Exercise 3.1.9

Proof.

1.

($A=X\setminus B$) By Exercise 3.1.4 it suffices to show that $A\subseteq X\setminus B$ and vice versa.

$A\subseteq X\setminus B$

Let $x\in A$ be arbitrary. We have $x\in A\cup B$ by Axiom 3.4. Since $x\in A\cup B,$ we have $x\in X$ by theorem assumption.
Since $x\in A,$ we have $x\notin B$ since $A\cap B=\varnothing$ (if $x\in B,$ we would have $x\in A\cap B,$ which would contradict $A\cap B=\varnothing$).
Since $x\in X,$ $x\notin B,$ we have $x\in X\setminus B.$

$X\setminus B\subseteq A$

Let $x\in X\setminus B$ be arbitrary. By the definition of the set difference, we have $x\in X$ and $x\notin B.$
Since $x\in X,$ by theorem assumption we have $x\in A\cup B.$ by Axiom 3.4 this means that $x\in A$ or $x\in B.$ But $x\in B$ is false, and therefore, $x\in A.$ Thus, we’re done.

2.

($B=X\setminus A$) By Exercise 3.1.4 it suffices to show that $B\subseteq X\setminus A$ and vice versa.

$B\subseteq X\setminus A$

Let $x\in B$ be arbitrary. We have $x\in A\cup B$ by Axiom 3.4. Since $x\in A\cup B,$ we have $x\in X$ by theorem assumption.
Since $x\in B,$ we have $x\notin A$ since $A\cap B=\varnothing$ (if $x\in A,$ we would have $x\in A\cap B,$ which would contradict that $A\cap B=\varnothing$).
Since $x\in X,$ $x\notin A,$ we have $x\in X\setminus A.$ Since $x\in B$ implies $x\in X\setminus A,$ we have $B\subseteq X\setminus A.$

$X\setminus A\subseteq B$

Let $x\in X\setminus A$ be arbitrary. By the definition of the set difference, we have $x\in X$ and $x\notin A.$
Since $x\in X,$ by theorem assumption we have $x\in A\cup B.$ by Axiom 3.4 this means that $x\in A$ or $x\in B.$ But $x\in A$ is false, and therefore, $x\in B.$ Thus, we’re done.