There are a lot of theoretical results in Real Analysis, Calculus, Topology, etc. that depend on or use this result below. This has been a question that has bothered me for a long time and seems to me very similar to Euclid's Fifth Postulate in Geometry. Negating the fifth postulate has led to different types of geometry which is why I am curious about the theoretical nature of how real numbers behave when negating the least upper bound property (i.e. given below).

The Completeness Axiom states every non-empty set $S$ of real numbers that has an upper bound also has a least upper bound. Sometimes this is a theorem that relies on other axioms which I find also terribly non-intuitive. Like why does this property have to work? It doesn't seem like a typical evident axiom like so many other theoretical objects are in math.

What I was wondering is what happens when we negate the completeness axiom? In other words, there is some non-empty set $S$ of real numbers that has an upper bound but lacks a least upper bound. I am curious as to what happens. So, my question is what happens when we negate the completeness axiom?

  • $\begingroup$ Since part of the definition of $\mathbb{R}$ itself is completeness, it sounds - if you want to think about "completeness-free topology" or similar - like you want to look for a system which proves that there are no complete linear orders (with more than one point) in the first place. Does that seem right? $\endgroup$ May 2, 2021 at 19:14
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    $\begingroup$ You can get some intuition by looking at the set of rationals. $\endgroup$ May 2, 2021 at 19:14
  • $\begingroup$ Yes, that sounds right to me. What happens to systems like this? I know the definition of $\mathbb{R}$ depends on completeness but what if that was just gone. What happens to the behavior of real numbers? Certain other properties of real numbers could still hold. I am curious about that. $\endgroup$
    – W. G.
    May 2, 2021 at 19:18
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    $\begingroup$ Like would the fundamental theorem of algebra hold? What other results would and wouldn't? $\endgroup$
    – W. G.
    May 2, 2021 at 19:28
  • $\begingroup$ FTA would not hold of course (assuming you mean in $\mathbb{Q}+i\mathbb{Q}$). $\endgroup$ May 2, 2021 at 19:31

2 Answers 2


By the real field axioms, you have the $0$, the $1$, $1>0$, and because you need all the inverse elements for addition and multiplication (except for the 0), you have at least the $(\mathbb{Q},+,\cdot)$ field.

A possible characterization of the sets you are interested may be the sum of rational numbers with all rational, finite, linear combinations of elements from an arbitrary subset $X$ of $\mathbb{R}\setminus\mathbb{Q}$, and all the multiplicative and additive inverses of any number of combinations.

It is not so simple because if you choose, say, numbers of the form $r+r^\prime\times\pi$, you need also $(r+r^\prime\times\pi)^{-1}$ and so on.

One more possibly interesting point is the fact that there are Hamel basis for $\mathbb{R}$. So, if you want to characterize all subsets of $\mathbb{R}$ which fulfill all ordered field axiom but not the completeness, you would need to require that the set $X$ cannot include any Hamel basis, etc.

  • $\begingroup$ $\mathbb Q$ is an ordered field with no completeness, and it has a Hamel basis even without the axiom of choice. And of course any linear space has a Hamel basis (under choice). So, I'm not sure what you are trying to say in the last paragraph. $\endgroup$ May 2, 2021 at 20:10
  • $\begingroup$ $\mathbb{R}$ admits a Hamel basis $H$ over linear rational linear combinations. So if $H\subset X$, then the generated field is $\mathbb{R}$, which does fulfill the completeness axiom. You are interested in ordered fields who do not. $\endgroup$ May 2, 2021 at 20:58
  • $\begingroup$ $\mathbb Q$ also admits such a Hamel basis, so, still, I don't quite understand what you are trying to say. $\endgroup$ May 2, 2021 at 21:10
  • $\begingroup$ If you pick a set $X\subset\mathbb{R}$, and take the set $F(X)$ as all finite combinations of the form $r_0+\sum r_i x_i$ where all $r_i$ are rationals and $x_i\in X$, then the smallest ordered field including $F(X)$ does not fulfill completeness only if it is strictly contained in $\mathbb{R}$. If $H\subset X$, for any Hamel basis of $\mathbb{R}$ over the rationals, then $F(X)=\mathbb{R}$ and the smallest ordered field which includes $X$ is $\mathbb{R}$, which does fulfill the axiom of completeness, and therefore you are not interested in. $\endgroup$ May 2, 2021 at 21:23
  • $\begingroup$ OK, I understand. now that you were talking about characterising the non-completes subfields of $\mathbb R$. $\endgroup$ May 3, 2021 at 7:21

The completeness axiom in the axiomatization of the reals is not so much like Euclid's Fifth. The completeness axiom is, well, a property that assures completeness. Euclid's Fifth has nothing to do with completeness. Further, there are many non-isomorphic models for geometry with Euclid's Fifth (e.g., the Euclidean spaces). However, all models of the axioms of the reals (as a complete ordered field) are isomorphic. Negating Euclid's Fifth admits new types of geometry that make perfect sense geometrically. Negating the completeness axiom of the reals results in many models (e.g., the rationals) that are quite terrible for the purposes of analysis. In geometry there are many consequences of Euclid's Fifth as well equivalent formulations in terms of equally reasonable geometric notions (e.g., triangles having angles summing up to $\pi$ radians). In analysis the completeness axiom has a few equivalents, but they are rather set-theoretic.

So, to see some of the bad things that happen if you neglect the real's completeness can be seen from trying to do analysis in $\mathbb Q$. Say bye-bye to $\sqrt 2$ and many other important numbers like $\mathrm e,\pi $, etc. Of course, a trace of these numbers is still present, namely instead of $\sqrt 2$ you have sequences of rationals that approach $\sqrt 2$. Of course, any such sequence of rationals that approximate $\sqrt 2$ is as good as any other, so you'd really like to treat all these sequences as equivalent. And of course you'd like to repeat that process for any 'number' that is missing in your system but can be approximated from within the system. Oh, but some sequences don't approximate anything at all. How can we detect those that do? Of course, they all satisfy Cauchy's condition (spelled out only using rationals). So, let's take all the Cauchy sequences, mod out by the suitable equivalence relation, and, voilà, we've just constructed the reals.

In other words, from the perspective of analysis, you always want to work with a complete space simply since it's convenient. The only thing that will change if the space is not complete is that you may run against some entities that you can perfectly well approximate within your system, but that do not have a 'name' within the system. There is a hole in the space. The space is not complete. Completing it merely introduces names for all the holes, so that whatever you can approximate within your system is necessarily in the system. That's all the completion does. It's not really a property of the same nature as Euclid's Fifth is. Euclid's Fifth is saying something about how the geometry behaves. Completeness simply states that the model is rich enough to contain everything it approximates. If that was not the case, you can always apply a generic completion process. There is no Fifthication process in geometry.


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