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Let $A_\beta$, $\beta \in B$ be a family of pairwise disjoint events. Show that if $P(A_\beta) > 0 \ \ \forall \beta \in B$, then $B$ must be countable.

My work

Suppose $B$ is uncountable. If we can write $B$ as $B = \bigcup_{i=1}^\infty B_i$ for a countable number of sets $B_i$ such that each $B_i$ is countable, then we may write $$P\left(\bigcup_\beta A_\beta\right) = P\left(\bigcup_{i=1}^\infty\bigcup_{\beta \in B_i} A_\beta\right) = \sum_{i=1}^\infty\sum_{\beta \in B_i} P(A_\beta)$$

And since $P(A_\beta) > 0$ for all $\beta$, the innner sum does not vanishes and so the total sum diverges; but this is a contradiction since $\bigcup_\beta A_\beta \subset \Omega \implies P(\bigcup_\beta A_\beta) \le 1$

Now, does this make any sense? And is it true that every uncountable $B$ may be written in that form? If not, how to prove this statement?

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    $\begingroup$ The only sets that can be written as countable unions of countable sets are the countable sets. $\endgroup$ Dec 6, 2014 at 18:41
  • $\begingroup$ @AndresCaicedo I see.. thank you. Can you offer an hint on how to tackle the problem ? $\endgroup$
    – Ant
    Dec 6, 2014 at 18:43
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    $\begingroup$ Nitpick: "$P(A_\beta)>0\;\forall \beta\in B$" is not correct notation. Either write "$\forall \beta\in B:\;P(A_\beta)>0$" (or some punctuation variant thereof), or "$P(A_\beta)>0$ for all $\beta$". The symbol $\forall$ cannot go after the formula it quantifies. $\endgroup$ Dec 6, 2014 at 18:48
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    $\begingroup$ @AndresCaicedo Thank you for your help, I think I finally figured it out. $\endgroup$
    – Ant
    Dec 6, 2014 at 18:52
  • $\begingroup$ @HenningMakholm Thanks! Next time I'll bear that in mind $\endgroup$
    – Ant
    Dec 6, 2014 at 18:53

3 Answers 3

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Hint: Consider for each $n \in \Bbb{N}$ the set

$$ M_n := \{\beta \in B \mid P(A_\beta) \geq 1/n\}. $$

Show that each of these sets is in fact finite.

Why does this help you?

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  • $\begingroup$ Very nice. I'm going to accept the other answer because is more complete but thank you! :) $\endgroup$
    – Ant
    Dec 6, 2014 at 18:54
  • $\begingroup$ Sure, no problem :) $\endgroup$
    – PhoemueX
    Dec 6, 2014 at 18:55
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    $\begingroup$ "Why does this help you?" -- and what choice axiom do you need to use? ;-) $\endgroup$ Dec 7, 2014 at 3:10
  • $\begingroup$ @SteveJessop: Hehe, ok true :) But probably, this question would have caused more confusion than good for questions at this level. $\endgroup$
    – PhoemueX
    Dec 7, 2014 at 8:31
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    $\begingroup$ @PhoemueX: I gave you +1 though for using a weaker choice than anyone else here $\endgroup$ Dec 7, 2014 at 15:58
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One of the sets $\{\beta | P(A_\beta)>\frac{1}{n}\}$ would have to be uncountable if $B$ is uncountable and the assumption holds, by Pidgeonhole.

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  • $\begingroup$ Thank you for your help, I think I finally figured it out. $\endgroup$
    – Ant
    Dec 6, 2014 at 18:53
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Assume $B$ is uncountable. Define for all $k \in \mathbb{N}$ $$ B_k := \left\{ \beta \in B \mid P(A_\beta) \geq \frac{1}{k} \right\} $$ Now $$ B = \bigcup_{n=1}^{\infty} B_k $$ Because $B$ is uncountable, there must exist $k_0 \in \mathbb{N}$ s.t. the set $B_{k_0}$ is uncountable. Then $$ \sum_{\beta \in B} P(A_\beta) \geq \sum_{\beta \in B_{k_0}} P(A_\beta) \geq \sum_{\beta \in B_k} \frac{1}{k_0} \geq \sum_{n=1}^\infty \frac{1}{k_0} = \infty. $$

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  • $\begingroup$ perfect. Thank you! :) $\endgroup$
    – Ant
    Dec 6, 2014 at 18:55

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