Is it true, that $\mathcal{A}(K) = \mathcal{B}(\mathbb{R})$ for $K = \left\{\left[a,b\right]\::\:-\infty <a<b<\infty \right\} \subset 2^{\mathbb{R}}$?

Question

As far as I understand, a Borel Algebra is generated from a family of open sets. But here, $K$ is a family of closed sets.

This was in today's lecture, and I don't know if it shouldn't be $(a,b)$ instead of $\left[a,b\right]$ here

Answer 1

If you want to get an open interval $(a,b)$ from the collection $K$, you can find two sequences $\{a_n\}_{n=1}^{\infty}$ and $\{b_n\}_{n=1}^{\infty}$ such that $a<a_{n+1}<a_n<b_n<b_{n+1}<b$ for every positive integer $n$ and $a_n \to a$, $b_n \to b$ as $n \to \infty$. Observe that:

\begin{equation*} (a,b) = \bigcup_{i=1}^{\infty}[a_n,b_n] \end{equation*}