An uncountable example is obtained by considering dense linear orders without endpoints. Any two dense linear orders without endpoints are elemenatarily equivalent. But $\langle\mathbb{R},<\rangle$ and $\langle\mathbb{I},<\rangle$ with the usual order, $\mathbb{I}$ the set of irrational numbers, are non-isomorphic as the first is complete while the latter is not. There is no such countable example as the theory of dense linear orders without endpoints is $\aleph_0$-categorical.

I am interested in countable and finite examples of elementarily equivalent but non-isomorphic first-order structures.

  • 5
    $\begingroup$ Your uncountable example is needlessly complicated. Two infinite sets, with no relations other than equality, are elementarily equivalent. So just take two infinite sets of different cardinalities, say, one countable and one uncountable. $\endgroup$
    – bof
    Commented Jun 18, 2015 at 22:19
  • $\begingroup$ This is stupid, but no day may end without using it at least once. Therefore I take this opportunity to whisper Löwenheim-Skolem. Btw. It's a nice exercise to prove (as Alex mentioned before) that finite models of the same theory are isomorphic (consider the case of a finite language first). $\endgroup$ Commented Jun 23, 2015 at 15:04

1 Answer 1


You're not going to find any finite examples. For any finite structure $A$, every model of $\text{Th}(A)$ is isomorphic to $A$.

But there are many many countable examples. In fact, if $T$ is any complete theory which is not $\aleph_0$-categorical, then (by definition) it has countable models which are elementarily equivalent but not isomorphic.

Here are some concrete examples:

  • Let $L$ consist of countably many constant symbols $\{c_i\mid i\in\mathbb{N}\}$, and let $T$ be the theory asserting that $c_i\neq c_j$ for all $i\neq j$. This is a complete theory, and it has countably many nonisomorphic countable models, determined by how many elements are in the structure which are not named by constants. The choices are $0, 1, 2, \dots, \aleph_0$.
  • Similarly, the theory of algebraically closed fields of characteristic $0$ is complete. The fields $\overline{\mathbb{Q}}$, $\overline{\mathbb{Q}(t_0)}$, $\overline{\mathbb{Q}(t_0,t_1)}$, $\dots$, $\overline{\mathbb{Q}(t_0,t_1,t_2,\dots)}$ are all countable and elementarily equivalent but nonisomorphic.
  • Any two discrete linear orders without endpoints are elementarily equivalent (this can be proven using EF games). Any model of this theory looks like $L \times \mathbb{Z}$, ordered lexicographically, where $L$ is some linear order. Anytime $L$ is finite or countable, this model is countable, but nonisomorphic orders $L$ give nonisomorphic models $M_L$.
  • Consider the structure $\langle \mathbb{Q},<,\{c_q\}_{q\in\mathbb{Q}}\rangle$. This is $\langle\mathbb{Q},<\rangle$ with every element named by a constant. Its complete theory is the theory of dense linear orders in which the constants pick out a copy of $\mathbb{Q}$. This theory has continuum-many nonisomorphic countable models, since you're free to choose which countable collection of the continuum-many cuts in $\mathbb{Q}$ to fill.

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