I was looking at various proofs of the completeness of real numbers, following the intuition that there are no "gaps" in the real number line. (Yes, I do realize that that's not an especially rigorous way to look at this, but bear with me). In particular, you can never have a situation like this:

enter image description here

My argument for why you can never have a situation like this is that, using the synthetic approach to constructing the real numbers, the real numbers are axiomatically defined as an ordered field. The fact that the real numbers are complete can be proven using the closure of addition and multiplication in a field. In particular, let $a, b \in \mathbb{R}$ with $a < b$ and $$c = (a + b) * \frac{1}{2}$$

Then $c \in \mathbb{R}$ (because $a, b, \frac{1}{2} \in \mathbb{R}$ and the real numbers are closed over addition and multiplication). Also, $a < c < b$.

Note that I specifically did not define $c$ as $\frac{a + b}{2}$ to avoid having to prove anything about division. We already know that $\frac{1}{2} \in \mathbb{Q}$ because it's the ratio of two integers, so we can assume that $\frac{1}{2} \in \mathbb{R}$ without further explanation.

A case like the following ends up being more annoying to prove with my argument because there's not a "greatest" real number that's less than 2 or a "smallest" real number that's greater than 4:

enter image description here

but that's a different topic.

I think that this would prove that something like I have in the first diagram can't possibly exist because the above argument proves that there must be some real number in between 2 and 4 (i.e. the gap isn't really a gap).

My main question, then: this seems vaguely circular to me. Does this actually work, or am I subtly assuming what I'm trying to prove? Or is it really this easy to prove that the real number line can't actually look like the first diagram?

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    $\begingroup$ The rationals are an ordered field (closed under addition and multiplication etc) but are not complete. One needs say the LUB axiom to get completeness. $\endgroup$ – coffeemath Feb 6 '18 at 5:59
  • $\begingroup$ @coffeemath You make a good point. Does my diagram at least work for the restricted case of my first diagram, or is the proof just incorrect? Is there some way I can restrict this further to "ban" something like the first diagram from happening? $\endgroup$ – EJoshuaS Feb 6 '18 at 6:02
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    $\begingroup$ All the gaps in your pictures have a positive width. If you carefully formalize everything, the kind of argument you're giving should prove that no gap of this sort can exist. But it won't account for the kind of $0$-width gap that, e.g., $\sqrt{2}$ forms in the rationals. This is where more subtle arguments based on things like the least upper bound principle are required. $\endgroup$ – Micah Feb 6 '18 at 6:04
  • $\begingroup$ EJoshuaS-- Yes your argument rules out things like the first diagram. It also works for the second if you take instead any two rationals whose average falls in the gap. See also the comment of @Micah ... $\endgroup$ – coffeemath Feb 6 '18 at 6:08
  • $\begingroup$ @Micah So, my proof does, in fact, prove that there are no positive-width "gaps" for either the real or the rational numbers, but not for zero-width gaps? Is my proof non-circular for that restricted case? That actually helps a lot. $\endgroup$ – EJoshuaS Feb 6 '18 at 6:08

The idea which you are trying to formulate is called denseness/density. Using the closure properties of arithmetic operations in an ordered field one can show the field possesses density property. That is given two elements there is another lying between them. This is key to all the definitions (based on $\epsilon, \delta$) in calculus/analysis.

Completeness on the other hand is more elusive and not an algebraic concept. It rather enhances the order relation of the field in an interesting way and ensures that those $\epsilon, \delta$ definitions of calculus don't work in vacuum. Dedekind first comprehended the idea of completeness by thinking deeply about the meaning of an unbroken/connected straight line and reasoned that if the line is cut into two parts, there must be a point which makes this cut and it will belong to one and only one of the two segments produced after the cut.

Thus if the elements of an ordered field are partitioned into two classes such that each member of one class is less than every member of another class then there should be a number which separates these two classes ie there should be an element $\alpha$ in the field such that all elements less than $\alpha $ lie in one class and those greater than $\alpha$ lie in another class. This desirable property was called continuity by Dedekind and later it was replaced by the word completeness. Not every ordered field possesss it, but more importantly Dedekind used this idea to extend an ordered field into a complete ordered field.

Your assumption is that gaps in ordered field are necessarily of a positive size. This is not correct. By the property of density we just can't have gaps of positive size anyway.


You showed that nothing like what happens in the first diagram occurs, that's correct. Between any two real numbers, there's another real number. But the problem is this is true for any ordered field, including the rationals. And the rationals satisfy addition and multiplication, and they have a $\frac{1}{2}$ as well.

And yet the rationals have plenty of holes. E.g. $\sqrt{17}$ is irrational and yet there are plenty of rationals nearby $\sqrt{17}$. The thing that differentiates the rationals and the reals in this respect is that the reals have what's called the Least Upper Bound property which says: If a set $S$ is nonempty and bounded above, there is some number $a$ at least as large as every element of $S$, and no smaller number than $a$ can make this claim. That is, $S$ has a least upper bound.

This fixes the $\sqrt{17}$ problem, because if $S$ is the set of all rationals $r$ for which $r^2<17$, then $S$ is nonempty and bounded above (by $5$, for instance), so it has a least upper bound $a$. If $a^2<17$ then $\frac{5a+17}{a+5}>a$ and $\left(\frac{17+5a}{a+5}\right)^2<17$ too. Thus $\frac{17+5a}{a+5}\in S$, impossible since $a$ is an upper bound for $S$. And if $a^2>17$ then $\left(\frac{17+5a}{a+5}\right)^2>17>r^2$ for any $r$ in $S$ and $a>\frac{17+5a}{a+5}$, so $\frac{17+5a}{a+5}$ is a better bound, impossible. So $a^2=17$ and $\sqrt{17}$ is a real number.

Obviously this example was highly contrived, but it shows how to prove it in general. Find a hole you want to plug, let $S$ be the set of all numbers less than the hole, show it has an upper bound, get a least upper bound for free, show that this least upper bound fills in the hole. This works whenever you have a dense set leading up to the hole, and your proof helps to show that those are the only holes possible.

So the point is that you showed what you wanted to show correctly, but what you wanted to show is density of the reals, which isn't enough for completeness of the reals; you need a totally separate axiom from the usual field axioms to show that.


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