Prove that $\sigma$-algebra of subsets of $\mathbb{R}$ of the form $(a,\infty)$ contains all the intervals. I want to prove that if a $\sigma$-algebra of subsets of $\mathbb{R}$ contains intervals of the form $(a,\infty)$, then it contains all the intervals.
Proof:
Let $\mathscr{B}$ to denote the $\sigma$-algebra defined above. We have that since $(a,\infty) \in \mathscr{B}$, then since it contains the complement, we have that it also contains the intervals of the form $(-\infty,b]$, i.e. $$(-\infty,b] \in \mathscr{B}$$
In particular, we have that $(a,b) \in \mathscr{B}$. It's enough to prove that:
$$[b, \infty) \in \mathscr{B}$$
$$[a,b] \in \mathscr{B}$$
$$[a,b) \in \mathscr{B}$$
$$(a,b) \in \mathscr{B}$$
This is clear, since: 
$$[b, \infty) = \bigcap_{k=1}^{\infty} (b-\frac{1}{n}, \infty)$$
$$[a,b] = \bigcap_{n=1}^{\infty} (a-\frac{1}{n},b+\frac{1}{n})$$
$$[a,b) = \bigcap_{n=1}^{\infty} (a,b+\frac{1}{n})$$
$$(a,b) = \mathscr{B} \backslash [(-\infty,a]\cup[b,\infty)]$$
Is this proof correct?
 A: 
A new organized answer is required.

$\newcommand{\b}{\mathscr B}$Ok, I will give it. 
What is the structure of an interval? It is one of the following : $(a,b),[a,b),(a,b],[a,b]$, where one of $a,b$ could be $\pm \infty$.
Let us start with $(a,\infty) \in \b$ for all $a \color{blue}{(1)}$.
Without complementation, note that $[a,\infty) = \bigcap_{n=1}^\infty \left(a-\frac 1n,\infty\right)$, so $[a,\infty) \in \b$ for all $a\color{blue}{(2)}$.
Now, the complements of $\color{blue}{(1)}$ give $(-\infty,b] \in \b$ for all $b\color{blue}{(3)}$.
The complements of $\color{blue}{(2)}$ give $(-\infty,b) \in \b$ for all $b\color{blue}{(4)}$.
Thus, one sees that all cases where one endpoint is infinite is covered. This can be used to systematically process the rest of the intervals.

Now, 


*

*$(a,b) = \mathbb R \setminus ((-\infty,a] \cup [b,\infty))$.

*$[a,b) = \mathbb R \setminus ((-\infty,a) \cup [b,\infty)$.

*$(a,b] = \mathbb R \setminus ((-\infty,a] \cup (b,\infty))$.

*$[a,b] = \mathbb R \setminus ((-\infty,a) \cup (b,\infty))$.
Which shows that each of these intervals, for arbitrary $a \leq b \in \mathbb R$ belong to $\b$. 
