Class of structures isomorphic to $(\mathbb{Z},<)$ in infinitary logic $L_{\omega_1 \omega}$

Question

We want to define the class of all structures isomorphic to $(\mathbb{Z},<)$ in the infinitary logic $L_{\omega_1 \omega}$. Therefor we define strict order as usual:

• $\forall x,y,z ~~ (x<y \land y<z)\rightarrow x<z$
• $\forall x,y ~~ x<y \rightarrow \neg(y<x)$

We define that it has no end points:

• $\neg \exists x \forall y (x \neq y \rightarrow x<y)$
• $\neg \exists x \forall y (x \neq y \rightarrow x>y)$

Obviously, the successor relation $S$ is definable. Then we use infinitary logic for the first time to express that the order is not dense:

$\varphi^n(x,y):=S^n(x)=y$

$\Phi := \{\varphi^n(x,y) \mid n < \omega\}$

• $\forall x,y (x<y \rightarrow ~\lor\Phi)$

Thus, we express that between two element $x<y$ there can only be a finite number of elements in between. Is that a correct and complete characterization of the classes isomorphic to $(\mathbb{Z},<)$? Intuitively think it is but I do not know how to proof the correctness.

Answer 1

The fact that the order is nowhere dense is expressible in $L_{\omega\omega}$. We simply say that every point has a successor.

What you expressed is that the order is the countable union of finite intervals. Namely for every $x$ and $y$, if $x<y$ then by applying the successor some finitely many times from $x$ we arrive $y$.

Equally, but perhaps more insightfully, we could have said: $$\exists x\forall y\left(\bigvee\Phi(x,y)\lor\bigvee\Phi(y,x)\right)$$

Namely there is some point that every thing is generated by a finite application of the successor and co-successor.

Now suppose that $(M,\prec)$ is a structure which satisfies that $\prec$ is a linear order which is nowhere dense, and satisfies the property written above, fix some $x_\mathbb Z$ and $x_M$ which satisfy the role of $x$ in the sentence, and you have but a unique way to extend the function to an isomorphism now.

Answer 2

You're missing that $<$ is linear, and not merely a partial order. Otherwise, any structure where $<$ is interpreted as the empty relation will satisfy your axioms. So add $$\forall x, y (x = y\ \vee\ x < y\ \vee\ y < x)$$ Given an arbitrary $(M, \prec)$ structure satisfying these axioms, you can construct an isomorphism between it and $(\mathbb{Z}, <)$. Start by picking some $x \in M$, then define $f : \mathbb{Z} \to M$ by $f(n) = S^n(x)$. As an exercise you can show that this is an isomorphism. NB: for this to make sense, you'll need to prove (easily) that $S$ is a total, injective function.