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I'm stuck on the following problem (Source: Real Analysis for Graduate Students; Exercise 2.6; Bass):

Suppose $\mathcal A$ is a $\sigma$-algebra with the property that whenever $A \in \mathcal A$, there exists $B, C \in \mathcal A$ with $B \cap C = \emptyset$, $B \cup C = A$, and neither $B$ nor $C$ is empty. Prove that $\mathcal A$ is uncountable.

I think there is the added assumption that this is only true for $A \in \mathcal A$ having at least two elements, so it doesn't hold for singletons and the emptyset?

I tried to show this by way of contradiction and say that $\mathcal A = \{A_k\}$ is countable, but I didn't see how this could get me to a contradiction.

The other approach I tried was to look at $X = B_1 \cup C_1$ where $B_1 \cap C_1 = \emptyset$, then look at $B_1 = B_2 \cup C_2$ where $B_2 \cap C_2 = \emptyset$ and so on. If I look at the $C_k$'s they are pairwise disjoint and I either have a finite number of them or an infinite number of them in which case I've created a countable sequence of pairwise disjoint nonempty elements of $\mathcal A$. This path looked promising, but I couldn't see what to do next. Any ideas?

As a secondary question: I think this result is supposed to be used to show that if $\mathcal A$ is a $\sigma$-algebra with infinitely many elements, then it's uncountable, but I wasn't able to show that the property mentioned above (the one I'm trying to show) was satisfied in this case.

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    $\begingroup$ Just a nitpick: no $\sigma$-algebra satisfies the hypotheses as stated, since $A = \emptyset \in \mathcal{A}$ cannot be decomposed into disjoint nonempty sets. $\endgroup$
    – user169852
    Sep 12, 2014 at 4:17
  • $\begingroup$ I addressed that in the post: I think we needed the added assumption that $A$ must have at least two elements, because it wouldn't hold for singletons or the emptyset. $\endgroup$ Sep 12, 2014 at 4:20
  • $\begingroup$ Sorry, I hadn't read that far yet. :-) I'm reading the rest now. $\endgroup$
    – user169852
    Sep 12, 2014 at 4:21
  • $\begingroup$ After reading the rest of your post, and Quang Hoang's answer, I agree with the argument: show that the $\sigma$-algebra is infinite, and use the result cited by Quang Hoang to show that this implies that it is uncountable. But we do need the additional hypothesis that $A$ is not the empty set. Also, notice that your argument shows that in fact $\mathcal{A}$ cannot contain any finite sets other than the empty set, because if it did, we can keep subdividing it until we reach a singleton, which can't satisfy the hypotheses. $\endgroup$
    – user169852
    Sep 12, 2014 at 4:28
  • $\begingroup$ @RobertCardona If we add the condition that the property only need hold for $|A|\ge 2$, then the claim becomes false. The power set of any finite set has this modified property but it is not uncountable; so we should not add that assumption into the question. $\endgroup$
    – Erick Wong
    Oct 5, 2016 at 11:25

1 Answer 1

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From the hypothesis, it follows that $\mathcal{A}$ is infinite, which you proved. The rest (also the secondary questions) follows from here.

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  • $\begingroup$ No, the sequence $B_k$ doesn't stop. It's clear that $$B_1\supsetneq B_2\supsetneq B_3\supsetneq \cdots.$$ $\endgroup$ Sep 12, 2014 at 4:34
  • $\begingroup$ I understand now! Thanks! $\endgroup$ Sep 12, 2014 at 4:34
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    $\begingroup$ Well, I understand why $B_k$ doesn't stop and why I have an infinite number of them, but I don't understand your proof in the linked answer. What is an $F_2$ vector space? Why does it being infinite imply it is infinite dimensional? and why does that imply it contains an uncountable set? Do you have an approach that uses just the definition of $\sigma$-algebra? This problem came out of the second chapter of a book that just defined what a $\sigma$-algebra was which suggests (big assumption here) it can follow in a self-contained manner from the definitions. $\endgroup$ Sep 12, 2014 at 4:47
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    $\begingroup$ Ah! Have you not studied linear algebra? $\mathbb F_2$ is the field of two elements $0,1$ where $1+1=0$. In that case, try to prove that the sequence $\{C_k\}$ is infinite. Since they are pairwise disjoint, the collection of their (sub)unions should be uncountable. $\endgroup$ Sep 12, 2014 at 4:53
  • $\begingroup$ @QuangHoang Your latter argument is correct and necessary. There are $\mathbb F_2$-linear algebras with countably infinite cardinality. The $\sigma$-algebra property is needed to prove uncountability, and your answer should be edited to include this. $\endgroup$
    – Erick Wong
    Oct 5, 2016 at 11:30

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