[Math] Are there sigma-algebras of cardinality $\kappa>2^{\aleph_0}$ with countable cofinality

Question

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!

Answer 1

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.