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(First, I apologize if I display any fundamental misunderstanding of how set theory works.)

I had a question regarding the limitations of ZFC (assuming its consistency, of course.) Is there any constructable set whose cardinality cannot be determined? As I can't think of a better or more rigorous definition of "constructable," let's just say use "definable in ZFC;" i.e. the real numbers can be constructed, but a hypothetical set which would disprove the continuum hypothesis, while not forbidden by ZFC, cannot be constructed.

For a more rigorous (though slightly less obviously connected) question, do constructable sets A and B exist such that there is an injection from A to B, but it cannot be determined whether there is a bijection?

Bat Dejean
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  • Constructible. – Andrés E. Caicedo Nov 29 '14 at 16:32
  • For second question, one can show there is an injection from $\aleph_1$ to $c$, but the bijection question is the continuum hypothesis. – André Nicolas Nov 29 '14 at 16:33
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    Anyway, the following is an example: Let $A$ be empty if $\mathsf{ZFC}$ is consistent, and let it be ${0}$ otherwise. If you do not like this example (because we know $A$ is empty, even if $\mathsf{ZFC}$ cannot prove it), then change in the definition of $A$ the statement "$\mathsf{ZFC}$ is consistent" with "$\mathsf{CH}$ holds". – Andrés E. Caicedo Nov 29 '14 at 16:35
  • (The point is that it is a theorem of $\mathsf{ZFC}$ that $A$ exists.) – Andrés E. Caicedo Nov 29 '14 at 16:36
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    Related: http://math.stackexchange.com/questions/814072/are-there-any-infinite-sets-that-are-not-known-to-be-either-countable-or-uncount – Asaf Karagila Nov 29 '14 at 17:40
  • @AndresCaicedo Yea, that was bugging me too. My spell check didn't like constructible, so I just went with constructable.

    Maybe a way of phrasing it to get around weird counterexamples is only consider sets made from larger sets via a rule such that, for any explicitly describable (w/e that means) element of the larger set, either it being in our set or it not being on our set is a theorem of ZFC? Honestly, the reason I'm asking is there are a few examples where I've made a set and shown it's larger than $\aleph_0$ and at most the cardinality of the continuum, (cont)

     – Bat Dejean Nov 30 '14 at 16:50
  • so if its cardinality can be determined, it has to be the cardinality of the continuum (otherwise we'd contradict the undecidability of the CH.) This is, at least, nicer than trying to find a bijection between it and something else with the cardinality of the continuum, but of course it needs to be justifiable. – Bat Dejean Nov 30 '14 at 16:52
  • We can prove the continuum hypothesis for Borel sets, and just about any set we can "explicity" describe is Borel. (For instance, the set of points where a continuous function is differentiable, the set of points where a sequence of Baire-class 2 functions converge, the set of points where a Riemann integral is differentiable and its derivative coincides with the integrand...) – Andrés E. Caicedo Nov 30 '14 at 16:54

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