Then otherwise the sentence "It is not possible for someone to find a counter-example" would be a proof.

I mean, are there some hypotheses that are false but the counter-example is somewhere we cannot find even if we have super computers.

Sorry, if this is a silly question.

Thanks a lot.

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    What do you mean by can't be found? It's incomputable? There's no constructive proof of the existence of a counterexample? – Loki Clock Apr 15 '14 at 21:05
  • maybe Riemann Hypothesis is false but no one could find a counter-example until now. In fact, if a problem is undecidable, this would be the case, you could try to find a counter-example until the end of time – Integral Apr 15 '14 at 21:06
  • @LokiClock yes, i am asking for an incomputable counter example whose existence is proved. – ThePortakal Apr 15 '14 at 21:09
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    Related: http://math.stackexchange.com/questions/710950/prove-that-a-counterexample-exists-without-knowing-one – leonbloy Apr 15 '14 at 21:10
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    What evidence is there that the counterexample exists if it cannot be found, and how is that evidence convincing if it does not amount to producing the counterexample? – Kaz Apr 15 '14 at 22:36
  • Do you have an example of a hypothesis that is known to be false (by some proof other than showing a counterexample), but for which the counterexample is difficult to find? – Kaz Apr 15 '14 at 22:38
  • I'm not 100% sure about the tagging, by the way. [proof-theory] feels a bit off here. – Asaf Karagila Apr 16 '14 at 09:28
  • What about the statement "there is no incomputable object"? – DanielV Apr 16 '14 at 18:23
  • Isn't it what the Gödel Theorem is about? That if the theory is consistent, then there *exists* undecidable statements (counterexamples of decidability) but none of them can be proven undecidable, therefore they cannot be *found*. There are systems that are outside the applicability of the theorem, so the question is incomplete without context (e.g. Natural numbers? Euclidean geometry?) – alfC Apr 16 '14 at 20:08
  • It's perfectly possible for a counter-example to exist (and be proven to exist) but the counter-example cannot be 'found' nor named - for example, the specific counter-example number would require more bits than there are particles in the observable universe. – Peteris Apr 16 '14 at 22:55
  • @alfC: Then how do you explain the fact that the theory of algebraically closed fields of characteristics $0$ is complete? – Asaf Karagila Apr 17 '14 at 02:41
  • @AsafKaragila, I don't know what you are talking about and how is that related to my comment. Sorry. – alfC Apr 17 '14 at 05:08
  • @alfC: It is a theory which is both consistent and complete. It is even effective. – Asaf Karagila Apr 17 '14 at 06:21
  • @AsafKaragila, Thanks for the information. Geometry is also consistent and complete. That is why I said in my comment that the answer to the question depends on the theory. In some theories if there is a counter example to a statement then it can also be found and the original question can be answer to the negative. – alfC Apr 17 '14 at 06:32
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    @alfC: The theory has to be first-order, recursively enumerable, and interpret basic truths of arithmetic. Geometry, if I recall, is not first-order; algebraically closed fields can't interpret arithmetic. – Asaf Karagila Apr 17 '14 at 07:27
  • @AsafKaragila, I think Tarsky proved that. This points to relevant material: http://en.wikipedia.org/wiki/Alfred_Tarski#Mathematician – alfC Apr 17 '14 at 18:23
  • @alfC: Tarski, you mean. No, my confusion was with Hilbert's geometry which was second-order in nature. Tarski's axioms of geometry were in fact a first-order theory, but I think this theory is not complete. I'm not 100% sure, though. – Asaf Karagila Apr 17 '14 at 18:31
  • [Intuitionistic mathematics](http://math.andrej.com/2008/08/13/intuitionistic-mathematics-for-physics/) seems relevant here. The law of the excluded middle is neither accepted nor entirely rejected, but one can prove **no** counter-example exists in intuitionistic logic. Also, see [Penon infinitesimals](http://math.stackexchange.com/questions/713415/defining-penon-infinitesimals). – Shaun Apr 18 '14 at 09:41

8 Answers8


A standard example of this is the halting problem, which states essentially:

There is no program which can always determine whether another program will eventually terminate.

Thus there must be some program which does not terminate, but no proof that it does not terminate exists. Otherwise for any program, we could run the program and at the same time search for a proof that it does not terminate, and either the program would eventually terminate or we would find such a proof.

To match the phrasing of your question, this means that the statement:

If a program does not terminate, there is some proof of this fact.

is false, but no counterexample can be found.

