Showing that the set of closed subsets of $\mathbb{R}$ generates the Borel set

Question

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 $\sigma(\mathcal{C})$.

Answer 1

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 $\sigma(\mathcal{C})=\sigma(\mathcal{O})$.

Answer 2

We know that $\mathcal{B}(\mathbb{R})$ contains all the open sets of $\mathbb{R}$ so it contains all the complements of all the open sets which are the closed sets.

Thus $ \sigma(\mathcal{C}) \subseteq \mathcal{B}(\mathbb{R})$

Also every open set in the real line can be expresed as a countable disjoint union of open intervals.

For every open interval we have $$(a,b)=\bigcup_{n=1}^{\infty}[a+\frac{1}{n},b-\frac{1}{n}]$$

thus every open set is in $\sigma(\mathcal{C})$ and $\sigma(\mathcal{C})$ is a sigma algebra which contains all the open sets thus $\mathcal{B}(\mathbb{R}) \subseteq \sigma(\mathcal{C})$ because $\mathcal{B}(\mathbb{R})$ is the smallest sigma algebra which contains all the open sets.

So $\sigma(\mathcal{C})=\mathcal{B}(\mathbb{R})$

Obviously $\sigma(\mathcal{O}) \subseteq \sigma(\mathcal{C})$

and $\mathcal{B}(\mathbb{R})\subseteq \sigma(\mathcal{O})$ because $\sigma(\mathcal{O})$ is a sigma algebra which contains all the open sets thus it contains as a subset the Borel sigma algebra by the definition of it.

Therefore we have the inclusions:$$\mathcal{B}(\mathbb{R}) \subseteq \sigma(\mathcal{O})\subseteq \sigma(\mathcal{C})= \mathcal{B}(\mathbb{R})$$