Is there an infinite countable $\sigma$-algebra on an uncountable set Let $\Omega$ be a set.
If $\Omega$ is finite, then any $\sigma$-algebra on $\Omega$ is finite.
If $\Omega$ is infinite and countable, a $\sigma$-algebra on $\Omega$  cannot be infinite and countable.

What about if $\Omega$ is not countable ? Is it possible to find an uncountable $\Omega$ with a $\sigma$-algebra that is infinite and countable ?


Let $\mathcal B$ be an infinite $\sigma$-algebra on a set $\Omega$. Partition $\Omega$ into two disjoint nonempty sets $A_1,B_1\in\mathcal B$. At least one of $\mathcal B\cap\mathcal P(A_1)$ and $\mathcal B\cap\mathcal P(B_1)$ is infinite; otherwise, if $|\mathcal B\cap\mathcal P(A_1)|=m\lt\aleph_0$ and $|\mathcal B\cap\mathcal P(B_1)|=n\lt\aleph_0$, we would have $|\mathcal B|=mn\lt\aleph_0$. We may assume that $\mathcal B\cap\mathcal P(B_1)$ is infinite. Next, partition $B_1$ into two disjoint nonempty sets $A_2,B_2\in\mathcal B$ so that $\mathcal B\cap\mathcal P(B_2)$ is infinite. Continuing in this way, we get an infinite sequence $A_1,A_2,A_3,\dots$ of pairwise disjoint nonempty elements of $\mathcal B$. (Every infinite Boolean algebra contains such a sequence; we haven't used the $\sigma$ yet.) Since $\mathcal B$ is a $\sigma$-algebra, the union of each subsequence belongs to $\mathcal B$, showing that $|\mathcal B|\ge2^{\aleph_0}$.
