Why doesn't the definition of (model-theoretic) conservative extension need strengthening?

Question

In the wikipedia page dedicated to conservative extensions, we find the following sentence:

$T_2$ is a model-theoretic conservative extension of $T_1$ if every model of $T_1$ can be expanded to a model of $T_2$.

Question: What do we actually want to do with this definition?

The reason I ask is because from my naive viewpoint, it seems moderately deficient. In particular

1. We have no guarantee that for all $M_1 \models T_1$ there aren't multiple expansions $M_2$ such that $M_2 \models T_2$, nor even that all such expansions are isomorphic.

2. We have no guarantee that for all $M_2 \models T_2$, we can find $M_1 \models T_1$ such that $M_1$ can be expanded to $M_2$.

Answer 1

The current earlier wikipedia phrasing is unclear: it doesn’t explicitly state that a model-theoretic conservative extension is assumed to be an extension. I guess the author thought that was clear from the terminology. A clearer phrasing of the standard definition would be:

An extension $T_2$ of $T_1$ is model-theoretically conservative if every model of $T_1$ can be expanded to a model of $T_2$.

This fixes your deficiency (2). Given any model of $T_2$, its reduct to the language of $T_1$ will give a model of $T_1$, since $T_2$ is an extension of $T_1$.

On the other hand, your point (1) is not a deficiency. Typically, the language of $T_2$ is richer than the language of $T_1$, and a model of $T_1$ may indeed be extended in multiple non-isomorphic ways to a model $T_2$. The intuition behind conservativity is that the extended theory may introduce more notions to the language — more expressive power — just so long as it doesn’t prove any more theorems within the original language.

For instance, working in single-sorted first-order logic, take $T_1$ to be any theory with at least one constant in its language; now extend the language of $T_1$ by a single new constant symbol $c$, and let $T_2$ be the deductive closure of the theorems of $T_1$ within this expanded language. Then any model of $T_1$ can be extended to a model of $T_2$ in as many ways as it has elements; the constants of the original language ensure that any model of $T_1$ is non-empty.

If you are happy with the proof-theoretic definition of conservativity, then the theorem “model-theoretic conservativity implies proof-theoretic conservativity” (a straightforward consequence of the completeness theorem) should reassure you that the model-theoretic definition is not deficient.