0.0.0.1 Problem
A family of sets \(\mathcal{R}\subset 2^X\) is called a [[ring|ring]] if it is closed under finite unions and differences (i.e. if \(E_{1}, \dots, E_{n}\in\mathcal{R}\) then \(\bigcup_{i=1}^nE_{j}\in\mathcal{R}\) and if \(E,F \in\mathcal{R}\) then \(E \setminus F:=E \cap F^c\in \mathcal{R}\)). A ring that is closed under countable unions is called a \(\sigma\)-ring.
Prove the following results:
- Rings (\(\sigma\)-rings) are closed under finite (countable) intersections.
- Rings (\(\sigma\)-rings) are algebras (\(\sigma\)-algebras) when \(X\in\mathcal{R}\).
- For \(\sigma\)-ring \(\mathcal{R}\) then \(\mathcal{F}_{1}:=\{ E \subset X:E \in \mathcal{R}~\text{or}~E^c\in\mathcal{R} \}\) is a \(\sigma\)-algebra.
- For \(\sigma\)-ring \(\mathcal{R}\) then \(\mathcal{F}_{2}:=\{ E \subset X: E \cap F \in\mathcal{R},~\forall F \in\mathcal{R} \}\) is a \(\sigma\)-algebra.
0.0.0.2 Solution
\begin{proof}[Proof (1).] Assuming that \(E_{1},E_{2} \in\mathcal{R}\) for \(\sigma\)-ring \(\mathcal{R}\) we have that \[
E_{1}\cap E_{2}=(E_{1}\cup E_{2})\setminus \left\{ (E_{1}\setminus E_{2})\cup(E_{2}\setminus E_{1}) \right\} \in\mathcal{R},
\] since \(\mathcal{R}\) is closed under countable unions and differences. Inductively we can expand this to the \(n\) set case (and onwards to countably many sets) since intersections are associative, i.e. \[
\bigcap_{j=1}^n E_{j}=\bigcup_{j=1}^nE_{j}\setminus \bigcup_{(i,j)\in \{ 1, \dots , n \}^2}E_{i}\setminus E_{j}.
\] \end{proof} \begin{proof}[Proof (2).] We need only show that \(\mathcal{R}\) is closed under complements. Assuming \(X \in\mathcal{R}\) we have that \(X^c=\emptyset=X\setminus X\in\mathcal{R}\). For all other sets \(E \in \mathcal{R}\) we therefore have \(E^c=X\setminus E\in\mathcal{R}\).\end{proof}
\begin{proof}[Proof (3).]
Since \(\mathcal{R}\) is a \(\sigma\)-ring it is non-empty hence neither is \(\mathcal{F}_1\). Furthermore clearly \(\mathcal{F}_{1}\) is closed under compliments hence all we need is to show \(F_{1}\) is closed under countable unions. Assuming \(\{ E_{j} \}_{j=1}^\infty \subseteq \mathcal{F}_{1}\) then \(\forall j\) either \(E_{j}\in\mathcal{R}\) or \(E_{j}^c\in\mathcal{R}\).
\end{proof}