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 $kappage 2^{aleph_0}$ 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 $kappage 2^{aleph_0}$ 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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