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A standard homework in measure theory textbooks asks the student to prove that there are not countably infinite $\sigma$-algebras. The only proof that I know is via a contradiction argument which yields no estimate on the minimum cardinality of an infinite $\sigma$-algebra.

Given an a set $X$ of infinite cardinality $\kappa$, the $\sigma$-algebra of all co-countable subsets of $X$ is of cardinality $2^\kappa$ $\kappa^{\aleph_0}$. This example doesn't tell me whether there are $\sigma$-algebras of cardinality below $2^{\aleph_0}$, if I don't assume the Continuum Hypothesis.

My question is as the title says: Are there $\sigma$-algebras of every uncountable cardinality?

Edit: The combined answer with Stephen, Matthew proves that the cardinality of a $\sigma$-algebra is necessarily at least $2^{\aleph_0}$. Further, for each cardinality $\kappa\ge 2^{\aleph_0}$ with uncountable cofinality, the $\sigma$-algebra of countable (or cocountable) subsets of a set $X$ with cardinality $\kappa$, is of cardinality $\kappa$.

What is left is whether for $\kappa\ge 2^{\aleph_0}$ with $cf(\kappa)=\aleph_0$ are there $\sigma$-algebras of cardinality $\kappa$. (I changed the title to reflect this.)

Thanks Stephen, Matthew, Apollo, for the combined work!

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It seems to me that Matthew's answer contradicts your claim in the second paragraph, and that Matthew is right. – Harald Hanche-Olsen May 10 '10 at 13:23
    
@Harald: As Griffeth pointed out, it seems that Matthew has omitted in his calculation those subsets which are countably infinite. – user2529 May 10 '10 at 13:41
    
Dear Harald, Well, what about the sigma algebra of all countable or co-countable subsets of a countable set? Wouldn't Matthew's argument "prove" that it has countable cardinality? Best, Stephen – GS May 10 '10 at 13:43
    
Argh! I shall never again try to do mathematics when surrounded by loud conversation. I used to be able to do that, but now it seems I can't maintain my concentration in such circumstances. Oh well. – Harald Hanche-Olsen May 11 '10 at 3:05

A Boolean algebra is $\sigma$-complete if every countable subset has a least upper bound and a greatest lower bound. Every $\sigma$-algebra is a $\sigma$-complete Boolean algebra. Every (infinite) $\sigma$-complete Boolean algebra $B$ satisfies $|B|^{\aleph_0}=|B|$. (I am almost certain that a proof of this fact is in the Handbook of Boolean Algebras, Volume 1.) Since for every infinite cardinal $\kappa$, $\kappa<\kappa^{\text{cf}(\kappa)}$, there is no infinite $\sigma$-algebra of some size $\kappa$ of cofinality $\aleph_0$.

Something stronger is actually true: By a result of Koppelberg [Boolean algebras as unions of chains of subalgebras, Algebra Universalis, Vol. 7 (1977), 195-203], no $\sigma$-complete Boolean algebra is the union of a countable increasing chain of proper subalgebras. This also implies that the size of a $\sigma$-algebra cannot be of countable cofinality.

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up vote 3 down vote accepted

I'm really not used to thinking about this sort of question (which is why I'm giving it a shot...) but here goes.

Given a $\sigma$-algebra $A$ of subsets of a set $X$, assume that $A$ is infinite. Then $A$ is a poset, with inclusion as the order relation. Apply Zorn's lemma to the poset $P$ of all linearly ordered subsets of $A$ to conclude that there is a maximal linearly ordered subset of $A$; since $A$ is infinite this implies there is an infinite chain $S_1 \subset S_2 \subset \cdots$ of elements of $A$. The pairwise differences $S_{i+1}-S_{i}$ are pairwise disjoint and generate a $\sigma$-subalgebra of $A$ of size at least the size of the power set of $\mathbb{Z}_{\geq 0}$. So any infinite $\sigma$-algebra is at least that big.

Edit: Actually, this combined Matthew's argument below almost finishes the problem: if the cardinality of X is at least that of the power set of $\mathbb{Z}_{\geq 0}$, then the $\sigma$-algebra of countable or cocountable sets in $X$ has cardinality $X$ if $X$ is not of countable cofinality. So a cardinal number is the cardinality of a sigma algebra exactly if it is at least the cardinality of the continuum (no CH needed) and not of countable cofinality.

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So this would seem to suggest that it's not possible to get a cardinality between $\aleph_0$ and $2^\aleph_0$... – Matthew Daws May 10 '10 at 14:41
    
Right... but I don't know about which larger cardinals are achievable. – GS May 10 '10 at 14:42
    
A small nitpick: the set of negative integers is infinite and linearly ordered but it'll take quite a lot of effort to extract an infinite increasing chain out of it. – fedja May 11 '10 at 2:28
    
Dear fedja, Yes, but fortunately A is closed under taking complements! I have to admit that I was unsure (for at least 2 reasons) what the right level of detail was here. Best, Stephen – GS May 11 '10 at 9:19
    
Dear Matthew, Oops! I should have been paying more attention. I guess combining what we've written finishes off the problem (unless I've made a mistake somewhere...) Best, Stephen – GS May 13 '10 at 17:55

Edit: Yeah, this is nonsense! So really $|S|=|X|^{|\mathbb N|}=|X|^{\aleph_0}$.

Let X be a set of uncountable cardinality. Let S be the collection of countable or cocountable subsets of X. Then $X\in S$, clearly $S$ is closed under taking complements, and S is easily seen to be closed under countable unions.

Every co-countable set corresponds to a countable set, so the cardinality of S is equal to the cardinality of the collection of countable subsets of X, say C. We can write C as the countable union of subsets of X of size $1,2,3,\cdots$. But the collection of subsets of X of size n has cardinality |X| for any n, and so we conclude that S has size |X| as well.

So, it seems you can find a sigma algebra of any uncountable cardinality. Or did I make a mistake?

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I don't understand where the countably infinite subsets of $X$ fit into your countable union (they don't have size $n$ for any integer $n$). Am I being dense? – GS May 10 '10 at 13:30
    
So, in fact, we get cardinality $|X|^{\aleph_0}$. And, examining the argument, we see that ANY uncountable sigma-algebra is at least this, so has cardinal at least ${\alpha_0}^{\aleph_0} = 2^{\aleph_0}$. Thus, if CH fails, then the answer to the original question is NO. – Gerald Edgar May 10 '10 at 13:59
    
This argument doesn't work. The set of countable subsets of a cardinal $\kappa$ will be of size $\kappa^\omega$ which will be equal to $2^\omega \kappa$ for $\kappa$ with uncountable cofinality (and hence $>\kappa$ for any $\kappa<2^\omega$. For $\kappa$ of countable cofinality $\kappa^\omega>\kappa$. So in all of these cases your construction fails. However, for $\kappa$ of uncountable cofinality you will get a $\sigma$-algebra of cardinality $\kappa$. – Apollo May 10 '10 at 14:09
    
Yes, I agree, I did indeed make a mistake. Hopefully the edit shows this... – Matthew Daws May 10 '10 at 14:36
    
No, the conclusion is $|X|^{\aleph_0}$, not $2^{|X|}$. – Gerald Edgar May 10 '10 at 14:45

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