Singleton in $\sigma$-algebra on a subset of $\mathbb{N}$ I'm very new to set theory. Given the set
$$\mathcal{L}=\left\{\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\dots\right\}$$
Where $\mathcal{L}$ is on the form $\{,+1,+2\}$, $\in\mathbb{N}$ and $3|(+2)$. Show that $\{1\}\notin\sigma(\mathcal{L})$. My only argumentation is that the subsets of the set $\mathcal{L}$ are disjoint given the condition that $\in\mathbb{N}$ and $3|(+2)$, and so any intersection is empty.
 A: This question can be solved in a easy way from a broader result.

Lemma. Let $X$ be a set and $\mathcal{S}=\{A_i\}_{i\in I}$ a countable partition of $X$. Then $\sigma(\mathcal{S})= \{ \bigcup_{i\in K} A_i : K \subseteq I \}$

Proof:
It is easy to check that $\{ \bigcup_{i\in K} A_i : K \subseteq I \}$  is a $\sigma$-algebra. In fact,

*

*$\emptyset = \bigcup_{i\in \emptyset} A_i$;

*Since $\mathcal{S}$ is a partition of $X$, $X \setminus (\bigcup_{i\in K} A_i)=\bigcup_{i\in I \setminus K} A_i $;

*Given a countable family $\{\bigcup_{i\in K_j} A_i\}_{j \in J}$, then
$\bigcup_{j \in J}(\bigcup_{i\in K_j} A_i)= \bigcup_{i\in \bigcup_{j \in J} K_j} A_i$;

*Clealy $X=  \bigcup_{i\in I} A_i $
So $\{ \bigcup_{i\in K} A_i : K \subseteq I \}$  is a $\sigma$-algebra. Since $\mathcal{S} \subseteq \{ \bigcup_{i\in K} A_i : K \subseteq I \}$, we have that
$\sigma(\mathcal{S}) \subseteq \{ \bigcup_{i\in K} A_i : K \subseteq I \}$.
Now, since $\mathcal{S}=\{A_i\}_{i\in I}$ a countable partition of $X$, we have that $I$ is countable and so, every $K \subseteq I$ is countable. So, for every $K \subseteq I$, $\bigcup_{i\in K} A_i \in \sigma(\mathcal{S})$. So, $ \{ \bigcup_{i\in K} A_i : K \subseteq I \} \subseteq \sigma(\mathcal{S})$.
Thus we proved that $\sigma(\mathcal{S}) = \{ \bigcup_{i\in K} A_i : K \subseteq I \}$. $\square$
Now let us consider the question. Assuming $\Bbb N =\{1, 2, \dots \}$. We have that
$\mathcal{L}=\left\{\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\dots\right\}$ is a countable partition of $\Bbb N$. By the lemma above, any $D \in \sigma(\mathcal{L})$ is the union of elements of $\mathcal{L}$. So, it is easy to see that $1 \in D$ if and only if $2, 3 \in D$. So $\{1\} \notin \sigma(\mathcal{L})$
Remark: If we assume that $\Bbb N =\{0, 1, 2, \dots \}$, then consider the countable partition
$\mathcal{L_0}=\left\{\{0\},\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\dots\right\}$. Then, repeating the previous argument, we have $\{1\} \notin \sigma(\mathcal{L_0})$. Since $\mathcal{L} \subseteq \mathcal{L_0}$, it is immediate that  $\sigma(\mathcal{L}) \subseteq \sigma(\mathcal{L_0})$. So $\{1\} \notin \sigma(\mathcal{L})$.
A: It appears that the question employs the convention $\mathbb{N}=\{1, 2, 3, \dots\}$ which we also follow below.

The $\sigma$-algebra $\sigma(F)$ generated by a family $F$ of subsets of $\mathbb{N}$ is defined as the intersection of all $\sigma$-algebras that contain $F$. Therefore, we can show that $\{1\}\notin\sigma(\mathcal{L})$ by exhibiting a $\sigma$-algebra $\Sigma$ such that $\{1\}\notin\Sigma$ and $\mathcal{L}\subset\Sigma$.
For a subset $A\subset\mathbb{N}$, define
$$
A^3 = \{3k, 3k-1, 3k-2 | k\in A\}
$$
and note that $|A|$ is either infinite or divisible by $3$. Define
$$
\Sigma=\{A^3 | A\in\mathcal{P}(\mathbb{N})\}.
$$
Now, since $\mathcal{L}$ can be written as $\mathcal{L}=\{\{k\}^3 | k\in\mathbb{N}\}$ we see that $\mathcal{L}\subset\Sigma$. Moreover, since every element of $\Sigma$ is either infinite or its size is divisible by $3$, we see that $\{1\}\notin\Sigma$. We will show that $\Sigma$ is a $\sigma$-algebra.
First, since $\mathbb{N}^3=\mathbb{N}$ we see that $\mathbb{N}\in\Sigma$. Next, define $A^c := \mathbb{N}\setminus A$ to be the complement of $A$ in $\mathbb{N}$. It is easy to see that $(A^3)^c=(A^c)^3$ which implies that $\Sigma$ is closed under complements. Finally, if $A_1, A_2, \dots\in\mathcal{P}(\mathbb{N})$ then
$$
A_1^3\cup A_2^3\cup \dots = (A_1\cup A_2\cup\dots)^3
$$
which implies that $\Sigma$ is closed under countable unions.
Thus, $\Sigma$ is a $\sigma$-algebra that does not contain $\{1\}$ and of which $\mathcal{L}$ is a subset. Therefore, $\sigma(\mathcal{L})$ - which is the intersection of all $\sigma$-algebras containing $\mathcal{L}$ including $\Sigma$ - does not contain $\{1\}$ as an element.
