There is no definition of the essential notion of substructure (=submodel) in Shelah's introduction E56 to AEC, 1st Volume. Could someone please define this for me? I think that $$M \subseteq N$$ ($M$ is a submodel of $N$) iff
$M$ is a subset of $N$,
all constants $c$ from $N$ are in $M$,
3.$f(M^n)\in M$ for all $n$-ary function symbols $f$,
- I don't know what for relation symbols...And yet, must it be the case $M\neq \emptyset$ ?
You're close, the ideas aer there, but it's preferable to write things in a more detailed manner. Hodges defines it using embeddings in his Model Theory book, but I guess it's equivalent to the following:
Notice that above, there's no mention of the word model whatsoever. One speaks about models when he wants structures satisfying a certain set of axioms. For example, if you take the usual language of groups $L_{gr}=\{e,\cdot,{}^{-1}\}$ where $\cdot$ is a binary function symbol, $e$ a constant symbol and ${}^{-1}$ is a unary function symbol, then if you take $\mathcal Z=(\mathbb Z,1,\times,-)$ with:
then $\mathcal Z$ is a perfectly fine $L_{gr}$ -structure despite it not being a group, because it doesn't satisfy the theory of groups $T_{gr}$ axiomatized by:
When we speak of a model, we mean a structure that satisfies a certain theory!
Hodges doesn't define submodel in his book, whereas Chang and Keisler define it without defining what a substructure is in their book Model Theory. The definition they gave corresponds to the following:
You see in the definition above that it is intended that $\mathcal N$ is a model of $T$. If you don't specify on which theory we're working on, we may end up with a problem. For example, take the language $L=\{<\}$ and the $L$ -structure $\mathcal R=(\mathbb R,<)$. Undoubtedly $\mathcal Z=(\mathbb Z,<)$ is an $L$ -substructure of $\mathcal R$. Is it a submodel of $\mathcal R$? Well, if you mean $\mathcal R$ as a model of the theory of total orders, then sure it is! However, if you mean $\mathcal R$ as a model of dense total orders without endpoints, then it is not because $\mathcal Z$ is not dense.