Find all the $\sigma$ algebras of $\Bbb N$ I can find a lot of examples of this. For example I can take a generating set, and consider the $\sigma$ algebra generated by those sets. I can even consider a "basis" even with disjoint sets. For example the generating set {1},{2,3,4},{1,3} can be replaced by {1},{3},{2,4} doing the correct intersections, I can do this always in the case that I have a finite generating set, I don't know if this always happens , I think that that it's not the case, for example the $\sigma$ algebra generated by {2},{4},{6},.....  But I don't know , maybe we can do something with this generating sets.
Please help me!
 A: For any countable set $S$ and any $\sigma$-algebra $\cal F$ on $S$, $\cal F$ is generated by a partition of $S$, meaning that there is a set $\cal P$ of subsets of $S$ such that any pair of elements of $\cal P$ are disjoint, $\bigcup {\cal P}=S$, and
$$
{\cal F}=\left\{\bigcup T\mid T\subseteq {\cal P}\right\}.\qquad \qquad (*)
$$
To see this, define the equivalence relation $\sim$ on $S$ by
$$x\sim y \ \ \ \  \Leftrightarrow \ \ \ \ \ \text{for all } A\in{\cal F}, \ \ x\in A \text{ if and only if } y\in A.
$$
Pick any $x\in S$.  Then if $y\in S$ has $y\not\sim x$, there is either $A_{xy}\in\cal F$ with $x\in A_{xy}$ and $y\notin A_{xy}$, or $A_{xy}\in\cal F$ with $x\notin A_{xy}$ and $y\in A_{xy}$.  By taking the complement of $A_{xy}$ if necessary, you can assume that $x\in A_{xy}$ and $y\notin A_{xy}$.  Then, since $S$ is countable, the intersection $B_x:=S\cap\bigcap_{y\in S: y\not\sim x} A_{xy}$ is in $\cal F$.  By construction, $y\notin B_x$ if $y\not\sim x$, and $x\in B_x$.  Since $B_x\in\cal F$, by the definition of $\sim$, $y\in B_x$ whenever $y\sim x$.
Let $\cal P$ be the partition of $S$ generated by placing $x$ and $y$ into the same member of $\cal P$ just when $x\sim y$.  The last paragraph proves that, for all $x$, $B_x$ equals the member of $\cal P$ containing $x$ and $B_x\in\cal F$.  Since $x$ was an arbitrary element of $S$, this proves that ${\cal P}\subseteq {\cal F}$, and then, by taking countable unions, $\bigcup T\in{\cal F}$ for all $T\subseteq\cal P$.  In the other direction, by the definition of $\sim$, any element of $\cal F$ must be a union of equivalence classes of $\sim$, so it is of the form $\bigcup T$ for some $T\subseteq\cal P$.  This proves $(*)$.
