Prove DeMorgan's Theorem for indexed family of sets. Let $\{A_n\}_{n\in\mathbb N}$ be an indexed family of sets. Then:
$(i) (\bigcup\limits_{n=1}^\infty A_n)' = \bigcap\limits_{n=1}^\infty (A'_n)$
$(ii) (\bigcap\limits_{n=1}^\infty A_n)' = \bigcup\limits_{n=1}^\infty (A'_n)$
I went from doing simple, straightforward indexed set proofs to this, and I don't even know where to start.
 A: It’s not really any different from proving the finite versions: just show that each of the sets is a subset of the other. For (i), for instance, you want to show that 
$$\left(\bigcup_{n\ge 1}A_n\right)'\subseteq\bigcap_{n\ge 1}A_n'\tag{1}$$ and that
$$\bigcap_{n\ge 1}A_n'\subseteq\left(\bigcup_{n\ge 1}A_n\right)'\;.\tag{2}$$
To show $(1)$, assume that $x\in\left(\bigcup_{n\ge 1}A_n\right)'$; then $x\notin\bigcup_{n\ge 1}A_n$. This means that for every $n\ge 1$, $x\notin A_n$, which by the definition of complement means that $x\in A_n'$ for every $n\ge 1$. But that’s exactly what it means to say that $x\in\bigcap_{n\ge 1}A_n'$, so I’ve just proved $(1)$.
To prove $(2)$, assume that $x\in\bigcap_{n\ge 1}A_n'$. From the definition of intersection this means that $x\in A_n'$ for every $n\ge 1$, and hence that $x\notin A_n$ for every $n\ge 1$. This in turn means that $x\notin\bigcup_{n\ge 1}A_n$, i.e., that $x\in\left(\bigcup_{n\ge 1}A_n\right)'$, so $(2)$ is also true. Finally the truth of $(1)$ and $(2)$ ensures that 
$$\left(\bigcup_{n\ge 1}A_n\right)'=\bigcap_{n\ge 1}A_n'\;.$$
I’ll leave (ii) to you; you should be able to use much of what I did here as a model.
A: If $a\in (\bigcup_{n=1}^{\infty}A_{n})'$ then $a\notin A_{n}$ for any $n\in \mathbb{N}$, therefore $a\in A_{n}'$ for all $n\in \mathbb{N}$.  Thus $a\in \bigcap_{n=1}^{\infty}A_{n}'$.  Since $a$ was arbitrary, this shows $(\bigcup_{n=1}^{\infty}A_{n})' \subset \bigcap_{n=1}^{\infty}A_{n}'$.  The other containment and the other problem are similar.
A: $(i)$ let $ x \in (\cup_{n=1}^\infty A_n)' \Rightarrow x \notin A_n $ for some $n$. Therefore, $x \in A_n$ for some $n$. and hence, $x \in \cap_{n=1}^\infty (A'_n)$
Now, using same reasoning, prove the other direction: $$\cap_{n=1}^\infty (A'_n) \subseteq (\cup_{n=1}^\infty A_n)'$$ This would give you the desired result. Second part is analogous.
A: My good sir! We need to show:
$$\left(\bigcup_{n\ge 1}A_n\right)'\subseteq\bigcap_{n\ge 1}A_n'$$ and 
$$\bigcap_{n\ge 1}A_n'\subseteq\left(\bigcup_{n\ge 1}A_n\right)'.$$
The following are all biconditional statements. See, if $x∈\left(\bigcup_{n\ge 1}A_n\right)'$, that is, if x is in the complement of all these sets collectively, then x certainly isn't in the union of all the sets, so we see $$x∉\left(\bigcup_{n\ge 1}A_n\right).$$ Now, if x is not in the union of all these indexed sets, then most certainly it is also not in the intersect of all these sets, or: $$x∉\left(\bigcap_{n\ge 1}A_n\right).$$ Then for a final step, another logical statement: if x is not in the intersection of all these indexed sets, then it is in the intersection of everything else (draw 3-part Venn-Diagrams to convince yourself this) thus, $$x∈\left(\bigcap_{n\ge 1}A_n'\right).$$
