$\sigma$-algebra generated by the set $A:=\{n,n+1,n+2\}$ where $n\in \Bbb{N}-\{1\}$. I am taking a course on measure theory and I have some issues regarding $\sigma$-algebra.

Here is the exercise, we are taking place on $\Bbb{N}$ and I need to characterize the $\sigma$-algebra generated by the set $A:=\{n,n+1,n+2\}$ where $n\in \Bbb{N}-\{1\}$.

So I have I understand correctly the definition, the question is to determine the intersection of all $\sigma$-algebras containing $A$. By definition it's also a  $\sigma$-algebra so it's closed under countable set operations (union,intersection) and $\sigma(A)$ is included in $\mathcal{P}(\Bbb{N})$. But I really don't understand how can I describe the sigma-field generated by a set.
Can someone explain how can I proceed ? I am not asking a solution but an explanation of what I need to understand here.
 A: First, a disclaimer: the intersection of all $\sigma$-algebras containing $A$ is not a $\sigma$-algebra "by definition". It is a $\sigma$-algebra because there is a theorem saying that an intersection of $\sigma$-algebras is a $\sigma$-algebra.

That said, to solve your problem, I advise you to play around with what you can construct with sets from $A$. Let's call $S$ the sigma algebra, generated by $A$. 
For example, you know that all sets of the type $\{n, n+1,n+2\}$ are in $A$, therefore, they are in $S$. But you also know that $S$ is closed under intersections, so, for example, $\{n, n+1, n+2\}\cap \{n+1, n+2, n+3\} = \{n+1,n+2\}$ is also in $S$.
Using thought processes like this will allow you to get a good idea of what "needs" to be in $S$. Then, at some point, you will realize that the things that "need" to be in $S$ also form a $\sigma$-algebra.
At that point, you basically have some condition which describes some part of $S$:

All sets $X$ which satisfy this condition must be in $S$.

Once this condition is general enough to also satisfy the statement:

The set of all sets $X$ which satisfy this condition form a $\sigma$-algebra

You are done, because at that point, you can prove:


*

*All $\sigma$-algebras that include $A$ must also include $S$.

*$S$ is a $\sigma$-algebra.


From these two points, it follows that $S$ is the $\sigma$-algebra, generated by $A$.
