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It is consistent with ${\sf ZF}$ that the reals are the countable union of countable sets. Since any countable set is Borel, it follows that in any such pathological universe, let's call it $W$, every set is Borel; this has come out here before.

Since there is a countable basis for the Borel sets, it follows that in $W$ there is a countable family of sets of reals, such that the $\sigma$-algebra they generate is all of ${\mathcal P}({\mathbb R})$. I would like to see whether we can improve this slightly:

Suppose ${\mathbb R}=\bigcup_n A_n$, where each $A_n$ is countable. Can we further assert that, in addition, the $\sigma$-algebra the $A_n$ generate is ${\mathcal P}({\mathbb R})$? More precisely, can we modify the $A_n$ to a new family $A_n'$ of countable sets with union ${\mathbb R}$ and this generating property?

Asaf's answer shows I'm being (somewhat) sloppy in this formulation. So, does it make sense to ask for some intrinsic construction? (Something other than: We start with such and such countable family that generates the Borel sets, we code it into bits of the $A_n$, and then we are done. I confess I do not see immediately how to formalize "intrinsic", perhaps it makes no sense.)

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Andres, I think I must not understand the first question. If say $\{0,1\} \subseteq A_n$ for all $n$, then you can never separate $0$ and $1$ in the generated $\sigma$-algebra. – Clinton Conley Mar 24 '12 at 0:36
OK, now I understand. I thought you were asking whether the first sentence implied the second, not whether it was possible to satisfy both simultaneously. Good question! – Clinton Conley Mar 24 '12 at 0:49
Andres, I recently posted an answer to an question of mine,…, in which I cite a paper by Truss in which every set is Borel, but countable unions of countable sets are countable. Pretty weird, eh? Anyhow, I thought it might interest you to know that $\mathcal P(\mathbb R)$ can be countably generated even if $\mathbb R$ is not the countable union of countable sets. – Asaf Karagila Oct 19 '12 at 6:26
Thanks Asaf! Nice reference. – Andrés Caicedo Oct 20 '12 at 2:51
up vote 7 down vote accepted

We can cheat in the following way:

Suppose $\lbrace A_n\rbrace$ is a countable collection of countable sets, let $\lbrace B_n\rbrace$ be an enumeration of the countable collection of open intervals with rational end points.

Now consider the family $\lbrace A_n\cap B_m\mid n,m\in\omega\rbrace$. It is a countable family of countable sets, and every open interval $B_k$ is the countable union $\bigcup_n(A_n\cap B_k)$. Therefore the $\sigma$-algebra generated by this family is the entire power set of the real numbers.

The reason we cannot talk much about the general case is that given a countable family of countable sets $\lbrace A_n\rbrace$ we can make them disjoint via $A'_n=A_n\setminus\bigcup_{k < n} A_k$. The $A'_n$ are still countable, disjoint and have the same union - namely $\mathbb R$.

Now the $\sigma$-algebra the disjoint family generates is exactly $\lbrace\bigcup_{i\in I} A'_i\mid I\subseteq\omega\rbrace$. Since there can only be countably many singletons in this family we cannot separate all the points of $\mathbb R$ from one another and therefore we cannot generate $\mathcal P(\mathbb R)$.

So if the $A_n$ were already disjoint we could not have generated $\mathcal P(\mathbb R)$ without some sort of modification as in the first part.

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Yes, sure. But isn't there some intrinsic way of achieving this result? Perhaps this is as intrinsic as it gets... – Andrés Caicedo Mar 24 '12 at 1:34
Andres, I think it is. I was thinking what happens if we "disjointify" the family. The $\sigma$-algebra becomes just all the countable combinations of elements. – Asaf Karagila Mar 24 '12 at 8:23
A necessary and sufficient combinatorial condition (in this model) for $\{A_n\}$ to generate the full power set $\sigma$-algebra is for $\{A_n\}$ to separate points (i.e., for all $X \in [\mathbb{R}]^2$, there exists an $n$ such that $|A_n \cap X|=1$). With this in mind, Asaf's argument localizes in the sense that you can build $B_n \subseteq \mathbb{R}$ such that $\bigcup_{n < \kappa} A_n = \bigcup_{n < \kappa} A_n$ for $\kappa \in \{1,2, \ldots, \aleph_0\}$ and moreover $\{B_n\}$ separates points. [cont] – Clinton Conley Mar 24 '12 at 13:51
(Sorry, I meant $\bigcup_{n < \kappa} B_n = \bigcup_{n < \kappa} A_n$ above.) However, every way I see of performing this alteration uses either an external separating family for $\mathbb{R}$ or induces a simultaneous enumeration of the $A_n$s (which of course is no good in this model). But this is probably just a failure of imagination. – Clinton Conley Mar 24 '12 at 13:56

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