Prove that the $\sigma$ - algebras are equal I want to show that $\sigma$-algebras on $\mathbb{R}$ generated by $(a,b), \ (a,b], [a, b), [a,b], (-\infty, a), (-\infty, a], (b, +\infty), [b, +\infty)$ for $a,b \in \mathbb{R}$ and $a,b \in \mathbb{Q}$ are all equal.
Here is what I've come up with so far:
($\mathcal{M} $ - sigma-algebra) 
Would it suffice to say that $\sigma$-algebras generated by $(a,b), \ (a,b], [a, b), [a,b], \ a,b \in \mathbb{R}$ are the same because 
$(a,b) \in \mathcal{M} \ \Rightarrow \mathbb{R} \setminus (a,b) = (-\infty, a] \cup [b, +\infty) \in \mathcal{M}$, so $[a,b] = \bigcap_{n \in \mathbb{N_+}} (a-\frac{1}{n}, b+\frac{1}{n}) = \bigcap_{n \in \mathbb{N_+}} (\mathbb{R} \setminus (-\infty, a-\frac{1}{n}] \cup [ b+\frac{1}{n}, +\infty) )=$
$ = \mathbb{R} \setminus(\bigcup_{n \in \mathbb{N_+}}  (-\infty, a-\frac{1}{n}] \cup [ b+\frac{1}{n}, +\infty) ) \in \mathcal{M}$
Similarly, $[a,b) = \bigcap_{n \in \mathbb{N_+}} (a-\frac{1}{n}, b)$, $(a,b] = \bigcap_{n \in \mathbb{N_+}} (a, b +\frac{1}{n})$.
When it comes to $(-\infty, a), (-\infty, a], (b, +\infty), [b, +\infty)$ for $a,b \in \mathbb{R}$, would it be enough to say that $(-\infty, a) = \bigcup_{n \in \mathbb{N}}(-n, a)$ and etc?
In case of $a,b \in \mathbb{Q}$ would it be all right to say that $\mathbb{Q}$ is dense in $\mathbb{R}$ so for every $r \in \mathbb{R}$ we will find a sequence $\{q_n\}_{n \in \mathbb{N}}$ that converges to $r$, so for example $(r, b) = \bigcup_{n \in \mathbb{N}}(q_n, b)$ ?
Could you tell me if my approach is right?
Please, help.
Thank you.

The easiest way is to show that any $\sigma$-algebra containing rational intervals must contain all real intervals.  
You've got the right idea: the rationals are a countable dense subset of the real numbers, so you can always find a sequence of rational numbers converging to any real number, which you can use to construct countable unions and intersections to make all the types of interval you want.  
