Show that $F$ is a $\pi$-system. Let $S$ = $\mathbb{N}$ and let $F$ = $\{{\{n, n +1, \ldots \} : n \in \mathbb{N} \} }$. 
(a) Show that $F$ is a $\pi$-system.
(b) Show that $\sigma(F)$ = $\mathcal{P}$$(\mathbb{N})$
Definition of a $\pi$-system: A collection $E$ $\in$ $\mathcal{P}$$(S)$ is called a $\pi$-system if for all A,B $\in$ $E$ one has A $\cap$ B $\in$ $E$.
For a):
The main problem is that I don't know how to 'see' the set $F$. 
Is $F$ the set containing all singletons {$n, n+1, \dots\}$ because in that case $(n + k) \cap(n+m)$ with $k, m \in \mathbb{Z}$ would be $\emptyset$ right? And I don't see how the empty set is included in $F$.
Or is $F$ the set containing all sequences with $\{n, n+1, \ldots\}$ because then $\{n, n+1, \ldots\} \cap \{k, k+1, \ldots\}$ would be $\{m, m + 1,\ldots\}$ with $m$ = max$\{n,k\}$ which is in $F$. 
for b) I would like a hint.
 A: First about the imagination of $F$. It is a set of subsets of $\mathbb{N}$:
Clearly you can write 
$$F = \{\{1,2,3,...\},\{2,3,4,...\},\{3,4,5,...\},...\}.$$ and
$$F = \{\mathbb{N},\mathbb{N}\backslash \{1\},\mathbb{N}\backslash \{1,2\},\mathbb{N}\backslash \{1,2,3\},... \}.$$
Secondly, neither all singletons nor the empty set are contained in $F$, but this is also not necessary for $F$ to be a $\pi$-system.
And there is no need to add elements, which are sets themselves, to $F$. This has to be done if you consider $\sigma(F)$.
So, let's talk about (a). For $F$ to be a $\pi$-system, we need to show, according to this that $F$ is not empty, but thats clear, since $\mathbb{N} \in F$. Secondly we need to show, that for some $A,B\in F$ also $A \cap B$ is in $F$. So let $A$ and $B$ be elements of $F$, then according to the second representation above (try to see it from your representation as well!), we can write $A = \mathbb{N}\backslash \{1,2,...,a\}$ for some $a\in \mathbb{N}$ and analogously for B. Then $A \cap B = \mathbb{N}\backslash \{1,2,...,\max\{a,b\}\}$. Thus $A \cap B \in F$.
Hint for (b): Try to show that the singletons are contained in $\sigma(F)$ and make use of a well-known theorem for $\sigma$-algebras and generators, stating that two generated $\sigma$-algebras coincide, if their generators are contained in the other $\sigma$-algebra.

For (b) we observe that it is enough to show, that all singletons $\{n\}_{n \in \mathbb{N}}$ are contained in $\sigma(F)$. But according to the axioms for $\sigma$-algebras, you know that the set-operator $\backslash$ is a valid operator (I hope it is clear how I mean that.). Because then you can write $$\{n\} = \underbrace{\{n,n+1, ...\}}_{\in F}\backslash\underbrace{\{n+1,n+2, ...\}}_{\in F}$$ and therefore conclude: $\{n\}\in F$. And this should be sufficient.

