Timeline for A Fraïssé class without the strong amalgamation property.
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2013 at 1:50 | vote | accept | surf_that_curl | ||
| Apr 13, 2013 at 1:45 | vote | accept | surf_that_curl | ||
| Apr 13, 2013 at 1:46 | |||||
| Apr 12, 2013 at 17:44 | history | edited | Ali Enayat | CC BY-SA 3.0 |
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| Apr 12, 2013 at 15:05 | history | edited | Ali Enayat | CC BY-SA 3.0 |
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| Apr 12, 2013 at 14:12 | history | edited | Ali Enayat | CC BY-SA 3.0 |
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| Apr 12, 2013 at 10:54 | comment | added | Emil Jeřábek | Also, in varieties of Heyting algebras, amalgamation, strong amalgamation, and superamalgamation are all equivalent, and they hold iff the corresponding logic has the interpolation property. Classical logic has interpolation, hence Boolean algebras have strong amalgamation, and this should also work for the class of finite algebras as the variety is locally finite. | |
| Apr 12, 2013 at 10:34 | comment | added | Emil Jeřábek | It seems to me that the stabilizer of a finite subset of the countable atomless Boolean algebra fixes only the finite subalgebra generated by the set, hence by Joel’s correction of the statement, it should have strong amalgamation. Am I wrong? | |
| Apr 12, 2013 at 1:00 | history | edited | Ali Enayat | CC BY-SA 3.0 |
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| Apr 12, 2013 at 0:58 | comment | added | Ali Enayat | @Joel: the theorem was (implicitly) stated for relational structures, and your proposed fix is the right one when there are function symbols around. I will edit. | |
| Apr 11, 2013 at 23:59 | comment | added | Joel David Hamkins | Probably you should mean to say that the stabilizer of a finite set should fix only the induced substructure generated by that set... | |
| Apr 11, 2013 at 23:53 | comment | added | Joel David Hamkins | Ali, is this stated correctly? After all, if there is a constant symbol in the language, for example, then every automorphism will fix that point, even if it is not among some other finite set. But we don't expect that merely having a constant symbol in the language ensures the failure of strong amalgamation. | |
| Apr 11, 2013 at 23:41 | history | answered | Ali Enayat | CC BY-SA 3.0 |