generating classes for Borel sigma-field (on $\mathbb R$)? I'm learning measure theoretic probability, and am a bit confused about the generating classes of the Borel $\sigma$-field. By definition, the Borel $\sigma$-field $\mathcal B$ on the real line is generated by the class of open sets in $\mathbb R$.  In David Pollard's book ("A User's Guide to Measure-theoretic probability"), it's shown that $\mathcal A_1=\{(-\infty, x]: x\in \mathbb R\}$ is also a generating class, i.e. $\mathcal B=\sigma(\mathcal A_1).$  I would like to confirm the following:
Q1: Let $\mathcal A_2=\{(-\infty, x): x\in \mathbb R\}$.  Since $(-\infty, x]=\bigcap_n(-\infty, x+\frac{1}{n})\in \sigma(\mathcal A_2)$, it follows that $\sigma(\mathcal A_1)\subset \sigma(\mathcal A_2)$.  Conversely, $(-\infty, x)=\bigcup_n(-\infty, x-\frac{1}{n}]\in \sigma(\mathcal A_1),$ so $\sigma(\mathcal A_2)\subset \sigma(\mathcal A_1)$. Hence $\sigma(\mathcal A_1)= \sigma(\mathcal A_2)$.  So $\mathcal A_2$ is also a generating class, right?
Q2: Let $\mathcal A^Q_1\triangleq \{(-\infty, q]:q\in \mathbb Q\}.$  Clearly, $\sigma(A_1^Q)\subset \sigma(A_1)$. 
 But the converse is also true, since $(-\infty, x]=\bigcap_n(-\infty, a_n]$ for any rational sequence $\{a_n\}$ with $a_n>x$ and $a_n\to x$, c.f. $[0,1]=\bigcap_n \big[0, 1+\frac{1}{n}\big)=\bigcap_n \big [0, 1+\frac{1}{n}\big]$?  Therefore, $\mathcal B=\sigma(\mathcal A_1^Q)$ ?
Q3: Similarly, $\mathcal A^Q_2\triangleq \{(-\infty, q):q\in \mathbb Q\}$ is also a generating class?  (Obviously $\sigma(A_2^Q)\subset \sigma(A_2)$.  But we also have $(-\infty, x)=\bigcup_n(-\infty, b_n)$ for any rational sequence $\{b_n\}$ with $b_n<x$ and $b_n\to x,$ implying the converse.)
In other words, $\mathcal B=\sigma(\{(-\infty, x]\})=\sigma(\{(-\infty, x)\})$, where $x$ may be any real number or any rational number.
Q4:  In general, for any class $\mathcal A$ of sets in $\mathbb R$, let $\mathcal A^c=\{E^c: E\in \mathcal A\}$.  Then $\sigma(\mathcal A)=\sigma(\mathcal A^c),$ right?  (For if $E \in \mathcal A,$ then $E^c\in \mathcal A^c$. So $E=(E^c)^c\in \sigma(\mathcal A^c)$, and $\sigma(\mathcal A)\subset \sigma(\mathcal A^c)$.  And vice versa.)
Q5: As a result, $\mathcal B$ may also be generated by the classes $\{(x, \infty)\}$ or $\{[x, \infty)\}$, where $x\in \mathbb R$ or $x\in \mathbb Q$.
To recap, $\mathcal B$ may be generated by any of the following classes: $\{(-\infty, x]\}$, $\{(-\infty, x)\}$, $\{(x, \infty)\}$ or $\{[x, \infty)\}$, where $x\in \mathbb R$ or $x\in \mathbb Q$.
Are the above arguments correct?  I'd appreciate it if someone can point it out if I'm mistaken.  Thanks a lot!
 A: Answer on Q1,Q2,Q3.
Your "conversely" proof on Q1 is enough to prove that $\sigma(\mathcal A_1)=\sigma(\mathcal A_2)$ (and then  $\sigma(\mathcal A_1)=\mathcal B$ on base of $\sigma(\mathcal A_2)=\mathcal B)$ but you can catch things also by proving directly that $\sigma(\mathcal A_1)=\mathcal B=\sigma(\mathcal\tau)$ where $\tau$ denotes the topology on $\mathbb R$, and in a sense that is enlightening in my view. 
It must be shown that every open set is element of $\sigma(\mathcal A_1)$. 
We have $(a,b]=(-\infty,a]^c\cap(-\infty,b]\in\sigma(\mathcal A_1)$.
For open set $U\subseteq\mathbb R$ it is not really difficult to prove the essential equality: $$U=\bigcup\{(a,b]\mid a,b\in\mathbb Q\wedge(a,b]\subset U\}\tag1$$
and consequently:$$U\in\sigma(\mathcal A_1)$$
This because the RHS of $(1)$ is a countable union of sets in $\sigma(\mathcal A_1)$.
Note that only intervals $(a,b]$ are used with $a,b\in\mathbb Q$. 
This fact confirms that also collections like $\mathcal A_1^{\mathbb Q}$ are generating. Secondly we can also do it with intervals $[a,b)$ so $\mathcal A_2^{\mathbb Q}$ will generate also.
Actually instead of $\mathbb Q$ we can use any countable and dense set $D\subset\mathbb R$.

The answers on Q4 and Q5 are both: "yes".
