Exercise 9.1.6

Proof. First let $⟨B_i\mid i\in I⟩$ be mutually disjoint sets where $\left|B_i\right|=\left|A_i\right|$ for every $i\in I$. It then follows that ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}\left|A_i\right|=\left|{\bigcup }_{i\in I}{B}_i\right|$ by definition. For each $i\in I$ we can choose a bijection $f_i$ from $A_i$ to $B_i$ by the Axiom of Choice since $\left|A_i\right|=\left|B_i\right|$. We construct a function $f:{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i\to {\bigcup }_{i\in I}{B}_i$. For each $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ we have that $x\in A_j$ for some $j\in I$. We choose one such $j$, which requires the Axiom of Choice, and set $f(x)=f_j(x)$. Clearly $f(x)=f_j(x)\in B_j$ so that then $f(x)\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$, which shows that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ can be the codomain for $f$.

We show that $f$ is injective. So consider any $x_1$ and $x_2$ in ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ where $x_1\ne x_2$. Then we have chosen unique $j_1$ and $j_2$ where $f(x_1)=f_{j_1}(x)$ and $f(x_2)=f_{j_2}(x)$. If $j_1=j_2$ then we have $f(x_1)=f_{j_1}(x_1)\ne f_{j_1}(x_2)=f_{j_2}(x_2)=f(x_2)$ since $f_{j_1}=f_{j_2}$ is injective and $x_1\ne x_2$. If $j_1\ne j_2$ then $f(x_1)=f_{j_1}(x_1)\in B_{j_1}$ whereas $f(x_2)=f_{j_2}(x_2)\in B_{j_2}$. Since $j_1\ne j_2$ and $⟨B_i\mid i\in I⟩$ are mutually disjoint, it follows that $f(x_1)\ne f(x_2)$. Since this is true in both cases and $x_1$ and $x_2$ were arbitrary, it shows that $f$ is injective.

Since $f$ is injective we, we have

as desired. □