I realize this is possibly a copy of another question (The Axiom of Choice and definability) but I would appreciate an explanation with less set theoretical explanations if that is at all possible.

I recently took the opportunity to look up the axiom of choice (AC). Wikipedia has both a formal and an informal definition. Based on my shortcomings in set theory I have to rely on the informal version: “given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object from each bin”.

It seems that this can be used to define an endless series of decimals such as an arbitrary number x between 0 and 1, in term corresponding to an arbitrary positive real number through such a function as f(x)=1/x-1, i.e. it can be used to define any real number (assuming a real number is an infinite decimal series).

A beauty spot is the cases of 0.99999… and 1.00000… - discussed in a number of questions on the site. It points to several issues:

I argue, albeit in a roundabout way, that for a choice involving the AC to have any value then the number received through the choice must also allow for existing mathematical operations, i.e. 2*0.11111…=0.22222… as discussed in another question. (“You're also assuming that you can multiply 0.9999... by 10 and get 9.9999... (or, rather, that arithmetic with infinite decimals works normally), which is not at all unreasonable to assume. – Isaac Jul 20 '10 at 20:59”) i) 0.99999…must be equal to 1.00000… for the same reason. ii) The fact that 0.99999…=1.00000…can challenge the one-to-one correlation between the real numbers, as we view as being without these doublets. However it can be disregarded as

I was assured in a recent question, in that removing a countable set such as 0,99999…, between the endless decimal series and 0.099999…and 0.0099999… etc. from a uncountable set would not change the one-to-one correlation to numbers without this anomaly. (I was referred to i.a. the Cantor–Bernstein–Schroeder theorem).

Reasoning in this manner I perceive the AC to be identical of postulating - or at least allowing for the definition of – real numbers. My question is whether or not this is the case.

Edit much later: I should add that I primarily am looking at the problems of defing invividual real numbers, so called pointwise definition, rather than definition of the whole set of reals. I didn't realize at the time that these are two different issues.

  • $\begingroup$ Mikael, with respect to "defining individual real numbers" you might be interested in reading my blog post about naming real numbers and the inherent difficulty with this notion. $\endgroup$
    – Asaf Karagila
    Dec 6, 2015 at 11:30
  • $\begingroup$ That is not a very easy post to read for amateurs. I assume that the axiom oh choice is critical here and that the arguments wouldn't work withour it. $\endgroup$ Dec 10, 2015 at 22:03
  • 2
    $\begingroup$ If everything was so easy, these sort of fallacious arguments wouldn't have been so common. You have to earn your knowledge with hard work, and hours of sweat and confusion. That's just how life works. $\endgroup$
    – Asaf Karagila
    Dec 10, 2015 at 23:37
  • $\begingroup$ It's cleaner to solve the dual representation problem by doing it the other way around, i.e. removing all of the finite sequences and using only the infinite ones. That way, you can use each finite sequence's "evil twin" to represent each dyadic/decadic rational as a corresponding real number (dyadic or decadic depending on which set of sequences you're using). $\endgroup$
    – Spencer
    Oct 12, 2021 at 16:29

4 Answers 4


The axiom of choice has nothing to do with the definition of the real numbers.

We define the real numbers either as Dedekind cuts over the rational numbers, which is simply a pair $(A,B)$ of subsets of the rational numbers which have some properties. Or we can define an equivalence relation on sequences of rational numbers and consider the equivalence classes as the real numbers themselves.

Then we can prove, again without any appeal to the axiom of choice, all sort of things about the real numbers. We can prove basic mathematical facts like convergence or divergence of some geometric series, the definition of $e$ and $\pi$, and $\sqrt2$. We can even prove that none of these are rational.

Without the axiom of choice we can even prove the Cantor-Bernstein theorem, and we can even prove that every set of real numbers whose complement is countable, has the same cardinality as the real numbers.

No, so far the axiom of choice isn't even playing a role. It does play a role if you want to prove other properties of the real numbers, such as "the real numbers are not the countable union of countable sets" (which is not to say they are countable!). But that doesn't seem to be what you're asking specifically.


A standard construction of the real numbers (going from the natural numbers, to the integers, to the rationals, and finally the reals as, say, Cauchy sequences of rationals) can be done entirely unproblematically without the Axiom of Choice. At each stage you define, without Choice, an appropriate equivalence relation, and the next number system is just the equivalence classes over this relation.

