There is a theorem in my book (Measure, Integral, and Probability by Marek Capinski and Peter E. Kopp) that is stated as such:
Theorem
If instead of the family of all intervals we take all open intervals, all closed intervals, all intervals of the form $(a,\infty)$ (or of the form $[a,\infty),(-\infty,b),\;or\;(-\infty,b]$), all open sets, or all closed set, then the $\sigma$-field generated by them is the same as $\mathfrak{B}$, which denotes the Borel set.
The definition of Borel set given is:
First, define $\mathfrak{F_{\alpha}}$ to be $\sigma$-fields and $\alpha \in A$ (A can be arbitrary). Put $$\mathfrak{F}=\bigcap_{\alpha \in A}\mathfrak{F}_{\alpha}$$ Next, let $I$ be the set of all intervals. We finally define a Borel set as: $$\mathfrak{B}=\bigcap\{\mathfrak{F}\supset I\}$$
Now the issue I'm having is that in the proof, they define a set $$\mathfrak{C}=\bigcap\{\mathfrak{F}\supset OI\}$$ where $OI$ stands for open intervals. They wish to show that $\mathfrak{B=C}$. I understand the first argument, where they show that $\mathfrak{C \subset B}$, but then, when showing that $\mathfrak{B \subset C}$, I get confused. Here is what they said:
Consider intervals $[a,b),[a,b],(a,b]$ (where know that that intervals of the form $(a,b)\in \mathfrak{C}$ by construction): $$[a,b)=\bigcap_{n=1}^{\infty} (a_n-\frac{1}{n},b_n)$$ $$[a,b]=\bigcap_{n=1}^{\infty} (a_n-\frac{1}{n},b_n+\frac{1}{n})$$ $$(a,b]=\bigcap_{n=1}^{\infty} (a_n,b_n+\frac{1}{n})$$ $\mathfrak{C}$ as a $\sigma$-field is closed with respect to countable intersection, so it contains the intervals listed above. The argument for unbounded intervals is similiar. Hence, the proof is complete.
So here's my question: is this valid because for all closed intervals (or half-closed,half-open intervals) since we can always find open intervals for which they are contained in?
Yes. First you define $\mathfrak{C}$ as the intersection between all $\sigma$-algebras containing only the open sets. Then, since $\mathfrak{C}$ is a $\sigma$-algebra, it is closed under countable intersections. So if we can write for example a closed interval as a countable intersection of open sets (which is done in the proof), it must also be in $\mathfrak{C}$.