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I am somehow confused about this exercise.

First of all, a definition: A system of subsets $\mathcal{R}\subset\mathcal{P}(\Omega)$, where $\Omega$ is a nonempty set $\Omega$, is called $\alpha$-system, if it does satisfy the following conditions:

  1. $\Omega\in\mathcal{R}$
  2. $A\in\mathcal{R}\Rightarrow A^c\in\mathcal{R}$
  3. For each sequence of subsets $A_1\subset A_2,...$ of elements from $\mathcal{R}$ it is true that $\bigcup_{n=1}^\infty A_n\in\mathcal{R}$

My goal is to prove, that for each subset $\mathcal{E}\subset\mathcal{P}(\Omega)$ it is true that there exists $\alpha(\mathcal{E})$, which the smallest $\alpha$-system which contains $\mathcal{E}$.

My thoughts: If I take a subset $\mathcal{E}\subset\mathcal{P}(\Omega)$, then I just have to

  1. add $\Omega$ and the empty set $\emptyset$ (if they are not already in $\mathcal{E}$)
  2. add each complement
  3. and make sure, that each union of a sequence of subsets $A_1\subset A_2,...$ of elements from $\alpha(\mathcal{E})$ is in $\alpha(\mathcal{E})$.

Concerning point three I have no ideas how to approach it. Any inspiration?

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    Thank you @all for the hints. I think I got it now. If someone want to move his comment to an answer, I will gladly accept it.2012-04-23

1 Answers 1

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We have to show that an arbitrary intersection of $\alpha$-systems $\mathcal R_i$, $i\in I$, still is a $\alpha$-system. We will denote it by $\mathcal R$

  1. Since $\Omega\in\mathcal R_i $ for all $i$, $\Omega\in\mathcal R$.
  2. Let $A\in\mathcal R$ and $i\in I$. Then $A^c\in\mathcal R_i$ (because $\mathcal R_i$ is a $\alpha$-system), and we conclude that $A^c\in\mathcal R_i$.
  3. Let $\{A_k\}$ an increasing sequence of elements of $\mathcal R$, and $A$ the union of the elements of this sequence. Then $A\in\mathcal R_i$ for all $i\in I$.

Hence for $\mathcal E\subset \mathcal P(\Omega)$, we can take the intersection of all the $\alpha$-systems containing $\mathcal E$. It's an $\alpha$-system and it's by construction the smallest among those which contain $\mathcal E$.

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    I don't know, and I've never seen this before.2012-08-05