I am going through the proof of one of the theorems from Jacod-Protter and unable to understand this. Theorem 2.1: The Borel $\sigma$-algebra of $\mathbb{R}$ is generated by intervals of the form $(-\infty,a]$ where $a\in \mathbb{Q}$
Proof: Let $\mathcal{C}$ = all the open intervals of $\mathbb{R}$. Then we have $\sigma(\mathcal{C})$ = the Borel $\sigma$-algebra of $\mathbb{R}$.
Assume $\mathcal{D}$ to be all the sets of the form $(-\infty,a]$.
Take a set $(a,b)\in\mathcal{C}$ and let $(a_n)_{n \geq1}$ be a decreasing sequence decreasing to $a$ while $(b_n)_{n \geq1}$ be a strictly increasing sequence increasing to  $b_n$.
Then it can be seen that $(a,b) = \cup_{i=1}^{\infty}(a_n,b_n] = \cup_{i=1}^{\infty}((-\infty,b_n] \cap (-\infty,a_n]^c)$
As $\sigma(\mathcal{D})$ is a $\sigma$-algebra generated by $\mathcal{D}$, then sets of the form $(-\infty,a]$ and $(-\infty,a_n]^c$ will belong to $\sigma(\mathcal{D})$ and so will the countable union and intersections of sets of these type $\implies \cup_{i=1}^{\infty}((-\infty,b_n] \cap (-\infty,a_n]^c) \in \sigma(\mathcal{D})$
Hence, $\mathcal{C}\in\sigma(\mathcal{D})$.
The next line in the book says: Therefore $\mathcal{C}\in\sigma(\mathcal{D})$, whence $\sigma(\mathcal{C})\subset\sigma(\mathcal{D})$
My question: As $\sigma(\mathcal{C})$ is the smallest $\sigma$ algebra containing $\mathcal{C}$, that means $\mathcal{C}\in\sigma(\mathcal{C})$. Now, how can we  infer from this that $\sigma(\mathcal{C})\subset\sigma(\mathcal{D})$? What am I missing or where am I going wrong?
 A: There may be a typo, as the correct conclusion is $\mathcal C \subset \sigma(\mathcal D)$. The proof demonstrates that each open interval $(a,b) \in \mathcal C$ may be written as a countable union of intersections of elements of $\mathcal D$ and their complements; as $\sigma(\mathcal D)$ is closed under each of these operations, it follows that $(a,b) \in \sigma(\mathcal D)$ and therefore $\mathcal C \subset \sigma(\mathcal D)$.
The last step is an appeal to the definition. Indeed, for a set $A$ the algebra $\sigma(A)$ is defined to be the smallest (i.e., minimal with respect to the $\subset$ relation) sigma algebra which contains $A$. Another way to state this is
$$\sigma(A) = \bigcap_{A \subset \mathcal F} \mathcal F$$
where the intersection is taken over all the sigma algebras $\mathcal F$ which contain $A$.
Consequently, if $A \subset \sigma(B)$ then $\sigma(A) \subset \sigma(B)$ for any set $B$. The algebra $\sigma(B)$ is an example of a sigma algebra containing $A$, therefore $\sigma(A)$ must be a subset of $\sigma(B)$ (and possibly coincide with it).
Said another way, $A \subset B \implies \sigma(A) \subset \sigma(B)$, therefore $A \subset \sigma(B) \implies \sigma(A) \subset \sigma(\sigma(B)) = \sigma(B)$.
