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I am trying to prove that $K_2$, the densely ordered set without a smallest or a biggest element is complete.

Now I arrived at the step where I want to proof that it is impossible for any wf $B$ that both $$ \nvdash _{K_2} B \mbox{ and } \nvdash_{K_2} \lnot B $$ To prove this, I assume that neither $\nvdash _{K_2} B \mbox{ and } \nvdash_{K_2} \lnot B$ . Then I know that $K_2 \cup \{B\}$ and $K_2 \cup \{\lnot B\}$ are both consistent. So we know (by Skolem-Lowenheim) that there exist denumerable models $M_1$ and $M_2$ such that $\models_{M_1} B$ and $\models_{M_2} \lnot B$.

I know that $K_2$ is $\aleph_0$-categorical. So $K_2$ has at least one normal model of cardinality $\aleph_0$ (indeed, this model can be denoted by the rational numbers), and any two normal models of $K_2$ with cardinality $\aleph_0$ are isomorphic.

Now the part where I get stuck comes: I want to prove that $M_1$ and $M_2$ are isomorphic, by the $\aleph_0$-categoricity of $K_2$ (to obtain a contradiction). But $M_1$ and $M_2$ are models for different theories ($K_2 \cup \{B\}$ and $K_2 \cup \{\lnot B\}$), and they are not models for $K_2$. Or are they? What is the correct way to proceed?

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    $\begingroup$ Of course they are models of $K_2$. A structure $M$ is a model of a theory $T$ if $M\models\varphi$ for all $\varphi\in T$ (that is the definition). What sentences $M$ satisfies apart from those in $T$ doesn't play a role. $\endgroup$ – martin.koeberl Mar 18 '17 at 15:01
  • $\begingroup$ Off course.. Any theorem that is true in a model for $K_2$ is also true in a model for $K_2 \cup \{ B \}$ if $B$ is not in $K_2$.. $\endgroup$ – Sven Mar 18 '17 at 15:03
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Remember that if $\Delta\subseteq \Gamma$, then every model of $\Gamma$ is also a model of $\Delta$. For example, both $S_{17}$ and $\mathbb{Z}/6\mathbb{Z}$ are models of the theory of groups, even though the first satisfies "$\exists x, y(x*y\not=y*x)$" and the second does not.

In your case both $M_0$ and $M_1$ are models of $K_2$ (note that your first sentence makes a small mistake, by implying that $K_2$ is a structure as opposed to a theory!). In particular, since they are each countable and $K_2$ is $\aleph_0$-categorical, we have $M_0\cong M_1$. The fact that $M_0$ and $M_1$ disagree on some sentence outside of $K_2$ (namely, $B$) doesn't affect the fact that each of them is a model of $K_2$.

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