Are there sigma-algebras of cardinality $\kappa>2^{\aleph_0}$ with countable cofinality? 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!  
 A: 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.
A: 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. 
A: 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?
