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
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 25, 2016 at 20:55 | vote | accept | Dino Rossegger | ||
| Aug 25, 2016 at 16:06 | comment | added | Dino Rossegger | Thanks, I figured out that every complete theory not satisfying the Schröder Bernstein property has such models but this is a nice example. | |
| Aug 25, 2016 at 16:04 | vote | accept | Dino Rossegger | ||
| Aug 25, 2016 at 20:52 | |||||
| Aug 25, 2016 at 15:52 | history | answered | Ehud Meir | CC BY-SA 3.0 |