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

$$A$$ 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 $$a so that $$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.

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.

• Thanks a lot for your help, really appreciate it. – KingDingeling Feb 18 at 14:23
• You are welcome. – drhab Feb 18 at 14:24

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 $$a so that $$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.

• Thank you for your help, very kind. I will check your points. – KingDingeling Feb 18 at 14:24