$\sigma$-algebra generated by all countable and co-countable sets Let $S=[0,1]$. 
Let $\Sigma$ be a $\sigma$-algebra generated by all countable and co-countable subsets of $S$.
I know that a set of all countable and co-countable subsets itself is a $\sigma$-algebra.
But what happens if this set generates a new $\sigma$-algebra $\Sigma$? 
Will $\Sigma$ be bigger, will it be $\mathcal{P}(S)$?
 A: The sigma algebra generated by co-countable and countable sets it is equal to the sigma algebra of the countable and co-countable sets,by definition of the sigma algebra generated by a family of sets.(which is the smallest with respect to set  inclusion.
A: If I understand correctly, first we take $\mathcal P(S)$, and then the set $\Sigma$ of all countable and cocountable sets. 
Then you already know that $\Sigma$ is a $\sigma$-algebra, and ask what is $\sigma(\Sigma)$, the generated $\sigma$-algebra.
And the answer is that $\sigma(\Sigma) =\Sigma$, since $\Sigma$ is the smallest $\sigma$-algebra containing $\Sigma$. 
A: As it has already been answered that $\Sigma$ will exactly be the sigma algebra of countable and co-countable sets, I will give a brief explanation of why is that the case?
As you have already pointed out that the set of all countable and co-countable sets form a sigma algebra. Let's call this sigma algebra $\Sigma_0.$ And, let the sigma algebra generated by countable and co-countable sets be denoted by $\Sigma.$ 
It is obvious that $\Sigma_0\subseteq \Sigma.$ But, by definition of $\Sigma$ we have that if any sigma algebra $\mathcal{F}$ contains all countable and co-countable sets then $\Sigma\subseteq \mathcal{F}.$ In particular, for $\mathcal{F}=\Sigma_0,$ we get that $\Sigma\subseteq \Sigma_0.$ It thus follows that $\Sigma=\Sigma_0.$
