Let $\Omega$ be a finite set and $\mathcal{A}$ be a $\sigma$-Algebra.
I want to show, that there exists one and only one single partition $\{\pi_1 ,\dots , \pi_n\}$ such that
- $\pi_i \in \mathcal{A}$ for every index $i\in\{1,\dots,n\}$
- For all $B\in\mathcal{A}$ and for all $\ell\in\{1,\dots,n\}$: $\pi_\ell\cap B\in\{\emptyset,\pi_\ell\}$
Here are my thoughts about that. The only way I can think of to get hold of $\{\pi_1 ,\dots , \pi_n\}$ is to break $\Omega$ as follows:
- Take an element $B\in\mathcal{A}$.
- Take all elements $C_i\in\mathcal{A}$, which have an intersection with $B$.
- Now build every possible intersections of the set $B$ with every count of the sets $C_i$.
- Chose $B \setminus \bigcup_i C_i$ as a an element of $\{\pi_1 ,\dots , \pi_n\}$
- Every intersection of the 3. step which has no other intersection from step 3. as a subset is an element of $\{\pi_1 ,\dots , \pi_n\}$ too.
This is the construction I came up with. But it seems really cumbersome. Is there a more intuitive way? (Or perhaps I should ask: Is this approach correct after all?)