Timeline for $\mathcal A\equiv\mathcal B\implies \mathcal A\cong\mathcal B$ for finite $\mathcal L$-structures where $\mathcal L$ is an infinite signature
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 11, 2016 at 15:35 | vote | accept | urpzilmöräqÜ | ||
| May 11, 2016 at 15:35 | comment | added | urpzilmöräqÜ | Now I understand :-D Thank you! Prof. Hamkin's answer and the additional explanations of Prof. Blass and Emil Jeřábek were very helpful! | |
| May 11, 2016 at 15:05 | comment | added | Emil Jeřábek | @urpzilmöräqÜ You don’t really need to talk about formulas here. If a bijection fails to be an isomorphism, it is because it fails to preserve some relation or function symbols from the signature, by the definition of isomorphism. So, fix one such symbol for each bijection, and voila, you have a finite sublanguage $L'$ such that the reducts of $\mathcal A$ and $\mathcal B$ to $L'$ are still nonisomorphic. | |
| May 11, 2016 at 14:50 | comment | added | Andreas Blass | @urpzilmöräqÜ Note that Joel wrote about atomic formulas, not atomic sentences. In your example, $i$ would not preserve satisfaction of the atomic formula $R(x_1,\dots,x_n)$ (where the $x_i$ are distinct variables, not constant symbols). So this atomic formula gives a "reason" why $i$ is not an isomorphism. | |
| May 11, 2016 at 14:48 | comment | added | urpzilmöräqÜ | For example, let us suppose that $R^{\mathcal A}(a_1, \dots, a_n)$ holds but $R^{\mathcal B}(i(a_1),\dots,i(a_n))$ does not hold, where $i$ denotes a bijection $A\to B$. Assuming that we have no constant symbols for some of the following elements: $a_1,\dots, a_n$, what would be the atomic formula that expresses the reason why $i$ is not an isomorphism? | |
| May 11, 2016 at 14:48 | comment | added | urpzilmöräqÜ | @AndreasBlass: I think that there is not necessarily an atomic formula expressing the reason why a bijection from $A$ to $B$ fails to be an isomorphism, because it is possible that there are some $a\in A$ such that there is no constant symbol $c\in\mathcal L$ with $c^{\mathcal A} = a$. | |
| May 11, 2016 at 14:28 | comment | added | Andreas Blass | @urpzilmöräqÜ As far as I can see, Joel is using the same meaning of "atomic formula" as the wikipedia article that you linked to. Why do you think these formulas don't suffice as reasons for non-isomorphism? In fact, one could limit further to atomic formulas that are of one of the forms $R(x_1,\dots,x_n)$ for some predicate symbol $R$ and $F(x_1,\dots,x_n)=y$ for some function symbol $F$. | |
| May 11, 2016 at 14:24 | comment | added | urpzilmöräqÜ | What do you mean by an "atomic formula"? I guess you do not mean the acceptation which is explained in en.wikipedia.org/wiki/… since the reason why a bijection from A to B fails to be an isomorphism cannot necessarily be expressed by an atomic formula (where "atomic formula" means what is explained in this article). Also, could you please explain why we can reduce to the case of a finite sublanguage if there are only finitely many bijections? | |
| May 11, 2016 at 14:01 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |