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Many logicians and philosophers believe that all sentences expressible in the language of Peano Arithmetic have determinate truth-values, even though no nice formal system can capture all of these truths. It is more controversial whether the Continuum Hypothesis has a determinate truth value. I am wondering whether the determinacy of arithmetic could potentially have consequences for the determinacy of the Continuum Hypothesis as expressed in the following two mathematical questions below. Using the fact that any arithmetic sentence (in the language of Peano Arithmetic) can be translated into a corresponding sentence in the language of set theory, we have:

(1) Is it the case that for every model M of ZFC, there exists a model N of ZFC such that M and N agree on all arithmetic truths (suitably interpreted in the language of set theory) yet disagree on the truth value of the Continuum Hypothesis?

If the answer to (1) is "No", what is the answer to the following much stronger question:

(2) Is it the case that for every model M of ZFC, any model N of ZFC that agrees with M on all arithmetic truths also agrees with M on the truth value of the Continuum Hypothesis?

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  • $\begingroup$ I may be wrong, but I believe that Skolem's paradox should imply that the determinacy of CH is no more controversial that the determinacy of PA. $\endgroup$
    – gamma
    Commented Nov 5, 2014 at 4:10
  • $\begingroup$ I'm aware of Skolem's paradox, but I don't see why it has the implication: If PA is determinate, then CH is determinate. (Is that what you are trying to say? Or are you saying the weaker claim that the above conditional is made plausible by the paradox? And if so, why?) $\endgroup$
    – Taro
    Commented Nov 5, 2014 at 4:13
  • $\begingroup$ @Nick: I see nothing about Skolem paradox here. The Skolem paradox is that a countable model of set theory has uncountably many reals, but it isn't really a problem since the model and the universe (i.e. us) disagree on the notion of "uncountable". $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 5, 2014 at 4:15
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    $\begingroup$ @Nick: I'm too tired to laugh. I just got out of bed 15 minutes ago. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 5, 2014 at 4:25
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    $\begingroup$ Incidentally, Joel David Hamkins has an interesting paper questioning whether arithmetical truth is really that definite. You might want to check it out: jdh.hamkins.org/… $\endgroup$
    – Nagase
    Commented Nov 8, 2014 at 19:00

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The answer to (1) is positive.

If you have a model of $\sf ZFC$, then you can just as well assume it is countable; otherwise you can take a countable elementary submodel, and these would have to agree on arithmetic truths. Now that you have a countable model of $\sf ZFC$ you can force with $Add(\omega^M,\omega_2^M)$ to create a model which have the same ordinals where the continuum hypothesis is false, or with $Add(\omega_1^M,1)$ which makes the continuum hypothesis true again.

Either forcing doesn't change the arithmetic truths of the model, since those are absolute between models with the same ordinals. So we can flick the continuum hypothesis like a switch, on and off, without changing anything about the first-order nature of the natural numbers.

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