Of course, if Choice fails you might get some strange, even counterintuitive, happenings (e.g., the reals being a countable union of countable sets (but still uncountable!), or every set of reals being Lebesgue measurable). But nowhere in the construction is there any appeal to the Axiom of Choice.

  • 2
    $\begingroup$ Some amount of choice is needed to show that, say, the Dedekind reals and Cauchy reals are isomorphic. $\endgroup$
    – Zhen Lin
    Sep 10, 2013 at 19:18
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    $\begingroup$ @ZhenLin: How so? $\endgroup$
    – Asaf Karagila
    Sep 10, 2013 at 19:20
  • 1
    $\begingroup$ Every Cauchy real defines a Dedekind real in the obvious way, but to get a Cauchy real from a Dedekind real appears to require DC, no? $\endgroup$
    – Zhen Lin
    Sep 10, 2013 at 19:21
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    $\begingroup$ @ZhenLin: Not at all. Write the rational numbers as $q_n$, now suppose that $(A,B)$ is a cut. Using the enumeration you can find a sequence which is strictly increasing and unbounded within $A$. Then you can show it is Cauchy. Then you're about done. $\endgroup$
    – Asaf Karagila
    Sep 10, 2013 at 19:23
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    $\begingroup$ Ah, yes, I forgot that $\mathbb{Q}$ is well-orderable. $\endgroup$
    – Zhen Lin
    Sep 10, 2013 at 19:28

To answer the question "Can the axiom of Choice be used to define the real numbers?" the answer is yes.

As you suggested, the "bins" in the axiom of choice explanation can be the familiar infinite decimal representation of the real numbers. You can, if you like, define the real-2 numbers as any integer with an infinite decimal expansion as per the axiom of choice. The next step is to formally show that these real-2 numbers are in fact equivalent to the standard definition real numbers, for example by showing a 1-to-1 correspondence to equivalence classes of Cauchy sequences, or to Dedekind cuts.

Now as to whether you need the Axiom of Choice to define real numbers, which is the question most have answered, this is clearly a no, you just need Cauchy sequences of rational numbers.

But how to generate all Cauchy sequences so as to generate all real numbers?


It seems to me that you need the Axiom of Choice in order to get most of the Cauchy sequences whose equivalence classes constitute the real numbers.

Consider the set of all rational sequences. There are many which can be defined using a formula, i.e. an algorithm. For example $(1, 1, 1, 1, 1, \dots)$ has the formula $x_n = 1 \ \forall n$. Similarly you can define a rational sequence whose squares converge to $2$ (and thus you get a rational Cauchy sequence whose equivalence class will be named $\sqrt 2$) and thus you have defined that particular real number. It seems to me you can do that to construct all roots of polynomials (i.e. algebraic numbers), but what about all the transcendental numbers? $\Bbb e$ is easy enough. But what about just any old real number which lies somewhere on the number line. How do you define a sequence that converges to it? How can you specify the terms of such a sequence? It seems that you need the Axiom of Choice to do this. Because you're choosing an infinite number of elements from $\Bbb Q$, without any algorithm / formula.

The point is that to say that infinite sequences (other than ones that can be constructed algorithmically) exist requires the Axiom of Choice.

It's all very well to say you don't need AoC for constructing the reals because you just need cauchy sequences - but the Cauchy sequences themselves (most of them - i.e. the ones that can't be specified by using some algorithm) need the Axiom of Choice.

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    $\begingroup$ This is wrong on so many levels. $\endgroup$
    – Asaf Karagila
    Dec 4, 2015 at 19:54
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    $\begingroup$ 1) The set $C$ of all Cauchy sequences of rational numbers exists. 2) If $\sim$ is an equivalence relation on $C$, then the set of equivalence classes of $C$ under $\sim$ exists. If you agree that both 1) and 2) are provable without choice, then you must also agree that the set of real numbers exists, because real numbers are exactly the equivalence classes of certain relation on $C$. $\endgroup$
    – Wojowu
    Dec 4, 2015 at 20:09
  • $\begingroup$ Why is this wrong? Can you explain? How can you say that the set of all Cauchy sequences of rational numbers exists without assuming some random way of defining most of them (the AoC). Thanks. $\endgroup$
    – SSD
    Jan 15, 2017 at 17:34

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