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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 $\mathcal P(\mathbb R) = \Sigma(P')$?

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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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As 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. – Andrés E. Caicedo Mar 23 '12 at 6:18
To your comment, use the fact that the Borel algebra is generated by a countable basis and that every set is Borel. – Asaf Karagila Mar 23 '12 at 8:10
Thank you, Andres. Your answer is really helpful. – Joe Zhou Mar 23 '12 at 11:02
@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. – Asaf Karagila Mar 23 '12 at 11:34
@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. – Andrés E. Caicedo Mar 23 '12 at 14:00

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