# Thread: Probability question involving sequences of mutually exclusive events.

1. ## Probability question involving sequences of mutually exclusive events.

Let $\{A_1, A_2, A_3, ...\}$ be a sequence of events of a sample space $S$. Find a sequence $\{B_1, B_2, B_3, ...\}$ of mutually exclusive events such that for all $n \geq 1$, $\bigcup_{i=1}^m A_i = \bigcup_{i=1}^n B_i$.

2. Originally Posted by Zennie
Let $\{A_1, A_2, A_3, ...\}$ be a sequence of events of a sample space $S$. Find a sequence $\{B_1, B_2, B_3, ...\}$ of mutually exclusive events such that for all $n \geq 1$, $\bigcup_{i=1}^m A_i = \bigcup_{i=1}^n B_i$.
What about $B_i := A_i\backslash \bigcup_{j?

3. Thanks, Failure. You wouldn't be able to explain how you got your answer would you? I don't exactly understand the question... which means I don't really understand your answer either.

4. Originally Posted by Zennie
Thanks, Failure. You wouldn't be able to explain how you got your answer would you? I don't exactly understand the question... which means I don't really understand your answer either.
You want a sequence $B_i$ of mutually exclusive events (i.e. disjoint sets) that has the same union as the sequence $A_i$.
So let's start by setting $B_1 := A_1$. Now for $B_2$ we can take $A_2$ but in order to keep it disjoint from $B_1$ we take away $A_1$. - So $B_1$ and $B_2 := A_2\backslash A_1$ are disjoint, agreed?

Similarly, for $B_3$ we can take $A_3$, but in order to keep it disjoint from $B_{1,2}$ must take away their union, or, what amounts to the same thing $A_1\cup A_2$. We choose $B_3 := A_3\backslash (A_1\cup A_2)$.

So generally, we would like to define $B_i := A_i\backslash \bigcup_{j.

Observe that $B_i \subseteq A_i$, for all i, by definition. Also, if $i_1 we see that

$B_{i_1}\cap B_{i_2}\subseteq A_{i_1}\cap \left(A_{i_2}\backslash \bigcup_{j

Finally, to show that $\bigcup_i B_i = \bigcup_i A_i$.

First, from $B_i\subseteq A_i$ for all i it follows immediately that

$\bigcup_i B_i \subseteq \bigcup_i A_i$

Second, suppose that $x\in \left(\bigcup_i A_i\right) \backslash \left(\bigcup_i B_i\right)$. In that case there must exist a smallest $i_0$, such that $x \in A_{i_0}$.
But then we have that $x\in A_{i_0}\backslash \bigcup_{j and therefore $x\notin \left(\bigcup_i A_i\right) \backslash \left(\bigcup_i B_i\right)$, a contradiction. Thus there can be no such x, hence we have that $\bigcup_i B_i \supseteq \backslash \bigcup_i A_i$