Alex Becker
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    I disagree that this can be concluded from that statement. What if for every program that never terminates there's a proof of this fact, but no unified way of producing a proof, or no single program which can verify that it terminates? – Loki Clock Apr 15 '14 at 21:13
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    @LokiClock I have avoided technical details here, but there always is such a way. Simply enumerate all proofs and check whether they are correct proofs of termination. The standard argument in computability theory is on step $2k$ run the $k$th step of the test program, and on step $2k+1$ check the $k$th proof to see if it is correct (proof checking is known to be computable). – Alex Becker Apr 15 '14 at 21:18
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    I might be completely off base, but I think a big stumbling block to the halting problem is that if you had a candidate program $p$ that solved the halting problem, it would also have to be able to determine if it would halt if it ran using itself as an input. Yikes. – corsiKa Apr 15 '14 at 21:48
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    -1 What is your statement a counter-example to? Certainly we can give a constructive proof that there are programs which do not terminate. Here is one: While True print "keeps on going and going..." – Andrew Kelley Apr 15 '14 at 21:50
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    @corsiKa That's not the issue as any program which solves the halting program would be required to halt. – Alex Becker Apr 15 '14 at 21:50
  • @AndrewKelley That all programs which do not terminate, provably do not terminate. – Alex Becker Apr 15 '14 at 21:52
  • Alex, I think it is slightly more subtle. Doesn't the halting problem state that there is no program which can decide in all cases? Does it also say that there exists a program, say $\mathbf{P}_0$ such that, there does not exist a proof that $\mathbf{P}_0$ does not terminate? I think this is similar to the word problem in group theory: There is no algorithm that works for all groups. But more than that, there even exists a specific group for which no algorithm exists. My objection is the same as that of @Loki. – Andrew Kelley Apr 15 '14 at 22:12
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    @AndrewKelley The statement "There exists a program, say P0 such that, there does not exist a proof that P0 does not terminate?" follows from the statement "there is no program which can decide in all cases". This is because a candidate program could just try all proofs. – Alex Becker Apr 15 '14 at 22:16
  • @AlexBecker How is the existence of a proof for every non-terminating program's non-termination equivalent to the assignment of the proof to each such program? A program which can go through all proofs - assuming there are countable many non-terminating programs - would require such an assignment. Assuming the axiom of choice, yes, we can choose a representative proof from each term of the product of all non-terminating programs' set of proofs of nontermination, and if the number of programs is countable, the axiom of countable choice. – Loki Clock Apr 15 '14 at 22:39
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    @LokiClock No choice is necessary. The key is that proofs are *computably enumerable*. For each program, we need only go through all possible proofs one by one until we find one that works. I never said it was an *efficient* way to do it, but that's the standard argument in computability theory. – Alex Becker Apr 15 '14 at 22:59
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    @Loki: the number of programs is necessarily countable, since a program in a Turing-complete language is equivalent to a finite sequence of characters. Alex: one cannot check for the existence of a proof by checking the set of "all possible proofs"; if such a proof does *not* exist, then this "proof-checking" program is non-terminating. – Kyle Strand Apr 15 '14 at 23:01
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    @KyleStrand That's why you spend alternating steps checking whether the program terminates. Under the (false) assumption that either a program terminates or such a proof exists, our strategy eventually causes the program to terminate or finds a proof. Of course, this contradicts the halting problem so we have a contradiction. – Alex Becker Apr 15 '14 at 23:04
  • @Alex I'm a bit lost now. "... you spend alternating steps checking whether the program terminates" already contradicts the halting principle. – Kyle Strand Apr 15 '14 at 23:09
  • @KyleStrand I mean on step $2k$, you run step $k$ of the program. So if the program terminates at step $n$, our hypothetical method finds out at step $2n$. – Alex Becker Apr 15 '14 at 23:17
  • @KyleStrand I suppose if there exists a bijection from a subset of the natural numbers to the proofs of non-termination, there must be some least number in the subset, and then the function whose value on each program is the proof whose value is that least natural number works as a choice of proof for each program. As long as you can test whether a number is in the image of the inverse bijection from the proofs, you can find the lowest number in the image. – Loki Clock Apr 15 '14 at 23:45
  • Alex: That wasn't the part I was confused about. You can't algorithmically check whether a program terminates. @Loki: Turing shows we can't perform such a test algorithmically in a finite number of steps. – Kyle Strand Apr 16 '14 at 02:28
  • @KyleStrand Now I'm confused. Of course you can algorithmically verify that a terminating program terminates. By assumption for any nonterminating program there is a proof of this fact. So try to verify that it terminates while also looking for a proof that it does not. – Alex Becker Apr 16 '14 at 02:29
  • Your assumption is simply incorrect. Have you read Turing’s *Entscheidungsproblem* paper? – Kyle Strand Apr 16 '14 at 03:28
  • @KyleStrand The answer said that assumption was false, the issue was the answer didn't say how to conclude that from the halting problem's undecidability. – Loki Clock Apr 16 '14 at 03:48
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    I'm pretty sure Alex Becker is completely correct here. In general, the undecidability of a certain decision problem guarantees that there exists a specific instance of the problem for which no proof of the correct decision exists in ZFC. This is because there exists a "universal prover" computer program that enumerates all possible consequences of the axioms of ZFC, thereby proving all provable statements. If all instances of a decision problem were provably solvable, then the universal prover would be an algorithm to decide it. – Jim Belk Apr 16 '14 at 03:52
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    In particular, since the halting problem is undecidable, it follows that there exists a computer program which never halts, but whose non-halting cannot be proven by the axioms of ZFC. (A computer program that halts can always be proven to do so by the axioms of ZFC -- the proof just consists of the sequence of internal states of the program from when it starts to when it halts.) – Jim Belk Apr 16 '14 at 03:54
  • Yes, Alex Becker is correct here, for the reasons he explained. Maybe the confusion stems from the fact that "terminate" here means "terminate on a blank input" (or a given input) rather than "terminate on all inputs" (in the former case it's trivial to verify that a *terminating* program terminates, in the second case it's undecidable). – david Apr 16 '14 at 06:53
  • Hm, I think I just got confused by the re-iteration of (intentionally false) premises and mistook them for (incorrect) statements of fact. Sorry. – Kyle Strand Apr 16 '14 at 17:02
  • There exists a set of laws of physics for which the halting problem can be solved. I can't prove the set is consistent but I can reference a proof that appears equivalent to the capacity to build a paradox machine does not prove a proposed set of laws of physics inconsistent. – Joshua Apr 16 '14 at 23:41
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    I disagree with this answer for a reason which is (I think) different from previous comments. For the statement to be precise, we must fix a notion of "proof". Say "proof" means "proof in $\mathsf{ZFC}$". But then a counterexample can be found: The program searching for a proof of $\bot$ in $\mathsf{ZFC}$ never halts, but there is no proof (in $\mathsf{ZFC}$) of this fact. Anyone who believes $\mathsf{ZFC}$ to be consistent must believe that this is a counterexample. (Of course, there is no proof in $\mathsf{ZFC}$ that it is a counterexample, but this seems to me to be beside the point.) – Trevor Wilson Apr 17 '14 at 00:40
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    @Trevor furthermore if you believe ZFC is inconsistent then "If a program does not terminate, there is some proof of this fact" is actually true since there is a proof of anything (and in this case it no longer contradicts the halting problem). So either ZFC is inconsistent and the statement is true or ZFC is consistent and the statement is false and a counterexample _can_ be found. And the same applies to any theory not just ZFC, so while interesting I agree this isn't a good answer to the question. – Dan Brumleve Apr 22 '14 at 20:59

