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The Continuum Hypothesis hypothesises

There is no set whose cardinality is strictly between that of the integers and the real numbers.

Clearly this is either true or false - there either exists such a set, or there does not exist such a set.

Paul Cohen proved that the Continuum Hypothesis cannot be proven or disproven using the axioms of ZFC.

If we find a set whose cardinality lies strictly between that of $\mathbb{N}$ and $\mathbb{R}$, then we are done, we have disproven it. But it has been proven that we cannot disprove it, thus by contrapositive, we cannot find such a set. If we cannot find such a set, then we can only conclude that such a set does not exist (if it did exist, there must be a non-zero probability that we would find it, so given enough time we would - contradiction. □)

So I have proven that the Continuum Hypothesis is true - there does not exist such a set. But this is a contradiction because it has been proven that we cannot prove it either. So where did I go wrong?

Thanks!

Superbus
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    "if it did exist, there must be a non-zero probability that we would find it, so given enough time we would" - wrong. – user2357112 Apr 15 '14 at 03:15
  • Possibly a duplicate, definitely related: http://math.stackexchange.com/questions/189471/why-is-the-continuum-hypothesis-not-true/ – Asaf Karagila Apr 15 '14 at 05:24
  • Provability and validity are not same concept. – Hanul Jeon Apr 15 '14 at 10:42
  • @tetori that statement very much intrigues me. Can you point me to some reliable resources which further expand on this statement? – Andrew Falanga Apr 15 '14 at 17:21
  • @AndrewFalanga I think some mathematical logic textbooks explain that point, but I don't know it for certain. – Hanul Jeon Apr 16 '14 at 05:56
  • In first-order logic, these two concepts are same by soundness theorem and completeness theorem. However, in another logic, there is no certain these two concepts are same. – Hanul Jeon Apr 16 '14 at 05:57

6 Answers6

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Cohen's result is that from a certain set of axioms ($\mathsf{ZFC}$) we cannot prove the continuum hypothesis. Gödel's result is that from the same set of axioms, we cannot refute the continuum hypothesis. This only means that the set of axioms under consideration is not strong enough to settle this question. If you manage to exhibit a set of intermediate size, your argument necessarily uses axioms beyond those in $\mathsf{ZFC}$, or is not formalizable in first-order logic. Similarly if you manage to show that there is no such set. It is worth pointing out that there are standard axioms beyond those in $\mathsf{ZFC}$ that settle the continuum problem. These axioms are not universally accepted yet, but this illustrates that, as explained, their results are not absolute, but relative to a very specific background theory.

Andrés E. Caicedo
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    These axioms are not universally accepted yet this suggests that there is a tendency that they will be so. Is this the current state of affairs? – Git Gud Apr 15 '14 at 03:06
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    Not quite. There are reasonable and fairly well-understood extensions of $\mathsf{ZFC}$ in the form of large cardinal axioms. These extensions are used routinely by the majority of set theorists, and many consider them part of the standard theory. Unfortunately, these additional axioms do not settle $\mathsf{CH}$ either. Beyond large cardinals, however, we are far from consensus. A well studied set of extensions takes the form of strong reflection principles, that decide $\mathsf{CH}$ (negatively). There are also extensions in other directions that prove $\mathsf{CH}$. (Cont.) – Andrés E. Caicedo Apr 15 '14 at 03:15
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    I find the picture of the universe with strong reflection principles compelling for a variety of reasons, but this does not mean I quite believe they are part of the true theory of the universe. In fact, the current multiverse proposals suggest that there is no one true universe of sets, and so issues such as the absolute status of $\mathsf{CH}$ are somewhat moot. These proposals are fairly recent, so we need time to appreciate them better and decide more objectively on their inclusion in our standard picture. – Andrés E. Caicedo Apr 15 '14 at 03:20
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    Thank you, this is very interesting. – Git Gud Apr 15 '14 at 03:22
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    I've heard it said that accepting axioms in set theory is analogous to modifying the parallel postulate in geometry - no one set is "correct", merely different theories. – RghtHndSd Apr 15 '14 at 03:22
  • @rghthndsd There is a hint of realism in all this. We are trying to describe an independent reality. Sure you can consider a mix of axioms, but they might not describe that reality. The goal is to describe it. – Git Gud Apr 15 '14 at 03:25
  • @rghthndsd Some set theorists indeed take this approach. Others disagree vehemently. For instance, most set theorists would argue that $V\ne L$, regardless of the fact that $V=L$ is consistent with $\mathsf{ZFC}$. – Andrés E. Caicedo Apr 15 '14 at 03:30
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"But it has been proven that we cannot disprove it" is not true. What has been proven is that you can't prove or disprove it within ZFC, or as you put it, "using the axioms of ZFC." You could easily prove it -- with a one-line proof, in fact! -- by adjoining to ZFC a new axiom, namely the Continuum Hypothesis.

