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?