ be any element in the set $A \cap \emptyset$. This means that $x$ is in the intersection of $A$ and $\emptyset$. This means that $x \in A$ and that $x \in \emptyset$. However, because $\emptyset$ is the empty set and doesn't contain any elements in it, our assumption that $x \in A \cap \emptyset$ is false; specifically, $A \cap \emptyset$ doesn't contain any elements in it. But we know that the set that contains no elements is called $\emptyset$, so $A \cap \emptyset \subseteq \emptyset$.
   • [Showing that $\emptyset \subseteq A \cap \emptyset$] Consider the set $\emptyset$, which doesn't contain any elements in it. The set $A \cap \emptyset$ contains all the elements of $A$ and of $\emptyset$ in it, which, because the set $\emptyset$ is empty, also contains no elements in it. As such, $\emptyset \subseteq A \cap \emptyset$.
   Because we showed that $A \cap \emptyset \subseteq \emptyset$ and $\emptyset \subseteq A \cap \emptyset$, we conclude that $A \cap \emptyset = \emptyset$ by the definition of set equality.◼