Timeline for Is the union of a chain of elementary embeddings elementary?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2019 at 15:38 | vote | accept | Joel David Hamkins | ||
| Sep 11, 2019 at 10:14 | comment | added | Joel David Hamkins | @GerhardPaseman, I had tried that initially, thinking it would be like the usual elementary-chain arguments. But things break down in the backwards direction of the existential case. Just because the target has a witness with $\varphi(x,j(b))$, how do you get a witness with $\varphi(x,b)$? In fact, you can't in general, because of the counterexamples that we now know about from Alex and Martin. | |
| Sep 10, 2019 at 23:49 | history | became hot network question | |||
| Sep 10, 2019 at 22:43 | answer | added | Alex Kruckman | timeline score: 19 | |
| Sep 10, 2019 at 21:51 | comment | added | Alex Kruckman | @JoelDavidHamkins Thanks. Yes, the copies of $\mathbb{Z}$ come after the copies of $\mathbb{N}$ in the order. I was waiting to post an answer until I came up with a counterexample to the main question - but if I don't manage to do that, I'll convert my comment into an answer. | |
| Sep 10, 2019 at 21:13 | comment | added | Joel David Hamkins | Alex, very nice example! (I guess you intend that the Z copy is on top.) Why not post the example as an answer? | |
| Sep 10, 2019 at 20:48 | answer | added | Goldstern | timeline score: 9 | |
| Sep 10, 2019 at 20:45 | comment | added | Gerhard Paseman | Can't you do some form of skolemnization of a target sentence wiith respect to some M_n, show j acts nicely on individuals witnessing the skolemnization by showing it acts nicely on a large enough submodel, to get the sentence to be preserved by j? (Seeing Alex's comment now suggests "No".) Gerhard "Or Are Elementary Chains Needed?" Paseman | |
| Sep 10, 2019 at 20:40 | comment | added | Alex Kruckman | The generalization is not true: For example, you could take $M_n = (\mathbb{N},<)$ for all $n$, and $N_n = (\mathbb{N}\sqcup \frac{1}{n}\mathbb{Z},<)$. Then $M_n \preceq N_n$ for all $n$, but $\bigcup_{n} M_n \not\preceq \bigcup_n N_n$, since the former is $(\mathbb{N},<)$ and the latter contains a copy of $\mathbb{Q}$, hence isn't discrete. I think there should be a counterexample to the original question as well, but I haven't come up with one yet. Of course, requiring $M_n = N_n$ means that any funny business in the union of the codomains happens in the union of the domains as well... | |
| Sep 10, 2019 at 16:22 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 160 characters in body
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| Sep 10, 2019 at 15:46 | history | asked | Joel David Hamkins | CC BY-SA 4.0 |