I apologise for this question to be quite long... it looks wordy since I'm trying to express the process of my thinking.

I'm recently using the appendix at the end of chapter 1 in Rudin's Principle of Mathematical Analysis to study the construction of $\mathbb{R}$ from $\mathbb{Q}$.

I fall into a somewhat circular argument $\color{red}{(\ast)}$ when I try to visualize cuts as finite lengths on an infinite straight line (by finite length I mean the segment from a reference point "0" to some point on the line).

I'm trying to visualize cuts as lengths on the line because I think "length" is the intuitive concept that is central to the story: people in the old days first found out that there exist some certain lengths that could not be represented by the elements of $\mathbb{Q}$ (namely the diagonals of squares), then this became the reason for cuts to come.

So, the first question I asked myself is:

How could one be sure that, after defining the cuts, every finite length on the line is indeed included in $\mathbb{R}$?

My answer to this is: I can now pick any one point on the line, denote it as $x$, and I can associate a set $r$ which consists of all the rational points to the left of the point $x$ so that $r$ is a cut. And I'm now representing $x$ by $r \in \mathbb{R}$.

Then, I asked myself the second question:

How could I know that such a cut $r$ is actually unique to my length $x$? That is, how could I ensure that my cut $r$ will not represent $x$ anymore even if I move my point $x$ to the right on the line for just a tiny tiny bit?

For this question, let's denote the original $x$ as $x_0$ and the $x$ after moving as $x_1$. My idea goes like: if we can always find a rational number between $x_0$ and $x_1$ (no matter how close they are), then $x_1$ must be represented by a different cut (let's denote it as $s$) because by the definition of cut, $r$ is now a proper subset of $s$.

[Please note also that I did not justify my premise above using the property that $\mathbb{Q}$ is dense in $\mathbb{R}$, because I still don't know if $x_0$ and $x_1$ are two distinct reals at the moment]

So, at this point, I made two important assumptions:

  1. Any two points on the line must either be the same point or be different. So, as long as I move the point, it becomes a different one to the original.

  2. The settings for the lengths on the line make them isomorphic to the cuts. i.e. assume that some proper, possibly customary, addition, multiplication and order which operates on the lengths are defined, and are preserved by some bijection between the set of all lengths and the the set of all cuts.

Using those two assumptions, we now have the two different points on the line representing by two different cuts.

For assumption 1, I think it is rather like an axiom.

And Finally this leads to the humble question I wrote in the title:

How do we prove assumption 2? That is, how can we construct an isomorphism between the set of all real numbers defined by (Dedekind's) cuts and the set of all finite lengths (including zero length) on an infinite straight line?

For now, there are two angles that I tried to solve this question in general:

  • As it is a fact that any two ordered fields with the least-upper-bound property are isomorphic, I was trying to prove that the set of all lengths on the line forms an ordered field with the least-upper-bound property under some good definitions. But I'm stucking somewhere in the middle.

  • Assume that there exist two points on the line $x_0$ and $x_1$ as described above where $x_1$ is to the right of $x_0$, and there exists some cut $r$ that catches them both. Then, the length between $x_0$ and $x_1$ must be finite (due to assumption 1), and I don't see a reason why any rational cuts wouldn't be surrounded by such "blank zone" or why those blank zones would have various lengths (I don't think my naive understanding to the concept of length would let me say more on these). By mathematical induction, there are infinitely many rationals between any two different rationals, and thus we would have some finite constant multiplied by infinity equals some finite quantity, which seems logically incorrect. But to me, such "proof by contradiction" is far not rigorous.

By the way, I have a feeling that constructing $\mathbb{R}$ from $\mathbb{Q}$ by means of Cauchy sequences might be a better approach to this question, I haven't read such construction yet because I believe that cuts should be equally good.

