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In mathematics the existence of a mathematical object is often proved by contradiction without showing how to construct the object.

Does the existence of the object imply that it is at least possible to construct the object?

Or are there mathematical objects that do exist but are impossible to construct?

Rodrigo de Azevedo
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Kasper
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8 Answers8

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Really the answer to this question will come down to the way we define the terms "existence" (and "construct"). Going philosophical for a moment, one may argue that constructibility is a priori required for existence, and so ; this, broadly speaking, is part of the impetus for intuitionism and constructivism, and related to the impetus for (ultra)finitism.$^1$ Incidentally, at least to some degree we can produce formal systems which capture this point of view (although the philosophical stance should really be understood as preceding the formal systems which try to reflect them; I believe this was a point Brouwer and others made strenuously in the early history of intuitionism).

A less philosophical take would be to interpret "existence" as simply "provable existence relative to some fixed theory" (say, ZFC, or ZFC + large cardinals). In this case it's clear what "exists" means, and the remaining weasel word is "construct." Computability theory can give us some results which may be relevant, depending on how we interpret this word: there are lots of objects we can define in a complicated way but provably have no "concrete" definitions:

  • The halting problem is not computable.

  • Kleene's $\mathcal{O}$ - or, the set of indices for computable well-orderings - is not hyperarithmetic.

  • A much deeper example: while we know that for all Turing degrees ${\bf a}$ there is a degree strictly between ${\bf a}$ and ${\bf a'}$ which is c.e. in $\bf a$, we can also show that there is no "uniform" way to produce such a degree in a precise sense.

Going further up the ladder, ideas from inner model theory and descriptive set theory become relevant. For example:

  • We can show in ZFC that there is a (Hamel) basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$; however, we can also show that no such basis is "nicely definable," in various precise senses (and we get stronger results along these lines as we add large cardinal axioms to ZFC). For example, no such basis can be Borel.

  • Other examples of the same flavor: a nontrivial ultrafilter on $\mathbb{N}$; a well-ordering of $\mathbb{R}$; a Vitali (or Bernstein or Luzin) set, or indeed any non-measurable set (or set without the property of Baire, or without the perfect set property); ...

  • On the other side of the aisle, the theory ZFC + a measurable cardinal proves that there is a set of natural numbers which is not "constructible" in a precise set-theoretic sense (basically, can be built just from "definable transfinite recursion" starting with the emptyset). Now the connection between $L$-flavored constructibility and the informal notion of a mathematical construction is tenuous at best, but this does in my opinion say that a measurable cardinal yields a hard-to-construct set of naturals in a precise sense.

$^1$I don't actually hold these stances except very rarely, and so I'm really not the best person to comment on their motivations; please take this sentence with a grain of salt.

Noah Schweber
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There exists a Hamel basis for the vector space $\Bbb{R}$ over $\Bbb{Q}$, but nobody has seen one so far. It is the axiom of choice which ensures existence.

Dietrich Burde
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    Another consequence of the axiom of choice: there is a well order in $\mathbb R$. But no one knows how it is. – Dog_69 Jun 05 '18 at 11:42
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    These might just be cases where we haven't discovered the construction yet. I assume the construction hasn't been proved impossible (or you'd have said so). – Barmar Jun 05 '18 at 17:06
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    @Barmar Actually, results from descriptive set theory do rule out "explicit" constructions. – Noah Schweber Jun 05 '18 at 17:12
  • @Noah Yes, although one can always force a $\Delta^2_2$ well-ordering without adding reals, so I guess this gives us a concrete limit to what we should mean by "explicit" within the realm of descriptive set theory. – Andrés E. Caicedo Jun 05 '18 at 18:10
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    @AndrésE.Caicedo I have no problem saying that $\Delta^2_2$ is not explicit, but you're right that that's important to mention. – Noah Schweber Jun 05 '18 at 18:17
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I am not sure if "constructible" in this context is related, but I think this example also counts.

