# Detail of the proof that the cardinality of a $\sigma$-algebra containing an infinite number of sets is uncountable

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.

## Problem

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.

## Proof

[[ 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.

## Questions

• 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)
• 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). – Pedro Sánchez Terraf Aug 21 '16 at 22:02
• @PedroSánchezTerraf, thank you for your correction, I try to amend the first part of the proof. – 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 $\mathcal{B}\subseteq\mathcal{A}$.