Is there an infinite countable $\sigma$-algebra on an uncountable set

Let $\Omega$ be a set.

If $\Omega$ is finite, then any $\sigma$-algebra on $\Omega$ is finite.

If $\Omega$ is infinite and countable, a $\sigma$-algebra on $\Omega$ cannot be infinite and countable.

What about if $\Omega$ is not countable ? Is it possible to find an uncountable $\Omega$ with a $\sigma$-algebra that is infinite and countable ?

• Isn't $\{\emptyset,\Omega\}$ a finite $\sigma$-algebra for arbitrary $\Omega$? – Hagen von Eitzen Oct 22 '14 at 18:14
• @HagenvonEitzen I edited my question accordingly. I'm looking for infinite countable $\sigma$-algebras. – Gabriel Romon Oct 22 '14 at 18:16

3 Answers

Suppose that $\lvert \Omega\rvert\ge\aleph_0$, and $\mathscr M\subset\mathscr P(\Omega)$ is a $\sigma-$algebra. We shall show that: $$\textit{Either}\,\,\,\, \lvert\mathscr M\rvert<\aleph_0\quad or\quad \lvert\mathscr M\rvert\ge 2^{\aleph_0}.$$ Define in $\Omega$ the following relation: $$a\sim b\qquad\text{iff}\qquad \forall E\in\mathscr M\, (\,a\in E\Longleftrightarrow b\in E\,).$$ Clearly, "$\sim$" is an equivalence relation in $\Omega$, and every $E\in\mathscr M$ is a union of equivalence classes. Also, for every two different classes $[a]$ and $[b]$, there are $E,F\in\mathscr M$, with $E\cap F=\varnothing$, such that $[a]\subset E$ and $[b]\subset F$.

Case I. If there are finitely many classes, say $n$, then each class belongs to $\mathscr M$, and clearly $\lvert \mathscr M\rvert=2^n$.

Case II. Assume there are $\aleph_0$ classes. Fix a class $[a]$, and let $\{[a_n]:n\in\mathbb N\}$ be the remaining classes. For every $n\in\mathbb N$, there exist $E_n,F_n\mathscr\in M$, such that $[a]\subset E_n$, $[a_n]\subset F_n$ and $E_n\cap F_n=\varnothing$. Clearly, $[a]=\bigcap_{n\in\mathbb N} E_n\in\mathscr M$, and thus $\lvert \mathscr M\rvert=2^{\aleph_0}$.

Case III. If there are uncountably many classes, we can pick infinite countable of them $[a_n]$, $n\in\mathbb N$, and disjoint sets $E_n\in\mathscr M$, with $[a_n]\subset E_n$, (using the Axiom of Choice), and then realise that the $\sigma-$algebra generated by the $E_n$'s has the cardinality of the continuum and is a subalgebra of $\mathscr M$.

• The use of "some $E$" is very confusing here. – Asaf Karagila Oct 22 '14 at 18:37
• I have fixed it a little bit. – Yiorgos S. Smyrlis Oct 22 '14 at 18:42
• Yeah, that's clearer. – Asaf Karagila Oct 22 '14 at 18:42
• Case II, why each class also belongs to $\mathcal{M}$? I don't see it obviously. And case III, how to chose the disjoint sets $E_n$? – Xiang Yu Jan 3 '16 at 7:29
• @XiangYu See be updated answer. – Yiorgos S. Smyrlis Jan 3 '16 at 14:20

No. The same proof is used as in the case where $\Omega$ is countable and uncountable.

If $\cal B$ is an infinite $\sigma$-algebra, then $\cal B$ has at least the cardinality of the continuum.

The proof is as follows, since $\cal B$ is infinite, it has a countable subset $\{A_i\mid i\in\Bbb N\}$. If this countable set is a $\subseteq$-chain (without loss of generality, it's increasing) take $B_i=A_i\setminus\bigcup_{j<i}A_j$, and this is an infinite family of pairwise disjoint sets. Otherwise, it's not a chain, and without loss of generality doesn't contain an infinite chain either, and by similar a induction (although now it's gonna be slightly dirtier, you might have to skip a few indices, between $B_i$ and $B_{i+1}$) create an infinite family $B_i$ of pairwise disjoint non-empty sets.

