prove $\bigcup_{n=1}^\infty A_n=\bigcup_{n=1}^\infty B_n$ prove$\bigcup_{n=1}^\infty A_n=\bigcup_{n=1}^\infty B_n$ if $A_i$ is a arbitrary set and $B_1=A_1$ and $B_n=A_n-\bigcup_{i=1}^{n-1} A_i$ for $n\ge 2$.
I have no ideas for solving it there was another question befor this that wanted to show:
$A_n\cap[A_n\cup B_n]=A_n$ that was easy to solve because $A_n\cup B_n=B_n$ and then $B_n \cap A_n=A_n$.I wrote this because I think it may helpful.
Thanks for your answers. 

I hope you can see why this is intuitively clear: you are essentially constructing the $B_n$'s so that they are disjoint and their union is that of the $A_n$'s.
To prove the claim formally: $\cup_{n=1}^\infty B_n \subset \cup_{n=1}^\infty A_n$ is clear. Let $x \in \cup_{n=1}^\infty A_n$. Consider the set $S_x = \{n: x \in A_n\}$. By the well-ordering principle $S_x$ has a least element, say $n_0$. Then $x \not \in A_n$ for all $n<n_0$ so $x\in B_{n_0}$. From here we get the right inclusion.
