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I'm making my way through Thomas W Hungerfords's seminal text "Abstract Algebra 2nd Edition w/ Sets, Logics and Categories" where he makes the statement that the Continuum Hypothesis (There does not exist a set with a cardinality less than the reals and no set strictly greater than the natural numbers.) is neither true or false.

This is utterly baffling to me, If it's possible to construct a set between $\mathbb{N}$ and $\mathbb{R}$ then this statement is demonstrably false, but if not then the statement is true.

This seems to be a straitforward deduction, but many with a more advanced understanding of the topic matter believe CH to be neither.

How can this be?

ŹV -
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    A related previous question: ["Impossible to prove" vs "neither true nor false"](http://math.stackexchange.com/q/154398/856) –  Sep 01 '12 at 00:12
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    This type of thing makes me like intuitionistic logic more. – Tunococ Sep 01 '12 at 00:14
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    "If it's possible to construct a set between $\mathbb{N}$ and $\mathbb{R}$ then this statement is demonstrably false, but if not then the statement is true." The predominant attitude among modern mathematicians is that math doesn't have to be constructive: that we can prove things exist without constructing them explicitly, and that such a proof is satisfactory. It then becomes possible to have things that you can prove exist, and can also prove can't be constructed explicitly; this is the situation, e.g., with ultrafilters. Independence of CH from ZFC says we can't rule this out for CH. –  Sep 02 '12 at 14:54

7 Answers7

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Set theory is much more complicated than "common" mathematics in this aspect, it deals with things which you can often prove that are unprovable.

Namely, when we start with mathematics (and sometimes for the rest of our lives) we see theorems, and we prove things about continuous functions or linear transformations, etc.

These things are often simple and have a very finite nature (in some sense), so we can prove and disprove almost all the statements we encounter. Furthermore it is a good idea, often, to start with statements that students can handle. Unprovable statements are philosophically hard to swallow, and as such they should usually be presented (in full) only after a good background has been given.


Now to the continuum hypothesis. The axioms of set theory merely tell us how sets should behave. They should have certain properties, and follow basic rules which are expected to hold for sets. E.g., two sets which have the same elements are equal.

Using the language of set theory we can phrase the following claim:

If $A$ is an uncountable subset of the real numbers, then $A$ is equipotent with $\mathbb R$.

The problem begins with the fact that there are many subsets of the real numbers. In fact we leave the so-called "very finite" nature of basic mathematics and we enter a realm of infinities, strangeness and many other weird things.

The intuition is partly true. For the sets of real numbers which we can define by a reasonably simple way we can also prove that the continuum hypothesis is true: every "simply" describable uncountable set is of the size of the continuum.

However most subsets of the real numbers are so complicated that we can't describe them in a simple way. Not even if we extend the meaning of simple by a bit, and if we extend it even more, then not only we will lose the above result about the continuum hypothesis being true for simple sets; we will still not be able to cover even anything close to "a large portion" of the subsets.


Lastly, it is not that many people "believe it is not a simple deduction". It was proved - mathematically - that we cannot prove the continuum hypothesis unless ZFC is inconsistent, in which case we will rather stop working with it.

Don't let this deter you from using ZFC, though. Unprovable questions are all over mathematics, even if you don't see them as such in a direct way:

There is exactly one number $x$ such that $x^3=1$.

This is an independent claim. In the real numbers, or the rationals even, it is true. However in the complex numbers this is not true anymore. Is this baffling? Not really, because the real and complex numbers have very canonical models. We know pretty much everything there is to know about these models (as fields, anyway), and it doesn't surprise us that the claim is true in one place, but false in another.

Set theory (read: ZFC), however, has no such property. It is a very strong theory which allows us to create a vast portion of mathematics inside of it, and as such it is bound to leave many questions open which may have true or false answers in different models of set theory. Some of these questions affect directly the "non set theory mathematics", while others do not.


