Is the power set of the real line, $\mathcal P(\mathbb R)$, countably generated, i.e., is there a countable subclass $P'\subseteq \mathcal P(\mathbb R)$ such that $\mathcal P(\mathbb R) = \Sigma(P')$?
Is the power set of $\mathbb R$ a countably generated $\sigma$-algebra?
1 Answers
Given a collection $\Sigma_0$ of subsets of a set $X$, define recursively a transfinite sequence of length $\omega_1$ (the first uncountable ordinal) by setting (for $\alpha<\omega_1$) $$ \Sigma_{\alpha+1}=\Sigma_\alpha\cup\{X\setminus A\mid A\in\Sigma_\alpha\}\cup B_\alpha, $$ where $B_\alpha$ consists of all those subsets of $X$ that are a countable union of sets in $\Sigma_\alpha$. Also, set $$ \Sigma_\lambda=\bigcup_{\alpha<\lambda}\Sigma_\alpha $$ for $\lambda\le\omega_1$ a limit ordinal.
Then $\Sigma_{\omega_1}$ is the $\sigma$-algebra $\Psi$ generated by $\Sigma_0$. This is easily proved:
- By induction, each $\Sigma_\alpha$ is a subset of $\Psi$.
- On the other hand, $\Sigma_{\omega_1}$ is closed under countable sequences, because any countable collection of sets in $\Sigma_{\omega_1}$ actually belong to $\Sigma_\beta$ for some $\beta<\omega_1$; this is simply the fact that a countable union of countable sets is countable, which translates into the fact that, given a countable collection of sets in $\Sigma_{\omega_1}$, they all appear at a countable stage, and the supremum of these stages is still countable.
This means that $\Sigma_{\omega_1}$ is a $\sigma$-algebra, so it coincides with $\Psi$.
(It doesn't matter here but, on the other hand, you may really need to go all the way up to $\omega_1$ to get this. For example, if $\Sigma_0$ is the collection of open subsets of ${\mathbb R}$, then you do not reach the $\sigma$-algebra of Borel sets at any countable stage.)
Ok. By induction, if $\Sigma_0$ is countable, then each $\Sigma_\alpha$ for $\alpha$ countable has size at most $|{\mathbb N}^{\mathbb N}|=|{\mathbb R}|$, and $\Sigma_{\omega_1}$ therefore has size at most $|{\mathbb R}|\times\omega_1=|{\mathbb R}|$. This means you cannot reach the whole power set of the reals this way.
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1As a technical aside, this argument uses the axiom of choice in two ways: First, that a countable union of countable ordinals is countable, and second, that $\omega_1\le|{\mathbb R}|$. It is consistent with the axioms of set theory, excluding choice, that the reals are a countable union of countable sets, say ${\mathbb R}=\bigcup_n A_n$. This implies that *every* set of reals is a countable union of countable sets, $D=\bigcup_n (A_n\cap D)$. But I do not quite see whether we can extend this to get without choice that the whole power set of the reals could be countably generated. – 2012-03-23
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2To your comment, use the fact that the Borel algebra is generated by a countable basis and that every set is Borel. – 2012-03-23
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0Thank you, Andres. Your answer is really helpful. – 2012-03-23
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0@Joe: If this answer helped you solve your problem, you can (and should in this case) accept Andres' answer by clicking the transparent check mark underneath the up/down arrows to the left of the answer. – 2012-03-23
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0@AsafKaragila Yeah, of course. I was being stubborn, mostly, trying to see directly whether this can be obtained from the sets $A_n$ rather than a posteriori. – 2012-03-23
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0@Andres: Perhaps $A_n$'s with the addition of $\{q\}$ for $q\in\mathbb Q$? – 2012-03-23
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0@Asaf: Yes, I checked the up-arrow. – 2012-03-25
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0@Andres: Thank you so much for your answer to my last question. I have raised in the Question Section another question: Is there a Lebesque measurable choice function? Your response will be most appreciated. – 2012-03-25
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0@Joe: You can accept the answer, not by the up-arrow but the $\large\checkmark$ symbol right below the down-arrow. :-) – 2012-03-25