Exercise 9.1.7

Proof. Suppose that $⟨A_j\mid j\in J⟩$ are sets where $\left|A_j\right|={\kappa }_j$ for each $j\in J$. It then follows that

Now, for any $i\in I$, set $B_i={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J_i}{A}_j$ so that

by definition.

Now we construct a bijection $f$ from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$ to ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$. So consider any $a\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$ so that $a=⟨a_j\mid j\in J⟩$ where $a_j\in A_j$ for every $j\in J$. Now, for each $i\in I$, set $a_i^{\prime }=⟨a_j\mid j\in J_i⟩$, noting that clearly $j\in J={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{J}_i$ for each $j\in J_i$ so that $a_j$ has been defined as in the range of $a$. We then have that $a_j\in A_j$ for $j\in J_i$ so that $a_i^{\prime }\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J_i}{A}_j=B_i$. So set $b=⟨a_i^{\prime }\mid i\in I⟩$ so that clearly $b\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$, and set $f(a)=b$. Since $f(a)=b\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ for any $a\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$, we have that $f$ is indeed a function from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$ into ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$.

We claim first that $f$ is injective. So consider any $\alpha$ and $\beta$ in ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$ where $\alpha \ne \beta$. It then follows that $\alpha =⟨{\alpha }_j\mid j\in J⟩$ and $\beta =⟨{\beta }_j\mid j\in J⟩$ where both ${\alpha }_j$ and ${\beta }_j$ are in $A_j$ for any $j\in J$. Now, since $\alpha \ne \beta$ it follows that there is a $j_0\in J$ such that ${\alpha }_{j_0}\ne {\beta }_{j_0}$. Also there is an $i_0\in I$ such that $j_0\in J_{i_0}$ since $j_0\in J={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{J}_i$. Now let ${\alpha }_i^{\prime }=⟨{\alpha }_j\mid j\in J_i⟩$ and ${\beta }_i^{\prime }=⟨{\beta }_j\mid j\in J_i⟩$ for $i\in I$. We then have that ${\alpha }_{i_0}^{\prime }\ne {\beta }_{i_0}^{\prime }$ since $j_0\in J_{i_0}$ and ${\alpha }_{j_0}\ne {\beta }_{j_0}$. Clearly then $f(\alpha )=⟨{\alpha }_i^{\prime }\mid i\in I⟩$ and $f(\beta )=⟨{\beta }_i^{\prime }\mid i\in I⟩$ by definition so that $f(\alpha )\ne f(\beta )$ since $i_0\in I$ and ${\alpha }_{i_0}^{\prime }\ne {\beta }_{i_0}^{\prime }$. This shows that $f$ is injective since $\alpha$ and $\beta$ were arbitrary.

We also claim that $f$ is onto. Consider any $b\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ so that $b=⟨b_i\mid i\in I⟩$ where each $b_i\in B_i$ for $i\in I$. So, for any $i\in I$, we have that $b_i\in B_i={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J_i}{A}_j$ so that $b_i=⟨a_{\mathit{𝑖𝑗}}\mid j\in J_i⟩$ where each $a_{\mathit{𝑖𝑗}}\in A_j$ for $j\in J_i$. Now we construct a function $g$. So consider any $j_0\in J$ so that there is a unique $i_0\in I$ such that $j_0\in J_{i_0}$, where the uniqueness clearly follows from the fact that $⟨J_i\mid i\in I⟩$ are mutually disjoint. Then simply set $g(j_0)=a_{i_0{j}_0}\in A_{j_0}$ so that clearly $g\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$. If we then set $a_i^{\prime }=⟨g(j)\mid j\in J_i⟩=⟨a_{\mathit{𝑖𝑗}}\mid j\in J_i⟩=b_i$ for all $i\in I$, then $f(g)=⟨a_i^{\prime }\mid i\in I⟩=⟨b_i\mid i\in I⟩=b$. Since $b$ was arbitrary this shows that $f$ is indeed onto.

It may not have been obvious, but the uniqueness of $i_0\in I$ for any $j_0\in J$ (such that $j_0\in J_{i_0}$) when constructing $g$ was critical for this proof. To see why, suppose that for some $j_0\in J$ there are distinct $i_1$ and $i_2$ in $I$ such that $j_0\in J_{i_1}$ and $j_0\in J_{i_2}$. Then it could very well be that $a_{i_1{j}_0}\ne a_{i_2{j}_0}$ (though they would both be in $A_{j_0}$) and we would have to choose one to be $g(j_0)$. Supposing we choose $g(j_0)=a_{i_1{j}_0}$ then we would have $a_{i_2}^{\prime }=⟨g(j)\mid j\in J_{i_2}⟩$ so that $a_{i_2}^{\prime }\ne b_{i_2}$ since $a_{i_2}^{\prime }(j_0)=g(j_0)=a_{i_1{j}_0}\ne a_{i_2{j}_0}=b_{i_2}(j_0)$. If we had set $g(j_0)=a_{i_2{j}_0}$ instead then, by the same argument, we would have $a_{i_1}^{\prime }\ne b_{i_1}$. Clearly in either case this would break the proof since we would have $f(g)=⟨a_i^{\prime }\mid i\in I⟩\ne ⟨b_i\mid i\in I⟩=b$. In fact, in this case there would be no $g\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$ such that $f(g)=b$ since we cannot choose a value for $g(j_0)$ for which $a_i^{\prime }=b_i$ for all $i\in I$.

Returning from our digression, we have shown that $f$ is a bijection from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{j\in J}{A}_j$ to ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ so that

Therefore we have

as desired. □