$\color{red}{(\ast)}$ : I think what is (probably) being circular here is as follows (and this is what really confuses me):

From the definition of cut, it is clear that any two distinct cuts cannot represent the same length on the line. But now I want to prove that any two distinct lengths on the line cannot be represented by the same cut. Symbolically, I want to prove that if the two lengths on the line are denoted as $x_0$ and $x_1$ where $x_0 \neq x_1$, and their associated cuts are denoted as $r_0$ and $r_1$ respectively, then $r_0 \neq r_1$. By the definition of the order defined on cuts, to prove $r_0 \neq r_1$ I need to show that one is the proper subset of another. This means that I need to find a rational in between. However, to conclude that there exists a rational between two cuts it requires the two cuts not to be equal in the first place.

Could anyone please tell me how to get rid of this circularity?

There may be ambiguities in multiple places in this question as I am still being confused about it right now... so anyone who notices why I'm confused is more than welcomed to write an answer even if the answer is unrelated to my arguments above :)

Thank you for your patience :)

  • 4
    $\begingroup$ How exactly are you defining the line, if not as the set of Dedekind cuts ordered appropriately? $\endgroup$ Dec 7, 2020 at 2:49
  • $\begingroup$ Suggested topics for reading: Linearly-ordered topological spaces (a.k.a ordered spaces), ordered groups (a.k.a. fully ordered groups), ordered fields.... And well-orders (a.k.a well-orderings). $\endgroup$ Dec 7, 2020 at 15:36
  • $\begingroup$ @DanielWainfleet Thank you for pointing these out :) may I confirm that I can find those things in abstract algebra books and in general topology books, right? $\endgroup$
    – J-A-S
    Dec 7, 2020 at 16:14
  • $\begingroup$ Yes, certainly..................... $\endgroup$ Dec 8, 2020 at 18:18

1 Answer 1


We can show that $(i)$ any two nontrivial connected ordered groups are isomorphic and $(ii)$ the set of Dedekind cuts, with the usual ordering and addition, forms a nontrivial connected ordered group. So we are forced to conclude that the Dedekind cut model faithfully captures the "Platonic line" as long as we grant that the Platonic line is connected and that we can add and subtract lengths appropriately.

To me, both connectedness and nontrivial groupiness are totally fundamental to my conception of the line, so this solves the problem completely.

  • $\begingroup$ Hi, thank you very much for your clear answer! Do you have any recommendation(s) on resources that teach (i) in details? I'm not that familiar with group theory (or is that topology?) right now $\endgroup$
    – J-A-S
    Dec 7, 2020 at 3:01
  • $\begingroup$ @J-A-S I'll add some details. For connectedness we don't need to go all the way to topology - we just need its order-theoretic sense. A linear order $L$ is connected iff whenever $A$ and $B$ are nonempty disjoint sets with $A\cup B=L$ and every element of $A$ is $<$ every element of $B$, then either $A$ has a maximum or $B$ has a minimum. Basically, connectedness says that the line doesn't 'break apart' anywhere. (The topological notion of connectedness is more technical.) $\endgroup$ Dec 7, 2020 at 3:03
  • $\begingroup$ Thanks! May I also ask, given two structures that are isomorphic, could we have multiple isomorphisms between them or must the isomorphism be unique? And what if the given two are ordered structures? $\endgroup$
    – J-A-S
    Dec 7, 2020 at 13:03
  • $\begingroup$ @J-A-S Sorry, I just saw this! In general there may be many isomorphisms between different structures. Indeed, we can phrase this in terms of a single structure: we're really looking at when a structure has non-trivial automorphisms (the trivial automorphism being the identity). Lots of naturally-occurring structures do have nontrivial automorphisms, while some others don't (structures which don't are called rigid). E.g. the field of complex numbers has one "simple" nontrivial automorphism (conjugation) and many "complicated" nontrivial automorphisms if we assume the axiom of choice. $\endgroup$ Jan 6, 2021 at 22:51
  • $\begingroup$ It should be surprising, in my opinion anyways, that the field of real numbers is rigid. I'd go further and say that that's a point in its favor in terms of "foundational status," although there are plenty of people who disagree with me about that. $\endgroup$ Jan 6, 2021 at 22:52

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