Show these unions are equal Let $A_1, A_2, ... $ be a sequence of subsets of $X$. Let $B_1 = A_1$ and define: \begin{equation}
B_n = A_n \backslash \bigcup_{i=1}^{n-1} A_i
\end{equation}
Show that $B_1, B_2, ...$ are pairwise disjoint (already did this, thanks to the latest post), and that 
\begin{equation}
\bigcup_{i=1}^n B_i = \bigcup_{i=1}^n A_i 
\end{equation}
And also show that: 
\begin{equation}
\bigcup_{i=1}^\infty B_i = \bigcup_{i=1}^\infty A_i 
\end{equation}
I showed the first part in a way, which I don't find too 'aesthetic', here's how:
\begin{equation}
\bigcup_{i=1}^n B_i = \bigcup_{i=1}^n (A_i \backslash \bigcup_{j=1}^{i-1} A_j) 
= A_1 \cup (A_2 \backslash A_1) \cup (A_3 \backslash (A_2 \cup A_1)) \cup ... \cup (A_n \backslash (A_{n-1} \cup .... \cup A_1)) = \bigcup_{i=1}^n A_i 
\end{equation}
This is not too pretty in my opinion, but I think it's pretty clear as the subtraction and unions "cancel out". If there's a better solution, please let me know! :-)
For the second part, I have no idea what to do, as the infinite part confuses me. I feel like I could do the same, but yet again, no... 
 A: The infinite part is actually quite simple. Your proof of the finite part is fine, so I shall deal only with the former.
Suppose that $x \in \displaystyle\bigcup_{i=1}^\infty B_i$. Then, $x \in B_n$ for some $n$, so that $x \in \displaystyle\bigcup_{i=1}^n B_i$. Now, given that $\displaystyle\bigcup_{i=1}^n B_i = \displaystyle\bigcup_{i=1}^n A_i$, we have that $x \in A_i$ for some $i \leq n$, hence $x \in \displaystyle\bigcup_{i=1}^\infty A_i$. Therefore $\displaystyle\bigcup_{i=1}^\infty B_i \subseteq\displaystyle\bigcup_{i=1}^\infty A_i$. The reverse containment follows similarly.
For the finite part, suppose that $x \in \displaystyle\bigcup_1^n B_i$. Then $x \in B_j$ for some $j$. But then $B_j \subset A_j$ since $B_j$ is $A_j$ minus some elements, which is still a subset of $A_j$. Hence, $x \in A_j$, so $x \in \displaystyle\bigcup_1^n A_i$.
For the reverse, if $x \in \displaystyle\bigcup_1^n A_i$, then $x \in A_j$ for some $j$. Let $S$ be the set of all numbers $j$ less than $n$ such that $x \in A_j$. $S$ has a least element, say $j_0$. Now, by $j_0$ being the least element in $S$, $x \in A_{j_0}$ and $x \notin A_k$ for any $k < j_0$. This means that $x \in A_{j_0} \backslash (\cup_1^{j_0-1} A_k)$, but then this is $B_{j_0}$, so that $x \in B_{j_0}$, hence $x \in \displaystyle\bigcup_1^n B_i$. This completes the other direction, and they are equal.
A: A more formal proof of the finite part could look like this: 
Clearly $B_i \subset A_i$ for all $i$, which implies $\bigcup \limits_{i = 1}^n B_i \subset \bigcup \limits_{i = 1}^n A_i$. So see the reverse inclusion, let $x \in \bigcup \limits_{i = 1}^n A_i$. Then there exists a smallest index $j$ so that $x \in A_j$. But this means that $x \in A_j$ and $x \notin A_i$ for $i < j$, so $x \notin \bigcup \limits_{i = 1}^{j - 1} A_i$. But this means $x \in B_j \subset \bigcup \limits_{i = 1}^n B_i$.
