if \(|\Omega|=n<\infty\) then \(|\mathcal{P}(\Omega)|=2^n\) which is finite.

As for the \( \sigma\)-algebra part.

\(\mathcal{P}(\Omega)\) is nonempty since it contains the empty set.

If we let \(X\in\mathcal{P}(\Omega)\) then \(X^c=\Omega\backslash X\subset\Omega\). and since \(\mathcal{P}(\Omega)\) contains all subsets of \(\Omega\) we have \(X^c\in\mathcal{P}(\Omega)\)