$\sigma$-algebra on subsets of $\mathbb{N}$ Consider the subsets of the natural numbers
$$A=\left \{ \left \{ 1,2,3 \right \},  \left \{ 2,3,4 \right \},  \left \{ 3,4,5 \right \}, \left \{ 4,5,6 \right \},... \right \}$$
and
$$B=\left \{ \left \{ 1,2,3 \right \},  \left \{ 4,5,6 \right \}, \left \{ 7,8,9 \right \},\left \{ 10,11,12 \right \},... \right \}$$

Show that the $\sigma(A)=\mathcal{P}(\mathbb{N})$ and that the singleton $\left \{ 1 \right \}\notin\sigma(B)$

I am extremly new to sets. I recognize that set $A$ is of the form $\left \{ k,k+1,k+2 \right \}, k\in \mathbb{N}$ and $B$ is of the form $\left \{ k,k+1,k+2 \right \}, k\in \mathbb{N}$ and $3|(k+2)$. I also know the number of all the subsets of a given set $S$ with cardinality $n$ is $|P(S)|=2^n$.
 A: Hint
To prove that $\{1\} \notin \sigma(B)$, notice that for any $s \in \sigma(B)$, $1 \in s \iff 2 \in s$.
A: For $A$:
$(i).$  For $3\le n\in \Bbb N$ we have $$\{n\}=\{n-2,n-1,n\}\cap\{n-1,n,n+1\}\cap\{n,n+1,n+2\}\in\sigma (A).$$ $(ii).$ So $C=\{n\in\Bbb N:n\ge 3\}\in\sigma (A)$ by countable union. So $\{1,2\}=\Bbb N$ \  $C\in\sigma (A).$ So $\{2\}=\{1,2\}\cap \{2,3,4\}\in\sigma (A).$
$(iii).$ So $D=\{2\}\cup C\in\sigma (A).$ So $\{1\}=\Bbb N$  \ $D\in\sigma (A).$
For $B:$
Let $C=B$ / $\{\{1,2,3\}\}.$ Let $D=\sigma (C)\cup \{x\cup \{1,2,3\}:x\in\sigma (C\}.$ Then $D$ is a $\sigma$-algebra with $D\supset B.$ But if $E$ is any $\sigma$-algebra with $E\supset B$ then $E\supset \sigma (C)$ and $\{1,2,3\}\in E$ so $E\supset D.$ Therefore $$D=\sigma (B).$$ Now if $y\in D$ there exists $x\in \sigma (C)$ with $y=x\subset \Bbb N$ \ $\{1,2,3\}$ or $y=x\cup \{1,2,3\}$, and in either case $y\ne\{1\}.$
OR observe that $B$ is a pairwise-disjoint countable family with $\cup B=\Bbb N,$ so $\sigma (B)=\{\cup F:F\in\mathcal P(B)\}.$ And for any $F\in\mathcal P(B),$ if $\{1,2,3\} \in F$ then $\{1,2,3\} \subset\cup  F,$ or if $\{1,2,3\}\not \in F$ then $\{1,2,3\}$ is disjoint from $\cup  F,$ so in either case $\cup F\ne \{1\}.$
A: Item 1. Assuming $\Bbb N =\{1,2,\dots \}$. To prove that $\sigma(A)=\mathcal{P}(\mathbb{N})$, just note that for any $n \in \Bbb N$,
$$ \{n\}= \{n, n+1, n+2\} \setminus \{n+1, n+2, n+3\} \in \sigma(A)$$
Since $\Bbb N$ is countable, it follows that  $\mathcal{P}(\mathbb{N}) \subseteq  \sigma(A)$. So we have $\sigma(A)=\mathcal{P}(\mathbb{N})$.
Remark: If we assume $\Bbb N =\{0,1,2,\dots \}$, just add to the above argument that
$$ \{0\} = \Bbb N \setminus \{1,2,\dots \} = \Bbb N \setminus \bigcup_{S \in A} S \in \sigma(A)$$
So, the same way, we have $\sigma(A)=\mathcal{P}(\mathbb{N})$.
Item 2. This item 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 second item of the question. Assuming $\Bbb N =\{1, 2, \dots \}$. We have that
$B=\left\{\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\dots\right\}$ is a countable partition of $\Bbb N$. By the lemma above, any $D \in \sigma(B)$ is the union of elements of $B$. So, it is easy to see that $1 \in D$ if and only if $2, 3 \in D$. So $\{1\} \notin \sigma(B)$
Remark: If we assume that $\Bbb N =\{0, 1, 2, \dots \}$, then consider the countable partition
$C=\left\{\{0\},\{1,2,3\},\{4,5,6\},\{7,8,9\},\{10,11,12\}\dots\right\}$. Then, repeating the previous argument, we have $\{1\} \notin \sigma(C)$. Since $B \subseteq C$, it is immediate that  $\sigma(B) \subseteq \sigma(C)$. So $\{1\} \notin \sigma(B)$.
