I'm stuck with the second exercise of chapter 9 from Jech and Hrbacek Introduction to set theory. It states:
If $J_i\,(i\in I)$ are mutually disjoint sets and $J=\bigcup_{i\in I}J_i$, and if $\kappa_j\,(j\in J)$ are cardinals, then $$\prod_{i\in I}\left(\prod_{j\in J_i}\kappa_j\right)=\prod_{j\in J}\kappa_j$$
Any help will be very appretiated.
Advanced greetings.
OK, I'll give it a go. It’s fun because there are lots of indices .
Define $G$ on $\Pi_{j\in J}^\ \kappa_j$ by $G:f\mapsto H_f$ where $H_f$ is a function on $I$ defined by $H_f:i\mapsto f|_{J_i}$. Now $f|_{J_i}$ is on $J_i$ and for $m\in\ J_i, f|_{J_i}(m)=f(m)\in\kappa_m$ as $f\in\Pi_{j\in J}^\ \kappa_j$. Hence $f|_{J_i}\in\Pi_{j\in J_i}^\ \kappa_j$. And $H_f$ is on $I$ and $H_f(i)\in\Pi_{j\in J_i}^\ \kappa_j$ so $H_f\in\Pi_{i\in I}^\ (\Pi_{j\in J_i}^\ \kappa_j)$ and $G:\Pi_{j\in J}^\ \kappa_j\rightarrow\Pi_{i\in I}^\ (\Pi_{j\in J_i}^\ \kappa_j)$.
Now we just need to show $G$ is 1-1 and ran $G=\Pi_{i\in I}^\ (\Pi_{j\in J_i}^\ \kappa_j)$. Suppose $f_1,\ f_2\in\mathrm{\Pi}_{j\in J}^\ \kappa_j$ and $G(f_1)=G(f_2)$. Then $H_{f_1}=H_{f_2}$. So $f_1|_{J_i}=f_2|_{J_i}$ for all $i\in I$. Now suppose $j\in J$. Then $j\in J_i$ for some $i\in I$. Hence $f_1|_{J_i}(j)=f_2|_{J_i}(j)$. And $f_1|_{J_i}(j)=f_1(j)$ and $f_2|_{J_i}(j)=f_2(j)$ and so $f_1(j)=f_2(j)$. Hence $f_1=f_2$ and $G$ is 1-1.
Now suppose $H\in\Pi_{i\in I}^\ (\Pi_{j\in J_i}^\ \kappa_j)$. Suppose $j\in J$. Then $j\in J_i$ for some $i$. Define the function $f$ on $J$ by $f:j\mapsto(H(i))(j)$ where $i$ is chosen so that $j\in J_i$. Then $G(f)=H$. Hence $\Pi_{i\in I}^\ (\Pi_{j\in J_i}^\ \kappa_j)\subseteq$ ran $G$. And we know $G:\Pi_{j\in J}^\ \kappa_j\rightarrow\Pi_{i\in I}^\ (\Pi_{j\in J_i}^\ \kappa_j)$ so ran $G\subseteq\mathrm{\Pi}_{i\in I}^\ (\mathrm{\Pi}_{j\in J_i}^\ \kappa_j)$ and so ran $G=\Pi_{i\in I}^\ (\Pi_{j\in J_i}^\ \kappa_j)$. Hence $\mathrm{\Pi}_{j\in J}^\ \kappa_j\approx\mathrm{\Pi}_{i\in I}^\ (\mathrm{\Pi}_{j\in J_i}^\ \kappa_j)$.
To show these sets are equal, you need to show if $\#A_i=\#B_i$ then $\#(\mathrm{\Pi}_{i\in I}A_i)=\#(\mathrm{\Pi}_{i\in I}B_i)$, which you should have already done.