$\sigma$-field generate by one point sets I have the following problem.
Let $\Omega$ be a non-empty set and let $\mathcal{C}$ be all one point subsets. 
Show that 
\begin{align}
\sigma(\mathcal{C})=\{A \subset \Omega : A  \text{ is countable or } A^c  \text{ is  countable}   \}
\end{align}
where $\sigma(\mathcal{C})$ is a $\sigma$-algebra generated by $\mathcal{C}$.
Lets define $\mathcal{B}=\{A \subset \Omega : A  \text{ is countable or } A^c  \text{ is  countable}   \}$. 
I was able to show that  $\sigma(\mathcal{C}) \subset \mathcal{B}$.
However, the reverse direction $\mathcal{B} \subset \sigma(\mathcal{C})$ is difficult (at least for me). 
Thank you
 A: To show $\mathcal{B} \subseteq \sigma(\mathcal{C})$, let $E \in \mathcal{B}$.  Then either $E$ is countable or $E^{c}$ is countable.
But if $E$ is countable, then $E = \{ x_{1}, x_{2}, \dots \}$ where $x_{i} \in \Omega$.  But this means $E = \bigcup \limits_{i = 1}^{\infty} \{ x_{i} \}$, and since $\{ x_{i} \}$ is in $\sigma( \mathcal{C})$ for each $i$ (since $\{ x_{i} \} \in \mathcal{C}$ for each $i$), then the countable union $\bigcup \limits_{i = 1}^{\infty} \{ x_{i} \}$ is in $\sigma(\mathcal{C})$ (since $\sigma(\mathcal{C})$ is a $\sigma$-algebra and thus closed under countable unions), and so $E \in \sigma(\mathcal{C})$.
If, however, $E^{c}$ is countable, then by the same argument above, we get that $E^{c} \in \sigma(\mathcal{C})$, and since $\sigma(\mathcal{C})$ is a $\sigma$-algebra (and thus closed under complements), this gives that $E \in \sigma(\mathcal{C})$, as desired.
So, we've shown that given $E \in \mathcal{C}$, if it is countable, it is in $\sigma(\mathcal{C})$, and if its complement is countable, $E$ is still in $\sigma(\mathcal{C})$, which implies $\mathcal{B} \subseteq \sigma(\mathcal{C})$.
