Borel $\sigma$-Algebra generated by open intervall with rational end point

Question

$A$ is the $\sigma$-Algebra generated by $(p,\infty)$ with $p\in \mathbb{Q}$. I need to show that $(a,b)\in A$ for $a,b\in \mathbb{R}$ and $a<b$ so that $A$ is a Borel $\sigma$-Algebra on $\mathbb{R}$.

Proof (what I tried):

We now that rational numbers are dense in the reals, so for a we can find a series ('from the right') with $$\lim_{n\to\infty} q_n = a$$ with $q_n>a$. Then $$\bigcup\limits_{n=1}^{\infty} (q_n,\infty)=(a,\infty)$$ This is a countable Union of sub-sets of $A$ so it is itselfe in $A$ and we get $(a,b)\in A$. That is true for every open set in $\mathbb{R}$. Since every open set in $\mathbb{R}$ can be written as a a countable union of intervals from $A$, $A$ contains the Borel $\sigma$-Algebra of $\mathbb{R}$. Because we said $q_n \downarrow a$ it is the smallest $\sigma$-Algebra containing the open intervall by definition. So it is the Borel $\sigma$-Algebra of $\mathbb{R}$.

I hope this is not a duplicate and not too bad, I would appreciate corrections, help and tips.

Answer 1

You proved that for every $a\in\mathbb R$ the set $(a,\infty)$ is an element of $\mathcal A$, but your conclusion that $(a,b)\in\mathcal A$ is not justified.

You can solve this by proving that also sets like $[a,\infty)$ are elements of $\mathcal A$ on a similar way with $q_n$ approaching from the left and taking $\bigcap_{n=1}^{\infty}(q_n,\infty)=[a,\infty)$. This in the knowledge that a $\sigma$-algebra is also closed under countable intersections.

Then also $(-\infty, b)=[b,\infty)^{\complement}$ and $(a,b)=(a,\infty)\cap(-\infty,b)$ are elements of $\mathcal A$.

If that is done then you can indeed claim that all open sets - as countable unions of open intervals - are elements of $\mathcal A$ because a $\sigma$-algebra is closed under complements and countable intersections.

This ensures that the Borel-$\sigma$-algebra on $\mathbb R$ if it is equipped with its usual order topology is a subcollection of $\mathcal A$.

Conversely we know that this Borel-$\sigma$-algebra will contain all sets $(p,\infty)$ where $p\in\mathbb Q$ since these sets are open. This ensures that $\mathcal A$ which is generated by these sets is a subcollection of the Borel-$\sigma$-algebra.

Answer 2

There are a couple of things that are a little inaccurate in your proof. That is, the general proof is fine, but it has a couple of issues.

Issue 1:

You claim that

I need to show that $(a,b)\in A$ for $a,b\in\mathbb R$ and $a<b$ so that $A$ is a Borel $\sigma$-Algebra on $\mathbb R$

however, in your proof, you don't do this. That is, you prove that $A$ is a borel sigma algebra on $\mathbb R$, but you didn't prove that by first proving $(a,b)\in A$ for $a,b\in\mathbb R$. This, to a reader of the proof, can be confusing. In a proof, if you announce you will prove $X$ by doing $Y$, you should then actually do $Y$.

Issue 2:

You claim that

Because we said $q_n \downarrow a$ it is the smallest $\sigma$-Algebra containing the open interval by definition.

However, this claim is quite nonsensical. I don't really know what you were trying to prove here. Since $A$ is generated by a subset of the Borel $\sigma$-algebra, it is obvious that $A$ must also be a subset of the Borel $\sigma$-algebra.