# What kind of numbers are inside a generating open interval of the Borel $\sigma$-algebra? [closed]

If it is enough to have all open intervals (a,b) with end points $$a$$ and $$b$$ belonging to the rational numbers, a < b, in order to generate a Borel $$\sigma$$-algebra on $$\mathbb{R}$$. Asked here: About the open intervals generating a Borel $\sigma$-algebra on $\mathbb{R}$

What kind of numbers do you need to have between $$a$$ and $$b$$? Only rational numbers or real numbers? And why?

## closed as unclear what you're asking by Andrés E. Caicedo, Lord Shark the Unknown, Shailesh, Cesareo, JaviApr 18 at 10:22

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• Not sure I understand the question. Let's write $(a,b)_{\mathbb{Q}} = \{ t \in \mathbb{Q} : a < t < b\}$. Are you asking whether the $\sigma$-algebra generated by $\{(a,b)_{\mathbb{Q}} : a,b \in \mathbb{Q}\}$ is the Borel $\sigma$-algebra of $\mathbb{R}$? The answer is no. – Nate Eldredge Apr 17 at 22:53
• @NateEldredge My doubt was whether $t \in \mathbb{Q} : a < t < b$ belongued to the rational numbers or the real numbers. I thought that open intervals with rational endpoints had to contained only rational numbers. So for what you said it has to contain real numbers in order to generate the σ-algebra on R? – roy212 Apr 17 at 23:16
• By open intervals $(a,b)$ with rational end points what is meant is the set of all real numbers lying between the rational numbers $a$ and $b$. – Kavi Rama Murthy Apr 17 at 23:27
• Yes. Since we are considering Borel sigma algebra of $\mathbb R$ it is understood that $(a,b)$ is interpreted ads real numbers between $a$ and $b$. If you want to consider only rational numbers you have to specify it, as done by Nate Eldredge. – Kavi Rama Murthy Apr 17 at 23:38
• Only in the same sense that any time you use the number 1 you are doing number theory. The question you are asking does not qualify. – Andrés E. Caicedo Apr 18 at 0:24

Following Nate Eldredge, I'll write "$$(a,b)_{\mathbb{Q}}$$" for $$(a,b)\cap\mathbb{Q}$$ (with $$a reals; note that we don't need $$a,b\in\mathbb{Q}$$ themselves for this to make sense). And I'll reserve "$$(a,b)$$" for the full interval of real numbers, as usual.

Now the key point is that each $$(a,b)_\mathbb{Q}$$ is countable. This implies that every element $$X$$ of the $$\sigma$$-algebra generated by such intervals is either countable or co-countable (= has countable complement). Specifically, let $$\mathfrak{S}$$ be the set of all sets of reals which are either countable or co-countable; it's easy$$^1$$ to check that $$\mathfrak{S}$$ is a $$\sigma$$-algebra, and it clearly contains each $$(a,b)_{\mathbb{Q}}$$.

• Note $$1$$: I'm not saying,incidentally, that $$\mathfrak{S}$$ is the $$\sigma$$-algebra so generated, merely that it contains it. Indeed, they're not the same, and it's a good exercise to find something in $$\mathfrak{S}$$ that isn't in the $$\sigma$$-algebra generated by the $$(a,b)_\mathbb{Q}$$s.

• Note $$2$$: More generally, it's usually the case that the $$\sigma$$-algebra generated by a collection of "small" sets consists entirely of "small" or "co-small" sets; e.g. every element of the $$\sigma$$-algebra generated by the null sets is either null or co-null, every element of the $$\sigma$$-algebra generated by the meager sets is either meager or co-meager, etc.

But there are plenty of Borel sets which are neither countable nor co-countable - for example, the interval $$(0,1)$$.

$$^1$$OK, let's sketch that here.

Complements are immediate: if $$A$$ is countable (respectively co-countable) then $$A^c$$ is co-countable (resp. countable).

Countable unions are just complements of countable intersections of complements, so the previous line shows that if $$\mathfrak{S}$$ is closed under countable intersections it's also closed under countable unions.

So suppose $$A_i\in\mathfrak{S}$$ for $$i\in\mathbb{N}$$. We want to show that $$X=\bigcap_{i\in\mathbb{N}}A_i$$ is either countable or co-countable. There are two cases:

• If some $$A_i$$ is countable, then we can conclude that $$X$$ is countable; do you see why?

• So suppose each $$A_i$$ is co-countable. The complement of $$X$$ is the union of the complements of the $$A_i$$s, which is to say a countable union of countable sets; what does this tell you about $$X$$?

• Indeed, the $\sigma$-field generated by all $(a,b)_\mathbb{Q}$ is even smaller than your $\mathfrak{S}$: it consists precisely of the subsets of $\mathbb{Q}$ and their complements. So for example the set $\{1/3, \pi\}$, while countable, is not in it. – Nate Eldredge Apr 18 at 1:45
• @Noah Schweber Thank you very much for your detailed answer. If I undertood it correctly, as the resulting $\sigma$-algebra only have countable and co-countable sets, it can't be a Borel $\sigma$-algebra on $\mathbb{R}$.is that right? About your questions I'm thinking about it. Thanks for all the comments too – roy212 Apr 18 at 1:46
• @NateEldredge Yes, of course, I didn't mean to imply that they were the same; fixing for clarity. – Noah Schweber Apr 18 at 1:47
• @roy212 I don't know what a Borel $\sigma$-algebra is, I just know what the Borel $\sigma$-algebra is - namely, the smallest $\sigma$-algebra containing all the open intervals. And since open intervals aren't in general countable or cocountable, $\mathfrak{S}$ is indeed far from the Borel $\sigma$-algebra (and so a fortiori your $\sigma$-algebra is as well). – Noah Schweber Apr 18 at 1:49
• By "your $\sigma$-algebra, do you mean the one made up of rational numbers only? – roy212 Apr 18 at 1:54