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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?

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

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    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Apr 17 at 22:53
  • $\begingroup$ @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? $\endgroup$ – roy212 Apr 17 at 23:16
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    $\begingroup$ 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 $b$. $\endgroup$ – Kavi Rama Murthy Apr 17 at 23:27
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    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Apr 17 at 23:38
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    $\begingroup$ 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. $\endgroup$ – Andrés E. Caicedo Apr 18 at 0:24
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Following Nate Eldredge, I'll write "$(a,b)_{\mathbb{Q}}$" for $(a,b)\cap\mathbb{Q}$ (with $a<b$ 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 $X$?

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  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge Apr 18 at 1:45
  • $\begingroup$ @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 $\endgroup$ – roy212 Apr 18 at 1:46
  • $\begingroup$ @NateEldredge Yes, of course, I didn't mean to imply that they were the same; fixing for clarity. $\endgroup$ – Noah Schweber Apr 18 at 1:47
  • $\begingroup$ @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). $\endgroup$ – Noah Schweber Apr 18 at 1:49
  • $\begingroup$ By "your $\sigma$-algebra, do you mean the one made up of rational numbers only? $\endgroup$ – roy212 Apr 18 at 1:54

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