The length of rational and irrational number's sets.

I knew the length of rational number's set is zero.

I understood from the following fact: the rational number's set is very fine. However it is not continuous since an infinitely many number of irrational numbers can exist between any two rational numbers. Therefore, rational number's set can be seen as discrete points, that is, length is zero.

However, I do not guess the length of irrational number's set. Can someone let me know the result and process of thinking?

• Your "Therefore..." is wrong. It's also true that between any two irrational numbers there are infinitely many rationals. This has very little to do with the fact that the Lebesgue measure of the rationals is $0$ while the Lebesgue measure of the irrationals is $\infty$. – Robert Israel Apr 11 '16 at 1:17
• Yes, after seeing answers, I noticed my notion is wrong! :($\\$I need to think again newly. – Danny_Kim Apr 11 '16 at 1:19
• Your reason for it not being continuous is not sufficient -- after all, an infinitely many irrational numbers can exist between any two nonequal irrational numbers as well. – Gregory Magarshak Apr 11 '16 at 2:50
• I'd refrain from using the word "length", but rather "measure". Clearly the lebesgue measure is very natural and present everywhere, but it's not the "only" measure – Ant Apr 11 '16 at 9:40

The irrationals have infinite length. Since the length of $\mathbb R$ is infinite, and the length of $\mathbb Q$ is zero, we must have that the length of $\mathbb R \setminus \mathbb Q$ is infinite.

• Ah, thank you. I got that $|\mathbb{Q}|=0, |\mathbb{R\setminus Q|}=\infty, |\mathbb{R}|=\infty$. – Danny_Kim Apr 11 '16 at 1:15
• similarly, the length of reals in $[a, b]$ is $b-a=||[a,b]||$ – MichaelChirico Apr 11 '16 at 4:36

You shouldn't be using the term "length"; for sets that are not unions of intervals (with more than one point), there is no notion of length that is easy to define. There is a notion "Lebesgue measure" that ultimately does give a notion similar to "length", but it is rather more advanced than the set-theoretic notions of being countable (like $\Bbb Q$) or not (like $\Bbb R$).

The situation of ordering of rational and irrational numbers is in fact rather counter-intuitive, as it marks a sharp deviation from what happens in the finite case. Imagine you want to make a fence across a certain opening. You can put two fence posts at the ends, and connect them with a wire; there is one more fence post then there are pieces of wire. Now you can add some fence posts in between; each time you add a fence post, you also divide one piece of wire into two, so it is easy to see (by induction) that at any time you will have one more fence post then piece of wire between them. (Some people tend to forget this small difference of one, and an error in computer programming that is due to this oversight even has been named after this.)

Anyway, no matter how long you go on subdividing, there are always nearly equally many fence posts as there are intervals between then, with the fence posts staying just one ahead of the intervals. However when in one fell swoop you put in (countably) infinitely many fence posts (like one at every rational number) then a surprising thing happens: then number of intervals has become much greater than the number of fence posts. Also intervals are in general no longer delimited by given a pair of fence posts; rather they are determined by which fence posts are to their left, and which are to their right (technically this is called a Dedekind cut). In fact, supposing the set of locations of fence posts is dense, not only has the length of the intervals become zero, each interval actually contains just a single point. And furthermore supposing the fence posts are precisely at the rational numbers, the intervals correspond precisely to the irrational numbers.

It doesn't make any sense to say that $\mathbb{Q}$ is continuous. Continuity is a property of functions, not sets. I believe what you are referring to is the function $f : \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 1$ when $x \in \mathbb{Q}$ and $f(x) = 0$ when $x \not\in \mathbb{Q}$. This function is not continuous, and indeed the key reason why is that any interval in $\mathbb{R}$ contains both rational and irrational numbers.

• Wow, you got it... I am seeing the function $f(x)=1$ when $x\in\mathbb{Q}$. Thank you, and you deleted your previous sentence, which is a countable set is measure zero. Is it wrong? – Danny_Kim Apr 11 '16 at 1:14
• No it's not wrong I just realized your question was about the measure of the irrationals, not $\mathbb{Q}$. – Ethan Alwaise Apr 11 '16 at 1:21
• – Carsten S Apr 11 '16 at 8:40
• Oh... my mistake haha – Ethan Alwaise Apr 11 '16 at 15:32

To understand this better, you can use the theorem that $\mathbb{Q}$ is countable and see this building. You can also read the theorems of Cantor.

• Thank you, can you recommend a name of textbook? Is it a part of set theory? or real analysis? Actually, my real analysis book does not contain about this notion. – Danny_Kim Apr 11 '16 at 1:28
• Functional Analysis - Walter Rudin – Hendrik Matamoros Apr 11 '16 at 1:30
• Thank you :) I will see that – Danny_Kim Apr 11 '16 at 1:30
• You're welcome, you can also check this: Thomas Jech, Set Theory. – Hendrik Matamoros Apr 11 '16 at 1:33
• For a light intro to infinite sets see Stories About Sets by Vilenkin. Any introductory textbook on set theory will include the countability of Q. – DanielWainfleet Apr 11 '16 at 5:04