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?
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| Aug 25, 2016 at 20:56 | comment | added | Noah Schweber | +1. Note that while this is much simpler than the other answers, it cannot be easily modified to produce an example involving countable structures. | |
| Aug 25, 2016 at 16:38 | history | answered | Ramiro de la Vega | CC BY-SA 3.0 |