I know that the theory of dense linear orders without endpoints is $\aleph_0$-categorical, and looking for two uncountable non isomorphic models of same cardinality, I found many examples, but nothing about what I initially thought: simply two copies of $\mathbb{R}$, one following the other (i.e. $2$ $\times$ $\mathbb{R}$ with lexicographical order) and $\mathbb{R}$ itself, are they isomorphic models of DLO?
Are $\mathbb{R}$+$\mathbb{R}$ and $\mathbb{R}$ isomorphic models of DLO
model-theory
Related Solutions
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.
Best Answer
$\mathbb{R}$ and $\mathbb{R} + \mathbb{R}$ are not isomorphic as linear orders. Suppose $f \colon \mathbb{R} + \mathbb{R} \to \mathbb{R}$ is order preserving. Let $A$ be the first copy of $\mathbb{R}$ in $\mathbb{R} + \mathbb{R}$ and $B$ be the second copy. Then $f(A)$ is bounded above in $\mathbb{R}$ by every element of $f(B)$, but $\sup f(A)$ can't be in $f(A)$ or $f(B)$, so $f$ isn't surjective.