# is there a countable generator for the (co-)countable $\sigma$-algebra?

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 $$\mathcal A$$?

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.