Timeline for Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property?
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| Oct 8 at 13:41 | comment | added | Adam Bartoš | Is there a reference for the fact that countable substructures of an ω-saturated Fraïssé limit have AP? | |
| Sep 29, 2023 at 16:00 | answer | added | Rob Sullivan | timeline score: 3 | |
| May 10, 2020 at 5:35 | history | edited | YCor |
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| Feb 17, 2020 at 10:31 | history | edited | YCor | CC BY-SA 4.0 |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jul 30, 2011 at 9:16 | vote | accept | Itaï BEN YAACOV | ||
| Jul 29, 2011 at 17:40 | answer | added | Ali Enayat | timeline score: 9 | |
| Jul 29, 2011 at 13:20 | comment | added | Ali Enayat | @Itai: yes, we can always transform a functional language to a relational one, but I was worried about substructures, which as you know behave very differently in a relational language; but I think I can now handle that hurdle and produce a relational counterexample that I can post later today once I go over it one more time. | |
| Jul 29, 2011 at 10:39 | comment | added | Itaï BEN YAACOV | @Ali: Yes of course this is of interest. Once you have a counter-example in a functional language, can you not add relations for the graphs of all terms and get a relational one? | |
| Jul 28, 2011 at 20:02 | comment | added | Ali Enayat | @Itai: If we change your question to allow function symbols in the language (and define the notion of substructure accordingly) then I can think of a counterexample, but probably this is not of interest to you. In the counterexample, M is a countable rec. sat. model of PA. If this is of interest, I can elaborate. | |
| Jul 28, 2011 at 13:37 | history | edited | Itaï BEN YAACOV | CC BY-SA 3.0 |
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| Jul 27, 2011 at 17:03 | history | edited | Itaï BEN YAACOV | CC BY-SA 3.0 |
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| Jul 27, 2011 at 15:19 | comment | added | James Freitag | Maybe it would be helpful to note some of the necessary properties of a counterexample. The class has AP, but should not have what I have heard called the strong AP, in which one can amalgamate the $B_i's$ so that $g_1(B_1) \cap g_2(B_2)=g_0 f_0 (A)=g_1 f_1 (A).$ If the class has this property, then it seems like a compactness argument would work. | |
| Jul 27, 2011 at 11:00 | answer | added | Emil Jeřábek | timeline score: 1 | |
| Jul 27, 2011 at 10:12 | history | edited | Itaï BEN YAACOV | CC BY-SA 3.0 |
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| Jul 27, 2011 at 10:04 | history | asked | Itaï BEN YAACOV | CC BY-SA 3.0 |