How can one show that a set is empty? For example, let the set $A = \{1,2,3\}$ and let the set $B = \{4,5,6\}$. I want to show that $A \cap B = \emptyset$.
I know that the typical approach to showing set equality would be, for this case, to show that $A \cap B \subseteq \emptyset$ and that $\emptyset\subseteq A \cap B $. Is this the correct approach?
For a set to be empty, its enough to show that no element belongs to it. If $x \in A \cap B$ as defined, then $x$ belongs to both $A$ and $B$. But if $x \in A$, $x = 1, 2$ or $3$, and that contradicts the fact that $x \in B$. Hence $x \not \in A \cap B$ and the intersection is empty.
Showing the first condition only ($A\cap B\subseteq \emptyset$) suffices. The empty set is anyway a subset of any set, so you don't have to show this.
Suppose $A\cap B$ is not empty. Then, by definition, there exists some $x$ such that $x\in A$ and $x\in B$. In your example, the first condition implies that either $x=1$, $x=2$, or $x=3$. Suppose that $x=1$ Since $x\in B=\{4,5,6\}$, it must mean that $1=4$, or $1=5$, or $1=6$. Any of these is clearly absurd. You can show similarly that the cases in which $x=2$ and $x=3$ lead to analogous absurdities. We have a contradiction, whence there must be no element in $A\cap B$. So, $A\cap B$ must be empty.
