Dedekind complete model

Question

Prove that there is no first order theory $T$ in $ L=\{<\}$ such that for every linearly order set $A$, $A$ is a model of $T$ iff $A$ is Dedekind complete.

Answer 1

Suppose $T$ is a first-order theory in the language $L=\{<\}$ such that $(A,<)\models T$ if and only if $(A,<)$ is Dedekind complete.

Let $T_0$ be the theory of dense linear orders without end points (i.e., the set of all sentences $\phi$ deduced from the axioms establishing that $<$ is a dense linear order without end points). It is known that $T_0$ is complete as it is $\omega$-categorical (every countable model is isomorphic to $(\mathbb{Q},<)$.

Let $T_1=Th(\mathbb{R},<)$. Since $(\mathbb{R},<)$ is Dedekind complete, we have $T\subseteq T_1$. On the other hand, since $T_0$ is complete and $(\mathbb{R},<)$ is also a dense linear order without end points, we have that $T_1=T_0$. Hence $T\subseteq T_0$, which implies that $(\mathbb{Q},<)$ is Dedekind complete, a contradiction.

Answer 2

I am using the following definition of Dedekind complete:

Definition: $(A,<)$ is Dedekind complete if every non-empty $S\subseteq A$ that is bounded above has a supremum in $A$.

Suppose $T$ is a first-order theory in $L=\{<\}$ such that for every linear order $\mathcal{A}=(A,<)$, $\mathcal{A}\models T$ if and only if $\mathcal{A}$ is Dedekind complete.

If you are familiar with ultraproducts, a possible answer is this one:

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Consider the structures $\mathcal{A}_n=(\mathbb{R},<)$ for every $n\in\mathbb{N}$, and let $\mathcal{U}$ be a non-principal ultrafilter on $\mathbb{N}$. Finally, consider $\mathcal{A}:=\prod_{\mathcal{U}}\mathcal{A}_n$, the ultraproduct of the structures $(\mathcal{A}_n:n\in\mathbb{N})$ with respect to $\mathcal{U}$. By Los' theorem, since $\mathcal{A}_n\models T$ for all $n\in\mathbb{N}$ (as $(\mathbb{R},<)$ is Dedekind complete), we have that $\mathcal{A}\models T$, and so $\mathcal{A}$ is Dedekind complete.

Consider the set given by $S=\{\overline{c}_k:k\in\mathbb{N}\}$ where $\overline{c}_k=\left[1-\frac{1}{k},1-\frac{1}{k},\ldots\right]_{\mathcal{U}}\in \mathcal{A}$. Notice that $\mathcal{A}\models \overline{c}_i<\overline{c}_j\leq 1$ for all $i<j$, and so $S$ is an increasing sequence which is bounded above. Let $\overline{a}$ be the supremum of $S$ in $\mathcal{A}$, with $\alpha=[a_n]_{\mathcal{U}}$. By Los' theorem, as $\overline{c}_k<\overline{a}$, we have that for every $k\in\mathbb{N}$, the set $X_k=\{n\in\mathbb{N}: 1-\frac{1}{k}<a_n\}\in\mathcal{U}$.

For every $n\in X_k$, let $b_{n,k}\in\mathbb{R}$ such that $1-\frac{1}{k}<b_{n,k}<a_n$, and finally put $\overline{b}=[b_{n,n}]_{\mathcal{U}}$. Then, for a fix $k\in\mathbb{N}$, $1-\frac{1}{k}<b_{n,k}\leq b_{n,n}<a_n$ whenever $n\in X_k\cap [k,+\infty)=X_k\cap (\mathbb{N}\setminus \{1,\ldots,k\})\in\mathcal{U}$, and by Los' theorem, $\overline{c}_k<\overline{b}<\overline{a}$ for all $k\in\mathbb{N}$.

In other words, $\overline{b}$ is an upper bound of the set $S$, with $\overline{b}<\overline{a}=\sup S$, a contradiction.

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Answer 3

Here is a slight variation of Wore's argument: Suppose that $T$ would characterize Dedekind complete linear orders.

Let $T^*$ be $T$ together with $\operatorname{DLONE}$ - the theory of dense linear orders without endpoints. Now $T^*$ characterizes Dedekind complete dense linear orders without endpoints and is consistent with infinite models (as $(\mathbb R;<)$ is such a model).

Hence, by Löwenheim-Skolem, it has a countable model $(X; \prec)$. It follows, by the $\aleph_0$-categoricity of $\operatorname{DLONE}$, that $(X; \prec) \cong (\mathbb Q; <) \models T^*$. Contradiction!