Consider the rigid relation ($RR$) principle, i.e.

"every set admits a rigid binary relation", that is,"that for every set $A$ there is a binary relation $R$ on $A$ such that the structure $(A,R)$ is rigid, meaning that it has no nontrivial automorphisms" ("that is, no bijection function $\pi$: $A$$\rightarrow$$A$ such that $aRb$ iff $\pi$$(a)$$R$$\pi$$(b)$, other than the identity function"). (Hamkins and Palumbo, arXiv: 1106.4635v1 [mathLO]).

In that paper, Hamkins and Palumbo prove that $AC$$\Rightarrow$$RR$, and that $ZF+{\lnot}AC+RR$ and $ZF+{\lnot}AC+{\lnot}RR$ are relatively consistent with $ZF$.

Question 1: What role (if any) does $RR$ play in the proof of the Kunen inconsistency?

Question 2: Since $ZF+{\lnot}RR$ is relatively consistent with $ZF$, is $NGB+{\lnot}AC+{\lnot}RR$ ($RR$ appropriately defined for $NGB$--call it $RR_{NGB}$) relatively consistent with $NGB$?

Question 3: Can the Kunen inconsistency be proved in $NBG+{\lnot}AC+{\lnot}RR_{NGB}$?

  • 1
    $\begingroup$ For question 3, I would have expected that you inquire whether the Kunen inconsistency can be proved in $\text{NBG}+\text{RR}$, that is, by weakening the use of $\text{AC}$ in the usual proofs to use merely $\text{RR}$ instead. This version of the question would amount to: can you prove the Kunen inconsistency using only the rigid relation principle instead of the axiom of choice? Is that what you had had in mind, or did you really intend to ask question 3 as you have stated it? $\endgroup$ Apr 6 '15 at 13:58
  • $\begingroup$ @JDH: Actually, I had both in mind, but asked the related question because,if I understand correctly, Kunen used the Erdos-Hajnal theorem to prove that in KM, the only elementary embedding j:$V$$\rightarrow$$V$ is the identity (Thm 1 of Kunen (1971)), which suggested to me that $RR$ is certainly needed in the proof. I asked question 3 because I thought that if the Kunen inconsistency could be proved from $NGB+{\lnot}AC+{\lnot}RR$, it would be a very nontrivial result. $\endgroup$ Apr 6 '15 at 17:02
  • $\begingroup$ My knowledge of the subject is quite narrow, but out of curiosity, I'd like to ask @JDH about the relationship between RR and the prime ideal axiom. It seems that at the time of publication of the mentioned paper, it was unknown whether one implies the other (or both). Is it still the case ? $\endgroup$ Apr 7 '15 at 20:53
  • $\begingroup$ @MathieuBaillif, Unfortunately, this question is still open. Why not give it a shot? There may still be some easy pickings about it... $\endgroup$ Apr 8 '15 at 2:18
  • $\begingroup$ Thanks for your answer. I'd love to give it a shot, but the time I have at disposal (and my scarce knowledge) makes it quite unlikely, unfortunately. $\endgroup$ Apr 8 '15 at 9:45

Claim: NGB + $\exists j\colon V\to V$ is equiconsistent with NGB + $\exists j\colon V\to V + \neg RR$.

In order to prove the left to right direction, we will start with a model of NGB + $\exists j\colon V\to V$ and take the symmetric extension that as described in Hamkins and Palumbo's paper. By the results of this paper, in the symmetric extension there is a set that witnesses $\neg RR$. Note that the poset, as well as the groups and the filter, that are used in the symmetric extension are all very low in the cumulative hierarchy (say in $V_{\omega + \omega}$). Therefore, they are fixed by the embedding $j$.

Let $W$ be the symmetric extension and let's extend $j$ to $\tilde{j}\colon W\to W$. We construct $\tilde{j}$ as in the proof of Levy-Solovay's theorem. For a hereditarily symmetric name $\dot{x}$, let's define $\tilde{j}(\dot{x}^G) := j(\dot{x})^G$.

Since $j$ does not move the forcing notion, $\mathbb{P}$, the groups of automorphisms of the forcing, $\mathcal{G}$, and the filter, $\mathcal{F}$, $j(\dot{x})$ is a hereditarily symmetric name (being hereditarily symmetric is a first order property with parameters $\mathbb{P}, \mathcal{G}, \mathcal{F}$), and therefore $j(\dot{x})\in W$. $\tilde{j}$ is elementary, by exactly the same argument of Levi-Solovay: using the forcing theorem (every statement in the generic symmetric extension is forces by some condition in the generic filter), and the fact that $j$ doesn't move the forcing or the conditions, $p\Vdash \varphi(\dot{x})$ iff $j(p) = p\Vdash \varphi(j(\dot{x}))$, by the elementarity of $j$ in $V$.

