Timeline for A model with $\kappa$ many automorphism and a rigid element.
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 23, 2014 at 18:03 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
another example
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| May 23, 2014 at 14:33 | comment | added | Emil Jeřábek | It is consistent with ZF that there exists a set that does not carry any rigid structure in a countable language, see mathoverflow.net/questions/6262 (it is stated there for a single binary relation, but the argument for ZFA applies to any well-orderable language, and as long as one bounds its cardinality, this transfers to ZF by the Jech–Sochor embedding theorem). | |
| May 23, 2014 at 14:17 | comment | added | Eric Wofsey | Actually, regardless of whether AC is needed to get a rigid structure, it is needed to get a group structure. | |
| May 23, 2014 at 14:17 | vote | accept | Ioannis Souldatos | ||
| May 23, 2014 at 14:17 | comment | added | Ioannis Souldatos | Emil Jerabek and EricWofsey: Thank you both for the answers. I will check Emil's answer as it came first. | |
| May 23, 2014 at 14:08 | comment | added | Eric Wofsey | Yes, but existence of well-orderings of arbitrary cardinality does. | |
| May 23, 2014 at 14:08 | comment | added | Emil Jeřábek | Rigidity of well orders does not require choice, actually. | |
| May 23, 2014 at 14:06 | comment | added | Eric Wofsey | The easiest example of a rigid structure of arbitrary cardinality I can think of is a well-ordering. Is there an example that does not require Choice? | |
| May 23, 2014 at 13:49 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
it does not need to be commutative, actually
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| May 23, 2014 at 13:34 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
added 106 characters in body
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| May 23, 2014 at 13:22 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
added 106 characters in body
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| May 23, 2014 at 13:18 | comment | added | Asaf Karagila♦ | That was an idea I had in mind, but the only examples of a rigid field I managed to think of were $\Bbb R$'s subfields. Is there an easy proof that there are rigid fields of every cardinality? | |
| May 23, 2014 at 13:05 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |