Consider the following two first-order theories:

  1. A theory $T$ with countably infinit many constant symbols $c_i,i=1,2,...$ and for each $i,j=1,2,3,...$ with $i\not= j$ an axiom $c_i\not= c_j$.
  2. A theory $T'$ without constant symbols but that for any $n\in\Bbb N$ contains an axiom that states that there are more than $n$ different elements.

Both imply that only infinite models are allowed and for me it seems $T$ and $T'$ are indistinguishable from a model theoretic point of view. But here (in the comments below my question) I was told, that there is a big difference:

One of the theories has only one countable infinit model, while the other one has more.

It was not stated for which holds which, but I assume the first one is assumed to be the one with the unique countable model. To be honest, I cannot think about any two countable models of $T'$ which are not isomorphic because the theories do not contain any relations or functions that can prevent these models from being so.

Can you please point out such two models for me and explain my misunderstanding?


You have it backwards. Your theory $T$ has (countably) infinitely many countable models up to isomorphism, while $T'$ only has one.

A model of $T$ looks like $\{c_i\mid i\in \mathbb{N}\}\cup X$, where $X$ is a set not containing any of the $c_i$ (so the union is a disjoint union). We have $M_X\not\cong M_Y$ whenever $X$ and $Y$ have different cardinalities, since an isomorphism must take $c_i$ to $c_i$ for all $i$ and then provide a bijection between the remaining elements (between $X$ and $Y$). And $M_X$ is countable whenever $X$ is finite or countable. So the countable models of $T$ are classified by the size of $X$, with possibilities $0,1,2,\dots,\aleph_0$.

On the other hand, a countable model of $T'$ is just a countably infinite set with no extra structure, and any bijection between two such models is an isomorphism.


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