I understand that a set whose members can, in principle, be enumerated (by having a formula) can be considered as a well-defined set. Therefore, set of all even numbers, multiples of 3, and so on constitute what a well-defined set, for its members are well-defined. I personally also refer to this as an identified set. Here I am saying that it makes "sense" to identify this set i.e. to assert its definite existence (in set theory) because there is a formal method to enumerate/identify its members.

Now, what about the reals, ? There is no method to enumerate its members (Cantor's diagonalization). Why is it not contradictory/absurd to speak of set R if one cannot formally identify it? We know what its members are integers, rational, irrational numbers...but relying on what we mutually understand appears more like handwaving. To support its formal existence, what concrete definitions have been provided by mathematicians/logicians?


You ask, "To support its formal existence, what concrete definitions have been provided by mathematicians/logicians?"

The real numbers can be defined in several equivalent ways.

One way is to start with the set of infinite sequences of rational numbers. Some of these sequences are Cauchy sequences, which means that the elements get arbitrarily close to each other as the sequence progresses. We can define the real numbers as Cauchy sequences of rational numbers, under an equivalence relation where two sequences are considered equivalent if their difference tends to zero as the sequence progresses.

Another way is by using Dedekind cuts of the rational numbers. A Dedekind cut of the rational numbers is a partition of the rational numbers into two sets (A, B) where all elements of A are less than all elements of B, and A has no greatest element. The "real number" is imagined to be the element right in the middle, larger than any element of A, and no larger than any element of B. We can define the real number as the Dedekind cut (A, B) and give rules for addition, subtraction, multiplication, and division of Dedekind cuts.


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    The only answer that actually answers the OP's question.
    – Alex
    Dec 29 '20 at 14:56

This is a misconception, you don't need to be able to enumerate the elements of a set.

In naive set theory (which has its problems but is useful to explain the set concept here), a set is defined by some property. Every object that satisfies the property is contained in the set. The only thing you can ask a set is whether some object is contained in it. If you use axiom of choice, you can also ask for some element that is contained.

A set is not numbered. For a set S, any enumeration is a bijective function f: N -> S, but such an enumeration doesn't exist for every set.

In summary: A set is not defined by some function that generates all its members, it's defined by a classification function which tells us whether an element belongs to the set or not.

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    Consider R/Q, the set of equivalence classes of the reals under the relation that x ~ y if x - y is rational. Let V be a choice set consisting of exactly one real from each equivalence class. Then there is no classification function that tells you whether 1/2, say, is in V. Such a set has pure existence, we know nothing about it at all except that it's uncountable and contains exactly one real from each equivalence class. See en.wikipedia.org/wiki/Vitali_set
    – user4894
    Dec 29 '20 at 5:37
  • @user4894 Ok, then maybe we can just say: A set is a set. It may or may not contain stuff. How can I fix this sentence? Most sets are defined by some property, at the very least the property of being in that particular set. There is an excluded middle: Something is in a set or it is not, but you are right it might not always be possible to tell whether something is in a set (but if I had an oracle for membership then I would have a function :-P ). In proofs we normally say: If an object is in Set S (meaning it has some property), then if follows that xyz.
    – kutschkem
    Dec 30 '20 at 15:44
  • I agree that any set has a classification function that says whether something is or isn't in the set, but that doesn't seem to help much since we can't know what the function is in most cases. There are more sets whose membership is random than there are sets whose members are defined by a rule. Noncomputable numbers again.
    – user4894
    Dec 30 '20 at 20:33

Most mathematicians are happy to use ZFC set theory or one of it's equivalents. These set theories support the "normal" real numbers.

There are, however, mathematicians such as the intuitionists and constructivists who might be said to be "suspicious" of ZFC set theory for reasons not unlike what those you have expressed in your question. In developing their mathematics, they do not use the axiom of choice, or use a modified version of it. There are also schools that attempt to avoid the power-set axiom.

There are theorems in "classical" mathematics that are not provable in these systems. They are, in a sense weaker. On the other hand, there are things that are provable in "classical" mathematics, such as the Banach-Tarski paradox, which may strain one's sense of how things "ought to be". In the end, however, it is impossible to prove that ZFC is either consistent or inconsistent due to Godel's incompleteness theorems.


Wikipedia entry on Constructivism

Wikipedia entry on Constructive Set Theory / Intuitionism

Wikipedia entry on Banach-Tarski Paradox

Stanford Encyclopedia of Philosophy entry on Constructivism

SEP entry on Intuitionism

  • Thanks for answer. Can you point out relevant resources/topics which expose these issues?
    – Ajax
    Dec 27 '20 at 17:07
  • I added some links to the answer. You can google Constructivism, Intuitionism, etc. Some of the principal names in Constructivism/Intuitionism include Brouwer and Martin-Lof. Dec 27 '20 at 17:20
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    ZFC isn't required to construct the real numbers, ZF is sufficient. So philosophical objections to choice don't affect the absurdity of the reals.
    – user4894
    Dec 28 '20 at 4:22
  • @user4894 The "absurdity" of the reals was a contemporary debate in the 1870's. It is no longer a matter of debate among professional philosophers and mathematicians. See cardinality of the coninuum.
    – J D
    Dec 28 '20 at 18:04
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    @JD (... cont) I might put down my thoughts about all this in a response to this question, but frankly I'm no expert. In the light of countable models of set theory and in light of neo-intuitionism, I hope you would recognize that the questions raised by the OP, no matter how inartfully phrased, are valid and of contemporary interest.
    – user4894
    Dec 28 '20 at 18:45

There are very well defined numbers that are demonstrated to be nor integers, nor a fraction of integers (rational numbers, who are "rational" because you can define them as a "ratio" of 2 integers, btw): Pi, e, square root of 2, sin(1)... From this we can derive an infinity of other numbers with the same property (2Pi, 1/Pi, etc...)

So the set of real numbers exists, undoubtedly.

Then you have the non polynomial functions, who give irrational results for integer inputs, like sin(1), so again an infinity of well identified yet transcendental numbers.

Since it seems strange that a functions like sin would be undefined for irrational inputs between two rationals (because stuff spin continuously, as far as we know), the existence of real number is in fact pretty intuitive.

  • First things first, that's not how it works. Second, we can prove a function is continous on ℝ only by assuming that every Cauchy Sequences converge in ℝ. In which case, we are taking a real number to be an equivalence class of Cauchy Sequences. I.e. we are appealing to axiom of specification and existence of ℚ. It's not as naive as you put it. Dec 29 '20 at 4:16
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    Are you ok? No one said sin is not continous on ℝ. The problem is you are assuming it is without knowing how we can prove that it is so. Sin is continous on ℝ because the real numbers (depending on how you foundationalize it Cauchy, dedekind, axiomatization) exist. Open any advance calculus/analysis text book and you will find this stated. Dec 29 '20 at 5:05
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    You convinced me that the countably infinite set of computable numbers exists. Pi, sqrt(2), e, and so forth. How about the noncomputable ones? Aren't they ontologically suspect, even a little? I'm no constructivist but I do see their point.
    – user4894
    Dec 29 '20 at 5:20
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    Even if the real numbers do not "exist" they are still a very useful construct. We could hardly imagine mathematics without them. Myself, I take all mathematical objects (including the integers) as counterfactual, and all theorems as true statements about what would follow if those objects existed.
    – causative
    Dec 29 '20 at 6:39
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    @armand What In the world is this nonsense!? "If they are continuous there is something there, between the rationals. And since we already have a set of numbers that are not rational, they fit right there inside." Rofl. I am so done with Philosophy SE. Utter joke. Dec 29 '20 at 7:29

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