Exercise 9.1.8

Proof. Suppose sets $⟨A_i\mid i\in I⟩$ and $⟨B_i\mid i\in I⟩$ where $\left|A_i\right|={\kappa }_i$ and $\left|B_i\right|={\lambda }_i$ for all $i\in I$. Then, by the definition of the cardinal product, we have that

Now, for any $i\in I$, we also have

so that we can choose an injective $f_i:A_i\to B_i$ (with the help of the Axiom of Choice).

We then construct a function $f$ from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ to ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$. So for any $a\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ we have that $a=⟨a_i\mid i\in I⟩$ where each $a_i\in A_i$. Thus $a_i\in A_i=\mathrm{dom}(f_i)$ for each $i\in I$ so that $f_i(a_i)\in B_i$. We then define $f(a)=⟨f_i(a_i)\mid i\in I⟩$ so that clearly $f(a)\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ and hence $f$ is a function into ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$.

We claim that $f$ as defined above is injective. To this end, consider any $\alpha$ and $\beta$ in ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ where $\alpha \ne \beta$. It then follows that $\alpha =⟨{\alpha }_i\mid i\in I⟩$ and $\beta =⟨{\beta }_i\mid i\in I⟩$ where both ${\alpha }_i$ and ${\beta }_i$ are elements of $A_i$ for each $i\in I$. Since $\alpha \ne \beta$ we must have that there is an $i_0\in I$ such that ${\alpha }_{i_0}\ne {\beta }_{i_0}$. It then follows that $f_{i_0}({\alpha }_{i_0})\ne f_{i_0}({\beta }_{i_0})$ since $f_{i_0}$ is injective. Therefore clearly $f(\alpha )=⟨f_i({\alpha }_i)\mid i\in I⟩\ne ⟨f_i({\beta }_i)\mid i\in I⟩=f(\beta )$. This shows that $f$ is injective since $\alpha$ and $\beta$ were arbitrary.

Finally, since $f$ is an injective function from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ to ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$, we have that

as desired. □