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Let $\mathbb{S} = \mathbb{R} / \mathbb{Z}$ be the $1$-dimensional sphere and consider the $\sigma$-algebra $$\mathcal{A}= \{A \subset \mathbb{S} : A \text{ countable or } \mathbb{S}\setminus A\text{ countable} \}$$on $\mathbb{S}$ (by countable I mean finite or countably infinite). A generator is for example the set $$\big\{ \{x\} : x \in \mathbb{S} \big\}.$$My question is: is there a countable set of sets generating $\mathcal A$?

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No. Suppose that $\mathcal{B}\subset \mathcal{A}$ is a countable, generating set and let $\mathcal{B}_1$ consist of the sets in $\mathcal{B}$ that are countable, and $\mathcal{B}_2$ those that are co-countable. Note that if we define $\mathcal{B}_2^C:=\{B^C : B\in \mathcal{B}_2\}$, then $\mathcal{A}=\sigma(\mathcal{B}_1\cup\mathcal{B}_2)=\sigma(\mathcal{B}_1\cup\mathcal{B}_2^C)$. However if we then define the countable set $\mathbb{S}_0:=\bigcup\limits_{B\in \mathcal{B}_1\cup\mathcal{B}_2^C} B$ we see that the $\sigma$-algebra $\mathcal{A}_0$ consisting of all subsets of $\mathbb{S}_0$ and their complements actually contains $\mathcal{B}_1\cup\mathcal{B}_2^C$. Hence $\mathcal{A}=\sigma(\mathcal{B}_1\cup\mathcal{B}_2^C)\subseteq \mathcal{A}_0$. But this is impossible, since for any $s\in \mathbb{S}\backslash\mathbb{S}_0$, $\{s\}\in \mathcal{A}$, but $\{s\}\notin \mathcal{A}_0$.

Interestingly, $\mathcal{A}$ is a much smaller $\sigma$-algebra than the one generated by the open sets (which is countably generated), so this is a good example of how "countable generatability" isn't perserved under taking sub $\sigma$-algebras.

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