Power Set \(\sigma\)-algebra

Theorem

Given any set \(S\), the power set of \(S\) is a sigma-algebra on \(S\).

Proof

First note that \(\varnothing \in \mathcal{P}(S)\) since \(\varnothing \subseteq \mathcal{P}(S)\).

Let \(A \subseteq S\). Then \(A^{c} = S - A \subseteq S\) since by definition \(S - A\) is \(S\) with some subset removed.

Finally, consider a countable sequence of subsets of \(S\):

\[ \{A_i\}_{i \in \mathbb{N}}\]

Now let \(x \in \bigcup_{i \in \mathbb{N}} A_i\). Hence by the definition of the union \(x \in A_i\) for some \(i \in \mathbb{N}\). However then by the definition of the subset, we have that \(x \in S\).

Therefore

\[ \bigcup_{i \in \mathbb{N}} A_i \subseteq S\]

and hence

\[ \bigcup_{i \in \mathbb{N}} A_i \in \mathcal{P}(S).\]