Not all the regular polygons are constructible. The Gauss–Wantzel theorem stated that,

A regular $n$-gon is constructible with compass and straightedge iff

  • $n$ is a Fermat prime (A prime in form of $2^{2^m}+1$)
  • $n=2^k$
  • $n$ is the product of a power of $2$ and distinct Fermat primes.

So, for instance, regular heptagon ($7$-gon) exists but it is not constructible.

Mythomorphic
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    +1 for an interesting reading of the problem, even though it's probably not what the OP had in mind. The heptagon can't be constructed with Euclidean tools but can with a more general definition of constructible, as in @NoahSchweber 's answer. – Ethan Bolker Jun 05 '18 at 12:19
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It may be worth mentioning that Errett Bishop developed his version of constructive mathematics precisely because of his disillusionment with his earlier research in complex analysis which he felt involved objects that don't admit any construction. His last work in classical mathematics was the article

Bishop, Errett. Differentiable manifolds in complex Euclidean space. Duke Math. J. 32 1965 1–21

which was and continues to be influential. After 1965, he published about a dozen articles, all dealing with constructive mathematics.

Mikhail Katz
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As others already said, it dependes on your notion of constructible.

Can you construct infinite objects? I think most people agree that you can construct any natural or integer or rational number. But the whole set is infinite. Can you construct it? On the other hand, most real numbers are "infinite" themselves. Can you construct $\sqrt{2}$? What about $\pi$ or $e$?

Things get complicated. Once you have a precise definition for $\pi$, you can trivialy construct it with this steps:

  1. Take $\pi$.
  2. Done.

Is that satisfactory?

Constructive Mathematics (see also links therein) attempt(s) to answer your question. Unfortunately, once you deviate from traditional mathematics, you face the reality of it: a human endeavour. I.e. multiple interpretations, different logics, proof systems, axioms sets, notations... So there are also multiple approaches to constructive mathematics...

Pablo H
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  • I'd love to see a readable and usable (practical) introduction to constructive mathematics. Say, up to basic linear algebra, group theory and analysis. – Pablo H Jun 05 '18 at 17:33
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    See Bishop's [Constructive Analysis] (https://www.springer.com/gp/book/9783642649059). – mrp Jun 05 '18 at 19:56
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    Given a Euclidean length of $1$, a ruler and compass construction of a straight length of $\sqrt{2}$ is rather easy, as is a curved length of $\pi$ – Henry Jun 07 '18 at 11:30
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I come from computer science and would like to define "construct" like this:

A real decimal number is constructible if you can write a program that prints out digits of the number forever. A decimal point must be printed at some time. The program must be of finite fixed size but can use more and more memory without limit as further digits are computed.

Given this definition, it is possible to describe a number that exists, but is impossible to construct.

This is related to the Halting Problem. Like in any version of the HP, we assign a number to each possible computer program in a certain language. The Halting Problem Theorem then states that there is no finite program that can decide whether another program will halt or get caught in an infinite loop.

In this version, we then asks for the real number between $0$ and $1$ where digit number $n$ after the decimal point is $1$ if program number $n$ halts and $0$ otherwise.

Any given program $n$ will either halt, or it will not. This means the number is well defined and exists.

However, writing an actual program to compute it would violate the HPT, so that is impossible.

Stig Hemmer
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  • Of course, you would struggle to find the computer with unlimited memory on which to run it but, even if you could get one, you would not violate the HPT, the program would just continue to run until something (e.g. end of the universe) stopped it. – badjohn Jun 06 '18 at 10:40
  • For this particular example, "write all $0$s", then for each $n$ spawn a thread [universal Turing machine] for $n$, so that when (if) program $n$ finishes, the $n$-th $0$ is replaced by a $1$. An infinte-memory computer has infinite cores, right? – Pablo H Jun 06 '18 at 14:55
  • It seems like my definition is not entirely water-proof. However, fixing it would drown the answer in technical details, so I won't – Stig Hemmer Jun 07 '18 at 08:45
  • @PabloH You can't go back and change stuff that you have already written, that is the crucial part here. – Arno Jun 07 '18 at 12:58
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    The standard term for what Stig described is "computable" or "recursive". The confusion then arises from the distinction between "computable" and "computably enumerable": for the former, the program must specify after some finite amount of time that the nth digit has been computed, whereas for the latter, the digit could remain 0 for arbitrarily long and we can never be sure that it won't be updated to a 1 at some later time. – Robin Saunders Jun 07 '18 at 15:04
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    Addendum: in constructive mathematics, the halting problem might not have a well-defined solution for all programs, and so it is consistent that your non-constructible number "does not exist". One formal version of this principle is the formal Church's thesis, which explicitly asserts that the only real numbers which exist are those which are computable (and not merely computably enumerable). – Robin Saunders Jun 07 '18 at 15:07
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Taking yet another definition of the terms, a 6 dimensional surface trivially exists, but I'm not sure what one would construct it from in the real world, so the object exists but cannot be (physically and real-world) constructed

