This following statement is from a book:
Let $C$ denote all open intervals. Since every open set in
$\mathbb{R}$ is the countable union of open intervals, we have
$\sigma(C)=$ the Borel $\sigma$-algebra of $\mathbb{R}$.
I need help understanding this. First, when it is said that $C$ denotes all open intervals, does it literally mean all imaginable open intervals $(a,b)$ in $\mathbb{R}$? If so, isn't the number of those intervals infinite and uncountable?
Second, can you give me a literal example of what is meant by the statement that every open set in $\mathbb{R}$ is the countable union of open intervals? I assume that an "open set" differs from an "open interval" in $\mathbb{R}$ by possibly having missing points or intervals in the middle?
Finally, how does one exactly go from every open set being the countable union of open intervals to $C$ generating the Borel sigma-algebra? This is not explained in the book at all.
Best Answer
Yes, $C$ is a set containing uncountably many elements, where each element is an open interval, of one of the three forms $(a,b)$ or $(-\infty,b)$ or $(a,\infty)$.
Every open interval is an open set. Any union of open sets is an open set. So, in particular, any countable union of open intervals $\bigcup_{n=1}^{\infty}I_n$ is an open set. An important fact about $\mathbb{R}$ is that every open subset of $\mathbb{R}$ can be expressed as a countable union of open intervals $\bigcup_{n=1}^{\infty}I_n$. (In fact, it's possible to do this with pairwise disjoint open intervals, in which case the representation is unique, but we don't need that here.) See here for some proofs of this fact: Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]
We already have the notation $C$ for the set of all open intervals, and $\sigma(C)$ for the $\sigma$-algebra generated by $C$.
Let us introduce the notation $O$ for the set of all open sets, and $B = \sigma(O)$ for the $\sigma$-algebra generated by $O$. By definition, this is the Borel $\sigma$-algebra.
Now, what does it mean that $\sigma(C)$ is the $\sigma$-algebra generated by $C$? It means three things: (1) $\sigma(C)$ is a $\sigma$-algebra; (2) $C \subseteq \sigma(C)$; (3) $\sigma(C)$ is the smallest $\sigma$-algebra containing $C$, meaning that if $\Sigma$ is another $\sigma$-algebra with $C \subseteq \Sigma$, then $\sigma(C) \subseteq \Sigma$.
Similarly, the Borel $\sigma$-algebra $B$ is the $\sigma$-algebra generated by the open subsets $O$, in other words, the smallest $\sigma$-algebra containing $O$.
Every open interval is an open set, and $B$ contains all open sets, so $C \subseteq B$. Therefore, $B$ is a $\sigma$-algebra containing $C$. But $\sigma(C)$ is the smallest $\sigma$-algebra containing $C$, so $\sigma(C) \subseteq B$.
On the other hand, every open set is a countable union of open intervals, and $\sigma(C)$ contains all countable unions of open intervals, so $\sigma(C)$ contains all open sets in $\mathbb{R}$. In other words, $O \subseteq \sigma(C)$. But $B$ is the smallest $\sigma$-algebra containing $O$, so $B \subseteq \sigma(C)$.
Summarizing the conclusions of the two preceding paragraphs, we have $\sigma(C) \subseteq B$ and $B \subseteq \sigma(C)$, so $B = \sigma(C)$.