Every $\sigma$-ring has a countable generator

Question

I am confused because the statement of the theorem suggests that for every set $E$ in $S(\textbf{E})$, there exists a countable subclass $\textbf{D}$ of $\textbf{E}$ so that $E\in S(\textbf{D})$; but the proof suggests that there exists a countable subclass $\textbf{D}$ of $\textbf{E}$ so that for all $E\in S(\textbf{E})$, $E\in S(\textbf{D})$, i.e. $S(\textbf{D})=S(\textbf{E})$.

I believe the former claim easily: For any $E\in S(\textbf{E})$, $E$ is given by countably many unions and set differences of sets in $\textbf{E}$; let $\textbf{D}$ be the class of those sets to get the result.

I believe the latter claim too (since I worked through the proof), but am surprised by one consequence: The $\sigma$-ring of a countable class can be uncountable. This doesn't seem right, since we're taking countable unions and set differences of a countable set.

Could someone clarify? In particular, am I reading the proof wrong? I would like to see an example of an uncountable $\sigma$-ring that is generated by a countable class.

Thanks a bunch in advance!

P.S. Text is Halmos Measure Theory. $\sigma$-ring is same thing as $\sigma$-algebra except if $\textbf{E}$ is a class of subsets of $X$, we don't require $X\in \textbf{E}$.

Answer 1

You are misinterpreting the proof. The proof is taking the union of all $\sigma$-subrings of $S$ which are generated by any countable subclass of $\mathbf{E}$, not one particular fixed countable subclass. That is, it is taking the union of all the elements of the set $$\{S(\mathbf{D}):\mathbf{D}\subseteq\mathbf{E}\text{ and }\mathbf{D}\text{ is countable}\}.$$

This union contains every element of $\mathbf{E}$, since you can take $\mathbf{D}$ to contain any particular element of $\mathbf{E}$. And it is a $\sigma$-ring: given countably many elements of the union, say they are elements of $S(\mathbf{D}_1),S(\mathbf{D}_2),\dots$ for some countable subsets $\mathbf{D}_1,\mathbf{D}_2,\dots$. Then if you define $\mathbf{D}=\bigcup_n\mathbf{D}_n$, $\mathbf{D}$ is countable and all of your elements are elements of $S(\mathbf{D})$, and hence their union (or the difference of any two of them) is also an element of $S(\mathbf{D})$.

With your interpretation, the result is not true. For instance, let $\mathbf{E}$ consist of all singleton subsets of $\mathbb{R}$. If $\mathbf{D}$ is any fixed countable subset of $\mathbf{E}$, every element of $S(\mathbf{D})$ cannot contain any elements other than the elements of the countably many sets in $\mathbf{D}$. In particular, any element of $\mathbf{E}\setminus\mathbf{D}$ is not an element of $S(\mathbf{D})$.

However, it is true that the $\sigma$-ring generated by a countable set can be uncountable. For a simple example, let $\mathbf{D}$ be the set of all singleton subsets of $\mathbb{N}$. Every subset of $\mathbb{N}$ is a countable union of singletons, so $S(\mathbf{D})=P(\mathbb{N})$ is uncountable. In fact, it turns out that any infinite $\sigma$-ring is uncountable.