56

The Continuum Hypothesis say that there is no set with cardinality between that of the reals and the natural numbers. Apparently, the Continuum Hypothesis can't be proved or disproved using the standard axioms of set theory.

In order to disprove it, one would only have to construct one counterexample of a set with cardinality between the naturals and the reals. It was proven that the CH can't be disproven. Equivalently, it was proven that one cannot construct a counterexample for the CH. Doesn't this prove it?

Of course, the issue is that it was also proven that it can't be proved. I don't know the details of this unprovability proof, but how can it avoid a contradiction? I understand the idea of something being independent of the axioms, I just don't see how if there is provably no counterexample the hypothesis isn't immediately true, since it basically just says a counterexample doesn't exist.

I'm sure I'm making some horrible logical error here, but I'm not sure what it is.

So my question is this: what is the flaw in my argument? Is it a logical error, or a gross misunderstanding of the unprovability proof in question? Or something else entirely?

RothX
  • 1,593
  • 12
  • 20
  • 5
    The flaw is in your understanding. That you cannot construct a counter example in ZF does not mean that within ZF, you can construct a proof of CH. Confusion between the object language and the metalanguage, the language used to talk about ZF, is the flaw. – William Elliot Sep 26 '17 at 01:44
  • @WilliamElliot I can believe that you can't construct a different proof using ZF, my confusion is how the fact that you can't construct a counterexample isn't a proof on its own. – RothX Sep 26 '17 at 01:52
  • @martin.koeberl I don't think my question is very similar to that one. I tried to edit it, but I feel I am already as clear as I could be. Why do you think my question seems similar to that one? – RothX Sep 26 '17 at 01:57
  • 8
    As Asaf's answer gets at, you seem to be assuming that there is One True Set Theory and that CH is innately either true or false; but what the proofs construct is _different_ set theories which settle the question differently. – Steven Stadnicki Sep 26 '17 at 02:19
  • 9
    It might help to think of the parallel postulate of (Euclidean) geometry; neither it nor its negation can be proven from the other postulates because there are models in which each is valid. – Steven Stadnicki Sep 26 '17 at 02:32
  • 1
    There is no theorem of ZF (nor of ZFC) that you can't construct a counter-example to $CH$ (unless $1=0$ happens to be a theorem of ZF.) The main result is that if ZF (or ZFC) is consistent then neither $CH$ nor $\neg CH$ are theorems of ZF (or ZFC) – DanielWainfleet Sep 26 '17 at 02:39
  • 1
    @RothX. That fact was proven outside ZF about what can and cannot be proven inside ZF. That fact is not a theorem of ZF. ZF talks only about sets; it does not talk about statements, proofs nor well formed formulas. That is what the metalanguage talks about. – William Elliot Sep 26 '17 at 02:56
  • 1
    Interestingly enough, the Goedel sentence for PA is a statement of the form "There is no number $n$ that has property $P(n)$" and indeed, since it is undecidable in $PA$, it is true, very much along the lines you state here. However there is a big distinction. $P(n)$ is a computable property here so we know that $PA$ would actually be able to prove it if it existed (although PA does not know this!). That a set has no bijection with the naturals or reals is not such an easy thing to prove that we should be confident ZFC could do it were it true. – spaceisdarkgreen Sep 26 '17 at 02:57
  • 1
    @spaceisdarkgreen: Also important is the fact that PA has a notion of "true" predicated on the existence of a canonical model. The same cannot be said about ZFC. – Asaf Karagila Sep 26 '17 at 03:01
  • This question reminds me of [Wildberger's comment](http://web.maths.unsw.edu.au/~norman/papers/SetTheory.pdf): "If you have an elaborate theory of ‘hierarchies upon hierarchies of infinite sets’, in which you cannot *even in principle* decide whether there is anything between the first and second ‘infinity’ on your list, *then it’s time to admit that you are no longer doing mathematics.*" – Wildcard Sep 26 '17 at 04:39
  • @RothX. Because that fact is not a theorem of ZF, so you cannot use it to prove CH within ZF . True and provable are two different concepts. – William Elliot Sep 26 '17 at 06:09
  • 10
    @Wildcard: What a silly comment... Let me know when you figure out how many numbers are between the fifth and sixth Ramsey numbers. – Asaf Karagila Sep 26 '17 at 07:49
  • 1
    @Wildcard "If you have an elaborate theory of 'lines' in which you cannot *even in principle* determine the number of lines parallel to a given line which pass through a given point not on the line, *then it is time to admit that you are no longer doing mathematics.*" ? – John Coleman Sep 26 '17 at 09:41
  • 1
