Exercise 9.1.2

Proof. Suppose that $⟨A_i\mid i\in I⟩$ are mutually disjoint sets such that $\left|A_i\right|={\kappa }_i$ for all $i\in I$. Similarly, suppose that $⟨B_i\mid i\in I⟩$ are mutually disjoint sets such that $\left|B_i\right|={\lambda }_i$ for all $i\in I$. It then follows by definition that

Now, we have $\left|A_i\right|={\kappa }_i\le {\lambda }_i=\left|B_i\right|$ so that there is an injective function $f_i:A_i\to B_i$ for all $i\in I$. With the help of the Axiom of Choice, we can choose one of these functions for each $i\in I$ and form the system of functions ${\left\{f_i\right\}}_{i\in I}$.

We claim that ${\left\{f_i\right\}}_{i\in I}$ is a compatible system of functions. To see this, consider any $i_1$ and $i_2$ in $I$. If $i_1=i_2$ then consider any $x\in \mathrm{dom}(f_{i_1})\cap \mathrm{dom}(f_{i_2})=A_{i_1}\cap A_{i_2}=A_{i_1}\cap A_{i_1}=A_{i_1}$. Then clearly $f_{i_1}(x)=f_{i_2}(x)$ since $i_1=i_2$ and $f_{i_1}=f_{i_2}$ is a function. On the other hand, if $i_1\ne i_2$, then we have that $\mathrm{dom}(f_{i_1})\cap \mathrm{dom}(f_{i_2})=A_{i_1}\cap A_{i_2}=\varnothing$ since $⟨A_i\mid i\in I⟩$ are mutually disjoint and $i_1\ne i_2$. Hence it is vacuously true that $f_{i_1}(x)=f_{i_2}(x)$ for all $x\in \mathrm{dom}(f_{i_1})\cap \mathrm{dom}(f_{i_2})$ since there is no such $x$. Since $i_1$ and $i_2$ were arbitrary, this shows that ${\left\{f_i\right\}}_{i\in I}$ is a compatible system (see Definition 2.3.10). It then follows from Theorem 2.3.12 that $f={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{f}_i$ is a function with domain ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}\mathrm{dom}(f_i)={\bigcup }_{i\in I}{A}_i$.

Though perhaps it may seem obvious, we show formally that $f(x)=f_i(x)$ for any $x\in A_i$ (for any $i\in I$). So consider any such $i\in I$ and $x\in A_i$. Then $(x,f(x))\in f={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{f}_i$ so that there is a $j\in I$ where $(x,f(x))\in f_j$. Suppose for a moment that $j\ne i$ so that $x\in \mathrm{dom}(f_j)=A_j$. Since $x\in A_i$ and $x\in A_j$ but $i\ne j$, this contradicts the fact that $⟨A_i\mid i\in I⟩$ are mutually disjoint. Hence it must be that $j=i$ so that $(x,f(x))\in f_j=f_i$. From this of course it follows that $f_i(x)=f(x)$ as desired.

We also claim that $f(x)\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ for any $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ so that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ can be the codomain of $f$. This is easy to show: consider any $x\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ so that there is an $i\in I$ where $x\in A_i$. It then follows that $f(x)=f_i(x)\in B_i$ since $f_i$ is a function from $A_i$ to $B_i$. Therefore we clearly have $f(x)\in {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$. This shows the result since $x$ was arbitrary.

We also claim 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 there are $i_1$ and $i_2$ such that $x_1\in A_{i_1}$ and $x_2\in A_{i_2}$. If $i_1=i_2$ then $f(x_1)=f_{i_1}(x_1)\ne f_{i_1}(x_2)=f_{i_2}(x_2)=f(x_2)$ since $f_{i_1}=f_{i_2}$ is injective. If $i_1\ne i_2$ then $f(x_1)=f_{i_1}(x_1)\in B_{i_1}$ whereas $f(x_2)=f_{i_2}(x_2)\in B_{i_2}$. Since $\left\{B_i\mid i\in I\right\}$ are mutually disjoint and $i_1\ne i_2$ it follows that $f(x_1)\ne f(x_2)$. Therefore $f$ is an injective function from ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i$ to ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{B}_i$ so that

as desired. □