If {$A_i$} and {$B_i$} are families of sets then

• Prove-($\bigcup\limits_{i}A_i$) $\bigcap$($\bigcup\limits_{j}B_j$)=$\bigcup\limits_{i,j}(A_i \bigcap B_j)$
• Prove-($\bigcap\limits_{i}A_i$) $\bigcup$($\bigcap\limits_{j}B_j$)=$\bigcap\limits_{i,j}(A_i \bigcup B_j)$
• If {$I_j$} is a family of sets with domain J,say;$\Bbb K$=$\bigcup\limits_{j}I_j$,and let {$A_k$} be the family of sets with domain $\Bbb K$ then prove $\bigcup\limits_{k\in\mathbb K } A_k$=$\bigcup\limits_{j\in\mathbb J}$($\bigcup\limits_{i\in\mathbb I_j}A_i$)

These exercises are from Halmos' Naive set theory text.Actually these arose while dealing with some theorems related to Measure theory.I'm not very good in set theory.I tried these via ven diagram,but unable to visualize them.

I need help in visualising the above problems through venn diagrams also in proving them analytically.

• A venn diagram with potentially infinitely many circles must look very confusing... Try proving double containment. Prove that the left side is a subset of the right side and vice versa. – JMoravitz Feb 22 '17 at 19:34
• You can prove these equations considering elements of sets and quantifiers. – ninjaaa Feb 22 '17 at 19:34
• I've removed inappropriate tags - descriptive set theory and measure theory are specific branches of mathematics unrelated to this question, and the set theory tag is reserved for questions at a higher level. – Noah Schweber Feb 22 '17 at 19:39

$x\in\left(\cup_iA_i\right)\cap\left(\cup_iB_i\right)\Leftrightarrow x\in\left(\cup_iA_i\right)$ and $x\in\left(\cup_iB_i\right)\Leftrightarrow x\in A_i$ for some $i$ and $x\in B_j$ for some $j\Leftrightarrow x\in A_i\cap B_j$ for some $i,j\Leftrightarrow x\in\cup_{i,j}A_i\cap B_j$. The second and third problems are solved similarly (most problems like this are, just a brute force showing that if $x$ is in one set it's in the other and vice versa).
When proving the equality of sets $S = T$, you usually
• let $s \in S$ and prove that $s \in T$, which means $S \subseteq T$, and
• let $t \in T$ and prove that $t \in S$, which means $T \subseteq S$, and
• since $S \subseteq T \subseteq S$, it follows $S=T$.
Let's do this for your first one and you can try the rest yourself. So we want to show that $$\left(\bigcup_{i}A_i\right) \cap \left(\bigcup_{j} B_j\right) = \bigcup_{i,j}\left(A_i \cap B_j\right)$$ so let $x \in \left(\bigcup_{i}A_i\right) \cap \left(\bigcup_{j} B_j\right)$. Therefore, $x \in \bigcup_{i}A_i$ and $x \in \bigcup_{j} B_j$. Hence, $\exists m,n$ such that $x \in A_m$ and $x \in B_n$. Thus, $x \in A_m \cap B_n$, and so $x \in \bigcup_{i,j}\left(A_i \cap B_j\right)$ as desired.
Now assume $x \in \bigcup_{i,j}\left(A_i \cap B_j\right)$. Then, $\exists m,n$ such that $x \in A_m$ and $x \in B_n$. (can you finish this?)