# Generating set of countable, co-countable sigma algebra on $\mathbb{R}$

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
$$\Sigma = \sigma (\{A_n \:|\: n\in\mathbb{N}\})?$$

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.

• Please describe what you've tried so far! – Xoque55 Dec 17 '15 at 1:37
• @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. – Raghava G D Dec 17 '15 at 1:54
• 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! – Xoque55 Dec 17 '15 at 2:25
• @Xoque55 :Sure, Thanks. – Raghava G D Dec 17 '15 at 2:30
• 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. – user149792 Dec 18 '15 at 14:13

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 $\Omega$.