Exercise 8.1.17

Proof. Since $B$ is infinite, it follows Theorem 8.1.5 that there is a unique ordinal $\alpha$ such that $\left|B\right|={\aleph }_{\alpha }$, noting that this requires the Axiom of Choice. Thus there is a bijection $f$ from $B$ onto ${\omega }_{\alpha }$. Clearly then, since $A\subseteq B$, we have that $f[A]\subseteq {\omega }_{\alpha }$. Moreover, clearly $f$ is a bijection from $A$ to $f[A]$ so that $\left|f[A]\right|=\left|A\right|<\left|B\right|={\aleph }_{\alpha }$. It therefore follows from Exercise 7.2.6 that $\left|{\omega }_{\alpha }-f[A]\right|={\aleph }_{\alpha }$. We now claim that $f[B-A]={\omega }_{\alpha }-f[A]$.

$\text{(⊆)}$ So consider any $y\in f[B-A]$ so that there is an $x\in B-A$ such that $y=f(x)$. Since clearly $x\in B$, we have that $y=f(x)\in {\omega }_{\alpha }$ since $f$ maps $B$ onto ${\omega }_{\alpha }$. Now suppose that also $y\in f[A]$ so that there is a $z\in A$ where $y=f(z)$. Then we have that $y=f(x)=f(z)$ so that $x=z$ since $f$ is injective. Hence we have $x=z\in A$ but also $z=x\notin A$ since $z=x\in B-A$, which is a contradiction. So it must be that $y\notin f[A]$ so that indeed $y\in {\omega }_{\alpha }-f[A]$. Therefore $f[B-A]\subseteq {\omega }_{\alpha }-f[A]$ since $y$ was arbitrary.

$\text{(⊇)}$ Now consider any $y\in {\omega }_{\alpha }-f[A]$ so that $y\in {\omega }_{\alpha }$ and $y\notin f[A]$. Since $f$ is onto ${\omega }_{\alpha }$, there is an $x\in B$ such that $y=f(x)$. Moreover, since $y\notin f[A]$ we can be sure that $x\notin A$. Therefore $x\in B-A$. Since $y=f(x)$ it therefore follows that $y\in f[B-A]$ so that ${\omega }_{\alpha }-f[A]\subseteq f[B-A]$ since $y$ was arbitrary.

Thus we have shown that $f[B-A]={\omega }_{\alpha }-f[A]$ so that clearly $f$ is a bijection from $B-A$ onto ${\omega }_{\alpha }-f[A]$. Hence we have that $\left|B-A\right|=\left|{\omega }_{\alpha }-f[A]\right|={\aleph }_{\alpha }=\left|B\right|$ as desired. □