Generating $\pi$-system for $\mathbb{Z}_+$ As we know from the definition a $\pi$-system is a collection where the intersection of two sets is again in that collection. In fact we can use $\pi$-systems to generate $\sigma$-algebra's. We know that a $\pi$-system generates a $\sigma$-algebra if the minimum $\sigma$-algebra containing the $\pi$-system is in fact the $\sigma$-algebra we want to generate. 
I'm curious to what the "simplest" $\pi$-system is that generates the power set of non-negative integers $\mathbb{Z}_+$. I think the $\pi$-system should look something like
$$ \mathcal{I}=\left\{\{0,\ldots,n\},n\in\mathbb{Z}_+\right\}.$$
Then this is in fact a $\pi$-system but I don't know how to make it rigurous that it generates the power set of non-negative integers ($\sigma(\mathcal{I})=2^{\mathbb{Z}_+})$ and that it is in fact the "simplest" one.
For a more advanced question I want to refer to an unanswered one $\pi$-systems for counting processes. Thanks for any help.

Notice that for $n\geqslant 1$,  $\{ n\}=\{0,\dots,n\}\setminus\{0,\dots,n-1\}$, which is an element of the $\sigma$-algebra generated by $\mathcal I$. We thus deduce that $\{0\}\in\sigma(\mathcal I)$. 
We have shown that $\{n\} \in\sigma(\mathcal I)$ for each $n\in\mathbb Z_+$. Since each element of the power set of $\mathbb Z_+$ is a countable union of sets of the form $\{n\}$, $n\in\mathbb Z_+$, we showed that $\sigma(\mathcal I)=2^{\mathbb Z_+}  $.
