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. 
 A: A start: The right-hand side is clearly a subset of the left-hand side, because $B_n\subseteq A_n$.
To show that the left-hand side is a subset of the right-hand side, pick an element $x$ of the left-hand side. Then there is a smallest $n$ such that $x\in A_n$. Use this to show $x$ is in the RHS.
A: 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.
A: 
Claim 1: $\bigcup_{n=1}^{\infty} B_n\subseteq \bigcup_{n=1}^{\infty} A_n$.

Proof: Pick any $x\in \bigcup_{n=1}^{\infty} B_n$. Then, there exists some $n^*\in\mathbb N$ such that $x\in B_{n^*}$. Since $B_n\subseteq A_n$ for every $n\in\mathbb N$ by construction, it follows that $x\in A_{n^*}\subseteq \bigcup_{n=1}^{\infty} A_n$. $\blacksquare$

Claim 2: $\bigcup_{n=1}^{\infty} A_n\subseteq \bigcup_{n=1}^{\infty} B_n$.

Proof: Suppose that $x\in \bigcup_{n=1}^{\infty} A_n$. Then, there exists some $n^*\in\mathbb N$ such that $x\in A_{n^*}$. Define $n^{\circ}$ as the least positive integer satisfying $x\in A_{n^{\circ}}$. Formally, $$n^{\circ}\equiv\min\{n\in\mathbb N\,|\,x\in A_n\}.$$ Note that the set over which the minimum is taken is not empty since it contains $n^*$.
If, on the one hand, $n^{\circ}=1$, then $A_{n^{\circ}}=B_{n^{\circ}}$, so that  $x\in B_{n^{\circ}}$.
On the other hand, if $n^{\circ}\geq 2$, then, by the definition of $n^{\circ}$, $x$ is not contained in $A_n$ for any $n\in\{1,\ldots, n^{\circ}-1\}$, but it is contained in $A_{n^{\circ}}$. Formally, $$x\in A_{n^{\circ}}-\left(\bigcup_{i=1}^{n^{\circ}-1}A_i\right)=B_{n^{\circ}}.$$ In either case, $x\in B_{n^{\circ}}\subseteq\bigcup_{n=1}^{\infty} B_n$. $\blacksquare$
A: Here is an easier way to make the main claim.  If $x\in\bigcup_n A_n$, by well-ordering of the integers, we know there is a smallest $n$ so that $x\in A_n$. Then $$x\in A_n - \bigcup_{k=1}^{n-1}A_k = B_n.$$
