It seems it's well known that if a sigma algebra is generated by countably many sets, then the cardinality of it is either finite or $c$ (the cardinality of continuum). But it seems hard to prove it, and actually hard to find a proof of it. Can anyone help me out?
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I am not sure I understand the statement of your question. The $\sigma$algebra on $[0,\infty)$ generated by all sets of the form $[0,n]$, $n\in\mathbb N$ is countable.\\You have mentioned Borel algebra in the title (but not in the body of you're question), so this is probably not what you want. – Martin Sleziak Oct 08 '11 at 16:02

7@Martin, that $\sigma$algebra doesn't look countable to me. Are you forgetting to close under complementation? – user83827 Oct 08 '11 at 16:14

1Thanks for the correction @ccc. – Martin Sleziak Oct 08 '11 at 16:54
2 Answers
Let's say the $\sigma$algebra on $X$ is generated by the sets $A_i \subseteq X$. For each subset $I$ of the natural numbers, consider the set $B_I = \bigcap_{i \in I} A_i \cap \bigcap_{i \notin I} (X \setminus A_i)$. For distinct sets $I$ and $J$, the corresponding sets $B_I$ and $B_J$ are disjoint. Now take cases: either only finitely many of the $B_I$ are nonempty, or infinitely many are. This will show that the $\sigma$algebra is either finite or has cardinality at least that of the continuum.
To show that the $\sigma$algebra cannot have cardinality strictly above that of the continuum is a bit more involved. I can't come up with an approach avoiding transfinite induction up the Borel hierarchy. Here's a sketch of what I have in mind:
We build an increasing family $S_\alpha$ of subsets of the power set of $X$, as $\alpha$ ranges over the countable ordinals. In the end, $\bigcup_{\alpha < \omega_1} S_\alpha$ will be a $\sigma$algebra of size at most continuum containing our countably many generators (in fact, it will be the $\sigma$algebra they generate, but that's just an added bonus). We start by setting $S_0$ to equal the (countable) set of generators. Given $S_\alpha$, we let $S_{\alpha+1}$ be the collection of subsets which can be written as countable unions of the form $\bigcup_i A_i \cup \bigcup_j (X \setminus B_j)$, where $A_i$ and $B_j$ are chosen from $S_\alpha$. Note that if $S_\alpha \leq 2^{\aleph_0}$, then $S_{\alpha+1} \leq 2^{\aleph_0}$ as well (since there are only continuum many choices of ways to write the union: this is essentially the cardinal equality $(2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0}$). For limit ordinals $\lambda$, let $S_\lambda = \bigcup_{\alpha < \lambda} S_\alpha$. This will again satisfy $S_\lambda \leq 2^{\aleph_0}$ provided each $S_\alpha$ in the union does.
Finally, we see $\bigcup_{\alpha<\omega_1} S_\alpha$ has cardinality at most that of the continuum, since $\aleph_1 \cdot 2^{\aleph_0} = 2^{\aleph_0}$. Moreover, it is closed under the $\sigma$algebra operations since any countable sequence of elements is accounted for in some $S_\alpha$ (with $\alpha < \omega_1$).
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1I don't have the references on hand, but surely this appears in either Kechris' or Srivastava's text on descriptive set theory (possibly as an exercise). In the meantime I will sketch the argument I have in mind in the answer. – user83827 Oct 08 '11 at 23:05

1Thanks! So it seems impossible to prove it without the axiom of choice? – Syang Chen Oct 09 '11 at 17:15

Well certainly some amount of choice is required, since (modulo large cardinals) it's consistent with ZF that the usual Borel $\sigma$algebra on the reals is equinumerous with $2^{2^{\aleph_0}}$ (see, e.g., Gitik's model in which everything is singular). I'm not sure whether, say, uncountable choice principles are really required, even though I tacitly used them in the sketch above. – user83827 Oct 09 '11 at 20:12

@Xianghong, ccc: Arnold Miller wrote a paper about long Borel hierarchies without the axiom of choice. He shows that it is possible to have (without large cardinals) the length at $\omega_2$ or at any successor of a limit ordinal in $(\omega_1,\omega_2]$, all this while having $\omega_1$ singular. He goes on to suggest that under some large cardinal assumptions we can get even higher. – Asaf Karagila Oct 15 '11 at 21:39