Moreover, the class $\dot{j} = \{(\dot{x}, j(\dot{x})) \mid \dot{x} \text{ is hereditarily symmetric name}\}$ (when writing the pairs as canonical names of pairs) is a hereditarily symmetrical name for a class (it is fixed by itself by any automorphism in the group, and every member of it is a pair of hereditarily symmetric names). So $\tilde{j}$ is a class of $W$.

  • $\begingroup$ It should be remarked that this absolutely does not generalize to $\sf ZF$. The reason is that here $j$ was a class in $V$; so it was easily extended to a class of the symmetric extension. On the other hand, in $\sf ZF$ we don't even know if the ground model is necessarily a class of the symmetric extension. So the fact that a class of names is stable under automorphisms does not necessarily means that it will be a class of the symmetric model. (In fact, this is probably the biggest open problems in symmetric extensions.) $\endgroup$
    – Asaf Karagila
    Apr 7 '15 at 20:50
  • $\begingroup$ Two remarks: (1) $fix(\dot x)$ is usually reserved for the pointwise stabilizer, and $\operatorname{sym}(\dot x)$ is more appropriate. But this doesn't matter. (2) You need to require in the last paragraph that $\dot x$ is hereditarily symmetric name, not just that its stabilizer is in the filter, otherwise it cannot possibly be a class of $W$ (since it will include objects not in $W$). $\endgroup$
    – Asaf Karagila
    Apr 7 '15 at 20:53
  • $\begingroup$ Let me nitpick again, and point out that the definition $j(\dot x^G)=j(\dot x)^G$ requires more justification: (1) $G$ is usually not in $W$, so you argue about a definition in $V[G]$. Then by the fact that the class itself is symmetric we have that it is a class of $W$ and it works out. (2) Again with hereditarily symmetric names. It's not sufficient just that $\operatorname{sym}(\dot x)=\operatorname{sym}(j(\dot x))$, you also need to ensure that all the names which appear in $j(\dot x)$ are already hereditarily symmetric. But this is a first-order property (with parameters) so it works too. $\endgroup$
    – Asaf Karagila
    Apr 7 '15 at 22:06
  • $\begingroup$ @Asaf: Yes, you're right. I edited the answer. $\endgroup$
    – Yair Hayut
    Apr 8 '15 at 8:12
  • $\begingroup$ @YairHayut: Your answer is very interesting but I need some clarification (forgive my ignorance). When you claim that $NGB+{\exists}j:V{\rightarrow}V$ is equiconsistent with $NGB+{\exists}j:V{\rightarrow}V+{\lnot}AC+{\lnot}RR$ does your model show that $j$ is nontirvial under ${\lnot}RR$? Also, since Prof. Hamkins, in his partial answer to me shows that $NGB$ is also equiconsistent with $NGB+{\lnot}AC+RR$, can you show that $NGB+{\exists}j:V{\rightarrow}V$ is equiconsistent with $NGB+{\exists}j:V{\rightarrow}V+ {\lnot}AC+RR$ with $j$ being only the identity in that model? Thanks. $\endgroup$ Apr 9 '15 at 11:19

Question 2 is an immediate consequence of the theorems in my paper with Justin.

Specifically, in that paper, we prove in theorem 5 that if there is a model of $\text{ZF}$, then there is a model of $\text{ZF}+\neg\text{AC}+\neg\text{RR}$. If one augments this model with the collection of definable classes (definable from parameters), then one arrives at a model of $\text{NGB}$, which has exactly the same sets as the original model, and therefore also $\neg\text{AC}+\neg\text{RR}$. So $\text{ZF}$ is equiconsistent with $\text{NGB}+\neg\text{AC}+\neg\text{RR}$. Doing the same thing with the model of theorem 6 shows that it is also equiconsistent with $\text{NGB}+\neg\text{AC}+\text{RR}$, as well as with $\text{NGB}+\text{global choice}$.

Meanwhile, I don't know whether one can prove the Kunen inconsistency using merely $\text{RR}$ in place of $\text{AC}$, but I find this to be a very interesting idea. (Note, however, that this is not actually what you asked, although I find it to be in the same spirit as what you asked.)

  • $\begingroup$ @JDH: That was implied by question 3. Does $RR$ imply the Erdos-Hajnal theorem used by Kunen in the original paper? $\endgroup$ Apr 6 '15 at 14:24
  • $\begingroup$ Your question 3 is about $\neg\text{RR}$, rather than $\text{RR}$, and this is the point of my comments about it. I don't know if we can get Erdos-Hajnal from $\text{RR}$. It would also suffice to get Ulam's theorem on stationary partitions, since this can be used to prove the Kunen inconsistency as in Woodin's argument (see jdh.hamkins.org/generalizationsofkuneninconsistency). $\endgroup$ Apr 6 '15 at 14:31

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