I have just recently started to read a book about mathematics and have stumbled upon an axiomatic definition of the real numbers. The real numbers are defined as a Dedekind-complete ordered field. While I understand the components of this definition, i.e the specifications of these axioms, I am absolutely lost as to the philosophical process in place. We define the real numbers by a set of properties that we assume to be true. If someone was to try and construct the set of real numbers using this definition, he would have to turn to the only set that has these properties. Which set is that? It is the set that we assume to have these properties. Well, which set is that? It is the set that we assume to have these properties and so on... I feel like without a concrete reference of some sort all we're left with is cyclical logic. Can anybody help me shed light on the philosophical process of this definition?

  • $\begingroup$ Try a philosopher perhaps? $\endgroup$ Feb 18 '18 at 12:04
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    $\begingroup$ All I'm interested in is making very basic sense of the process in this definition. Surely that would not require a philosopher. $\endgroup$ Feb 18 '18 at 12:15
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    $\begingroup$ Well your concern is genuine. You should better get familiar with the construction of real numbers as given by Dedekind to get the real feeling of the way they work. $\endgroup$
    – Paramanand Singh
    Feb 18 '18 at 17:21

It seems like a better question to me than some people seem to think. The question of which set we're talking about doesn't matter, but the question of how we know such a structure exists certainly matters. Lemme paraphrase what it seems to me the issue is, in language that people here will understand:

Q: I've read that the reals are a complete ordered field. Ok, one can easily show that any two (Dedekind-)complete ordered fields are isomorphic, so this characterizes the theory of the reals, fine. But how do we know that a complete ordered field exists? After all, if there's no such thing as a complete ordered field then the theory of complete ordered fields seems a little pointless.

A perfectly reasonable mathematical question; no need to snicker about the need for a philosopher. Luckily:

A: There are various methods one can use to show that a complete ordered field exists. One of the best known is via "Dedekind cuts", which you can read about in various places online.


Axiomatic definition like this doesn't tell you “this particular thing is the set of reals”, but rather “we call the reals anything that satisfies the conditions”. It just specifies the interface, not giving any particular implementation.

The point here is that not only some implementation exists, but that it is unique up to isomorphism. This is not part of the definition, it is a theorem that has to be proved. But it justifies the name “the reals” (i.e. using the definite article).

Actually, it doesn't matter which particular set realizes the reals, the structure is important. You can for example look at a simpler case of natural numbers and integers – do you know their actual set theoretic representation? (There are some natural choices, maybe one of them is considered canonical.) Does it really matter? I mean, the structure itself is important, and the fact that it can be realized as a set. But the particular realization is not so important.

Axiomatic definitions are more often used in the situation without uniqueness – vector spaces, metric spaces, topological spaces, Banach spaces – they are axiomatically defined structures/interfaces that abstract over particular instances, so you can formulate some theorems that hold in general.

  • $\begingroup$ Thank you for your comment! I understand now that the structure of this set is more important than any particular realization. And I understand there is esentially only one set that holds these properties. But how do we know which set holds these properties? How can I know that the set that contains every number represented by an infinite string of decimals actually holds these particular properties? $\endgroup$ Feb 18 '18 at 12:20
  • $\begingroup$ “How can I know?” – as always in math, you have to prove that. A simpler thing: can you actually prove that the set of all integers with addition satisfies the axioms of commutative groups? On one hand, rather obvious, on the other hand, you have to consider some actual construction of integers. $\endgroup$
    – user87690
    Feb 18 '18 at 13:16
  • $\begingroup$ @PeterHeinig: If you want to represent all reals as decimal strings, you have to also encode the sign, and then $0$ has also two representatives. $\endgroup$
    – user87690
    Feb 18 '18 at 13:17
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    $\begingroup$ @PeterHeinig what are the two decimal representations of $1/3$? I only know one (0.3333333...) $\endgroup$ Feb 18 '18 at 17:54

Your inference "If ... . he would have to turn to the only set that has these properties" is not a mathematical inference and rests on non-mathematical presuppositions. In particular, you simply presuppose that the axiom-system characterizes the real numbers strictly uniquely (up to equality in some universe of sets); this is not the usual point of view nowadays. At most 'uniqueness up to isomorphism of ordered fields' is usually considered, and for that you even forgot to mention a condition: the real numbers are the only complete ordered field, up to isomorphism of ordered fields.

A similar comment is in order in response to your question "Which set is that?" This question presupposes a view which is considered outdated: that any property (like e.g. 'satisfying all the axioms you are currently asking about') would define a set. G. Frege, among others, thought that something similar could be adopted as a principle of thought, but this turned out to be an untenably naive point of view. As user87690 has already hinted at, in usual set-theory there are infinitely-many sets which are isomorphic as fields to the real numbers.

The intuition you evince in " I feel like without a concrete reference of some sort all we're left with is cyclical logic"(usage) is quite relevant: the Axiom schema of specification arguably arose out of a similar intuition.

Re your "Can anybody help me shed light on the philosophical process of this definition?": the 'philosophical process' is that usually only the isomorphism type is considered essential for mathematical and scientific applications of $\mathbb{R}$, while the set-theoretic realization is considered a, as it were, set-and-forget(pun) issue of (model-)theoretic interest only. Relevant keywords with which you can learn more are 'axiomatic method' and 'model theory' (where a relevant topic to learn about is categoricity).


(pun) Pun intended. The model is usually taken to be a set which for most intents and purposes you then forget.

(usage) By the way, 'circular logic', or 'circular reasoning' is a much more usual collocation in English.

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    $\begingroup$ Thank you, you have referenced many things in your response which I have yet to familiarize myself with. I have a lot of reading to do. Thanks again! $\endgroup$ Feb 18 '18 at 12:54
  • $\begingroup$ Actually up too isomorphism there is only one complete ordered field. It's not hard to show that a complete ordered field must be Archimedean. $\endgroup$ Feb 18 '18 at 16:34
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    $\begingroup$ And saying "only the isomorphim type is considered essential, not the set-theoretic realization" seemms to gloss over what seems to me to be the point to the question: How do we know that there is a compete ordered field? $\endgroup$ Feb 18 '18 at 16:36
  • $\begingroup$ @DavidC.Ullrich There is a small issue in some books/contexts using "complete" for "Cauchy complete" (which doesn't imply Archimedean) and others using it for "Dedekind complete" (which does). But I agree with you wholeheartedly that the primary issue is likely "why should something satisfying those axioms exist?" $\endgroup$
    – Mark S.
    Feb 18 '18 at 16:47
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    $\begingroup$ Simple proof that any Dedekind-complete ordered field is Archimedean: Suppose not: $\epsilon>0$ but $\epsilon/le1/n$ for every $n\in\Bbb N$. Then $n\le1/\epsilon$ for every $n$, so $\Bbb N$ is bounded above. Hence $\Bbb N$ has a least upper bound. But it's clear that $\Bbb N$ does not have a least upper bound, because if $b$ is an upper bound then $b-1$ is a smaller upper bound. $\endgroup$ Feb 18 '18 at 16:55

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