Continuum hypothesis (CH) states that there can be no set whose cardinality is strictly between that of integers and real numbers. Godel, 1940 and Paul Cohen,1963 showed that CH can neither be proved nor be disproved.

How can we assert that a statement is true if it cannot be proven? What are the bases to make such assertions? I know this is a deep subject, and I confess that I don't have technical tools to understand everything on this topic. But I would like to have some understanding on why CH has to be true if it cannot be proven.

Do you have any other examples where statements are absolutely true (i.e with 100% certainty) but cannot be proved??

EDIT: Is "CH cannot be proved" the same as "proof will not exist ever in future"? Or is it that "CH cannot be proved" implies the insufficient human knowledge to prove it?

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    $\begingroup$ "...the hypothesis can neither be disproved nor be proved using the axioms of Zermelo–Fraenkel set theory." $\endgroup$
    – Pedro
    Commented Sep 15, 2013 at 4:51
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    $\begingroup$ Who says that $\mathsf{CH}$ is true? $\endgroup$ Commented Sep 15, 2013 at 4:53
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    $\begingroup$ OP you insinuate that "we assert that [CH] is true [when] it cannot be proven." Can you link to examples where established people argue CH's truth, and in what sense? You are probably laboring under a false impression. $\endgroup$
    – anon
    Commented Sep 15, 2013 at 4:59
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    $\begingroup$ @ILOVEMATH: When we say that "The continuum hypothesis is true for a family of sets" we mean to say that all sets in this family are either countable, or of size continuum. $\endgroup$
    – Asaf Karagila
    Commented Sep 15, 2013 at 5:49
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    $\begingroup$ You may want to read two nice (if somewhat involved) papers by Penelope Maddy, "Believing the Axioms", Parts I and II. J. Symbolic Logic 53 (1988), no. 2, 481-511 and J. Symbolic Logic 53 (1988), no. 3, 736-764. Part of what the articles discuss is what kind of evidence leads us to "accept" an "axiom" as true, and how one goes about deciding between conflicting statements that are proved independent of the standard axioms of set theory. $\endgroup$ Commented Sep 15, 2013 at 15:33

1 Answer 1


In mathematics, and in particular when talking about the incompleteness phenomenon, there is a grave danger of confusing the two terms "True" and "Provable".

Being "true" is a semantic property of a statement. Statements are true in a particular interpretation of the language, or a particular model of a theory. Whereas being provable is a syntactical property which means that there is a sequence statements which are axioms, or inferences from the axioms which is both finite, and its final statement is the statement we wanted to prove.

The completeness theorem tells us that $T$ proves a statement if and only if the statement is true in every model of $T$. So we sometimes abuse the meaning and say things like "Cantor's theorem is true in $\sf ZF$" when we mean that $\sf ZF$ can prove Cantor's theorem.

Sometimes, as in the case of arithmetic, there is an intended interpretation or a standard model for the theory. With the integers we have a standard model that we know to characterize. That is the model we care about. So in the context of Peano's axioms of arithmetic we say that $\varphi$ is true if it is true in the standard model, and provable if it is true in every model. However $\sf ZFC$ is far from $\sf PA$ in this context, and it does not have a standard or intended interpretation (in set theory the term "standard model" means something which is far more arbitrary than in $\sf PA$).

So finally we ask what does it mean for $\sf CH$ to be true? Since we don't have a standard model for set theory, we can't associate some "Platonic truth" to the statement in the language of set theory. Some will be true in some models, and some will be false in other models. The continuum hypothesis - and its negation - are both such statement. And when we say that the continuum hypothesis is not provable from the axioms of $\sf ZFC$, we mean that mathematically we have proved that this statement does not have a proof from the axioms. How did we do that? We showed that there models where it is true, and models where it is false (assuming there are models to begin with, of course).

So saying that $\sf CH$ is true, but not saying where is meaningless in the sense that it doesn't offer sufficient information to properly evaluate the claim. It is true in $L$, which is Godel's constructible universe (i.e. in every model of set theory satisfying the statement $V=L$ the continuum hypothesis is true). But it need not be true in other models of set theory (e.g. models of $\sf PFA$).


From the question's comments it shows that the question rose after reading a statement of the form "the continuum hypothesis is true for all practical purposes" and "the continuum hypothesis is true for Borel sets". The statement simply say that if we are only concerned about Borel sets (or some other indicated family of sets) then they are either countable, or have the cardinality of the continuum.

Historically when Cantor set to prove his continuum hypothesis, it was simply to find a bijection between open intervals and the real numbers; and using a clever method he showed that uncountable closed sets must have the cardinality of the continuum. Cantor expected that these proofs can carry on on some "complexity" of sets, and eventually cover all the sets. However the construction of taking complement and countable unions only gives us the Borel sets, and indeed shortly after leaving the Borel sets one can already run into classes which do not have to satisfy the continuum hypothesis.

Further reading:

  1. Why is the Continuum Hypothesis (not) true?
  2. Unprovable things
  3. Neither provable nor disprovable theorem
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    $\begingroup$ +5 for wonderfully lucid and judicious answer! $\endgroup$ Commented Sep 15, 2013 at 8:25
  • $\begingroup$ Thanks Peter. It means a lot coming form you! $\endgroup$
    – Asaf Karagila
    Commented Sep 15, 2013 at 10:05
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    $\begingroup$ As a layman I also found this answer very clear and exhaustive. $\endgroup$ Commented Sep 16, 2013 at 7:49

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