Set $A \subset$ $\sigma$-algebra $B$ implies $\sigma(A) \subset B$ (Jacod and Protter, proof of Theorem 2.1)

Question

I am studying from Jacod and Protter's Probability Essentials, 2nd ed. In particular, I am stuck on their proof of the following theorem:

Theorem 2.1 The Borel $\sigma$-algebra of $\mathbf{R}$ is generated by intervals of the form $(-\infty, a]$, where $a \in \mathbf{Q}$ ($\mathbf{Q}$ = rationals).

In particular, they claim that $A \subset B$, where $B$ is a $\sigma$-algebra, implies that $\sigma(A) \subset B$, where $\sigma(A)$ is the minimal sigma algebra generated by $A$.

It seems to me that $A \subset \sigma(A)$, so that $A \subset B \not\Rightarrow \sigma(A) \subset B$, i.e., in general $\exists a \in \sigma(A): a \not\in A$, and $a \in \sigma(A) \not\Rightarrow a \in B$.

Where does this claim come from?

Answer 1

$\sigma(A)$ is the intersection of all $\sigma$-algebras containing $A$. $B$ is one such $\sigma$-algebra and so...

Answer 2

By definition. We have that $\sigma(A)$ is the smallest $\sigma$-algebra containing the set $A$, i.e. $$\sigma(A) = \bigcap_{\substack{C \text{ $\sigma$-algebra}\\A \subseteq C}} C.$$ Now since $B$ is a $\sigma$-algebra containg $A$, we have that $$\bigcap_{\substack{C \text{ $\sigma$-algebra}\\A \subseteq C}} C \subseteq B$$ and the claim follows.