Timeline for Sufficient Condition for Defining $\in$
Current License: CC BY-SA 3.0
5 events
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| Apr 25, 2014 at 12:41 | comment | added | user45939 | It is an interesting conjecture. Would you please explain more about your intuition about this problem. Why does it seem unprovable within ZFC? Is there any large cardinal strength in existence of a rigid relation $\in'$ on $V$ with Kunen Inconsistency Property which cannot define $\in$? | |
| Apr 25, 2014 at 2:42 | comment | added | Joel David Hamkins | I am not sure whether in ZFC without any extra assumptions we can prove there is a relation $\in'$ such that $\langle V,\in'\rangle$ has no nontrivial elementary embeddings, such that $\in$ is not definable from $\in'$. | |
| Apr 24, 2014 at 21:02 | comment | added | Joel David Hamkins | Of course, your original argument about $j$ doesn't require that $j$ is internally available as a class, and it is a general model-theoretic fact that if one relation is not definable from another, then you can go to a saturated model where there will be automorphisms of the second relation that don't fix the first. That is, in a saturated model, you get a positive answer, using external embeddings. | |
| Apr 24, 2014 at 20:30 | comment | added | user45939 | It is a very nice answer. Thank you Prof.Hamkins. Beside finding a counterexample for suggested condition, it is an interesting quest to find a sufficient condition which implies the definability of $\in$ as a fundamental relation using another definable relation of set theory. It will be more interesting if such a sufficient condition is expressible in terms of self-elementary embeddings of universe which is closely related to Kunen inconsistency theorem for relations different from $\in$ and a possible corresponding large cardinal tree. | |
| Apr 24, 2014 at 19:54 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |