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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

  1. $\pi_i \in \mathcal{A}$ for every index $i\in\{1,\dots,n\}$
  2. 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:

  1. Take an element $B\in\mathcal{A}$.
  2. Take all elements $C_i\in\mathcal{A}$, which have an intersection with $B$.
  3. Now build every possible intersections of the set $B$ with every count of the sets $C_i$.
  4. Chose $B \setminus \bigcup_i C_i$ as a an element of $\{\pi_1 ,\dots , \pi_n\}$
  5. 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?)

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1 Answer 1

up vote 3 down vote accepted

Hint: For each $\omega\in\Omega$ define $\pi(\omega)$ to be the intersection of all sets in $\cal A$ that contain $\omega$.

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This definition helped a lot, thanks! –  Aufwind Nov 2 '11 at 10:51
    
Glad to help out. –  Byron Schmuland Nov 2 '11 at 11:47

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