Exercise 9.1.5

Proof. Suppose that $⟨A_i\mid i\in I⟩$ are mutually disjoint sets such that $\left|A_i\right|={\kappa }_i$ for all $i\in I$. Then by definition ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{\kappa }_i=\left|{\bigcup }_{i\in I}{A}_i\right|$. Also suppose that $B$ is a set such that $\left|B\right|=\lambda$.

We claim first that $B\times {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}{A}_i={\bigcup }_{i\in I}\left(B\times A_i\right)$. This is easy to show since, for any $x$ and $y$, we have

We then have

as desired. We note that $\left|{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}(B\times A_i)\right|={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i\in I}\left|B\times A_i\right|$ works since the sets $⟨B\times A_i\mid i\in I⟩$ are mutually disjoint. This is easy to see by considering $i$ and $j$ in $I$ where $i\ne j$. Then, if $(x,y)\in B\times A_i$ and also $(x,y)\in B\times A_j$, it follows that $y\in A_i$ and $y\in A_j$, which cannot be since $i\ne j$ and $⟨A_i\mid i\in I⟩$ are mutually disjoint. Hence it must be that there is no such ordered pair $(x,y)$ so that $B\times A_i$ and $B\times A_j$ are disjoint, which proves the result since $i$ and $j$ were arbitrary. □