# How to prove: For each $\mathcal{E}\subset\mathcal{P}(\Omega)$ exists an $\alpha(\mathcal{E})$?

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 $\mathcal{E}$)
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?

-
Your approach works, but requires transfinite induction. You have to repeat and repeat the operations. The simpler approach is noting that the powerst of $\Omega$ are $\alpha$-systems and that the intersection of arbitrary families of $\alpha$-systems is again an $\alpha$-system. So just take the intersection of all $\alpha$-systems containing $\mathcal{E}$. If you want to put this construction into context, take a look at Moore closures. – Michael Greinecker Apr 23 '12 at 9:24
@jspecter: Note that this is not a sigma algebra; here the countable unions are monotone increasing, whereas in a sigma algebra they can be arbitrary. – Najib Idrissi Apr 23 '12 at 10:05
@zulon, does this mean, the proposed solutions do not hold for an $\alpha$-system? – Aufwind Apr 23 '12 at 10:18
@Aufwind: It does hold, the methodology is exactly the same. All you need to prove is that the intersection of $\alpha$-systems is an $\alpha$-system. – Najib Idrissi Apr 23 '12 at 10:19
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. – Aufwind Apr 23 '12 at 10:22

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 $\mathcal E$.
Do you know what purpose $\alpha$-systems serve? I've never seen that notion before. It seems very closely related to a Dynkin system but it is weaker if I'm not completely mistaken. – t.b. Aug 4 '12 at 19:03