troubles proving $ \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=1}^{n-1} A_k\bigr)$ for sets $A_i$. How can we show
$$ \bigcup_{n\in\mathbb N}A_n = \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=1}^{n-1} A_k\bigr)$$
for any family of sets?
So let $x\in \bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=1}^{n-1} A_k\bigr)=\bigcup_{n\in\mathbb N}\bigl(A_n\cap\bigcap_{k=1}^{n-1} A_k^c\bigr)$. So $\exists n: x\in A_n\cap\bigcap_{k=1}^{n-1} A_k^c $ and therefore $x\in A_n \wedge x\in A_1^c\wedge\dots\wedge x\in A_{n-1}^c   $. So we get $x\in \bigcup_{n\in\mathbb N}A_n.$ 
What happens if $A_i$ isn't pairwise disjoint?
And how do you get the other way?
 A: Here's a (rather abbreviated) proof:
$$x\in \bigcup_{n \in \mathbb{N}}(A_n \setminus \bigcup_{k=1}^{n-1}A_k) \implies (\exists n \in \mathbb{N} : x\in A_n \setminus \bigcup_{k=1}^{n-1}A_k) \implies $$
$$\implies (\exists n \in \mathbb{N}:x\in A_n) \implies x \in \bigcup_{n\in \mathbb{N}}A_n$$
And for the other way:
$$ x \in \bigcup_{n\in \mathbb{N}}A_n \implies H:=\{k\in \mathbb{N}:x\in A_k\}\not= \emptyset$$
Then by the well ordering principle $H$ has a least element $n_0$; it follows that $x \in A_{n_0}$ and $x \notin \bigcup_{k=1}^{n_0-1} A_k$, so $$x\in A_{n_0} \setminus \bigcup_{k=1}^{n_0-1}A_k \implies x\in \bigcup_{n \in \mathbb{N}}(A_n \setminus \bigcup_{k=1}^{n-1}A_k)$$
I hope that is of use!
A: $x \in \bigcup_{n\in\mathbb N}A_n \Rightarrow x \in A_i$ for some $i \in \mathbb{N}$. Since we have at least $1$ such $i$ then we can take the minimal $j\in\mathbb{N}$ such that $x \in A_j$. Then $x \in A_j\setminus\bigcup_{k=1}^{j-1} A_k$. Hence in $\bigcup_{n\in\mathbb N}\bigl(A_n\setminus\bigcup_{k=1}^{n-1} A_k\bigr)$. This is one of the great features of natural numbers. Every non-empty subset of them has a minimal element.
