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(This question is inspired by deleted question, and the questioner who write the deleted question wrote new question.)

It is well-known that consistency of "ZF + DC + every set of reals are Lebesgue measurable" is equiconsistent with some large cardinal axiom. On the other hand, ZFC proves existence of non-Lebesgue measurable set. So if some large cardinal axiom is consistent with ZFC, then the consistency of "ZF + DC + every set of reals are Lebesgue measurable" is undecidable within ZFC.

But, above result requires consistency of large cardinal axiom. I want to find a such example which does not need a stronger axiom (in the sense that their consistency strength exceed that of ZFC.) Can we find a sentence $\varphi$ such that both the consistency of $\mathsf{ZF+\mathsf{Con}(ZF+\varphi)}$ and $\mathsf{ZF+\lnot\mathsf{Con}(ZF+\varphi)}$ is a theorem of ZFC+Con(ZFC)? Thanks for any help.

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Hanul Jeon
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1 Answers1

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No, this is not possible.

Assume there exists a model $M$ of ZF which is also a model of $Con(ZF + \varphi)$. This means that there is no integer in $M$ that codes a proof of $\lnot \varphi$ from $ZF$. Since every integer in "our" universe has a corresponding integer in $M$, none of "our" integers can code such a proof. Therefore $Con(ZF + \varphi)$ is true.

(EDIT: Alternatively, $M$ has within it what it believes to be a model $N$ of $ZF + \varphi$. But $N$ must then really be a model of $ZF + \varphi$.)

Thus $Con(ZF + Con(ZF + \varphi))$ implies $Con(ZF + \varphi)$. Therefore the situation you're speaking of can't arise.

EDIT: The statement in Tetori's problem may need to be changed as follows in order for my arguments to be conclusive. Thanks to GME for pointing out the error. However, even this formulation turns out to be trivially true for another reason.

Assume that ZFC + Con(ZFC) is consistent. Can there be a sentence $\varphi$ such that both of the following are theorems of ZFC + Con(ZFC)?

  1. Con(ZF + Con(ZF + $\varphi$)).
  2. Con(ZF + Con(ZF) + $\lnot$Con(ZF + $\varphi$)).

Then the answer is no, at least for the reasons above, as well as the fact that for any model of ZF, there is a model of ZFC with the same integers.

Tetori's original question had the following instead of 2.

2'. Con(ZF + $\lnot$Con(ZF + $\varphi$)).

2' is a consequence of Con(ZF) by Gödel's incompleteness theorem, so in this case, the question reduces to whether 1 can ever be a theorem of ZFC + Con(ZFC). But I don't think it can! Perhaps someone else knows the answer to this. So even the question as I reformulated it probably isn't all that interesting.

I think it probably would have been appropriate to replace ZFC + Con(ZFC) in the original question with ZFC + Con(ZFC + Con(ZFC)). My head is beginning to hurt, so I won't say anything else about this for now, but I think this might fix things.

user187373
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  • Why is it problematic that ZFC + $Con(ZFC)$ proves $Con(ZF + \phi)$? If $\phi = AC$, this would be fine. Thanks in advance! –  Oct 27 '14 at 08:26
  • Well, the consistency of $ZF + \lnot Con(ZF + \varphi)$ cannot then be true, contrary to the requirements of Tetori. – user187373 Oct 27 '14 at 12:15
  • Sorry for being dense, but why not? –  Oct 27 '14 at 12:27
  • Doesn't $Con(ZF + \neg Con(ZFC))$ follow from $Con(ZFC)$ by Godel's second? –  Oct 27 '14 at 12:35
  • @GME You're right. I've made "corrections" above. Careful thought needs to be given to the correct formulation of the original problem, and I don't have an immediate answer. – user187373 Oct 27 '14 at 19:03
  • I think you're right about 1. If $\phi = AC$, then by appealing to $L$, 1 implies $Con(ZFC + Con(ZFC))$. So if ZFC + $Con(ZFC)$ proved 1, it would prove $Con(ZFC + Con(ZFC))$ which is impossible if it is consistent. That was the first thought I had about this. –  Oct 27 '14 at 19:08