Exercise 3.1.5 (Equivalent formulations of subset relation)

Proof. To prove that those $3$ statements are logically equivalent, we use the following chain of implications (see Appendix A: Basics of mathematical logic):

• Suppose that $A\subseteq B$. By Exercise 3.1.4, it suffices to show that $A\cup B\subseteq B$ and $B\subseteq A\cup B$ hold simultaneously.

  $A\cup B\subseteq B.$

  Let $x\in A\cup B$ be arbitrary. By Axiom 3.4, $x$ is an element of $A$ or $x$ is an element of $B.$ In the first case, $x\in A$ implies that $x\in B$, since by theorem assumption $A\subseteq B.$ In the other case, $x\in B$ as well. Thus, we get $A\cup B\subseteq B.$

  $B\subseteq A\cup B.$

  Let $x\in B$ be arbitrary. Then, $x\in A\cup B,$ since by Axiom 3.4, $\forall <!--\; FUNCTION\; APPLICATION\; -->x:x\in A\cup B⟺x\in A\text{OR}x\in B.$

  Thus, $A\subseteq B⟹A\cup B=B.$

• Suppose that $A\cup B=B$. By Exercise 3.1.4, it suffices to show that $A\cap B\subseteq A$ and $A\subseteq B\cap A$ hold simultaneously.

  $A\cap B\subseteq A.$

  Let $x\in A\cap B$ be arbitrary. By Definition 3.1.23, $x$ is an element of $A$ and $x$ is an element of $B.$ In particular, $x$ is an element of $A$. Since our choice of $x$ was arbitrary, this follows for all $x\in A\cap B$.

  $A\subseteq A\cap B.$

  Let $x\in A$ be arbitrary. Suppose for the sake of contradiction that $x\notin A\cap B$. $x\notin A\cap B$ means that $x$ is not an element of $A$ or $x$ is not an element of $B$. $x\notin A$ cannot be true because this would contradict the original assumption. Therefore, $x\notin B.$ Since $x\in A,$ we have $x\in A\cup B.$ Since $A\cup B=B$ by assumption, we have that every $x$ of the set $A\cup B$ is also contained in $B$ by Definition 3.1.4. Thus, $x\in B$. But this is a contradiction, since as we showed above, $x\notin B.$ Thus, $x\in A\cap B,$ and therefore, $A\subseteq A\cap B,$ as desired.

• Finally, suppose that $A\cap B=A.$ Let $x\in A$ be arbitrary. We have $x\in A\cap B$ since $A\cap B=A$ by assumption. $x\in A\cap B$ means that $x$ is contained in $A$ and $x$ is contained in $B$. The fact that $x$ is an element of $B$ completes the proof.