This is a problem from Sohrab, Basic Real Analysis

I hope my proof is correct. I have a question about the conclusion and the reason of the hint from the author of the book. The central large paragraph can be left unread.


Show that, if $\mathcal{A}$ is a $\sigma$-algebra containing an infinite number of sets, then this (cardinal) number is uncountable. Hint: Start by showing that $\mathcal{A}$ contains a sequence $(A_n)_{n=1}^\infty$ of (non-empty) pairwise disjoint sets and use Problem 9.


[[ I have removed three paragraphs, which were wrong, because they distracted some reader from the real question. The three paragraphs aimed to obtain a sequence $(B_i)$ of non-empty pairwise disjoint sets in $\mathcal{A}$. This the topic of this question ]]

We have $\mathbb{N}\sim X\sim \{B_i\mid i\in\mathbb{N}\}$. The set of all unions of the $B_n$, let call it $\mathcal{B}$, is a subset of $\mathcal{A}$, since $\mathcal{A}$ is a $\sigma$-algebra. $\mathcal{B}$ is equivalent to $\mathcal{P}(X)$. Summarizing: $|\mathcal{A}|\geq|\mathcal{B}|=|\mathcal{P}(X)|=|\mathcal{P}(\mathbb{N})|$ and therefore uncountable (Corollary 1.4.24).

Correction to the proof

As @PedroSánchezTerraf pointed out, the choice of $B_n$ is, in general, wrong. Meanwhile I removed the wrong part of the proof.

I opened another question here, with a possible construction of the $(B_i)$, that should replace paragraphs 1, 2, and 3 of the proof above.


  • Is the last paragraph correct?
  • Why should I use Problem 9:
    • Problem 9: Let $(A_n)_{n=1}^\infty$ be a partition of a non-empty set $U$; i.e., the $A_n$ are nonempty, pairwise disjoint, and $\bigcup A_n=U$. Show that the set of all unions of the $A_n$ (including the ``empty union'' which we define to be $\varnothing$) is a $\sigma$-algebra.
  • [option] in another question (from this webpage) that I used to make the large paragraph of the proof, and that I am not able to find anymore, the answer used twice the concept on measurable. Why? It should not be needed (it is the topic of chapter 10 of my book, and I am still at chapter 1)
  • $\begingroup$ I would say that the third paragraph of the proof incorrect. That's because you have some care choosing the $x_i$, you might end with only one $B_i$ (for instance, if there is a minimal nonempty member $A$ of $\mathcal{A}$, and all of your $x_i$ lie there). $\endgroup$ – Pedro Sánchez Terraf Aug 21 '16 at 22:02
  • $\begingroup$ @PedroSánchezTerraf, thank you for your correction, I try to amend the first part of the proof. $\endgroup$ – PeptideChain Aug 22 '16 at 3:58

Your last paragraph is correct (except that I'm not sure what the variable $X$ is supposed to mean--I guess it's defined in the omitted earlier paragraphs?).

The hint to use Problem 9 is just wrong and totally irrelevant to the question. Maybe whoever wrote the hint was thinking that Problem 9 also stated that the set of all unions of the $A_n$ is the smallest $\sigma$-algebra which contains each $A_n$, which you could then use in your argument to say that $\mathcal{B}\subseteq\mathcal{A}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.