Are there $\sigma$-rings countable?

Question

I know that a $\sigma$-algebra $\mathfrak{A}$ is finite or non-countable. I was reviewing the proof, and I did not find it necessary that $X\in\mathfrak{A}$. So, Is this true for $\sigma$-rings?

Answer 1

Yes, the result holds for $\sigma$-rings as well, with few if any modifications to the proof (depending on exactly what your proof is). In fact, you can prove the $\sigma$-ring case directly from the $\sigma$-algebra case. If $\mathfrak{A}$ is an infinite $\sigma$-algebra, let $A_1,A_2,\dots$ be an infinite sequence of distinct elements of $\mathfrak{A}$ and let $Y=\bigcup_{n\in\mathbb{N}} A_n$. Then $Y\in\mathfrak{A}$, and $\mathfrak{B}=\mathfrak{A}\cap P(Y)$ is a $\sigma$-algebra on $Y$. Since $A_n\in\mathfrak{B}$ for each $n$, $\mathfrak{B}$ is infinite, and hence uncountable. Since $\mathfrak{B}\subseteq\mathfrak{A}$, this implies $\mathfrak{A}$ is uncountable as well.