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Assume that CH is false. Then, there exists a counterexample; that is, there exists some set $A$ such that there exist an injection $f\colon \Bbb N\to A$ and an injection $g\colon A\to\Bbb R$, but there are no surjections of those kinds. By providing such a counterexample, we could actually prove that CH is false. Since CH is proven to be independent of ZFC, this is impossible. Thus, a counterexample cannot exist, which means CH is true.

Is this argument flawed? I've seen a similar one brought up in the context of the Goldbach conjecture. But CH probably is not equivalent to any $\Pi_1^0$ sentence (proof?), so this reasoning doesn't apply, I think.

Elvis
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    A set $A$ with the desired property exists independently of CH: Take $A = \mathbb{N}$. Perhaps you mean something else? – Ben Steffan Nov 03 '24 at 23:27
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    How is this any different from a proof that all groups are commutative? If a group is not commutative, then there is an example of that, so we can prove that all groups are not commutative, but that's not provable from the axioms of a group. Therefore there can't be a counterexample. There all groups are commutative. – Asaf Karagila Nov 03 '24 at 23:55
  • There doesn't need to be a proof it's not equivalent a $\Pi_1^0$ sentence to answer the question (just needs to be the case that there is no proof that it is equivalent), but here's one way we can know that, anyway: CH's truth value can be toggled by forcing, whereas the truth value of first-order arithmetic statements cannot. – spaceisdarkgreen Nov 04 '24 at 00:07
  • @RobArthan I know, and I did not write that. – Elvis Nov 05 '24 at 14:00
  • @spaceisdarkgreen: I misunderstood the way that Elvis had stated CH. – Rob Arthan Nov 05 '24 at 19:57
  • @AsafKaragila but why does that analogy not apply to the case where GC is proven independent of PA, for example? Or to RH, whose independence would imply that it's true as well? – Elvis Nov 05 '24 at 23:45
  • @Elvis: When we say that "if RH is independent, then it is true", here truth is relative to the standard model of Peano Arithmetic. And independent means "from PA". This is because we have a notion of "default truth" for arithmetic which does not exist for set theory, or group theory (in the sense that statements in the language of groups are not considered true exactly when they hold in one specified group). – Asaf Karagila Nov 06 '24 at 22:36

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Yes, the argument is flawed. If CH is false, there exists a counterexample, but there is no guarantee that such a counterexample can be proven to be a counterexample. We might not be able to explicitly specify a counterexample, or even if we can specify it we might not be able to prove that it is a counterexample.

The differences between this and Goldbach are:

  1. Every natural number can be written down with a finite number of symbols. But here, only countably many subsets of $\mathbb R$ can be explicitly written down, because there are only countably many strings in a given alphabet.
  2. Given a natural number, we can check in a finite number of steps whether it is the sum of two primes. But here, there is no finite algorithm that can take a description of a set and compare its cardinality to $\aleph_0$ and $\mathfrak c$.
Robert Israel
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  • This is an answer to the question in the title. The body of the question badly needs attention. I think you should have pointed that out. – Rob Arthan Nov 04 '24 at 23:57