In Terence Tao's book Analysis I, he says

there are still an infinite number of “gaps” or “holes” between the rationals, although this denseness property does ensure that these holes are in some sense infinitely small.

I think a gap between the rationals should have zero length.

Supposing $A_{1}=\{a\in {\mathbb {Q}}:a^{2}<2{\text{ or }}a<0\}, {\displaystyle A_{2}=\{a\in \mathbb {Q} :a^{2}>2{\text{ and }}a>0\}} $, I define the "length" of the gap between $A_{1}$ and $A_{2}$ to be the greatest lower bound of $A =\left\{ a \middle| a = a_{2} - a_{1}, {\ a}_{1} \in A_{1}{,a}_{2} \in A_{2} \right\}$, so how to prove the greatest lower bound is $0$ ? especially using the density property of rational numbers to prove it?

Maybe I have a lack of understanding in the density property of rational numbers , so I am unable to give a proof to my question here.


It is possible to prove the greatest lower bound of $A =\left\{ a \middle| a = a_{2} - a_{1}, {\ a}_{1} \in A_{1}{,a}_{2} \in A_{2} \right\}$ is 0 in the rational number system, since if any other positive number $b$ is the greatest lower bound, we could always find a positive number $c=a_{2} - a_{1}<b$ . I think what is important here is that the question won't make much sense unless we prove the greatest lower bound of $A$ is 0 in a continuous number system ($\mathbb R$). We may use Archimedean property of $\mathbb R$ to prove the conclusion .


It's a good question! Denseness alone is not enough. Consider the set $P=\mathbb Q\setminus[1,2]$. This $P$ is also dense as an ordered set. (But it is not dense as a subset of $\mathbb R$.)

Now let $B_{1}=\{a\in P:a^{2}<2{\text{ or }}a<0\}, {\displaystyle B_{2}=\{a\in P :a^{2}>2{\text{ and }}a>0\}} $, and $B =\left\{ a \middle| a = a_{2} - a_{1}, {\ a}_{1} \in B_{1}{,a}_{2} \in B_{2} \right\}$. This is a recapitulation of your $A$; the only difference is that $\mathbb Q$ is replaced by $P$. You can see that $\inf B=1$!

So you cannot prove that $\inf A=0$ merely from the knowledge that $\mathbb Q$ is dense. You need something else: for example, that $\mathbb Q$ is closed under averages, which is after all how Proposition 4.4.3 is proved.

  • $\begingroup$ I only concern gaps in the rational number system(not other number system) and the length of each gap in it, rational number system is dense everywhere, I think Proposition 4.4.3 is a presentation of this property. $\endgroup$ – iMath Jul 28 '17 at 8:23
  • $\begingroup$ It seems I have found a way to prove using the density property of rational numbers. We may start with assuming the gap between the rationals is not zero-length, then it should have positive length, so that the gap itself is an interval, no matter closed or open, which has positive length and is free from rational,... $\endgroup$ – iMath Jul 28 '17 at 9:36
  • $\begingroup$ ...such interval would lead contradiction according to highlighted paragraph from here. $\endgroup$ – iMath Jul 28 '17 at 9:36
  • $\begingroup$ @iMath I would bet that your proposed proof does one of two things. Either it calls on some additional property of the rational numbers besides denseness, or it fails at some subtle juncture, because it attempts to prove too much. (In particular, it attempts to prove that $B=0$ as well, which is false.) Maybe you'll need to assume something about the endpoints on either side of the gap, which will generally be false. In any case, feel free to edit your question with the claimed proof, or enter it as an answer, and I'll take a look! $\endgroup$ – Chris Culter Jul 28 '17 at 17:15
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    $\begingroup$ The difference between $A$ and $B$ is coming from the difference between $\mathbb{Q} $and $P $. And it is this property: given any two distinct elements $a, b $ of $\mathbb{Q} $ and a positive rational $d$ we can find between them a finite number of elements $x_{1},x_{2},\dots, x_{n} $ in $\mathbb{Q} $ such $0<x_{i+1}-x_{i}<d$ for all $i=1,\dots,n-1$. Set $P$ does not possess this property. And this is certainly more than denseness. $\endgroup$ – Paramanand Singh Nov 13 '17 at 20:38

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