How can there be three model isomorphism classes for this extended theory of unbounded dense linear orders?

Question

I want to show that the theory $$ T := DLO \cup \{ c_i < c_{i+1} \mid i \in \omega \} $$ has exactly three non-isomorphic countable models, where $DLO$ is the theory of unbounded dense linear orders.

I struggle to see why there is not just one countable model of $T$ up to isomorphism: Suppose $M$ and $M'$ are any two countable models of $T$. Then $M$ and $M'$ are also countable models of $DLO$; so $M$ and $M'$ are countable unbounded dense linear orders. Thus, by Cantor's Theorem, $M$ and $M'$ are isomorphic to each other.

Where is my mistake? Thanks!

Answer 1

You've shown that the reducts are isomorphic, but not the structures themselves.

Suppose I have languages $\Sigma\subseteq\Sigma'$. Then to each $\Sigma'$-structure $\mathcal{M}$ I can associate its reduct to $\Sigma$; this is the $\Sigma$-structure $\mathcal{M}_\Sigma$ with the same domain as $\mathcal{M}$ and the same interpretation on all the symbols in $\Sigma$ (so basically I just throw away the extra symbols).

Now you've argued (correctly!) that if $\mathcal{A},\mathcal{B}\models T$ then $\mathcal{A}_{\{<\}}\cong\mathcal{B}_{\{<\}}$. But that's not enough to conclude that $\mathcal{A}\cong\mathcal{B}$. While it's certainly true that $\mathcal{M}\cong\mathcal{N}\implies\mathcal{M}_\Sigma\cong\mathcal{N}_\Sigma$, the converse fails badly. For an easy example, take $\Sigma=\emptyset$, $\Sigma'$ any language not consisting of only constant symbols, and consider two nonisomorphic $\Sigma'$-structures of the same cardinality.

The point is:

Isomorphisms between reducts need not preserve the "forgotten" structure - that is, the "$\Sigma'\setminus\Sigma$-part."

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Of course, this only points out the error in your argument - it doesn't explain why in fact $T$ has exactly three countable models up to isomorphism. Towards that, let me describe two nonisomorphic countable models of $T$, and leave finding the third as an exercise:

• In both $\mathcal{A}$ and $\mathcal{B}$, the underlying set will be $\mathbb{Q}$ and $<$ will be the usual ordering.

• In $\mathcal{A}$ we'll have $c_n=n$.

• In $\mathcal{B}$ we'll have $c_n=1-2^{-n}$.

Now, what's the third possibility? (HINT: it's sort of a mix between $\mathcal{A}$ and $\mathcal{B}$ ...)