Timeline for If $\mathcal{A} \equiv \mathcal{B}$ and $\mathcal{A} \not \cong \mathcal{B}$, is it possible that $\mathcal{A}$ and $\mathcal{B}$ are bi-embeddable?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2016 at 3:05 | comment | added | Danielle Ulrich | Chris Laskowski and John Goodrick (among others) have investigated this behaviour in much detail, see for example math.umd.edu/~laskow/Pubs/SB2revise.pdf | |
| Aug 31, 2016 at 2:57 | comment | added | Danielle Ulrich | @NoahSchweber: or just $Th(\mathbb{Z}, <)$: compare $\mathcal{A} = \mathbb{Z} \times \mathbb{Q}$, and $\mathcal{B} = \mathbb{Z} \times (\mathbb{Q} \cap [0, 1])$. | |
| Aug 25, 2016 at 20:55 | vote | accept | Dino Rossegger | ||
| Aug 25, 2016 at 20:52 | vote | accept | Dino Rossegger | ||
| Aug 25, 2016 at 20:55 | |||||
| Aug 25, 2016 at 20:48 | comment | added | Noah Schweber | Yet another example, which I don't think is worth another answer but is valuable to keep in mind: the usual example of a complete theory with three countable models gives an example (consider the two models where $\{c_i\}$ is bounded). | |
| Aug 25, 2016 at 16:42 | answer | added | Noah Schweber | timeline score: 4 | |
| Aug 25, 2016 at 16:38 | answer | added | Ramiro de la Vega | timeline score: 14 | |
| Aug 25, 2016 at 16:04 | vote | accept | Dino Rossegger | ||
| Aug 25, 2016 at 20:52 | |||||
| Aug 25, 2016 at 15:52 | answer | added | Ehud Meir | timeline score: 8 | |
| Aug 25, 2016 at 15:42 | review | First posts | |||
| Aug 25, 2016 at 16:44 | |||||
| Aug 25, 2016 at 15:42 | history | asked | Dino Rossegger | CC BY-SA 3.0 |