0

I'm trying to prove Boole's inequality

$$P\left(\ \bigcup_{i=1}^\infty A_i\right) \leq \sum_{i=1}^\infty P(A_i)$$

by constructing $\{B_1, B_2,..., B_n\}$ such that $F_0 = \emptyset$, $F_n =\bigcup_{i=1}^n A_i$ and $B_i = A_{i} - F_{i-1}$. I have showed by induction that for $n \geq 1$, we have

$$ \bigcup_{i=1}^n A_i = \bigcup_{i=1}^n B_i.$$

Since for all $i\neq j$, $B_i \cap B_j = \emptyset$, therefore $\{B_1, B_2,..., B_n\}$ is a partition of $F_n$.

I am stuck at the step where I must show that this is true for $\infty$:

$$ \bigcup_{i=1}^\infty A_i = \bigcup_{i=1}^\infty B_i.$$

RobPratt
  • 45,619
Heng Wei
  • 419
  • We have $x\in \bigcup_{i=1}^\infty A_i$ iff there exists $j\in \mathbb{N}$ such that $x\in A_j$. Now use your result for finite union. – Severin Schraven May 22 '23 at 21:53
  • But $j$ could be arbitrarily big no? – Heng Wei May 22 '23 at 21:56
  • Well, yes, but why would that bother us? You have shown the equality for any $n$ (as large as you like). – Severin Schraven May 22 '23 at 21:58
  • Then why can't I directly use the inclusion-exclusion identity that is true for $n$? – Heng Wei May 22 '23 at 22:07
  • What do you mean by that? – Severin Schraven May 22 '23 at 22:08
  • Is the inclusion-exclusion identity true for an infinite amount of sets? But I realize even if it's true, it's not obvious Boole's inequality would be true. Coming back to your original point, I don't see why an union of infinite $A_i$'s would be equal to an union of finite $A_i$'s. I understand the latter is a subset of the former, but I don't see why the opposite is true – Heng Wei May 22 '23 at 22:24
  • 2
    By definition we have $x\in \bigcup_{i=1}^\infty$ iff there exists some $j\in \mathbb{N}$ such that $x\in A_j$ (an element is in the infinite union iff it is some of the sets that we are taking the union over). However, we have then $$x\in A_j \in \bigcup_{i=1}^j A_i = \bigcup_{i=1}^j B_i \subseteq \bigcup_{i=1}^\infty B_i.$$ Thus, we get $$ \bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty B_i.$$ By interchanging the roles of $A_i$ and $B_i$ we get that the infinite unions are the same. – Severin Schraven May 22 '23 at 22:31
  • I think I’m struggling to understand the concept of infinity here. Since there are infinite sets, isn’t it possible that we never reach the set that contains $x$? – Heng Wei May 22 '23 at 23:07
  • 1
    No, that is not possible. The infinite union is defined (as written above) as "everything that is contained in any of the $A_i$". So if it is in the infinite union, then it must be in some of the $A_i$ and we are in great shape :) – Severin Schraven May 22 '23 at 23:10
  • 2
    Ok I understand now! Thank you! – Heng Wei May 22 '23 at 23:40
  • 1
    @AnneBauval yes – Heng Wei May 23 '23 at 07:02

0 Answers0