    @AsafKaragila, no one wonders whether there is or isn't a 5 1/2th Ramsey number, so the comparison doesn't really work, interesting as the question is. And John Coleman, I dare you to find a geometry that seriously calls itself a geometry and assumes **only** the first four Euclidean axioms. (The further difference being that geometry has an observable relevance to the real physical universe and infinite cardinalities do not.) But none of that is relevant to my statement that this question *reminds me* of Wildberger's comment—and it doesn't surprise me it's now closed as off-topic. :P – Wildcard Sep 26 '17 at 09:57
  • 2
    @Wildcard: Interesting. So you're claiming that the $n$th Ramsey number is exactly $n$? But sure, let's make it as explicit as humanly possible. When Mr. Wildberger, or you for that matter, find out exactly which natural numbers lie between the fifth and sixth Ramsey numbers, then I will hear more about how CH makes set theory "not mathematics". Or maybe I'll just ask about the interval between the sixth and seventh instead... who knows. – Asaf Karagila Sep 26 '17 at 10:14
  • 1
    @AsafKaragila, I think your response proves why this question is off-topic, as Wildberger's comment is likewise off-topic. There is a distinct parallel to be drawn between this question and Wildberger's comment. Which, you might notice, prompted me to state the question *reminds me* of his comment. I find your assumption of my agreement with his comment to be humorous. That comment itself is silly, although the fundamental point he makes in the broader context of his paper is not—but you don't do yourself any favors drawing nonsensical parallels. Is the $n$th infinite cardinality exactly $n$? – Wildcard Sep 26 '17 at 10:29
  • 3
    @Wildcard: I disagree. This is a question of someone not as familiar with logic. Wildberger is just... Let's not finish this comment. – Asaf Karagila Sep 26 '17 at 10:31
  • @Wildcard: Did you know there are (finite) systems of integer polynomial equations that you cannot, *even in principle*, determine whether they have any solutions? –  Sep 26 '17 at 16:01
  • @Hurkyl again a false parallel. Given a claimed solution, it is not only possible but trivial to check whether it is or is not a solution. So there's nothing "in principle" to stop you. Any given number either is or is not a solution no matter how tricky or complex the polynomial. There's no *general method* of solution for some types but again that's an irrelevant non-parallel. Please do tell me an example of such an equation if you truly meant your comment literally, though. – Wildcard Sep 26 '17 at 17:26
  • 2
    @Wildcard: There *is* a general method of solution for any type of Diophantine equation that can be proven to have (or not have) solutions -- simply check every possible string of symbols to see if it expresses a proof of solution or a proof there are none. The interesting phenomenon here is that even this algorithm will fail for some Diophantine equations. –  Sep 26 '17 at 17:48
  • @Hurkyl can you please give an example, or a link to such? – Wildcard Sep 26 '17 at 17:51
  • @Wildcard: This is the subject of the negative answer to Hilbert's tenth problem, and [Matiyasevich's theorem](https://en.wikipedia.org/wiki/Diophantine_set#Matiyasevich.27s_theorem). I have seen formulas explicitly written down in the past; they weren't enlightening, though. My google-fu is failing me currently. –  Sep 26 '17 at 17:58
  • 2
    I don't want to compete with the good answers here, but there's a short quippy answer: proof and truth value are not the same thing. We often forget this, because we're so good at proving things true or false, and if you prove it one way, its truth value is that way. However, if something is unprovable, that doesn't mean it is either true or false. It states nothing of the truthood of the predicate. – Cort Ammon Sep 26 '17 at 18:13
  • @Hurkyl, thanks for the link. You wrote "...even this algorithm will fail for some Diophantine equations." This is false. That naive algorithm will never fail, but it can only be used to prove that an equation *does* have a solution; if it hasn't halted yet, we don't know that there is or isn't a solution. A *solution* to a Diophantine equation is a well-defined object; given a *supposed* solution we can easily check whether it truly is one or not. There is no "solution" to infinite cardinalities as (a) they are not equations, and (b) relative cardinality is ill-defined for infinite sets. – Wildcard Sep 26 '17 at 22:04
  • Counter-question: As $CH$ can't be proven, we cannot find a counter example to $\neg CH$, i.e. an injection $\mathbb{R} \to \omega_1$. So actually $\neg CH$ is true? – Jonathan Schilhan Sep 27 '17 at 10:15
  • The first comment (by William Elliot) reminds me of the fact that something can be known to everyone, and yet not be common knowledge. (https://en.wikipedia.org/wiki/Common_knowledge_(logic)) – Mike Jones Sep 29 '17 at 14:36

5 Answers5

94

Here's an example axiomatic system:

  1. There exist exactly three objects $A, B, C$.
  2. Each of these objects is either a banana, a strawberry or an orange.
  3. There exists at least one strawberry.