I actually like this one:

There are uncountably many real numbers. However, given that all specifications of specific real numbers (be it by digits, by an algorithm, or even a description of the number in plain English) is ultimately given by a finite string of finitely many symbols, there are only countably many descriptions of real numbers.

A straightforward formalization (but not the only possible, nor the most general one) of that idea is to model the descriptions as natural numbers (think e.g. of storing the description in a file, and then interpreting the file as natural number), and then having a function from the natural numbers (that is, the descriptions) to subsets of the real numbers (namely the set of real numbers which fit the description). A description which uniquely describes a real number would, in this model, be a natural number which maps to a one-element subset of the real numbers; the single element of that subset is. Since there are only countably many natural numbers (by definition), they can only map to at most countably many one-element subsets, whose union therefore only contains countably many real numbers. Since there are uncountably many real numbers, there must be uncountably many numbers not in this set.

Therefore in this formalization, for any given mapping, almost every real number cannot be individually specified by any description. Therefore there exist uncountably many counterexamples to the claim "you can uniquely specify any real number".

Of course I cannot give a counterexample, because to give a counterexample, I'd have to specify an individual real number violating the claim, but if I could specify it, it would not violate the claim and therefore not be a counterexample.

Note that in the first version, I omitted that possible formalization. As I learned from the comments and especially this MathOverflow post linked from them, in the original generality the argument is wrong.

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    Actually, there are models of $\mathsf{ZFC}$ in which every real number (and indeed every set) is definable. (It is tempting to say that the statement "every real number is definable" is consistent with $\mathsf{ZFC}$, but this statement cannot be formalized: definability may fail to be definable.) The mistake is the assumption that there is a (partial) function mapping (putative) definitions to the numbers that they define. Although it is tempting to believe that there is an obvious such function (because it seems like we can define it) this is not necessarily the case. – Trevor Wilson Apr 17 '14 at 00:48
  • See Joel Hamkins' slides for more information on this subtle point: http://jdh.hamkins.org/files/2011/09/Vienna-2011-Pointwise-Definability-Talk.pdf for – Trevor Wilson Apr 17 '14 at 00:50
  • @TrevorWilson, surely we should be considering models of second-order set theory, i.e. $\mathrm{ZFC2}$ for this question. Otherwise, the model's real line needn't be the true real line relative to the ambient universe. – goblin GONE Apr 17 '14 at 02:16
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    Great answer! "There exists a real number whose description cannot ever be uttered." IMHO these numbers should be called Valdemort numbers. – Michael Apr 17 '14 at 02:19
  • @user18921 I'm not sure why we should consider such models. But anyway, the only reason I mentioned models was to illustrate a problem with the argument. The problem can still be explained without reference to models. The problem is: why should there be a function from definitions to the objects they define? (By Tarski's theorem on the undefinability of truth, the "obvious definition" doesn't work.) – Trevor Wilson Apr 17 '14 at 02:23
  • @TrevorWilson, I'm not sure I follow, so let me take a look at Hamkin's stuff and I'll get back to you. – goblin GONE Apr 17 '14 at 02:25
  • @user18921 It is a bit counterintuitive. He also wrote some answer on Math.SE or MathOverflow about this; I couldn't find it immediately, but possibly you would have better luck. – Trevor Wilson Apr 17 '14 at 02:27
  • @Trevor: He wrote it on MathOverflow. Keywords for a search should be "pointwise" and "definable", add `is:answer` to help removing possible questions that include both terms. – Asaf Karagila Apr 17 '14 at 02:48
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    Ah, here is Joel's MO post: http://mathoverflow.net/a/44129/1682 (thanks, Asaf.) – Trevor Wilson Apr 17 '14 at 02:52
  • So, ..., what is the smallest positive real number without a finite English description? (except for this one, obviously)? – Eric Towers Apr 17 '14 at 16:05
  • Interesting. I can't say I understand all of what was discussed, but the example with the ordinals in the linked MathOverflow answer is quite convincing that something is wrong with the argument (@EricTowers: There exist nonempty sets of positive real numbers without a minimum — most notably the set of positive real numbers itself —, so this is no argument; however one could impose a well-ordering on the reals by mapping to an ordinal of the right cardinality and make the argument with that ordering). I'm not sure what I now should do with my answer. Just delete it? Amend it saying it's wrong? – celtschk Apr 17 '14 at 18:07
  • Thinking again about my parenthetical comment in the previous comment: The well-ordering might itself be impossible to specify, therefore this trick probably doesn't work. Anyway, the fact that the very same argument can be applied to the ordinals and gives a contradiction already shows that the argument as such cannot be correct. – celtschk Apr 17 '14 at 18:22
  • I think the answer can be salvaged. The main problem, really, is that the question does not specify what "found" means. For many possible notions of "found" there is a local version of the argument: the statement "every real number can be found" has a counterexample, but no counterexample can be "found". For example if "found" means "computed" then we get something like the halting problem. – Trevor Wilson Apr 17 '14 at 19:51
  • @celtschk: I wasn't (intentionally) saying anything negative about your answer. Your answer reminded me of this standard example of a problem with naive set theory. For a variant, see the Interesting Number Paradox: http://en.wikipedia.org/wiki/Interesting_number_paradox – Eric Towers Apr 17 '14 at 22:53
  • @EricTowers For every positive real number there exists a smaller real number that cannot be expressed as a finite description. There is no smallest one. – trichoplax Apr 18 '14 at 09:22
  • @TrevorWilson: So the solution is to just demand a specific type of specification to be chosen, and then for this specific type of specification it is true? Or does the type of specification itself have to fulfil some requirements? – celtschk Apr 19 '14 at 05:29
  • @EricTowers: Actually the ordinal number version was exactly this interesting number paradox, with "natural" replaced by "ordinal" and "uninteresting" replaced by "undefinable". My point was that this paradox doesn't work with real numbers under the ordinary ordering, because other than non-empty sets of natural numbers, non-empty sets of real numbers (even bounded ones) don't need to have a minimum. – celtschk Apr 19 '14 at 05:34
  • The simplest sufficient condition I can think of for the argument to work is (1) the set of allowed definitions is countable, and (2) there is a surjection that maps each definition to the real number that it defines. Fortunately, most "restricted" or "bounded complexity" notions of definability have these properties. For example, definability over the structure $(\mathcal{P}(\mathbb{N}),\mathbb{N}; \in, +, \times, 0,1)$, which would include any definitions that are likely to arise in "ordinary" mathematics. – Trevor Wilson Apr 21 '14 at 06:34
  • @TrevorWilson: Is the post now OK? – celtschk Apr 21 '14 at 18:39
  • I don't think it's quite right. You way "Therefore in this formalization almost every real number cannot be individually specified in _any_ way" (emphasis added.) However, I think you need to fix a notion of "description" or "specification" _first_ (this can be formalized as a function, as you noted) and only then can we say that almost every real number cannot be individually specified in _that particular_ way. – Trevor Wilson Apr 21 '14 at 20:42
  • OK, I changed that. Now OK? – celtschk Apr 21 '14 at 20:46
  • I forgot the @TrevorWilson in the previous comment. – celtschk Apr 22 '14 at 21:10
  • I think it still has the same problem, but maybe I'm not reading it correctly. You say "for any given mapping, almost every real number cannot be individually specified by any description." It's not clear to me what "for any given mapping" has to do with the rest of the quoted sentence. Do you mean "almost every real number is not individually specified _via that mapping_ by any description"? – Trevor Wilson Apr 22 '14 at 21:49
  • Also, the following sentence "Therefore there exist uncountably many counterexamples to the claim 'you can uniquely specify any real number'" sounds suspiciously close to the statement that is refuted in Hamkins' answer. To me, the point is that for any particular notion of "specify" it's true, but if you leave it open-ended then it can't be formalized, and there is a meta-mathematical sense in which it's false. – Trevor Wilson Apr 22 '14 at 21:55

Yes. For instance,

There exists a non-measurable subset of $\mathbb{R}$ (with respect to Lebesgue measure).

