Let $\mathcal{C}$ be a monotone class of subsets of $\mathbb{R}^n$ containing all the open and closed sets.Prove that $\mathcal{C} \supset \mathbb{B}^n$ where $\mathbb{B}^n:$ topological Borel field.
My approach: Let $\mathcal{F}_0$: collection of all open and closed sets.
$\mathcal{C}_0$:minimal monotone class containing $\mathcal{F}_0$.
It is sufficient to show $\mathcal{C}_0$ is a field,because then it will also be a Borel field and hence contain $\mathbb{B}^n$.
Define $\mathcal{C}_3=\{E \in \mathcal{C}_0:E^c \in \mathcal{C}_0 \}$. I have shown that $\mathcal{C}_3$ is a monotone class and $\mathcal{C}_3 \supset \mathcal{F}_0$. Also, by minimality, $\mathcal{C}_0 \subset \mathcal{C}_3 \implies \mathcal{C}_0=\mathcal{C}_3$
So, $\mathcal{C}_0$ is closed under complementation.
But I can't show closure under intersection.
Anyone to help?
Actually it is enough for the monotone class to contain all open sets.
The most reasonable way to proceed, in my opinion, is first to show that the monotone class contains an algebra generating the Borel $\sigma$-algebra and then to employ the monotone class theorem.
That said, consider the standard algebra $\mathcal A$ of finite unions of rectangles of the form $$ \prod_{i=1}^n [a_i,b_i), $$ where $-\infty\le a_i<b_i\le \infty$ and $[-\infty,x):= (-\infty,x)$. It is contained in the monotone class containing all open sets, since $$ \prod_{i=1}^n [a_i,b_i) = \bigcap_{m\ge 1} \prod_{i=1}^n (a_i-1/m,b_i). $$ But $\mathcal A$ generates the Borel $\sigma$-algebra, so we get the desired statement by virtue of the monotone class theorem.