Why are the Borel sets of $\mathbb{R}$ equal to the monotone class generated by the open intervals? Let $\mathcal{B}(\mathbb{R})$ denote the Borel subsets of $\mathbb{R}$, and let $\mathcal{M}$ denote the monotone class generated by all open intervals $(a,b) \subset \mathbb{R}$. Since $\mathcal{B}(\mathbb{R})$ is also the $\sigma$-field generated by the open intervals, the monotone class theorem implies that $\mathcal{B}(\mathbb{R}) = \mathcal{M}$.
The Borel $\sigma$-field contains all open subsets of $\mathbb{R}$, and so the above implies that $\mathcal{M}$ must as well. However, I don't understand why this follows from the definition of $\mathcal{M}$. For example, how can I write $(0,1) \cup (1,2)$ as the countable union of an increasing sequence of open intervals?
Edit: to be clear, I understand that the fact that $\mathcal{M}$ is closed under countable monotone intersections and unions does not imply that all $A \in \mathcal{M}$ can be written as countable monotone intersection or union, but I'm still not really clear about how sets like the one above can arise in the monotone class.
To put this another way, how can I prove that $(0,1) \cup (1,2) \in \mathcal{M}$ without using the monotone class theorem?
Edit: the open intervals are not an algebra, e.g. they are not closed under finite unions, so I think the monotone class theorem isn't used correctly here.
 A: You are very right to be confused - the result as stated in wikipedia is incorrect. This is almost noticed at the talk page there, and see here for a correct statement of the theorem.
The issue is that we need to also include the ability to form finite set differences in order to get things like $(0,1)\cup(1,2)$. To see this, note the following:

Let $\mathfrak{C}$ be the class of all convex subsets of $\mathbb{R}$, that is, all $A\subseteq\mathbb{R}$ such that whenever $a,b\in A$ with $a<b$ we have $[a,b]\subseteq\mathbb{R}$. Then $\mathfrak{C}$ contains every open interval, is closed under countable increasing unions, and is closed under countable decreasing intersections.

At a glance, I believe that what happened is this: the term "monotone class" is generally used to refer to a class of sets closed under countable increasing unions and countable decreasing intersections. However, at least one source (see the talk page) uses "monotone class" to refer to Dynkin systems. The article then mixes this up.
Finally, to wrap things up, suppose $\mathfrak{D}$ is a Dynkin system containing each of the open intervals. For each $\epsilon<1$ we have $$X_\epsilon:=(0,2)\setminus (1-\epsilon, 1+\epsilon)\in \mathfrak{D}.$$ Now consider the countable increasing union $$Y=\bigcup_{n\in\mathbb{N}}X_{1\over n}.$$ This $Y$ is again in $\mathfrak{D}$, and is exactly $(0,1)\cup (1,2)$.
A: I think you misunderstood the definition  of $\mathcal M$. $\mathcal M$ is the smallest class of sets closed under increasing and decreasing unions and containing all open intervals. This does not mean that every set in it is  a countable increasing union of open intervals. In fact your set cannot be written in this form.
