11
$\begingroup$

I asked this question about two weeks ago on MSE and haven't gotten an answer, so I thought I would post the question here.

Do there exists sentences which are independent of ZFC, cannot be shown to be independent through some method of forcing, and do not increase the consistency strength of ZFC (e.g. so Large Cardinal Axioms are out)?

If there does exist such a sentence, I would love to know a concrete example. One with a combinatorial flavor would be ideal.

Edit: Feel free to increase the consistency strength of ZFC (say ZFC+ $\Delta$) in the "meta-"sense (i.e. work where you want to work). Does there exist a "non-forcible" independent sentence that does not increase the consistency strength of ZFC?

$\endgroup$
  • 1
    $\begingroup$ Arithmetical statements can be a way to go - iirc, forcing cannot change truth of any arithmetic statement. $\endgroup$ – Wojowu Feb 28 '15 at 21:08
  • $\begingroup$ @Wojowu: Do you know where I could find something on this? $\endgroup$ – Kyle Feb 28 '15 at 21:12
  • $\begingroup$ No, sorry - I know this just as a "well-known result", but I don't know a reference. $\endgroup$ – Wojowu Feb 28 '15 at 21:12
  • $\begingroup$ A related question: Consistency results using nonstandard models $\endgroup$ – Mohammad Golshani Mar 1 '15 at 5:57
  • $\begingroup$ What Wojowu said is an immediate result of Shoenfield Absoluteness Theorem. $\endgroup$ – Zoorado Mar 5 '15 at 9:56
13
$\begingroup$

The Gödel-Rosser sentence $R$ for $\text{ZFC}$ is an arithmetic assertion, such that $\text{ZFC}$ is equiconsistent with $\text{ZFC}+R$ and with $\text{ZFC}+\neg R$. So the Rosser sentence does not increase consistency strength. Since arithmetic assertions are preserved by forcing, one cannot use forcing directly to prove the independence of $R$.

Another example would be $\neg\text{Con}(ZFC)$, since $\text{ZFC}$ is equiconsistent with $\text{ZFC}+\neg\text{Con}(\text{ZFC})$, and so this doesn't increase consistency strength. (the theory $\text{ZFC}+\text{Con}(\text{ZFC})$, in contrast, does have strictly higher consistency strength). So this is an arithmetic assertion that is independent of $\text{ZFC}$, assuming that $\text{ZFC}$ is consistent, but this is not possible to prove in any direct way by forcing, since forcing does not affect arithmetic truth.

$\endgroup$
  • $\begingroup$ The Rosser sentence is a great example. The second one, not as much. $\endgroup$ – Asaf Karagila Feb 28 '15 at 23:11
  • $\begingroup$ @AsafKaragila What's wrong with the second one? $\endgroup$ – Will Sawin Mar 1 '15 at 0:11
  • 1
    $\begingroup$ @Will: It's sort of cheating, since $\operatorname{Con}(\mathsf{ZFC})$ does in fact increase the consistency strength of the theory, whereas its negation does not. It feels a bit like cheating. $\endgroup$ – Asaf Karagila Mar 1 '15 at 0:19
8
$\begingroup$

If we consider $ZF$ instead of $ZFC$, then we can say more. There are examples of such results which are obtained by Krivine, using his method of realizability. For example in the paper Realizability algebras II: New models of ZF+DC the following is stated:

Using the proof-program (Curry-Howard) correspondence, we give a new method to obtain models of $ZF$ and relative consistency results in set theory. We show the relative consistency of $ZF + DC$ + there exists a sequence of subsets of $\mathbb{R}$ the cardinals of which are strictly decreasing + other similar properties of $\mathbb{R}$.

As it is stated in the introduction of the paper:

These results seem not to have been previously obtained by forcing.

see also 50 years after forcing, the Curry-Howard correspondence gives new models of ZF

$\endgroup$
  • $\begingroup$ Not forcing, but easily with symmetric extensions which can be argued as part if forcing, or at least in the context of this question forcing and inner models. (And if we allow forcing over $\sf ZF$ directly, I'm not sure this is not obtainable with just forcing under some minor assumptions.) $\endgroup$ – Asaf Karagila Mar 1 '15 at 7:09
  • $\begingroup$ @AsafKaragila Yes, it might be possible to get this result using symmetric models (though I am not aware of it), and what I wrote above is taken from the abstract of the paper. But note that the model he constructs has more properties (where in the above abstract it is written as "other similar properties of $\mathbb{R}$"). $\endgroup$ – Mohammad Golshani Mar 1 '15 at 7:47
  • $\begingroup$ Also in the slide introduced above, it is mentioned that the above methods present the first new models of $ZF$, fifty years after Cohen’s forcing (of course I personally rather work with Cohen's forcing!!!). $\endgroup$ – Mohammad Golshani Mar 1 '15 at 7:49
  • $\begingroup$ I'm not quite sure what are "other similar properties of $\Bbb R$", but I didn't spend too much time looking through the paper (I did look at it, several times in the past as well). I'm not saying that it's not impressive, I'm just saying that it's a bit of cheating if you hide the symmetries in your construction and claim this to be a completely new result unobtainable by forcing. That being said, these are indeed impressive results and an interesting approach. $\endgroup$ – Asaf Karagila Mar 1 '15 at 8:04

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.