Prove these sets are pairwise disjoint

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. I was wondering how to show this. I started by showing: \begin{equation} B_2 = A_2 \backslash A_1 = A_2 \backslash B_1 \end{equation} So obviously $B_1$ and $B_2$ are disjoint. From here, I should probably use induction to show every step, but I can't even show that $B_3$ is disjoint from $B_2$ and $B_1$, so I doubt I'll be able to show they're all pairwise disjoint. Here's what I got when I attempted to show $B_3$ is disjoint from $B_2$ and $B_1$: \begin{equation} B_3 = A_3 \backslash (A_1 \cup A_2) = (A_3 \backslash A_1) \cap (A_3 \backslash A_2) = (A_3 \backslash B_1) \cap (A_3 \backslash A_2) \end{equation} So I managed to show $B_3$ is disjoint from $B_1$, but not $B_2$. Any help would be appreciated :-)

• If $j>i$, $B_i\subset A_i$ and $A_i\cap B_j=\emptyset$ by definition. Thus $B_i\cap B_j=\emptyset$. No need for induction. – zuggg Nov 16 '16 at 9:55

You need to show that $B_n \cap B_m = \emptyset$ when $m \ne n$. Assume wlog $m < n$. Then $$B_n = A_n \setminus \bigcup \limits_{i = 1}^{n - 1} A_i = A_n \cap \bigcap \limits_{i = 1}^{n - 1} A_i^c =A_n \cap \left(\bigcap \limits_{i = 1}^{n - 1} A_i^c\right) \cap A_m^c.$$
This implies $$B_n \cap B_m \subset B_n \cap A_m = A_n \cap \left(\bigcap \limits_{i = 1}^{n - 1} A_i^c\right) \cap A_m^c \cap A_m = \emptyset.$$
• This is basically just a more formal way to prove that $B_j \cap A_i = \emptyset$ for $i < j$. – Dominik Nov 16 '16 at 10:23
I hope this helps $\ddot\smile$