Fields and $\sigma$ - fields generated by given sets. Could you check if my reasoning is correct?
The field of subsets of $\mathbb{N}$ (let's denote it by $\mathcal{M}$) generated  by singletons is, I think,  the set of all subsets of $\mathbb{N}$, because 
$\emptyset, \mathbb{N} \in \mathcal{M}$  and 
if $\{n\} \in \mathcal{M}$ , then its complement $\mathbb{N} \setminus \{n\} \in \mathcal{M}$ , and 
if $\{n_1\},..., \{n_k\} \in \mathcal{M}$, then $\{n_1, ..., n_k\} \in \mathcal{M}$ and so $\mathbb{N} \setminus \{n_1, ..., n_k\} \in \mathcal{M}$. 
Doesn't this mean that $\mathcal{M} = 2^{\mathbb{N}}$? 
It would mean that this field of subsets of $\mathbb{N}$  same as $\sigma$-field of subsets of $\mathbb{N}$ generated by singletons.
Next, let's consider field $\mathcal{A}$ and $\sigma$-field $\mathcal{B}$ of subsets of $\mathbb{R}$ generated by $(n, + 
\infty), n\in \mathbb{Z}$. 
$(n, +\infty), (m, +\infty) \in \mathcal{A}, \ m>n \ \ \Rightarrow(n, m]\in \mathcal{A}$, because $(m, +\infty) \in \mathcal{A} \ \ \Rightarrow (-\infty, m] \in \mathcal{A}$
So here we get $\mathcal{A}=\{\emptyset, \mathbb{R}, (-\infty, m], (n,m], (n, +\infty) : m, n \in \mathbb{Z}\} = \mathcal{B}$
I would very much appreciate all your insight.
 A: Every finite subset of $\Bbb N$ will be an element of $\mathcal M,$ as will every subset of $\Bbb N$ having a finite complement. However, infinite subsets of $\Bbb N$ with infinite complements--such as $\{1,3,5,7,...\}$--will not be elements of $\mathcal M.$
Now, $\mathcal A$ will include all sets that are finite unions of sets of form $(-\infty,m],$ $(n,m],$ and $(n,\infty),$ where $m,n\in\Bbb Z.$ More explicitly, they will be of one of the following forms:


*

*$\emptyset$

*$\Bbb R$

*$(n,m],$ where $n<m$

*$(-\infty,m]$

*$(n,\infty)$

*$(n_1,m_1]\cup(n,\infty),$ where $n_1<m_1<n$

*$(-\infty,m]\cup(n_1,m_1],$ where $m<n_1<m_1$

*$(-\infty,m]\cup(n,\infty),$ where $m<n$

*$\bigcup\limits_{j=1}^k(n_j,m_j],$ where $n_1<m_1<…<n_k<m_k$

*$(-\infty,m]\cup\bigcup\limits_{j=1}^k(n_j,m_j],$ where $m<n_1<m_1<…<n_k<m_k$

*$(n,\infty)\cup\bigcup\limits_{j=1}^k(n_j,m_j],$ where $n_1<m_1<…<n_k<m_k<n$

*$(-\infty,m]\cup(n,\infty)\cup\bigcup\limits_{j=1}^k(n_j,m_j],$ where $m<n_1<m_1<…<n_k<m_k<n$


We have a similar situation with $\mathcal B,$ but it will also include sets of the following forms:


*

*$\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $n_1<m_1<n_2<m_2<…$

*$\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $m_1>n_1>m_2>n_2>…$

*$\bigcup\limits_{j=-\infty}^\infty(n_j,m_j],$ where $…<n_{-1}<m_{-1}<n_0<m_0<n_1<m_1<…$

*$(-\infty,m]\cup\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $m<n_1<m_1<n_2<m_2<…$

*$(n,\infty)\cup\bigcup\limits_{j=1}^\infty(n_j,m_j],$ where $n>m_1>n_1>m_2>n_2>…$

