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.

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.

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.