Timeline for Does every set admit a rigid binary relation? (and how is this related to the Axiom of Choice?)
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| May 8, 2020 at 8:37 | comment | added | YCor | @JoelDavidHamkins In the linked paper you prove the consistency with ZF of (X): there exists a set in which every binary relation is non-rigid. Two follow-up questions: (a) is is consistent with ZF+DC? (b) can it be promoted to (X'): there exists a set in which for every $n$, every $n$-ary relation is non-rigid?. Are (X) and (X') equivalent in ZF (or ZF+DC)? I'm asking as a comment but could ask a question if it's worth (no idea if it sounds obvious extensions of the argument, or if it's interestingly harder, or plainly false). | |
| Jun 25, 2011 at 18:39 | history | edited | Justin Palumbo | CC BY-SA 3.0 |
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| Jan 6, 2010 at 5:04 | comment | added | Justin Palumbo | Both are me; I was originally confused about how to make an account and lost access to the first one. It probably is a good idea for me to contact the admins and ask them to put them together, or maybe delete the old one. | |
| Jan 6, 2010 at 4:18 | comment | added | Joel David Hamkins | Thanks for the explanation, but your description makes it sound like the forcing construction. I'll have a look at Jech's description of the embedding theorem. Meanwhile, Justin, there seem to be two of you, one at mathoverflow.net/users/2426, and another at mathoverflow.net/users/2436. Which is really you? Perhaps the administrators can merge these identities. | |
| Jan 5, 2010 at 20:32 | comment | added | Justin Palumbo | About the ZFA proofs: Given a ZFA model with a permutation group G and associated normal filter F, the embedding theorem gives a symmetric forcing model where the permutations on A are now translated to very similar permutations on certain names for sets of ordinals. I am not sure when the reals in these models will have no well-ordering (probably in almost all nontrivial cases). In certain cases it definitely will, for example if the set of reals could be well-ordered, then its powerset could be linearly ordered and these embedding theorems can be used to give models where that fails. | |
| Jan 2, 2010 at 21:19 | vote | accept | Joel David Hamkins | ||
| Jan 2, 2010 at 21:19 | comment | added | Joel David Hamkins | I am accepting this answer, since it shows that some amount of choice is required. But it seems to remain open whether the question of every set having a rigid binary relation is actually equivalent to AC. It could be weaker than AC. | |
| Jan 2, 2010 at 21:17 | comment | added | Joel David Hamkins | This is great! Thanks very much. But now I am confused. I had always thought that there was a rough equivalence between the ZFA symmetric models via the Jech-Shore embedding theorem and the symmetric models obtained by the forcing technique (by looking at the class of symmetric names). For example, Cohen used the latter method to build a model of ZF with a non-wellorderable set of reals. But you are saying that the ZFA symmetric models only map their atoms in at a higher level? So do the ZFA proofs via Jech-Shore provide a model of ZF with no well-ordering of the reals? | |
| Dec 28, 2009 at 16:53 | history | answered | Justin Palumbo | CC BY-SA 2.5 |