0
$\begingroup$

To my understanding, the Continuum Hypothesis cannot be proven true or false under ZFC. But to me this is very ambiguous as to the actual provability of CH.

Does this mean that even with infinite attempts, no person or computer could ever define a set that disproves the Continuum Hypothesis? Do we know if that is even possible?

If we do not know, do we know if it is possible to prove whether or not we can know? How much do we know about what we can know about this problem?

I am not trying to ask strictly as an answer to that hypothetical. I am asking in order to understand more about what it specifically meant by the -non-provability of the Continuum Hypothesis.

$\endgroup$
6
$\begingroup$

Assuming the consistency of ZFC, there is a model of ZFC in which CH holds, and there is a model in which CH fails.

Think of this. There is a group which has infinite order, and there is a group which has finite order. Therefore the hypothesis that "$G$ is infinite" is independent of the axioms of group theory: there are models which fail the hypothesis, and there are models which satisfy the hypothesis.

That's the sense of "unprovable" we're talking about here.

$\endgroup$
1
$\begingroup$

Another example, also from Group Theory but easier: Is every Group abelian? The integers with ordinary addition are an abelian group, however the 2x2 Matrices with real entries and ordinary matrix multiplication is a non-abelian group.

So we found different models for the group axioms, in wich different first order sentences hold; namely the first order sentence asserting commutativity holds in one whereas not in the other structure. This seems very unproblematic to us, the group axioms alone do not determine groups completely, i.e. They do not completely determine wich sentences hold in groups.

Set Theory is to be a foundation for all of mathematics, and this ideal science is to be completely understandable, other then the chaotic empirical world, or so our philosophical prejudice goes. Now Gödel ofc showed this to not be the case. For the realm Of set theory no theory ought to provide a complete picture of all the sets.

In the case of groups we can add commutativity as an axiom and thus only study abelian groups, in the case of set theory we can just as well add CH as an axiom and henceforth study only models of our set picture that satisfie CH, but we can as well add not CH as an axiom. Simply put: Our pictureabout sets in general is not complete and will never be.

$\endgroup$
  • 1
    $\begingroup$ The first paragraph is false. $0$ has no multiplicative inverse, and similarly all the singular matrices. For the matrices to be a group under multiplication you need to consider only the collection of those with an inverse, first. $\endgroup$ – Asaf Karagila Apr 12 '16 at 17:15

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.