Some reading material:

  1. A question regarding the Continuum Hypothesis (Revised)
  2. Neither provable nor disprovable theorem
  3. Impossible to prove vs neither true nor false
Asaf Karagila
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    It's always fun when everyone writes a short answer and I find myself writing a short story! :-) – Asaf Karagila Sep 01 '12 at 00:59
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    This answer is illuminates a murky subject matter with exceptional clarity and wisdom. Thank you! – ŹV - Sep 01 '12 at 01:16
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    "...the real and complex numbers have a very canonical model. We know pretty much everything there is to know about these models..." except for whether they have any subsets with cardinality strictly between that of the integers and the reals. And where to find the zeros of the zeta function. – Gerry Myerson Sep 01 '12 at 01:21
  • @Gerry: I thought the rational numbers had cardinality strictly between the integers and the reals! :-) – Asaf Karagila Sep 01 '12 at 01:25
  • @AsafKaragila: The :-) suggests you know this, but the rationals are as countable as the integers. You can, for example, encode the bits of the numerator in the odd bits of an integer and the bits of the denominator in the even bits of the integer. This will cover all rational numbers. – Mark Hurd Sep 01 '12 at 07:52
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    @Mark: I think that writing a +17 answer about independence of CH from ZFC should hint that I am *probably* aware to that, among other things such as the `:-)`, however for the general point for those who might be unaware of this fact: thanks. – Asaf Karagila Sep 01 '12 at 07:57
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    @AsafKaragila: Yes, perhaps I should have introduced that differently :-) But as this is Math.SE, not MO, I think the clarification was worth it. – Mark Hurd Sep 01 '12 at 08:02
  • @zyx My intention was Borel sets. This is also true for analytic sets. Further than that in the hierarchy requires further assumptions. – Asaf Karagila Sep 01 '12 at 19:32
  • In ZF without choice, is CH phrased as you did, with "uncountable", or as "cardinality greater than $\aleph_0$"? – dfeuer Sep 20 '13 at 13:09
  • @dfeuer: The two ways are actually equivalent. To see this, if there are no Dedekind-finite sets of reals, then it's vacuously equivalent; if there is $X$ which is Dedekind-finite then $X\cup\Bbb N$ is a counterexample which is Dedekind-infinite. Note that the real numbers cannot be written as the sum of a Dedekind-finite set and a countable set (because removing a countable set from the real numbers doesn't change cardinality, whereas here we can find a countable set whose removal ends up with a strictly smaller set). – Asaf Karagila Sep 20 '13 at 13:35
  • @dfeuer: More generally, however, there are many different ways to formulate the continuum hypothesis without the axiom of choice. I prefer Cantor's original "there is no intermediate cardinality". See my answer [here](http://math.stackexchange.com/questions/404807/how-to-formulate-continuum-hypothesis-without-the-axiom-of-choice/404813#404813) for more examples and details. – Asaf Karagila Sep 20 '13 at 13:36
  • I'm taking my first Discrete and Foundational Math class this semester. I was presented this hypothesis. One thing I'm not entirely certain of: to what is "Continuum" referring? – Andrew Falanga Apr 21 '14 at 19:44
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    @Andrew: "Continuum" is a term for the real numbers. It has other, more general meanings, but in this context it simply means the real numbers. – Asaf Karagila Apr 21 '14 at 19:50
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    How does this answer answer the question? The question was how the continuum hypothesis could be neither true nor false, not how it could be neither provable nor disprovable. Since it's easy to prove in Zermelo-Fraenkel set theory that it's either true or false, maybe a better answer would explain the existence of non-Boolean systems where you can't prove that it's either true or false. This question couldn't have meant how can the continuum hypothesis be undecidable in such a non-Boolean system because there's no reason to be sure it's decidable. – Timothy Oct 25 '17 at 03:47
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    @Timothy: The term "true" is meaningless without a context. And since set theory does not have a default context like arithmetic, just saying something is true or false is meaningless. Sure, we often use "true in ZFC" to mean "provable from ZFC", but that is an abuse of language, and not a correct way of stating it. (Twice, in two different places I made such abuse of language in a talk I gave, and twice I was paused and asked for clarification.) – Asaf Karagila Oct 25 '17 at 05:47
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A number of mathematicians have definite opinions about the truth of CH, the majority I believe opting for false, Kurt Gödel among them. What there is agreement on, because it is a theorem, is that CH is neither provable nor refutable in ZFC. But that is quite a different assertion than "neither true nor false."

