$\sigma$-algebra generated by singletons of $X$ Let $X$ be an uncountable set and 
$$S=\{E \subset X: E \text{ or } E^c \text{is at last countable} \}$$
I want to prove that $S$ is equal to the $\sigma$-algebra generated by singletons of $X$. 
I think I need to write down that sigma algebra, and then by taking elements in each of these sets prove that these two are equal. But: 


*

*I don’t know how to create the sigma algebra generated by singletons. 

*Is there any other way for answering the question? 
 A: Let  $\mathcal{S}=\{\{x\}:x\in X\}$ be the family of all singletons of $X$ and $\sigma(\mathcal{S})$ the $\sigma$-algebra generated by $\mathcal{S}$. Since $S$ is $\sigma$-algebra and $\mathcal{S}\subset S$, we have $\sigma(\mathcal{S})\subset S$. Consequently, it only remains to prove that $S\subset\sigma(\mathcal{S})$. Let $A\in S$. If $A$ is countable, then $A\in \sigma(\mathcal{S})$, due to $A=\bigcup_{x\in A}\{x\}$. If $A$ is not countable, then $A^c$ is countable and hence $A^c\in \sigma(\mathcal{S})$. But $\sigma(\mathcal{S})$ is $\sigma$-algebra and, accordingly, $A\in \sigma(\mathcal{S})$. Therefore, in any case $A\in \sigma(\mathcal{S})$, which proves that $S=\sigma(\mathcal{S})$.
A: Note that each singleton is in $S$. So the $\sigma$-algebra generated by singletons (which is the smallest $\sigma$-algebra containing all singletons) is a subset of $S$ as $S$ itself is a $\sigma$-algebra (obviously $X,\phi \in S$. Also $S$ is closed under complement operations. Finally let $\{E_n\}_n\subseteq S$, then $\bigcup_{n=1}^{\infty} E_n \in S$ due to fact: if each $E_n$ is countable then  $\bigcup_{n=1}^{\infty} E_n$ is also so , and if one of $E_m$ is uncountable implies $E_m^c$ is countable so that $\bigcap_{n=1}^{\infty} E_n^c(\subseteq E_m^c)$ is countable i.e. $(\bigcap_{n=1}^{\infty} E_n^c)^c=\bigcup_{n=1}^{\infty} E_n\in S$. Hence $S$ is a $\sigma$-algebra.)
Now note that if $ E\in S$ then either $E$ or $E^c$ is countable i.e. either  $E$ or $E^c$  can be written as countable union of singletons i.e. either $E$ or $E^c$ is in the $\sigma$-algebra generated by singletons. Hence both  $E$ and $E^c$ is in the $\sigma$-algebra generated by singletons as any $\sigma$-algebra is closed under complement operations.
Therefore $S$ is equals to the $\sigma$-algebra generated by singletons.
