If $\nu : \mathcal{M} \rightarrow \mathcal{N}$ is an elementary embedding, then is it true that $\mathbb{M} \subseteq \mathbb{N}$?

Question

Exactly as the title suggests:

If $\nu : \mathcal{M} \rightarrow \mathcal{N}$ is an elementary embedding, then is it true that $\mathbb{M} \subseteq \mathbb{N}$?

Answer 1

An elementary embedding $\nu \colon \mathcal M \to \mathcal N$ between $\mathcal L$-structures is a map $\nu\colon M\to N$ on the underlying sets such that for every $\mathcal L$-formula $\varphi(\bar x)$, one has $$ \mathcal M \models \varphi(\bar m) \iff \mathcal N \models \varphi(\nu(\bar m)) .$$ In particular, $\nu$ is injective (prove it!). So maybe $\mathcal M$ is not a substructure of $\mathcal N$ (see André Nicolas's comment), but $\nu(\mathcal M)$ is. Of course, by $\nu(\mathcal M)$ I denote the $\mathcal L$-structure

• whose underlying set is $\nu(M)$,
• where a constant symbol $c$ has interpretation $\nu(c^\mathcal M)$,
• where a function symbol $f$ has interpretation $(n_1,\dots,n_k) \mapsto \nu\left(f^\mathcal M(\nu^{-1}(n_1),\dots,\nu^{-1}(n_k))\right)$,
• where a relation symbol $R$ has interpretation $\nu(R^\mathcal M)$.

Answer 2

Not just the general answer is negative as shown by others, it is also possible to have the opposite, namely $j\colon\cal M\to N$ an elementary embedding, but $\Bbb{N\subsetneq M}$.

For example, pick an order isomorphism from $\Bbb Q$ to $\Bbb Q\cap(0,1)$.