I've been studying some elementary set theory, semi axiomatically. And I've come across the continuum hypothesis $\mathsf{CH}$ which states that there is no cardinal between integers and reals. So if im not mistaken the negation of the hypothesis is that there is a subset of real numbers which is uncountable and it's not equipotent to real numbers set. I've also read that godel has shown that $\mathsf{CH}$ is consistent with the current set theory axiom system so that means that he indirectly showed that there is no subset of reals with such property. Because if we could find a subset of reals with such properties then $\mathsf{CH}$ would be inconsistent with the axiom system. So doesn't that mean that godel has indirectly proven that the $\mathsf{CH}$ is true?

On the other hand I've read that cohen has proven that we can't prove $\mathsf{CH}$ false either so it means using the axioms we can't prove that there is no such set neither can we prove that there is such a set. But from all that we know that with the current axioms like specification and power axioms we can't create a subset of reals with such properties.

So what's exactly unsolved about this?

  • 5
    $\begingroup$ "I've also read that godel has shown that The $\mathsf{CH}$ is consistent with the current set theory axiom system so that means that he indirectly showed that there is no subset of reals with such property." That claim is false. All we can extract from Godel's argument is that there is no subset of the reals which $\mathsf{ZFC}$ can verify has the property in question. Indeed, Godel's arguments do not (indirectly or otherwise) imply $\mathsf{CH}$. $\endgroup$ May 20 at 19:37
  • 2
    $\begingroup$ As a very rough analogy, consider the axioms for a group, which as you'll recall do not include commutativity of the group operation. To continue the analogy and ignoring obvious examples and history, we can imagine that mathematicians wondered whether we can prove commutativity, disprove commutativity, or is it possible that commutativity can both hold and fail (for different groups). The group analogy of Godel showed there exist commutative groups, so we know that commutativity doesn't contradict the other axioms. The group analogy of Cohen showed (continued) $\endgroup$ May 20 at 19:50
  • 1
    $\begingroup$ @Aria Assuming ZFC and only ZFC, it is provably impossible to prove the CH, and it is provably impossible to disprove (provide a counterexample for) the CH. It's not that such a subset both exists and doesn't. It is that we can't prove that such a subset exists and we can't prove that it doesn't. $\endgroup$
    – Arthur
    May 20 at 20:06
  • 2
    $\begingroup$ We haven't proven that there is no counterexample. Just that ZFC is too weak to construct one (or otherwise prove its existence). And in the other direction, ZFC is also too weak to prove that such a set can't exist. $\endgroup$
    – Arthur
    May 20 at 20:32
  • 1
    $\begingroup$ It's always good to try and fish out the red herring from an argument. The answer below suggests geometry, but you can also try something simpler, e.g. "$\sqrt2$ exists" and the axioms of a field; the axiom of a group and commutativity; the empty language and "there is exactly one thing" vs. "there are at least two things". All perfectly fine examples of independence. $\endgroup$
    – Asaf Karagila
    May 20 at 21:58

1 Answer 1


Of course none of these arguments take place in "the real world of sets". They take place in a certain system of axioms (called ZF, let's say).

Gödel's proof goes like this. Given any model of ZF: inside that model we can construct another model of ZF where CH holds. From this new model it follows that $\neg$CH cannot be proved from ZF. This is what it means to say "CH is consistent with ZF".

Something similar to this was done much earlier. Inside Euclidean plane geometry, we can construct a model of hyperbolic (non-euclidean) plane geometry: The Klein model. From this we know that the parallel postulate cannot be proved from the other axioms for Euclidean geometry.

Study this example to get an idea where your argument above goes wrong.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.