Timeline for Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property?
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11 events
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| Jul 27, 2011 at 17:53 | comment | added | James Freitag | I think I read this before your last edit. But, it was not at all clear what the $B_i$ were to someone reading the proof. Were they finite? Finitely generated. Of course they are countable, but it just wasn't clear, especially when you said to take them "as above" and the only above was Itai's $B_i's$ where they were finite (add to this the red herring finitely generated, and I think it was very unclear). I agree with what you wrote now. Or with Itai's point that essentially categoricity implies saturated implies countable amalgamation. | |
| Jul 27, 2011 at 17:43 | comment | added | Emil Jeřábek | @James: I missed your last comment. Why do I not seem to have proved it as written? | |
| Jul 27, 2011 at 17:40 | comment | added | Emil Jeřábek | You are right, I’m implicitly using finiteness of the language, which is not included in the original assumptions. I also clarified some of the wording, based on James's comments. | |
| Jul 27, 2011 at 17:39 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
finite language, some clarifications
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| Jul 27, 2011 at 17:31 | comment | added | James Freitag | Ok, that is what I thought. In the finite language case, I agree with what Emil is trying to prove (but he does not seem to have proved it as written). This comment of yours clears all of that up (via saturation). | |
| Jul 27, 2011 at 16:59 | comment | added | Itaï BEN YAACOV | My assumptions definitely do not imply $\aleph_0$-categoricity (I never said finite language). More generally, even, if $M$ is saturated then its countable substructures have AP, so a counter-example will have to be non saturated, and in particular non $\aleph_0$-categorical. | |
| Jul 27, 2011 at 15:11 | comment | added | James Freitag | Emil: Why are you talking about finitely generated substructures? The language is relational. Also, maybe you edited your post and I am missing something, but when you say to take $A,B_i,f$ "as above", do you mean as in Itai's post. Then the $B_i's$ are finite, so I don't understand what you are proving. | |
| Jul 27, 2011 at 15:06 | comment | added | James Freitag | I should add that for those who have less experience with model theory, the proof of $\omega$-categoricity in the finite relational language case goes like this: the finite language means that there are only finitely many $n$-types, like Emil mentions. Then two finite tuples with the same type are in the same $Aut(M)$ orbit. Then it is clear that the automorphism group acts oligimorphically. This implies $\omega$-categoricity (the equivalence of this condition to $\omega$-categoricity is called the Ryll-Nardzewski theorem). | |
| Jul 27, 2011 at 14:59 | comment | added | James Freitag | On the $\omega$-categoricity statement: Are you assuming a finite language here? Itai: Are you assuming the language is finite? Countable? No assumption? | |
| Jul 27, 2011 at 12:29 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
simplify the argument
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| Jul 27, 2011 at 11:00 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |