Proving that two interpretations are isomorphic implies the structures induced by them are isomorphic.

Question

Definition: Isomorphism of interpretations

Two interpretations $I,J$ are said to be isomorphic if there exists a function $F: U_I\to U_J$ that satisfies:

1. $F$ is bijective.
2. If $c\in C$ and $c_I,c_J$ its respective interpretations then $F(c_I)=c_J$.
3. $f^k\in F\implies F(f^k_I(t_1,..,t_k))=f_J^k(F(t_1),...,F(t_2))$.
4. $p^k\in P$ then $(a_1,...,a_k)\in P_I^k\iff (F(a_1),..,F(a_k))\in P_J^k$.

Proposition

Let $I, J$ be two isomorphic interpretations of $L$. Let $\xi(I)=(U_I,F_I,C_I,P_I)$ denote the structure induced by $I$. Then $\xi(I)\simeq\xi(J)$.

Example:

Let $L=(F=\{f^2\},C=\emptyset,P=\{=\})$. Let $\xi(I)=(\Bbb R_{>0},+),\xi(J)=(\Bbb R_{<0},+)$.

My proposition says that if $I,J$ are isomorphic, then, as $U_I,+$ is closed under addition ($f^2_I$), $U_J,+$ must be closed under addition ($f^2_J$) as well, which is indeed the case.

How could I prove this?

E: I think that my proposition is incomplete, but I'm not sure, should I add that $\xi(I)\simeq \xi(J)$ "up to expressibility"? Meaning that if there exists a property $P$ that $\xi(I)$ has, but not $\xi(J)$ but it can't be expressed in first order logic, then that shouldn't preclude the isomorphism of interpretations, right?

Answer 1

Since $f^2$ is a (binary) function, it's part of the definition of "model for L" that the universe of the model is closed under $f^2$ – there's nothing to prove.

If $f^2$ were replaced by a 3-place predicate $P^3_1$, then you'd have something to prove. In that case, a sentence of $L$ says "$P^3_1$ is a total function". That sentence holds in $\xi (I)$ iff the interpretation of $P^3_1$ in $\xi (I)$ is in fact a total function on $\mathbb{R_{> 0}}$, and then it also holds in the isomorphic $\xi (J)$, where it says the same thing about the corresponding relation on $\mathbb{R_{< 0}}$.

This generalizes: Suppose $\mathscr{L}$ is a first-order language, and $\mathsf{M}$ is a model appropriate for $\mathscr{L}$ -- $\mathsf{M} = (X, (f_i)_{i\in I}, (P_j)_{j \in j}, (c_k)_{k \in K}) $ has the appropriate signature (the correct number of functions and relations with correct arities, and appropriately many constants, etc.). Then all of the functions $f_i$ are total: $X$, the domain or universe of the model, is closed under every $f_i$.

A general statement of what you want to show might be the following: Suppose $\mathsf{M}$ and $\mathsf{M'}$ are isomorphic models of $\mathscr{L}$. If some relation $n+1$-place relation $P_j$ in $\mathsf{M}$ is a total function on $X$, then the corresponding relation $P'_j$ in $\mathsf{M'}$ is a total function on $X'$. (Here, $\mathsf{M}$ is as above and $\mathsf{M'} = (X', (f'_i)_{i\in I}, (P'_j)_{j \in j}, (c'_k)_{k \in K})$.)

Suppose $P_{j_0}$ is such an $n+1$-place relation which is a total function on $X$. There is a sentence $S$ of $\mathscr{L}$ expressing which is true in a model of iff the $j_0$-th relation is a total function. If $\mathsf{M} \models S$ ($S$ is true in $\mathsf{M}$), then $\mathsf{M'} \models S$, so $P'_{j_0}$ is a total function on $X'$.