Now it's easy to show that $\cal B$ has at least the cardinality of the continuum. If $D\subseteq\Bbb N$, take $B_D=\bigcup_{i\in D}B_i$. And it's quite easy to see that $D\neq D'$ implies that $B_D\neq B_{D'}$. $\quad \square$

A word on choice. $\tiny\textsf{(with regards to tomasz)}$

Note that this proof uses the axiom of choice. We used the fact that if $\cal B$ is infinite, then it has a countably infinite subset. In fact it is consistent that there is a $\sigma$-algebra which is infinite, but has no countably infinite subset. Of course this $\sigma$-algebra is not countable either.

If we are only interested in the answer to the original question, then the axiom of choice is not used anywhere. If there is a countably infinite subset, then the proof follows (note that all the choices above, except the $A_i$'s, are done by induction on the chosen sequence, so they are in fact AC-free); and if there is no countably infinite subset, then certainly $\cal B$ is not countable!

(One example of such $\sigma$-algebra that has no countably infinite subset, is the power set of an amorphous set; where amorphous means that every subset is finite or its complement is finite. Why is this a $\sigma$-algebra? From a countably infinite collection of subsets we can define an infinite co-infinite subset, in a way similar to the above induction from $A_i$ to $B_i$.)

• I was just about to post about this. I mean if the OP knows that for a countably infinite set $\Omega$ there are no countably infinite sigma algebras $\Sigma$. Then let $|\Omega|\ge \aleph_1$, and $\Sigma$, a sigma algebra such that $|\Sigma| = \aleph_0$. Let $\omega$ be a countably infinite subset of $\Omega$. Then $\Sigma \restriction \omega$ is a countably infinite sigma algebra on $\omega$, a contradiction to the OP's knowledge... Unless there are some details I'm missing – Rustyn Oct 22 '14 at 18:33
• Rustyn, but how can you assure that there is a countable subset of $\Omega$ that the restriction of $\Sigma$ to it is infinite? (That's a good approach, though, but I'd go for the approach showing that there is some quotient of $\Omega$ which is countable that the $\sigma$-algebra can be pulled to define a $\sigma$-algebra on that, instead.) – Asaf Karagila Oct 22 '14 at 18:36
• Oh @Asaf, Yeah that seems like the correct approach. My bad-- I was just spewing some naive-intuitiony stuff – Rustyn Oct 22 '14 at 18:37
• You're not saying about the fact that we're assuming axiom of choice? I'm disappointed. ;-) – tomasz Oct 22 '14 at 18:58
• @tomasz: I'm a bit sick, and tired. I should be sleeping now. But I've added to the answer, regardless of my condition. – Asaf Karagila Oct 22 '14 at 19:08

Let $\mathcal B$ be an infinite $\sigma$-algebra on a set $\Omega$. Partition $\Omega$ into two disjoint nonempty sets $A_1,B_1\in\mathcal B$. At least one of $\mathcal B\cap\mathcal P(A_1)$ and $\mathcal B\cap\mathcal P(B_1)$ is infinite; otherwise, if $|\mathcal B\cap\mathcal P(A_1)|=m\lt\aleph_0$ and $|\mathcal B\cap\mathcal P(B_1)|=n\lt\aleph_0$, we would have $|\mathcal B|=mn\lt\aleph_0$. We may assume that $\mathcal B\cap\mathcal P(B_1)$ is infinite. Next, partition $B_1$ into two disjoint nonempty sets $A_2,B_2\in\mathcal B$ so that $\mathcal B\cap\mathcal P(B_2)$ is infinite. Continuing in this way, we get an infinite sequence $A_1,A_2,A_3,\dots$ of pairwise disjoint nonempty elements of $\mathcal B$. (Every infinite Boolean algebra contains such a sequence; we haven't used the $\sigma$ yet.) Since $\mathcal B$ is a $\sigma$-algebra, the union of each subsequence belongs to $\mathcal B$, showing that $|\mathcal B|\ge2^{\aleph_0}$.

• Yes. That was my original plan, but I couldn't remember why you can keep the induction going. How silly of me. :-) – Asaf Karagila Oct 23 '14 at 1:45