Sigma algebra generated by the set of all singletons

Solution 1:

Let $$\mathscr C\equiv\{E\subseteq\mathbb R\,|\,E\text{ is countable or }E^c\text{ is countable}\}.$$ It is not difficult to check that $\mathscr C$ is a $\sigma$-algebra on $\mathbb R$. Let $\mathscr S$ denote the $\sigma$-algebra generated by singletons.

$\textbf{Proposition:}\phantom{---}$$\mathscr C=\mathscr S$.

$\textit{Proof:}\phantom{---}$ Clearly, $\{x\}\in\mathscr C$ for every $x\in\mathbb R$, so $\mathscr S\subseteq\mathscr C$, since $\mathscr C$ is a $\sigma$-algebra containing the singletons and $\mathscr S$ is, by definition, the smallest $\sigma$-algebra containing the singletons. Conversely, suppose that $E\in\mathscr C$. If $E$ is countable, then $E$ is a countable union of singletons. Therefore, $E\in\mathscr S$ (because $\mathscr S$ is $\sigma$-algebra containing the singletons). If $E^c$ is countable, then, by the same argument, $E^c\in\mathscr S$, from which it follows that $E\in\mathscr S$ (since $\sigma$-algebras are closed under complements). In conclusion, $\mathscr C\subseteq\mathscr S$. $\blacksquare$

Clearly, $(0,1)\notin\mathscr C=\mathscr S$, but $(0,1)\in\mathscr B_{\mathbb R}$. It follows that $\mathscr B_{\mathbb R}\neq\mathscr S$.

Solution 2:

Your proof for both parts is exactly correct as stated. There is no more formalization to be done. Just make sure to say that $( 0,1 )$ is uncountable.