Let's name the system $X$.

Vincent's Continuum Hypothesis (VCH): Every object is either a banana or a strawberry (i.e., there are no oranges).

Now, to disprove this in $X$, you would have to show that one of $A, B, C$ is an orange ("construct a counterexample"). But this does not follow from $X$, because the following model is consistent with $X$: A and B are bananas, C is a strawberry.

On the other hand, VCH does not follow from $X$ either, because the following model is consistent with $X$: A is a banana, B is a strawberry, C is an orange.

As you can see, there is no contradiction, because you have to take into account different models of the axiomatic system.

Vincent
  • 1,892
  • 11
  • 20
  • 21
    I love this answer for it's immediate accessibility for wider public. – Maciej Sep 26 '17 at 08:39
  • Great answer. It was so hard to pick which one to accept. This one is very useful, but it doesn't answer my question as directly as the one I accepted. – RothX Sep 26 '17 at 22:24
  • 6
    I'm pretty sure I've been searching for an example just like this for years and just never knew how to ask for it. – KDecker Sep 27 '17 at 13:01
  • 1
    This *must* be the canonical example of MSE going *bananas*. – Mikhail Katz Sep 27 '17 at 17:04
  • 1
    Is this actually how the real CH was proven independent of ZFC? Did we construct models of ZFC with CH and !CH? If so, what framework did we construct those models in? – ASKASK Oct 04 '17 at 08:09
55

I think the basic problem is in your statement that "In order to disprove it, one would only have to construct one counterexample of a set with cardinality between the naturals and the reals." Actually, to disprove CH by this strategy, one would have to produce a counterexample and prove that it actually has cardinality between those of $\mathbb N$ and $\mathbb R$.

So, from the fact that CH can't be disproved in ZFC, you can't infer that there is no counterexample but only that no set can be proved in ZFC to be a counterexample.

Andreas Blass
  • 65,963
  • 4
  • 76
  • 146
  • 4
    If you're the kind of person who believes there is a standard model of ZFC which is the "real" universe of sets, then I think this is exactly it: you believe that CH is either true or not, but ZFC alone does not tell you which. – Ben Millwood Sep 26 '17 at 12:28
  • 4
    I like this answer because it points out the specific problem in my logic rather than giving a general lesson on how axioms work. I assume that showing CH is not disprovable means showing there is no counterexample, when really you would show that no set can be proved to be a counterexample. – RothX Sep 26 '17 at 13:56
  • This [paper](https://arxiv.org/abs/0809.4144) seems related, but it is to difficult for me. What do you think about it? Where is it wrong? – Hjan Sep 26 '17 at 19:01
  • 5
    @Hjan, as a rule of thumb you can disregard anything claiming to disprove Cantor's theorem :P After a glance it looks like the claim "When N is large, counting the number of nodes is the same as counting the number of paths (to infinity) in the tree." in Proposition 2.2 is incorrect – Kai Rüsch Sep 26 '17 at 21:04
  • The real reason why it's impossible to derive CH from ZFC is that ZFC says nothing about continuumness and uncountability. One would need to define that smallest uncountably long string contains every countable substring to define CH. – rus9384 Sep 27 '17 at 12:41
  • @rus9384: So how do you explain the fact that V=L implies not just CH, but GCH? – Asaf Karagila Sep 28 '17 at 06:39
  • @AsafKaragila, so, if smallest uncountable string (if one does, they all do) contains every countable, we have GCH? – rus9384 Sep 28 '17 at 07:32
  • @rus9384: I have no idea what you're talking about. Or how that has to do with GCH. – Asaf Karagila Sep 28 '17 at 08:29
  • @AsafKaragila, what are V and L? – rus9384 Sep 28 '17 at 08:51
  • @rus9384: V is the common notation for the set theoretic universe; L is Gödel's constructible universe. The axiom V=L means that all sets are in L. This can be expressed via a single, finite, axiom. – Asaf Karagila Sep 28 '17 at 08:54
  • @AsafKaragila, don't we have that every real can be expressed as a string with point? – rus9384 Sep 28 '17 at 09:00
  • @rus9384: I don't see how that has to do with anything. – Asaf Karagila Sep 28 '17 at 11:00
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/66326/discussion-between-rus9384-and-asaf-karagila). – rus9384 Sep 28 '17 at 11:05
  • 1
    @rus9384: Let's not. Have a nice day. – Asaf Karagila Sep 28 '17 at 11:05
32

How would you prove that not all groups are abelian? Namely, how would you prove that the axioms of groups do not imply $\forall x\forall y(xy=yx)$?

Well. You'd find a model for the theory, namely a group, which is not abelian. At the same time, you can also show this is not disprovable by finding an abelian group.

So, does that mean that all groups are abelian, because the axiom $\forall x\forall y(xy=yx)$ is independent of the axioms of a group? No. It does not.

The independence of the continuum hypothesis from ZFC goes along the same lines. We can show there are models of ZFC where CH holds, and others where it fails. This is regardless to the techniques we use, which are clever and useful for so much more in set theory.

Asaf Karagila
  • 370,314
  • 41
  • 552
  • 949
  • So the idea is that ZFC describes a system where CH is true as well as a system where CH is false? – RothX Sep 26 '17 at 13:54
  • 1
    Well, there's some issue with "true" and "false" here. You should perhaps read https://math.stackexchange.com/questions/494099/why-is-ch-true-if-it-cannot-be-proved/494104#494104 where I wrote about that in length. – Asaf Karagila Sep 26 '17 at 14:04
  • 1
    @RothX That sounds correct, but I'd phrase it a bit more carefully. There are multiple different systems that are described by ZFC; and among those systems which are described by ZFC, there is at least one system where CH is true, as well as at least one system where CH is false. – Tanner Swett Sep 26 '17 at 17:21
  • @TannerSwett Yeah, I think that's pretty much what I was trying to say. I just need to get used to using mathematical language more. – RothX Sep 27 '17 at 11:38
8

The situation is really rather similar to that of non-Euclidean geometry.

From the first four Euclidean postulates, you can neither prove nor disprove the fifth (parallel) postulate. This has been proven only two millennia after Euclid wrote his Elements. If you assume the fifth postulate is true, you get Euclidean geometry; if you assume it is false, you get elliptic or hyperbolic geometry. The common ground to both is absolute geometry.

Similarly, from ZFC, you can neither prove nor disprove the continuum hypothesis. If you assume it is true, you get one kind of set theory; if you assume it is false, you get another. The common ground to both is ZFC.

Richard Bos
  • 81
  • 1
  • 2
    Correction: elliptic geometry is inconsistent with absolute geometry. If you replace the parallel postulate with its negation, you get hyperbolic geometry. – eyeballfrog Sep 26 '17 at 08:31
1

CH: "There is no set with cardinality between that of the reals and the natural numbers."

disprove CH by: demonstrating such a set

You can't demonstrate such a set. Therefore, you can't disprove it.

To prove it, you have to show that ALL sets have cardinality:

  1. less than or equal to natural numbers -or-
  2. greater than or equal to real numbers

You can't prove it either.

Therefore, you can neither prove nor disprove CH.

The basic problem is the unfounded assumption that you can prove something by simply showing that you can't disprove it. You'll have to prove that one first.

Actually, the continuum hypothesis itself demonstrates that you can't, by virtue of "you can't prove it" above.

allwind
  • 9
  • 2
  • `It isn't necessarily true, I think.` - to anyone with a background more in theoretical computer science, there is no "I think" here; reasoning about proving is not so far away from reasoning about computability, and assuming something is true only because you cannot prove that it is false is a very basic fallacy. – AnoE Sep 27 '17 at 17:01
  • fine. Removed "I think". But in response, I AM an algorithm, therefore "I think.." is still valid within the domain of theoretical computer science. But that is of course not the subject of this thread. – allwind Sep 27 '17 at 17:27
  • You can think whatever you want, and so can I, that's the beauty of it. :D (My comment was not related to a vote in either direction). – AnoE Sep 27 '17 at 17:50