(In other words, you need to pay closer attention to what it means to prove something within a specific formal system. In metalogic, theorems are relative to what axioms you start with. It makes no sense to speak of proving or disproving something, tout court.)

You'd also be rather hard-pressed to explain to a probabilist what probability measure allows you to say "if it did exist, there must be a non-zero probability that we would find it."

EDIT: you may find this paper by Solomon Feferman, entitled "Does mathematics need new axioms?", a fairly gentle introduction to some of the larger issues at stake here, which Andres has sketched in his answer.

symplectomorphic
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A nice analogy is this: from the axioms which define a ring, it is impossible to prove that multiplication is commutative. "Preposterous!", you might claim. "Take your ring to be the integers! Now, multiplication is repeated addition, and addition is commutative because we are working in a ring, and so perhaps by induction or maybe something simpler, multiplication is commutative!"

But if instead we use matrices as our ring, I can find you some rather obvious counterexamples. It turns out that the axioms which define a ring are not strong enough to prove or disprove the commutativity of multiplication—we can have rings where multiplication is commutative, and other rings where it is not. Similarly, we can construct worlds according to the axioms of ZFC in which the continuum hypothesis holds, and others in which it does not, hence proving that the ZFC axioms are not strong enough to prove the continuum hypothesis.

Mitchell Spector
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crf
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  • FWIW, this doesn't answer the question. The question was: "where did I go wrong?" not "explain what independence means." – symplectomorphic Apr 15 '14 at 03:16
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    The answer to "where did I go wrong" is that OP took "Clearly this is either true or false - there either exists such a set, or there does not exist such a set" to be true, and this demonstrates that that's not so obvious. – crf Apr 15 '14 at 03:17
  • Fair enough, but it might have been useful to say as much, since that was the question. – symplectomorphic Apr 15 '14 at 03:21
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Clearly this is either true or false - there either exists such a set, or there does not exist such a set.

This is the crux of the issue here. Gödel's completeness theorem tells us that a theory T proves a sentence $\varphi(\overline{x})$ if and only if $\varphi(\overline{x})$ holds in every model of T. The problem here is that if ZFC is consistent, then it has an enormous number of models, while you've implicitly asserted that there is some specific model, some structure, which is ZFC. This is simply not the case. How Cohen's proof works is that he shows that given some model of ZFC, you can both construct a model that has a new set that violates CH, or you can create a model that no longer believes it has such a set.

Travis Nell
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  • I think you are mixing things up a bit. The OP is taking the position that there is a true universe of sets, regardless of whether we have identified a theory that axiomatizes its main features. You are arguing that they are equating the theory of the universe with $\mathsf{ZFC}$, which does not quite seem to be what they are doing. Besides, "there is some model which is $\mathsf{ZFC}$" makes no sense. Models are structures, not theories. Perhaps you meant to say that "there is a model whose theory is precisely $\mathsf{ZFC}$". As you mention, this is indeed false. – Andrés E. Caicedo Apr 15 '14 at 03:26
  • True I conflated a theory with a model here. However I think this illustrates the real difficulty in the OP's question, which is saying that there being some "true" universe of sets says anything about what is provable in ZFC. – Travis Nell Apr 15 '14 at 06:21
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Another problem is that because just because a set exists doesn't mean you can find it.

Logan Tatham
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You have a mistake here: "if we cannot find such a set, then we can only conclude that such a set does not exist".

That's not right. The fact that you can't find a set does not mean that it does not exist. That's like the planets located beyond the observable universe: you can't prove if there are some people like us over there or not (or whatever), even showing a particular case, because you can't ever look at these planets, and this does not mean they do not exist.

The Gödel-Cohen demonstration states that starting from these axioms you will never be able to assert that 'there is at least a set -that I don't know which one it is- that has a strict intermediary cardinality', and that you'll never be able to assert that such a set cannot exist. That is: no 'deductive' prove can ever state that it exists or that it doesn't. So, there is no need to try this kind of demonstration. But the question about this hypothesis remains always open (that's why it is a hypothesis): there could exist such a set, or there could not. If it exists and you can write it, you just can't prove its cardinality with the only help of the Zermelo-Fraenkel axioms. It could exist but no one could prove it. Why not?

This would not necessarily apply starting from a different (and not equivalent) set of axioms. But the 'logic world' that create these axioms (or any other axiomatic system that includes them) is always uncomplete: not every true statement can be proven. The "provable statements" is a subset of the "true statements". There is also an "observable universe" limit in maths. That's what Gödel demonstrated.

Andrés E. Caicedo
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user143522
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