It requires the axiom of choice; so we could claim that it can't really be found. Any non-constructive proof would be another possible answer. See Wikipedia on constructive proofs.

Note the interesting, related question on MO: Are there non-constructive existence proofs without the axiom of choice?

Andrew Kelley
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    I don't get this answer: if you assume the axiom of choice then a non-measurable subset of $\mathbb R$ exists and can be found; if you don't then a non-measurable subset doesn't exist – dani_s Apr 15 '14 at 21:19
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    The axiom of choice guarantees the existence of this subset, but its not necessarily constructive, so does not follow from this that this subset can be found. – Integral Apr 15 '14 at 21:32
  • How is the construction of Vitali set non-constructive? – dani_s Apr 15 '14 at 21:47
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    @Integral Depends on what you mean by "found". – Alex Becker Apr 15 '14 at 21:51
  • @dani_s See [Wikipidia on Vitali sets](http://en.wikipedia.org/wiki/Vitali_set). You must use the axiom of choice. – Andrew Kelley Apr 15 '14 at 21:51
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    My objection is that mathematically the axiom of choice isn't more "special" than other set theory axioms. You could have just said something like "There exists an infinite set but it requires axiom of infinity so it can't be found" – dani_s Apr 15 '14 at 22:06
  • @dani_s Er...it's not? It's proven to be independent from ZFC. That's pretty "special." – Kyle Strand Apr 15 '14 at 23:03
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    In any case, the "construction" of the Vitali **sets** doesn't actually allow you to uniquely specify one **single** Vitali set in such a way that you could determine, for every real number in the unit interval, whether or not that number is in the set. Thus you aren't really "constructing" anything, and in fact you haven't even uniquely defined a particular subset of the reals. – Kyle Strand Apr 15 '14 at 23:07
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    @KyleStrand And axiom of infinity is independent from the other axioms of ZF. How is being independent from ZF more "special"?The point is that the answer looks like this: some object exists, but the proof of its existance and/or its construction requires some axiom, so it can't be "found". I'm guessing this is not what the OP was looking for... – dani_s Apr 15 '14 at 23:35
  • @dani_s I meant ZF, of course. Yes, the axiom of infinity is independent from the *other* axioms of ZF, but it's still part of ZF. The axiom of choice is much less universally accepted than the axiom of infinity (see, for instance, the work done with the axiom of determinacy). But you're still ignoring the larger point that the Vitali sets aren't actually "constructed"; their "construction" is merely a proof that they exist, but it fails to characterize a single set. – Kyle Strand Apr 16 '14 at 02:26
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    @dani_s: The axiom of infinity is in fact equivalent to the statement "$\omega$ is a set". So we do have a canonical construction of a set. On the other hand, we can prove that there is no canonical construction of a Vitali set. – Asaf Karagila Apr 16 '14 at 22:16
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    I don't disagree with this answer, but I think it is worth pointing out that it is consistent with $\mathsf{ZFC}$ that there is a _definable_ non-measurable subset of $\mathbb{R}$ (for example, this holds in Goedel's constructible universe $L$.) So if $V = L$ then a counterexample can be "found" in some sense. It is also possible (meaning consistent with $\mathsf{ZFC}$) that there is no definable non-measurable set of reals, and I suppose this scenario is what the answer refers to. – Trevor Wilson Apr 17 '14 at 00:57

The question has a bit of ambiguity in it.

What does it mean that a counterexample cannot be found? What does it mean to find something in mathematics?

We might want to talk about purely physical computation (by which I mean using our current technology and reasonably conservative projections thereof; not some hypothetical future in which we have new theorems, algorithms and hardware capable of factoring prime numbers larger than the volume of the observable universe instantly). In this case the word "find" means also verify without any doubt, as well.

In this case we can easily make claims like "Every prime number larger than $2^{2^{100000000000}}$ is of the form $2^n-1$". Of course we can prove that there exists a prime number larger than $2^{2^{100000000000}}$ which is not a Mersenne prime, but computationally speaking verifying that a number which is not a Mersenne prime is prime, is an immensely difficult task.

If the above feels a bit wide, you can also plug in all the other "relatively easily verifiable prime" into the list above. The point in the above example is that there is no efficient way of verifying whether or not a certain number is prime or not; and so to verify whether or not an arbitrary number larger than $2^{2^{100000000000}}$ is a prime number, we are expected to run a very lengthy computation. We can replace "prime number" by "solution to sufficiently complicated problem" just as well.

In contrast, we can easily verify if a given number is even or not. We just need to check one bit of its binary representation (or one digit of a decimal, or hexadecimal representation), or any other "sufficiently simple problem".