The theory ZFC captures many common intuitions about sets. It has been the dominant "set theory" for many years. There is no good reason that it will remain that forever.

André Nicolas
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    Doesn't one have to say, "neither provable nor refutable in ZFC, provided ZFC is consistent"? – Gerry Myerson Sep 01 '12 at 01:16
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    @GerryMyerson: Yes. But the axioms (at least of ZF) are true in the universe of sets, and the relative consistency of ZFC is a theorem. So I thought that in the context of a question about truth, the obligatory "if ZFC is consistent" could be dispensed with. – André Nicolas Sep 01 '12 at 01:20
  • Would you know (trusted estimates of) the proportion of mathematicians with *definite opinions about the truth of CH*, and amongst them, the proportion of those *opting for false*? – Did Sep 01 '12 at 14:23
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    Should probably not have said mathematicians. For set theorists, one can make lists. The pro-CH list would be short. – André Nicolas Sep 01 '12 at 17:58
  • WP has a nice summary of this: http://en.wikipedia.org/wiki/Continuum_hypothesis#Arguments_for_and_against_CH –  Sep 02 '12 at 14:46
  • @AndréNicolas Understood. (But please use the @ thing.) – Did Sep 03 '12 at 09:45
  • @BenCrowell Thanks for the link. The author of the WP page seems to use *mathematicians* for *mathematicians with definite opinions about the truth of CH*. – Did Sep 03 '12 at 09:46
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One can construct a model of set theory in which CH is true, and one can construct a model in which CH is false.

Gerry Myerson
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It is not possible to explicitly "construct" such a set and prove (using the ZFC axioms of set theory) that its cardinality is strictly between those of $\mathbb N$ and $\mathbb R$. That doesn't mean that no such set exists.

As a Platonist, I would not say that CH is "neither true nor false", rather that we do not know (and in a certain sense we cannot know) which it is. Truth and provability are very different things.

Robert Israel
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If you take the parallel axiom away from Euclidean geometry, you cannot prove (using the remaining axiom system) whether it is true or false. But even in a geometry without parallel axiom, you can have interesting results (see http://en.wikipedia.org/wiki/Absolute_geometry ).

Landei
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    Maybe add a sample of said *interesting results*. – Did Sep 01 '12 at 11:39
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    So you added a link to a WP page on Absolute geometry. Fine. But could you be more specific about **interesting results**? – Did Sep 02 '12 at 12:53
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    WP mentions the first 28 propositions from Euclid's Elements, the exterior angle theorem and the Saccheri-Legendre theorem. Maybe this isn't *terrible* interesting, but I guess good enough to get my point across. It's just an analogy, dude... – Landei Sep 02 '12 at 13:18
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I beleive that your confusion rise from bad definition of CH. It is not "there does not exist a set with a cardinality less than the reals and no set strictly greater than the natural numbers" as you stated it, but rather "there does not exist a set with a cardinality less than the reals AND strictly greater than that of the natural numbers.".

Eran
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(1)To make your wording more accurate, I assume you mean "there exists" when you state it as "It's possible to construct."

(2)Truth and Falsity in logic might have a different meaning than you think. Truth and Falsity here can only be talked within a scope, which is given by ZFC axiom system. Under the consistency assumption of ZFC, the system is incomplete, meaning that there are statements there not having a T/F answer.

(3) When people say CH "is neither true or false" in this context, they really just mean that such Truth or Falsity cannot be deduced from ZFC system. More precisely, they mean that under the consistency assumption of ZFC, if you add CH or its negation to ZFC, the system remains consistent, and therefore CH and its negation cannot be deduced from ZFC.

(4) In a larger system, it is possible to give a definitive answer to CH. As a trivial example, if we add CH to ZFC, then CH would be true.