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The problem is: prove that $\mathcal{A} \subset 2^{\Omega}$ is an algebra iff:

(i) $\Omega \in \mathcal{A}$

(ii) $\mathcal{A}$ closed under complements

(iii) $\mathcal{A}$ closed under intersections

The definition of algebra is: (ia) as above, (iia) $\mathcal{A}$ is \-closed, (iiia) $\mathcal{A}$ is $\cup$-closed.

No need to prove (ia). For (iiia) it is trivial to take a complement of the intersection and use property (ii) to show that $\mathcal{A}$ is also closed under unions. However, I am struggling a bit with showing that $\mathcal{A}$ is \-closed, which would conclude my proof.

This is what I attempted:

Using property (iii): $A, B \in \mathcal{A} \rightarrow A \cap B \in \mathcal{A}$,

but $A \cap B = A \backslash (A \backslash B) \in \mathcal{A}$, but this not the proof...

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Let $A,B\in\mathcal{A}$. We have to show that $A\setminus B\in\mathcal{A}$.

Due to (ii) we have $\Omega\setminus B\in\mathcal{A}$. And using (iii) it follows that $A\setminus B = A\cap (\Omega\setminus B)\in \mathcal{A}$.

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