If we talk about theoretically computation, it depends on what do you mean prove that a counterexample exists. In particular what do you mean by "prove"? More specifically, prove from what theory?

From $\sf PA$, the axioms of Peano arithmetic, we can develop a nice theory of computation and computability, and we can prove that many natural problems that can be solved, but not in a computable way. For example the ability to decide whether or not a sentence is true or false in the natural numbers.

In this context the term "supercomputer" can be interpreted, perhaps, as an oracle, or more specifically some "additional" function[s] that work in a way we don't necessarily know, and this helps us to solve problems that we couldn't solve before. But even then, we can prove there are always statements that are not computable (now in this stronger sense), but are provably false, so we can't "find" a counterexample.

We can extend this to mathematical questions that we can prove existence of certain objects, without our ability to give an explicit construction (read: definition) of an example. For this we usually move one up a notch and use set theory as our theory for the term "prove". Many objects, in particular infinite (and often uncountable) objects are researched in modern mathematics, and we can prove many things about these sets using an axiom known as the Axiom of Choice. This axiom is non-constructive in its nature, as it asserts the existence of certain objects, but doesn't provide us with a way to define them explicitly.

In this context we have so many examples. For example, the existence of non-measurable sets; a linear order of all the sets of sets of natural numbers; a free ultrafilter on the natural numbers.

To sign off this answer, let me point again, that this depends on what you mean "cannot be found" and what it means "prove to exist". In different contexts the answers will differ. And things which may be considered "definable" or considered "found" in one context might not be considered as such in another.

Asaf Karagila
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  • -1 Although I agree that the question is ambiguous, the [largest known prime](http://en.wikipedia.org/wiki/Largest_known_prime_number) is bigger than the proposed upper bound of $2^{10000}$. – Andrew Kelley Apr 16 '14 at 15:59
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    @Andrew: Fine. $2^{2^{100000000000}}$ it is, then. As long as you focus on the main idea of the answer, and not a petty detail, it's fine. – Asaf Karagila Apr 16 '14 at 16:00
  • Yeah, I agree; sorry about that. Since it was minor, I should have just offered an edit with my comment as an explanation. Your main point *is* explained well. +1 – Andrew Kelley Apr 16 '14 at 17:40
  • Well, no harm done. Thanks for the suggestion. – Asaf Karagila Apr 16 '14 at 18:18
  • As long as I'm nit-picking at the other answers in this thread, I'll state an objection to your argument regarding composite numbers. It seems to me that you might as well say that the statement "every number larger than $2^{2^{100000000000}}$ is odd" is false but we cannot find a counterexample. But of course we can find a counterexample: $2^{2^{100000000000}}+2$. The point is that we can express the counterexample in a compact notation. But this may be true for the "composite" version as well... – Trevor Wilson Apr 17 '14 at 01:07
  • ...it's not obvious to me that $2^{2^{100000000000}}+3^{100000} + 5$ is composite; maybe it is a counterexample? (Maybe it's not, but my point is that I don't see how you can dismiss the possibility of "finding" a counterexample like this.) – Trevor Wilson Apr 17 '14 at 01:08
  • (I agree with your main point that it depends on what we mean by "find". However, I think the definition of "find" that you use in the primality example is unreasonably narrow.) – Trevor Wilson Apr 17 '14 at 01:11
  • @Trevor: Well, "find" in the physical computational sense would require a verification. I'm not sure how you can computationally (even with a probabilistic proof) that your suggested number is prime. – Asaf Karagila Apr 17 '14 at 01:14
  • Perhaps with some clever argument. It seems to me that you are thinking of primality tests (deterministic or probabilistic) that apply to _all_ numbers. However, what if there is a compact notation that expresses a particular huge number, together with a short and clever proof that that particular huge number is prime? Why can't that happen? – Trevor Wilson Apr 17 '14 at 01:18
  • @Trevor: Of course that can happen. But you can't prove nor disprove that it can happen from this point in time. Since I cannot possibly foresee the future, I interpreted "physical computation" and "supercomputer" (in the first part of the answer) with current human knowledge. It might be that in the year $252525$ the remaining humans will find a way to tap into the Power Cosmic and use what we consider an infinite time computation without trouble. Of course this is a hyperbole, and probably less likely than your scenario. But my point remains: I can't foresee the future. – Asaf Karagila Apr 17 '14 at 01:28
  • I can't prove that my number is prime at the moment, but maybe that's just because I'm not a clever number theorist, not because I can't tap into cosmic power. My point is simply that you did not justify your statement that "computationally speaking we will never be able to find a counterexample." I think you need to explain why my "is odd" version is different from your "is composite" version. Presumably it is because primality is more complicated than evenness in some sense? Can you make this precise? – Trevor Wilson Apr 17 '14 at 01:32
  • @Trevor: Yes, exactly because of that. Let me think about it for a bit and see if I can find some reasonable way to wrap this up in an explanation. – Asaf Karagila Apr 17 '14 at 01:34
  • @Trevor: Perhaps now you can rest easily. (While we're at it, see my comment on the question regarding its tags.) – Asaf Karagila Apr 17 '14 at 02:07
  • There is also the "possibility" that we are working in an inconsistent theory, with some version of ultrafinitism being what is correct. The (silly) picture that may help understand what this could mean is one where $\mathbb N$ is "non-standard", and $\mathsf{PA}$ is inconsistent, but any proof is of non-standard length (the unorthodox ultrafinitist would say that it is unfeasibly large), so it exists, but in a concrete sense we cannot find it. – Andrés E. Caicedo Apr 17 '14 at 02:15
  • @Andres: I've been claiming for a long time that the universe is, if anything, an amorphous set. :-) – Asaf Karagila Apr 17 '14 at 02:19
  • I think your edit takes care of my objection, as long as you're only claiming that finding and verifying a counterexample is "probably" (in some fuzzy sense) intractable and not that you proved it to be intractable. I wouldn't be too surprised if you were wrong, but at least now the answer is in a form where I'd have to do some serious work to refute it (and of course I am too lazy) :-) – Trevor Wilson Apr 17 '14 at 02:49

