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!