Borel $\sigma$-algebra is equivalent to the smallest $\sigma$-algebra containing (a,$\infty$)

I'm trying to prove that the Borel $$\sigma$$-algebra is equivalent to the smallest $$\sigma$$-algebra containing (a,$$\infty$$). The approach I'm trying to use is to show that if A$$\subset$$B and B$$\subset$$A, then A=B. So far I think I have the first part:

(a, $$\infty$$) = $$\cup$$(a, n) and since each (a, n) is a finite open interval, then they are included in the Borel $$\sigma$$-algebra and since a $$\sigma$$-algebra is closed with respect to countable union, then the smallest $$\sigma$$-algebra containing (a, $$\infty$$) $$\subset$$ Borel $$\sigma$$-algebra.

Looking at (a,b) it seems obvious that it would be contained within (a,$$\infty$$), but I'm not sure if it is sufficient to just say that for the second part of this proof or if there is a formal way to show that (a,b) is contained in (a,$$\infty$$).

• $\sigma$-algebra are stable under finite intersection ! Commented Nov 24, 2020 at 6:17
• Your English seems to be quite good but some of the technical terms you use have no meaning. For example, there is no such thing as $\sigma$- algebra of the form $(a,\infty)$. Commented Nov 24, 2020 at 6:44

Let $$\mathcal A$$ be the smallest $$\sigma$$-algebra containing sets of the form $$(a, \infty)$$. It remains to show that the Borel $$\sigma$$-algebra $$\subset \mathcal A$$.
Every open subset of $$\mathbb R$$ can be written as countable union of open intervals. Thus we are done if we show that all open intervals belongs to $$\mathcal A$$.
Since $$(a, \infty) \in \mathcal A$$, we have $$(-\infty, a] \in \mathcal A$$. Thus $$(a, b]= (-\infty, b] \cap(a,\infty) \in \mathcal A$$ and consequently $$(a, b)= \cup (a, b-\frac{1}{n}]\in \mathcal A$$.
Since $$a,b \in \mathbb R$$ are arbitary, it follows that all open intervals belongs to $$\mathcal A$$. As discussed above this proves that the Borel $$\sigma$$-algebra $$\subset \mathcal A$$.