This answers your question, maybe in an indirect way. There are a lot of probabilistic proofs in which one proves only the existence of something without the possibility of finding it.

For example, in information theory, the random coding argument is used to show that there is a code which is capacity achieving however we really cannot find any specific code using the argument. Capacity achieving codes were discovered only recently.

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  • Given the tags, I think the OP wants situations in which a counter-example provably *cannot* be found. Is that the case here? – Jack M Apr 16 '14 at 05:34
  • In OP, it is stated that "we cannot find even if we have super computers" which I interpreted as computational barrier to find it. A random code has infinite length and so the exhaustive search needs infinite computational power. – Arash Apr 16 '14 at 08:32
  • Regarding the statement "provably cannot be found", I had the same feeling as Asaf Karagila (another answer in this page). So to be really precise OP should clarify issues stated in his answer. – Arash Apr 16 '14 at 08:35

There are two opposite answers, both correct on their own terms.

One is that in most situations that have distinct concepts of "finding an example" and "existence of an example", for which finding an example always implies an example exists (but no assumption is made about the reverse implication being valid or not in general), concrete instances can be shown where something "exists" but cannot be "found". For example:

  • seeing a dead body in some unusual place and circumstance, it might be easy to demonstrate with at least 99.9% certainty that there exists a murderer of that person, but finding the murderer with 99.9% certainty of the identification is far more difficult both logically and practically.

  • a prime factorization of a large number "exists", in the sense that theories of arithmetic have that statement and its generalizations as provable theorems, but one might not be able to "find" the prime factorization if that means writing it down, or computing it efficiently, from the description of the number. Tables of factorizations of special large numbers (such as Mersenne numbers, repunits, Fibonaccis, etc) include entries like C307 to mean a 307-digit factor that is known to be composite number but without a known factorization.

  • there "exists" a first-player win in finite two-player games such as $30 \times 30$ Tic Tac Toe or Chomp that are amenable to strategy-stealing arguments, but the resources to "find" and describe the winning strategy are vast compared to the resources needed to find the existence proof, and the strategy is not known.

  • there "exists" an optimal strategy for chess, or any finite two-player game with rules that forbid unlimited repetition of positions, but to "find" the strategy requires impossibly large resources compared to what is needed to show "existence".

  • the "exists" a shortest length of LISP computer code that will print out the Baghavadad Gita, but if that length is more than $1000$, no proof system that currently exists, such as formalized Zermelo-Fraenkel set theory, could "find" that program. Doing so would require writing down a program code, proving that it works by running it until termination, and a proof that the program is minimal-length, but no such proof exists in the proof system.

The other point of view, which is approximately the "intuitionist" or "constructivist" position, is that:

  • the idea of existence proofs without any power to construct examples, is meaningless, and

  • any healthy notion of proof will either define existence proof to mean finding an example, or have the property that an abstract existence proof can always be materialized into a proof that a specific example exists (e.g., Numerical Existence Property in intuitionistic proof systems).

In the finite examples, the supposed separation appears when there is an ability to efficiently describe and operate upon elements of a data structure (such as the positions and moves in the game tree of chess), and then extrapolating from that to an imaginary superpower to exhaustively search through the whole structure, and larger structures constructed from it (such as the space of strategies for black and for white). The problem in the infinite case is similar, where it is assumed that one can casually perform an infinite search to decide questions such as whether there "exists" a largest Fermat prime. If you reject the extrapolated abilities as no more than a pleasant fairy tale, it makes sense to reject the distinction between existence and the ability to find witnesses.

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The busy beaver function has many opportunities for this sort of thing. 'In an axiomatic system of ordinary mathematics, there is a true-but-unprovable sentence of the form "Σ(10↑↑10) = n", and there are infinitely many true-but-unprovable sentences of the form "Σ(10↑↑10) < n".'

