I am trying to prove that the countable, co-countable sigma algebra on $\mathbb{R}$ cannot be countably generated.
In more precise terms.
Let $\Sigma$ be the $\sigma$-algebra generated by countable subsets of $\mathbb{R}$, that is
$$ \Sigma = \sigma (\{A\subseteq \mathbb{R} \:|\: A \textrm{ is countable}\})$$
It is easy to see that $A\in \Sigma$ iff $A$ is countable or co-countable.
Question: Is there a countable family $\{A_n\}_{n\in\mathbb{N}}$ such, for all $n\in\mathbb{N}$, $A_n\in \Sigma$ and
$$ \Sigma = \sigma (\{A_n \:|\: n\in\mathbb{N}\})?$$
I think the answer is NO, and I am trying to prove it. Can someone please help me in proving this?
My attempt is to prove by contradiction. That is assuming that the countable generating set exists then to show that sigma algebra generated by this set would miss some singletons of $\mathbb{R}$. Since the given sigma algebra contains all singletons this leads to contradiction. I am following this approach because I know that the set of all singletons generate the given sigma algebra and they are uncountable.
Best Answer
Your idea to prove by contradiction is correct. Here are the details.
Suppose there is a countable family $\{A_n\}_{n\in\mathbb{N}}$ such, for all $n\in\mathbb{N}$, $A_n\in \Sigma$ and
$$ \Sigma = \sigma (\{A_n \:|\: n\in\mathbb{N}\})$$
For each $n\in\mathbb{N}$, define \begin{align} &B_n = A_n & \textrm{ if $A_n$ countable}; \\& B_n = A_n^c & \textrm{ if $A_n$ cocountable} \end{align}
Then we have, for all $n\in\mathbb{N}$, $B_n$ is countable and, it is easy to see that: $$ \Sigma = \sigma (\{A_n \:|\: n\in\mathbb{N}\})= \sigma (\{B_n \:|\: n\in\mathbb{N}\}) \tag{1}$$
Let $C=\bigcup_{n\in\mathbb{N}}B_n$. Since $C$ is a countable union of countable sets, we have that $C$ is countable.
Since, for each $n\in\mathbb{N}$, $B_n$ is a countable subset of $C$, we have $B_n\in \sigma(\{\{p\} \:|\: p\in C\})$ and so we have $$\sigma (\{B_n \:|\: n\in\mathbb{N}\})\subseteq \sigma(\{\{p\} \:|\: p\in C\}) $$
On the other hand, for each $p\in C$, $\{p\}\in \Sigma$ (because $\{p\}$ is obviously countable). So, considering $(1)$, for each $p\in C$, $\{p\}\in \sigma (\{B_n \:|\: n\in\mathbb{N}\})$, and we can conclude that $$\sigma(\{\{p\} \:|\: p\in C\}) \subseteq \sigma (\{B_n \:|\: n\in\mathbb{N}\})$$ and so we have $$\Sigma= \sigma (\{B_n \:|\: n\in\mathbb{N}\})= \sigma(\{\{p\} \:|\: p\in C\}) $$
Let $\Sigma_0= \{E \:|\: E\subset C\} \cup \{E\cup C^c \:|\: E\subset C \}$. It is easy to prove that $\Sigma_0$ is a $\sigma$-algebra, and for each $p\in C$, $\{p\}\in \Sigma_0$. So $$\Sigma= \sigma(\{\{p\} \:|\: p\in C\}) \subseteq \Sigma_0 \tag{2}$$
Now, note that, since $C$ is countable, $\mathbb{R}\setminus C\neq \emptyset$, that is, $C^c \neq \emptyset$. Let $q$ be any element in $C^c$. We have $\{q\}\in \Sigma$ (because $\{q\}$ is obviously countable) but $\{q\}\notin \Sigma_0$. Contradiction.
Remark 1: We can easily prove that $$\sigma(\{\{p\} \:|\: p\in C\}) = \Sigma_0$$ but all we need is the inclusion presented in $(2)$.
Remark 2: All we used from $\mathbb{R}$ is that it is uncountable. The proof above works for any uncountable space $\Omega$.