Finding the $\sigma$-algebra $\mathbf{X}$ = $\sigma$(S) generated by the collection of subsets S in these two questions. a) $X = \mathbb{R}$ , $S = \{(-\infty, n): n \in \mathbb{Z} \} $     (i.e. all intervals in the form $(-\infty, n)$).
b) $X = \mathbb{R}^2$ , $S = \{\{(x, y)\} : (x, y) \in \mathbb{R}^2\} $     (i.e all singleton sets in $\mathbb{R}^2$)
for part (a) I feel like the answer is the Borel Algebra on $\mathbb{Z}$ ? but I don't know how to show it.
and for part (b) I amn't sure where to start.
 A: We use the term "countable" to means finite or countably infinite.
a) Let  $S = \{(-\infty, n): n \in \mathbb{Z} \} $ and let
$S_f = \{[m, n): m,n \in \mathbb{Z} \textrm{ and } m\leqslant n \} $
Note that $\emptyset \in S_f$, since $\emptyset =[n,n)$.
Now let $\Sigma= \{\bigcup_{i\in \mathbb{N}} A_i: \textrm{for all }  i \in \mathbb{N}, A_i \in S \cup S_f \}$.
It is easy to see that $S_f \subseteq \sigma(S)$, and so,   $S \cup S_f \subseteq \sigma(S)$. So, countable unions of sets in  $S \cup S_f$ are also in $\sigma(S)$. It means $\Sigma \subseteq \sigma(S)$.
On the other hand, it is easy to prove that $\Sigma$ is a $\sigma$-algebra and, since $S\subseteq \Sigma$, we have that $\sigma(S) \subseteq \Sigma$.
So we have
$$\sigma(S)= \Sigma=\left \{\bigcup_{i\in \mathbb{N}} A_i: \textrm{for all }  i \in \mathbb{N}, A_i \in S \cup S_f \right \}$$
Remark: Since the set in $S$ are countable unions of sets in $S_f$, we can simplify the expression of $\sigma(S)$:
$$\sigma(S)= \Sigma=\left \{\bigcup_{i\in \mathbb{N}} A_i: \textrm{for all }  i \in \mathbb{N}, A_i \in S_f \right \}$$
b) Let  $S = \{\{(x, y)\} : (x, y) \in \mathbb{R}^2\} $. Let $\Sigma=\{A \subseteq \mathbb{R}^2 : A \textrm{ is countable or }  \mathbb{R}^2 \setminus A \textrm{ is countable} \}$.
It is easy to see that $\Sigma \subseteq \sigma(S)$.
On the other hand, it is easy to prove that $\Sigma$ is a a $\sigma$-algebra and all singletons are in  $\Sigma$. So  $\sigma(S) \subseteq \Sigma$. So we conclude that
$$\sigma(S) = \Sigma=\{A \subseteq \mathbb{R}^2 : A \textrm{ is countable or }  \mathbb{R}^2 \setminus A \textrm{ is countable} \}$$