So, for the description "a counter-example exists but it cannot be found", the equality with $n$ in the quotation is such a counterexample. And it "cannot be found" in the strong sense that has been discussed in the other answers -- the counterexample exists (it's just some number) but even if you had some finite, "useful" (This word in this context baffles me.) representation of that number, you would not be able to prove the equality. This $n$ would be the greatest counterexample to the inequality with $n$ above.

Eric Towers
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  • I can prove that $\Sigma(10\uparrow\uparrow10)=\Sigma(10\uparrow\uparrow10)$. – Mario Carneiro Apr 22 '14 at 19:04
  • @MarioCarneiro: Logically, yes; arithmetically, I'm not entirely convinced. ($\Sigma$ is not a computable function. http://en.wikipedia.org/wiki/Busy_beaver#Non-computability_of_.CE.A3 ) Additionally, neither the LHS nor the RHS is a "useful" representation of that number. (I still don't know what that word would mean here.) – Eric Towers Apr 22 '14 at 20:28
  • I'm not sure what the distinction between "logical" and "arithmetical" proofs is, but I think that what it means for a sentence to be "of the form $\Sigma(10 \uparrow \uparrow 10) = n$" is that the sentence is "$\Sigma(10 \uparrow \uparrow 10) = \underbrace{1+\ldots + 1}_n$" for some $n \in \mathbb{N}$. That is, the sentence doesn't contain the symbol $n$ at all, but contains $n$ many 1's. So then I suppose Mario's comment wouldn't qualify as "finding a counterexample" because the sentence it proves doesn't have a numeral ($\underbrace{1+\ldots + 1}_n$) on the right hand side. – Trevor Wilson Apr 22 '14 at 22:24
  • @TrevorWilson: It's logically true because the same string of symbols occurs on both sides of an equivalence relation. This variety of "true" has nothing to do with the interpretation of the symbols. Example: "tree bark = tree bark" is logically true with the usual logical interpretation of the symbol "=". However, there is no way to validate that a claimed value of $\Sigma$ (with argument $\geq 4$) is actually correct -- the function is uncomputable. So it's not even recursive, so even if you had a valid value, how would you compare that value with anything else? – Eric Towers Apr 22 '14 at 22:39
  • I know why $\Sigma(10 \uparrow \uparrow 10) = \Sigma(10 \uparrow \uparrow 10)$ is true; my first point was (or should have been) that I don't see any sense in which it is _not_ true. My second point is that the first point is irrelevant: the result mentioned in your second link is (I hope) using a specific meaning of "of the form", which I think is the one that I mention, and Mario's statement is not of this form. – Trevor Wilson Apr 22 '14 at 23:05
  • @Eric I agree with Trevor's assessment. I think it is necessary for the right side to be an actual numeral, that is, a string of $1$'s of some length, for the quote from wikipedia to apply. Note also that I can prove that $\Sigma(10\uparrow\uparrow10)=\Sigma(10\uparrow\uparrow10)+1-1$ and several other things too, since $\Sigma(10\uparrow\uparrow10)$ is provably an integer. Just because a number is uncomputable doesn't mean you can't work with it like anything else. – Mario Carneiro Apr 22 '14 at 23:08
  • @Mario: This is also my position -- there is no way to compare any finite string of "$1+1+1+\cdots +1$" with the LHS. See the article section immediately folowing "Non-computability of $\Sigma$", "$\Sigma$, complexity and unprovability". The relevant axiom systems aren't powerful enough to compare these objects. (This is why I keep saying that a "useful" representation of this value is incomprehensible to me.) – Eric Towers Apr 22 '14 at 23:13

Here is a very simple wrong hypothesis for which providing a counter-example is impossible:

Every natural number has been said (written, provided, imagined) by someone.

You can't show me a counter-example, but you can prove that there are a bit more numbers than we have ever imagined.

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  • This seems more of a semantic-philosophical example, rather than a mathematical example. – Asaf Karagila Apr 18 '14 at 10:02
  • You can mathematically calculate upper limit for amount of possible imagined numbers, for example by using Planck time and size or something like that. Whatever you do, it's quite finite. – Džuris Apr 18 '14 at 10:06
  • And then you can imagine a number just slightly larger. – Asaf Karagila Apr 18 '14 at 10:10
  • @Asaf Karagila, Juris mentioned "limit for amount of possible imagined integers", not a number itself. – VividD Apr 19 '14 at 08:47
  • How can you prove an upper limit on the number of numbers that have been imagined by people? Do you have a mathematical definition of what it means for a number to be imagined by a person? – Trevor Wilson Apr 22 '14 at 22:32
  • No, I don't. I already pointed out that I'd say you can only imagine a single number at once i.e. in a Planck time. However, if you want to make this precise, you have to use numbers that have been described and ask to provide a counter-example. It's not that hard to define a described number - $n$ is a described number iff such written or told description has ever existed that the natural number $n$ fits it and no other natural number fits it. – Džuris Apr 23 '14 at 08:09