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Let $A$ and $B$ be models of the theory of Dense Linear Orders without Endpoints such that $|B| \subset |A|$. I'm trying to prove that $B$ is an elementary submodel of $A$. Using Tarski-Vaught test I was able to prove the following:

Proposition 1: If $A$ and $B$ are $L$-structures, $|B| \subset |A|$ and for every finite sequence $b_1,\ldots,b_n$ of elements of $B$ and an element $a$ of $|A|$ there is an automorphism (elementary embedding of $A$ onto itself) $\Phi$ of $A$ such that $\Phi(b_i) = b_i$ for $1 \leq i \leq n$ and $\Phi(a) \in B$ then $B$ is an elementary substructure of $A$.

I was trying to use Proposition 1 to my specific problem and I actually have a pretty good idea of what the automorphism should look like, but I can't construct it. Given $b_1,\ldots,b_n$ elements of $|B|$ and an element of $|A|$, if $b_i < a < b_j$ for some $i$, $j$, then I can associate $a$ with $b \in (b_i, b_j)$. And having constructed $\Phi(x)$ for a finite numbers of elements $x_1,\ldots, x_n$, I can construct $\Phi(x_{n+1})$ just respecting the order of $x_1,\ldots, x_{n+1}$. My problem is: $|A|$ can have any cardinality, but this procedure works only for countable universes.

Any ideas on how to fix my attempt?

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  • $\begingroup$ Minor remark: I know that the result is an immediate consequence of Tarski-Vaught test, but I want to prove using the automorphism proposition I just stated. $\endgroup$ May 17 '15 at 21:33
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Take $A = \mathbb{R} \times \mathbb{Q}$ ordered lexicographically and $B = \{0\}\times\mathbb{Q}$, Then the hypotheses of your proposition don't hold, e.g., with $n = 1$, $b_1 = (0, 0)$ and $a = (1, 0)$, there is no automorphism of $A$ that fixes $b_1$ and maps $a$ into $B$.

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  • $\begingroup$ Yes, I meant $a = (1, 0)$ - I sometimes suffer from a form of dxsleyia when thinking about the lexicographic order in the plane. I have corrected that. Does it make sense now? The idea is that there are uncountably many elements between $b_1$ and $a$ in $A$, So no automorphism of $A$ can fix $b_1$ and map $a$ into the countable set $B$. $\endgroup$
    – Rob Arthan
    May 17 '15 at 22:17
  • $\begingroup$ You're right of course. There is no such thing as a denumeration of $\mathbb{Q}\times\mathbb{R}$. I'll think it over $\endgroup$ May 17 '15 at 22:20
  • $\begingroup$ I understand what you're saying but still am not convinced. Can you elaborate more? As I see, cardinality shouldn't be an issue, given that the image of the isomorphism is A itself $\endgroup$ May 17 '15 at 22:25
  • $\begingroup$ The automorphism must be one-to-one and order-preserving and must map the uncountable interval $(b_1, a)$ in $A$ to an interval $(b_1, \Phi(a))$ in $B$. As $B$ is countable, this is impossible $\endgroup$
    – Rob Arthan
    May 17 '15 at 22:32
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    $\begingroup$ Because everything in $A$ between $b_1$ (which is in $B$) and $\Phi(a)$ (which is required to be in $B$) is in $B$. $\endgroup$
    – Rob Arthan
    May 17 '15 at 22:41

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