@Asaf: Really? But ccc pointed out above that it is consistent with ZF. What extra assumption does Miller make? – Syang Chen Oct 21 '11 at 02:32

2@Xianghong: For the hierarchy to be longer than $\omega_2$ you would need both $\omega_1$ and $\omega_2$ to be singular. This is approximately a Woodin cardinal in consistency strength. – Asaf Karagila Oct 21 '11 at 09:02

@Xianghong: Also, Gitik model assumes ZF+a proper class of strongly compact cardinals. – Asaf Karagila Oct 21 '11 at 14:12

2When is induction done over *all* ordinals and when over countable ordinals only? – Rudy the Reindeer Jan 16 '12 at 14:36

@MattN. When we want to prove something about ordinals, we usually use induction over ALL ordinals. On the other hand, when we are proving something about a particular set using transfinite induction (and transfinite recursion), we use the set of all ordinals small than a particular ordinal, say $\gamma$. In this answer, we need to choose $\gamma$ to be large enough to reach all Borel measurable sets, yet small enough so that we add at most continuum many new subsets only at most continuum many times. – Jisang Yoo Jan 04 '14 at 15:29

We choose $\gamma$ to be the first uncountable ordinal because that's when $\cup_{\alpha < \gamma} S_\alpha$ becomes a sigma algebra, thereby reaching all Borel measurable subsets. This trick of choosing $\gamma$ to be an initial ordinal also appears in the construction of a subset of the plane crossing every line exactly twice, or other similar constructions. – Jisang Yoo Jan 04 '14 at 15:30

As written, the argument in the first paragraph doesn't work. Take for instance $X=\mathbb N$, the $\sigma$algebra $\mathcal P(X)$, and $A_n=\{n\}$. Then $B_I=\emptyset$ for all nonempty $I$, and no conclusion can be drawn. – Martin Argerami Jan 20 '21 at 04:28

@AsafKaragila May I ask where the axiom of choice is needed in the above proof? Thanks! – Jiu Jan 21 '22 at 10:44

1@Jiu: Numerous places. For example, $\aleph_1\cdot2^{\aleph_0}$ could be strictly larger than both; the cardinality of $\{A\subseteq\Bbb R\mid A\text{ is countable}\}$ could be larger than $2^{\aleph_0}$ which means that the argument about the cardinality of each "level" having size at most $2^{\aleph_0}$ also requires AC, and that the hierarchy stabilises at stage $\omega_1$ also requires that $\omega_1$ is a regular cardinal, which again requires a bit of AC. – Asaf Karagila Jan 21 '22 at 10:46

Is it enough to define $S_{\alpha+1}$ as the set that contains all countable union and complement of elements from $S_{\alpha}$? – Jiu Jan 21 '22 at 12:35
It is easy to prove that the $\sigma$algebra is either finite or has cardinality at least $2^{\aleph_0}$. One way to prove that it has cardinality at most $2^{\aleph_0}$, without explicitly using transfinite recursion, is the following. It is easy to see that it is enough to prove this upper bound for a "generic" $\sigma$algebra, e.g., for the Borel $\sigma$algebra of $\{0,1\}^{\omega}$, or for the Borel $\sigma$algebra of the Baire space $\mathcal{N} = \omega^{\omega}$. Note that $\mathcal{N}$ is a Polish space, so we can talk about analytic subsets of $\mathcal{N}$. Every Borel subset is an analytic subset of $\mathcal{N}$ (in fact, $A \subseteq \mathcal{N}$ is Borel if and only if $A$ and $X \setminus A$ are both analytic). So it is enough to prove that $\mathcal{N}$ has $2^{\aleph_0}$ analytic subsets. Now use the theorem stating that every analytic subset of $\mathcal{N}$ is the projection of a closed subset of $\mathcal{N} \times \mathcal{N}$. Since $\mathcal{N} \times \mathcal{N}$ has a countable basis of open subsets, it has $2^{\aleph_0}$ open subsets, so it has $2^{\aleph_0}$ closed subsets. So $\mathcal{N}$ has $2^{\aleph_0}$ analytic subsets.
The proof using transfinite recursion might be simpler, but I think the analytic subset description gives a slightly different, kind of direct ("less transfinite") view on the Borel sets, that could be useful to know.
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