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Does there exist a statement P in the language of first-order arithmetic, such that P is is a theorem of ZFC, and not P is a theorem of ZF + not C? In other words, can the question of whether the axiom of choice is true or not have an effect on what the truths of first-order arithmetic are? What about if we replace first-order arithmetic with higher-order arithmetic?

If there exists no such statement, then does there exist any statement X independent of ZF, for which there is a statement P in first-order arithmetic such that P is a theorem of ZF + X and not P is a theorem of ZF + not X? (Or replace ZF with ZFC.) Of course there are examples like Con(ZF) and not Con(ZF), but I'm looking for statements X such that neither ZF + X nor ZF + not X is obviously unsound. Or to put it another way, I want a statement which we're not obliged to believe or disbelieve if we accept the soundness of ZF (or ZFC).

Now the continuum hypothesis can be written as a statement of third-order arithmetic, and it's independent of ZFC. Does it and its negation have contradictory consequences for first-order arithmetic? Do they have contradictory consequences for second-order arithmetic?

Any help would be greatly appreciated.

Thank You in Advance.

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MO question: mathoverflow.net/questions/139376/… – Martin Sleziak Aug 14 '13 at 5:40
    
There is no way to merge this with the question on MO. At best, we could cross-link the questions. That's been done here, but I don't know if MO would go for that. – robjohn♦ Aug 15 '13 at 7:58
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If $P$ is a statement in first-order arithmetic then $P$ is absolute between $V$ and $L$. This means that $P$ is true in $V$ if and only if it is true in $L$. Since the axiom of choice is true in $L$, this means that the answer is negative. Similarly for the continuum hypothesis, which is always true in $L$.

The key theorem here is Shoenfield's absoluteness theorem. This theorem also holds for "simple enough" second-order statements.

As for the statement $X$ which might exist, note that $V=L$ decides most "natural looking" statements of set theory (at least those which don't require a greater consistency strength, like large cardinals). So the same logic applies here as well. If we do want to talk about large cardinals, then note that you can always consider $V_\kappa$ and $L_\kappa$ for the least $\kappa$ that these are models of $\sf ZF$. The natural numbers are there, and it is easy to see that the same statements are true in these smaller universes, which do not contain large cardinals, as in the full universe.

On the other hand, there are number theoretical statements which can be taken as a set theoretical axioms. For example $\newcommand{\con}{\operatorname{Con}}\con(\sf ZFC)$ and $\lnot\con(\sf ZFC)$ are such statements. If there is a model of $\sf ZFC$ in the universe, then $\con(\sf ZFC)$ is true, but otherwise it is false. Moreover if $\sf ZFC+\con(ZFC)$ is consistent then we can find a model of $\sf ZFC+\lnot\con(ZFC)$. Note, however, that in a model of $\sf ZFC+\lnot\con(ZFC)$ there are non-standard integers (standard ones cannot encode a proof of the contradiction from $\sf ZFC$).

Also relevant:

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What about less natural-looking statements? What about any statements at all whose truth or falsity doesn't follow from the truth of the axioms of ZF? Also, is there an analogue of Shoenfield's absoluteness theorem for the first-order theory of real numbers, i.e. the theory of real closed fields? Is it possible for different set theories to have different consequences for that? – Keshav Srinivasan Aug 14 '13 at 1:36
    
1. What sort of less natural-looking statements? I'm having hard time thinking about such statements. Perhaps you mean things like $\text{Con}(\sf ZFC)$ or something like that. It might be possible, however, I don't see really how. 2. The theory of real-closed fields is complete. I'm not sure if you need the axiom of choice to prove that (and model theory without choice can go a bit out of the window), but I don't see how two extensions of $\sf ZFC$ can disagree on that. But don't hold me for that last statement. – Asaf Karagila Aug 14 '13 at 1:49
    
1. Yes, I was thinking of things like Con(ZFC). Obviously Con(ZFC) and its negation are contradictory statements in first-order arithmetic, but we can easily say that Con(ZFC) is true because ZFC is sound, so that's a trivial example. So I was trying to find an example whose truth-value isn't a consequence of the soundness of ZFC. 2. Sorry, I forgot that the theory of real-closed fields was complete. What about if we changed "first-order theory of real numbers" to "second-order theory of real numbers"? Then can different set theories have contradictory consequences? – Keshav Srinivasan Aug 14 '13 at 2:05
    
Note that $\text{Con}(\sf ZFC)$ is unprovable from Peano's axioms. Clearly the number theoretic statement "$\sf ZFC$ is consistent" can be changed between some models of $\sf ZFC$ (if $\sf ZFC+\rm Con(\sf ZFC)$ is consistent then so is $\sf ZFC+\lnot\rm Con(\sf ZFC)$). But this is not a theorem of $\sf PA$. As for the second-order theory of the real numbers, it's not hard to formulate the continuum hypothesis, or the well-orderability of $\Bbb R$ using second-order logic. Clearly both depend on the set theory. – Asaf Karagila Aug 14 '13 at 2:13
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To complement Asaf's answer: The continuum hypothesis has no effect on statements of second order arithmetic. One can force $\mathsf{CH}$ without adding reals. This means that any model of $\lnot\mathsf{CH}$ has the same theory of the reals as a model of $\mathsf{CH}$, so $\mathsf{CH}$ can be eliminated from any proof of such a statement. – Andrés E. Caicedo Aug 14 '13 at 3:52

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