Stilez
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  • But you can easily construct one in the mathematical world, for example $x^2 + y^2 + z^2 + w^2 + u^2 + v^2 = 100$ describes a 6D equivalent of a sphere. You can plot all the coordinates on the sphere to arbitrary resolution, for example. You can project it down into 3D if you want. – user253751 Jun 06 '18 at 04:02
  • Yes. Mathematically. But construct and object can also have a literal meaning. A cube and a sphere are mathematical objects and can be literally constructed. A 6-sphere, so far, cannot, it can only be mathematically constructed, or at best projected instead. – Stilez Jun 06 '18 at 04:23
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    A sphere cannot be constructed physically, only an approximation to one. – user253751 Jun 06 '18 at 04:58
  • True - and on that case even more so. But I think an everyday meaning of "construct" allows for real-world scale imperfections. If I went to a carpenter/engineet and said ""here is a mathematical description of a 3d box 10cm on each side, please construct it", they would. But if I said "construct me a 6D box" or "construct me a perfect box", they wouldn't be able to – Stilez Jun 06 '18 at 06:02
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    Taking "construct" that literally would eliminate a vast number of mathematical objects. Irrational numbers would go, however accurate we get, we could never measure the diagonal of a unit square as exactly $\sqrt{2}$. Even larger integers would go since it is easy to name one larger than the number of atoms in the universe. – badjohn Jun 06 '18 at 10:42
  • You could still define $\sqrt{2}$ as a pair $(0,1)$ with a multiplication defined on it, similar as how complex numbers are defined. – Kasper Jun 06 '18 at 12:00
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Another sort of an example: there is a positive real $x\approx 1.16730397826142$ where $x^5-x-1=0$. The Intermediate Value Theorem proves that there is one. But it can't be constructed: we do not have the mathematical language to express it.

Rosie F
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  • Thanks, that is an example, I can really follow with my knowledge of mathematics. – Kasper Jun 07 '18 at 10:53
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    Surely we have the mathematical language to express it. "The real number $x$ for which $x^5 - x - 1 = 0$" is one way to express it. In what sense is that less mathematical language than, say "$\sqrt{2}$"? – JiK Jun 07 '18 at 12:36
  • @JiK But that merely restates the property that the number has. What construction has taken place? – Rosie F Jun 07 '18 at 13:04
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    But we can construct the number: you started constructing it yourself by listing the first few digits. We can use any root-finding algorithm (bisection, Newton, etc.) to get the rest of the number to arbitrary precision. Perhaps you mean that it can't be described in terms of roots and elementary operations. But that's like saying, "We don't have the mathematical language to express $\sqrt2$" on the basis that it can't be described as a ratio of whole numbers. – Théophile Jun 07 '18 at 14:57
  • I constructed an approximation to it, and root-finding algorithms can also be used to construct approximations to it, to arbitrary precision, as you say. But to do so is not to construct the number itself. – Rosie F Jun 07 '18 at 15:30
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    @RosieF What construction takes place in "$\sqrt{2}$"? That merely restates the property that that the number has. You can construct an approximation to it to arbitrary precision, but how is that constructing the number itself? – JiK Jun 07 '18 at 18:38