Exercise 7.8

Proof. First, for $x\in X^{\omega }$, clearly the function that maps $x$ to the set $\left\{x\right\}$ is an injective function from $X^{\omega }$ to $B$.

Now we construct an injection $f_1:B\to {(X^{\omega })}^{\omega }$. So consider any $S\in B$ so that $S$ is a countable subset of $X^{\omega }$. Then, by Theorem 7.1, we can choose a surjective function $g:{\mathbb{Z}}_{\text{+}}\to S$. Note that this does require the Axiom of Choice since we must choose such a surjection for each $S\in B$, and clearly $B$ is infinite. Since $S\subset X^{\omega }$, $g$ can be considered as a function from ${\mathbb{Z}}_{\text{+}}$ to $X^{\omega }$ so that $g\in {(X^{\omega })}^{\omega }$, though of course it would no longer necessarily be surjective with this range. So we simply set $f_1(S)=g$

To show that $f_1$ is injective consider $S,S^{\prime }\in B$ where $S\ne S^{\prime }$. Then, setting $g=f_1(S)$ and $g^{\prime }=f_1(S^{\prime })$, we have that $g({\mathbb{Z}}_{\text{+}})=S$ and $g^{\prime }({\mathbb{Z}}_{\text{+}})=S^{\prime }$ by definition. Since $S\ne S^{\prime }$, we have that $g$ and $g^{\prime }$ have the same domain but different image sets. Clearly this means that $f_1(S)=g\ne g^{\prime }=f_1(S^{\prime })$, which shows that $f_1$ is injective since $S$ and $S^{\prime }$ were arbitrary.

Hence $f_1$ is an injection from $B$ to ${(X^{\omega })}^{\omega }={(X^{{\mathbb{Z}}_{\text{+}}})}^{{\mathbb{Z}}_{\text{+}}}$. Now, from Lemma 2, we have that ${(X^{{\mathbb{Z}}_{\text{+}}})}^{{\mathbb{Z}}_{\text{+}}}$ has the same cardinality as $X^{{\mathbb{Z}}_{\text{+}}\times {\mathbb{Z}}_{\text{+}}}$ so that there is a bijection $f_2:{(X^{{\mathbb{Z}}_{\text{+}}})}^{{\mathbb{Z}}_{\text{+}}}\to X^{{\mathbb{Z}}_{\text{+}}\times {\mathbb{Z}}_{\text{+}}}$, which is of course also an injection. Finally, since ${\mathbb{Z}}_{\text{+}}\times {\mathbb{Z}}_{\text{+}}$ has the same cardinality as ${\mathbb{Z}}_{\text{+}}$ (by Corollary 7.4), it follows that there is an injection from ${\mathbb{Z}}_{\text{+}}\times {\mathbb{Z}}_{\text{+}}$ to ${\mathbb{Z}}_{\text{+}}$. Since also the identity function on $X$ is an injection, we have that there is an injection $f_3:X^{{\mathbb{Z}}_{\text{+}}\times {\mathbb{Z}}_{\text{+}}}\to X^{{\mathbb{Z}}_{\text{+}}}$ by Lemma 1. Then clearly $f_3\circ f_2\circ f_1$ is an injection from $B$ to $X^{{\mathbb{Z}}_{\text{+}}}=X^{\omega }$.

Since there is an injection from $X^{\omega }$ to $B$ and vice-versa, it follows that they have the same cardinality by Exercise 7.6 part (b) as desired. □