Given a sequence of events $E_1,E_2,....E_n$ (which may or may not be mutually exclusive) and another disjoint sequence of events $F_1,F_2,...F_n$.For any $n≥1$, $$\ \bigcup_{i=1}^n\textstyle{E\scriptstyle {i}}\textstyle{=}\bigcup_{i=1}^n\textstyle{F\scriptstyle {i}}$$ which means that the sequence $F_i$ has the following properties:

  • The $F_i$'s should be mutually exclusive and

  • the union of first $n$ $E_i$'s should be exactly equal to the union of first $n$ $F_i$ for all positive integers $n$.


2 Answers 2


Big hint and essentially the answer: you can take any sequence of sets $\left(E_i\right)_{i\in\mathbb{N}}$ and turn them into a disjoint sequence of sets $\left(F_i\right)$ by successively removing all previous members of the sequence from the current member, that is: $$F_i := E_i \backslash \bigcup\limits_{k <i}E_i \text{.}$$

The only thing left to proof is that $\forall n$: $\bigcup\limits_{i=0}\limits^{n} E_i = \bigcup\limits_{i=0}\limits^{n} F_i$, I'll leave that up to you.


Hint: Consider $$E_{n}-\bigcup_{n=1}^{n-1} Ei$$

Edit: I'll do the special case of a single union in gory detail, see if you can translate the language:

Let $F_1 :=E_1$ and$F_2:=E_2-E_1=E_2 \cap E_1^c$. Of course these are disjoint by construction.

Using Demorgan's laws, we also see that: $$F_1 \cup F_2=E_1 \cup(E_2 \cap E_1^c)=(E_2 \cup E_1) \cap (E_1^c \cup E_1)=E_2 \cup(E_1 \cap E_1^c)\cup E_1=E_2 \cup E_1.$$

  • $\begingroup$ it means I assumed that F1 =E1 and than by the mathematical induction ,the union of Ei from I=1 till n equal to the union of Fi from I=1 till n and shows that it hold for n+1's case ? $\endgroup$
    – steve chu
    Oct 1, 2016 at 10:05
  • $\begingroup$ @ccnotes edited so that perhaps it's more clear. You don't need induction (although that would work just fine.) This is the idea, although you don't need to use complements, they're just nicer to work with for me. $\endgroup$ Oct 1, 2016 at 12:10
  • $\begingroup$ but you just shows for the case n equal to one ,will it hold for the others n?Thanks $\endgroup$
    – steve chu
    Oct 1, 2016 at 12:53
  • $\begingroup$ that's what you have to do. It will hold as well, try to follow the reasoning. $\endgroup$ Oct 1, 2016 at 13:02
  • $\begingroup$ ok let me try to following your logic thanks $\endgroup$
    – steve chu
    Oct 1, 2016 at 13:07

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