Let $\mathcal{B}(\mathbb{R})$ be the Borel $\sigma$-algebra on $\mathbb{R}$. Let $\mathcal{C}=\{\text{all closed subsets of $\mathbb{R}$}\}$ and $\mathcal{O}=\{\text{all open subsets of $\mathbb{R}$}\}$. I want to show that $\sigma(\mathcal{C})=\sigma(\mathcal{O})\;(\,=\mathcal{B}(\mathbb{R})\,)$ by showing
- $\sigma(\mathcal{C})\subset\mathcal{B}(\mathbb{R})$
- $\mathcal{B}(\mathbb{R})\subset\sigma(\mathcal{C})$
The argument I'm trying to make is
- All open sets $\mathcal{O}$ are complements of closed sets. Since $\sigma(\mathcal{C})$ contains complements of closed sets, then $\mathcal{O}\subset\sigma(\mathcal{C})$.
- I'm stuck at trying to show that all sets in $\sigma(\mathcal{O})=\mathcal{B}(\mathbb{R})$ that are not in $\mathcal{O}$ are also in $\sigma(\mathcal{C})$.
Best Answer
As you said "All open sets $\mathcal{O}$ are complements of closed sets. Since $\sigma(\mathcal{C})$ contains complements of closed sets, then $\mathcal{O}\subset\sigma(\mathcal{C})$".
By definition, $\sigma(\mathcal{O})$ is the smallest $\sigma$-algebra containing $\mathcal{O}$. As $\sigma(\mathcal{C})$ is a $\sigma$-algebra it follows that $\sigma(\mathcal{O})\subseteq \sigma(\mathcal{C})$. Similarily, $\mathcal{C}\subseteq \sigma(\mathcal{O})$ and therefore by the same argument, $\sigma(\mathcal{C})\subseteq \sigma(\mathcal{O})$. It follows that $\sigma(\mathcal{C})=\sigma(\mathcal{O})$.