Exercise 9.1.12

Proof. First, suppose that $A$ is a set such that $\left|A\right|=\kappa$, and that $⟨B_i\mid i\in I⟩$ are mutually disjoint sets such that $\left|B_i\right|={\lambda }_i$ for each $i\in I$. Then, by the definitions of cardinal products and sums, we have

and

We now construct a bijective function F from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}^{B_i}$ to $A^{{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i}$, which clearly shows the desired result since we then have

So, for any $f=⟨f_i\mid i\in I⟩\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}^{B_i}$, we have that each $f_i$ is a function from $B_i$ to $A$. It then follows from the fact that $⟨B_i\mid i\in I⟩$ are mutually disjoint that ${\left\{f_i\right\}}_{i\in I}$ is a compatible system of functions. We then that $g={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{f}_i$ is a function from ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}\mathrm{dom}(f_i)={\bigcup }_{i\in I}{B}_i$ to $A$ by Theorem 2.3.12. Hence $g\in A^{{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i}$, and we of course set $F(f)=g$ so that $F$ is a function from ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}^{B_i}$ to $A^{{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i}$.

To show that $F$ is injective, consider any $f$ and $f^{\prime }$ in ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}^{B_i}$ where $f\ne f^{\prime }$. Let $g=F(f)$ and $g^{\prime }=F(f^{\prime })$. It then follows that there is an $i_0\in I$ such that $f_{i_0}\ne f_{i_0}^{\prime }$. Then, it has to be that there is a $b\in B_{i_0}$ where $f_{i_0}(b)\ne f_{i_0}^{\prime }(b)$. Clearly also $b\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ since $i_0\in I$ so that $b\in \mathrm{dom}(g)$ and $b\in \mathrm{dom}(g^{\prime })$ (since by the definition of $F$ we have that $g$ and $g^{\prime }$ are functions with domain ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$). Moreover, we have that $g(b)=f_{i_0}(b)$ and $g^{\prime }(b)=f_{i_0}^{\prime }(b)$, which again follows from the fact that $⟨B_i\mid i\in I⟩$ are mutually disjoint or equivalently that ${\left\{f_i\right\}}_{i\in I}$ is a compatible system. Hence we have $g(b)=f_{i_0}(b)\ne f_{i_0}^{\prime }(b)=g^{\prime }(b)$ so that $F(f)=g\ne g^{\prime }=F(f^{\prime })$. This shows that $F$ is injective since $f$ and $f^{\prime }$ were arbitrary.

To see that $F$ is onto, consider any $g^{\prime }\in A^{{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i}$. For each $i\in I$, define $f_i=g^{\prime }\upharpoonright B_i$, which is clearly a function from $B_i$ to $A$ so that $f_i\in A^{B_i}$. Then let $f=⟨f_i\mid i\in I⟩$ so that $F(f)=g={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{f}_i$, and clearly $f\in {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}^{B_i}=\mathrm{dom}(F)$. Now consider any $(x,y)\in g={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{f}_i$ so that there is an $i_0\in I$ where $(x,y)\in f_{i_0}=g^{\prime }\upharpoonright B_{i_0}$. Therefore $(x,y)\in g^{\prime }$ since obviously $g^{\prime }\upharpoonright B_{i_0}\subseteq g^{\prime }$. Consider next any $(x,y)\in g^{\prime }$ so that $x\in \mathrm{dom}(g^{\prime })={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ and $y=g^{\prime }(x)$. Then there is an $i_0\in I$ where $x\in B_{i_0}$ so that $(x,y)\in g^{\prime }\upharpoonright B_{i_0}=f_{i_0}$. Hence clearly $(x,y)\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{f}_i=g$ since $i_0\in I$. Therefore $g\subseteq g^{\prime }$ and $g^{\prime }\subseteq g$ so that $F(f)=g=g^{\prime }$, which shows that $F$ is onto since $g^{\prime }$ was arbitrary.

We have thus shown that $F$ is a bijection so that the result follows. □