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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.

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    $\begingroup$ Please describe what you've tried so far! $\endgroup$ – Xoque55 Dec 17 '15 at 1:37
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    $\begingroup$ @Xoque55 : I am trying to prove by contradiction. That is, assuming that there is a countable generating set I am trying to prove that I am sure going to miss some singletons in the sigma algebra generated by this set. Since singletons are in the countable, co-countable sigma algebra this leads to contradiction. I am trying this approach because I know that the singletons generate the countable, co-countable sigma algebra and they are uncountable. $\endgroup$ – Raghava G D Dec 17 '15 at 1:54
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    $\begingroup$ The reason I prompted you is so you could a description like that into your original question. Maybe you could edit all that into your question? Showing your efforts within a question is the best way to receive the best help! $\endgroup$ – Xoque55 Dec 17 '15 at 2:25
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    $\begingroup$ @Xoque55 :Sure, Thanks. $\endgroup$ – Raghava G D Dec 17 '15 at 2:30
  • $\begingroup$ Every countably generated $\sigma$-algebra has a minimal generator. You are asking whether there exists a countably generated $\sigma$-algebra with no minimal generators – the answer is no. See here. $\endgroup$ – user149792 Dec 18 '15 at 14:13
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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 $\Omega$.

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