Proving Th(($\mathbb{Z}$, <)) has continuum many countable models.

Question

I am tasked with proving that Th(($\mathbb{Z}, <))$ has continuum many models. For this we are given the following construction.

Let $\alpha \in \mathcal{C} = \{0,1\}^{\mathbb{N}}$. We define for each $\alpha$ the set $V_{\alpha}$: $$V_{\alpha} = \{q \in \mathbb{Q}\mid\exists n[2n \leq q \leq 2n+1]\ \lor\ \exists n[\alpha(n) = 1\ \land\ q = 2n + \frac{3}{2}]\}$$ Define for each such $V_{\alpha}$ the set $W_{\alpha} := V_{\alpha}\times \mathbb{Z}$.

Now consider the structures $(V_{\alpha}, <)$ and $(W_{\alpha}, <')$ with $<$ the usual ordering on $\mathbb{Q}$ and $<'$ the lexicographic ordering on $\mathbb{Q}^2$.

There are now three things to do:

1. Prove: $\forall\alpha\in\mathcal{C}\forall\beta\in\mathcal{C}[\alpha\neq\beta\to (V_{\alpha}, <) \ncong (V_{\beta}, <)]$.

2. Prove: $(V_{\alpha}, <) \cong (V_{\beta}, <)$ if and only if $(W_{\alpha}, <') \cong (W_{\beta}, <')$.

3. Prove: $(W_{\alpha}, <') \equiv (\mathbb{Z}, <)$. That is that both structures are elementary equivalent.

My question

I have difficulties proving 3. Thus far I tried using Ehrenfeucht-Fraïssé games but as of now no result yet. This is probably because $(W_{\alpha}, <')$ can be thought of as a plane and $(\mathbb{Z}, <)$ as a line. Successfully devising a winning strategy basically comes down to good choices when the first player chooses an element in $W_{\alpha}$. I know how to win if the games takes at most 3 moves from each player. Yet generalizing this has proven to be difficult.

Also I have some troubles with 1. I see why this is true; if $\alpha \neq \beta$ then either $V_{\alpha}$ or $V_{\beta}$ has one "successor" more. But translating this to a formal proof has yet to be made.

Can I get help on these problems? I thank you in advance.

Answer 1

This is not exactly the answer that the OP is looking for, but it may be interesting.

First an answer that assumes the continuum hypothesis.

For every $\alpha<\omega_1\smallsetminus\{0\}$ there is a model $\alpha\times \mathbb Z$. Here the relation $<$ is interpreted as the lexicographic order. These are $\omega_1$ non isomorphic countable models.

Now without continuum hypothesis.

For every sequence $(n_i)_{i\in\omega}$ of positive integers consider a model obtained "concatenating" the models $n_i\times\mathbb Z$ separated by a copy of $\mathbb Q\times\mathbb Z$. These are $2^\omega$ non isomorphic countable models.

Answer 2

Point 1. : let $\alpha \neq \beta \in \mathcal{C}$. Then there is $n \in \mathbb{N}$ such that $\alpha(n) \neq \beta(n)$. WLOG, $\alpha(n) = 0$ and $\beta(n) = 1$. Now take $q := 2 n + \frac{3}{2}$.

$q \notin V_\alpha$ but $q\in V_\beta$, hence $V_\alpha \neq V_\beta$.

Point 3. : The theory of $(\mathbb{Z}, <)$ is the theory of discrete linear orders without endpoints (which happens to be a complete theory). So, all you have to do is to check that $(W_{\alpha}, <')$ is a discrete linear order without endpoints.