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Let $T$ be a complete first order theory with a infinite model $M$. I want to show that every model $N$ of $T$ must be infinite.

Since $T$ is complete then for every sentence $\phi$ we can either derive $\phi$ or $\neg \phi$ from $T$.

By a corollary of the Upwards Löwenheim-Skolem theorem, if $T$ is a consistent theory which has an infinite model of cardinality $\kappa, \kappa \geq \vert L \vert$. In particular if $L$ is countable then $T$ has models of every infinite cardinal.

So we know that $T$ has a lot of infinite models but how can I be sure that no finite structures model $T$. How can I show this?

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Hint: suppose $M \models T$ is finite of size $n$. Can you come up with a first order statement that expresses that $M$ has size $n$? Then since $T$ is complete, that statement or its negation would be in $T$.

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Any complete theory with a finite model is categorical. So if $T$ a complete theory has an infinite model, it is not categorical, so it cannot have a finite model.

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It also works if $T$ is not necessarily complete.

Let's show that if $T$ an $\mathcal{L}$-theory has an infinite model then $T$ has infinite models of arbitrary large cardinal (i.e for all set $X$ there exists a model $\mathfrak{M}\models T$ such that $\vert M\vert \ge \vert X\vert$ where $M$ is the domain of $\mathfrak{M}$).

Let's add to $\mathcal{L}$ the following set of new constants $\{c_x : x\in X\}$. Then let's define $\mathcal{L}'=\mathcal{L}\cup \{c_x : x \in X\}$ where the $(c_x)_{x\in X}$ are pairwise distincts.

Let's consider $T' = T \cup \{c_x \neq c_y : x,y \in X ;x\neq y \}$ and prove that $T'$ is consistent. Indeed if $\mathfrak{N}' \models T'$ (in particular $\mathfrak{N}'\models T$) hence $\mathfrak{N}:= \mathfrak{N}'_{\vert \mathcal{L}} \models T$ and if we denote by $N$ the common domain of $\mathfrak{N}$ and $\mathfrak{N}'$ then the map $f : X \to N,\ x \mapsto c_x^{\mathfrak{N}'}$ is injective.

Let $T_0\subseteq T'$ finite. There exists $X_0$ a finite set such that $T_0 \subset T \cup \{c_x \neq c_y : x,y \in X_0 ;x\neq y \}$. Let $\mathfrak{M}$ an infinite model of $T$ (the existence is provided by the hypothesis), $g$ an injective map from $X_0$ to $M$ and $a \in M$. Then we define $\mathfrak{M}'$ an expansion of $\mathfrak{M}$ to $\mathcal{L}'$ which is a model of $T \cup \{c_x \neq c_y : x,y \in X_0 ;x\neq y \}$ and in particular of $T_0$ by setting $\mathfrak{M}'_{\vert \mathcal{L}}=\mathfrak{M}$, for all $x\in X_0, \ g(x)=:c_x^{\mathfrak{M}'}$ and for all $x\in X\setminus X_0$, $c_x^{\mathfrak{M}'}:=a$. Then $T'$ is finitely consistent hence consistent by compactness theorem.

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  • $\begingroup$ this seems to be a correct proof of the upward LS theorem but how does it have any relevance to the question, which is about showing the theory has no finite models (which does require completeness)? $\endgroup$ Nov 9, 2021 at 1:10
  • $\begingroup$ @spaceisdarkgreen With that, we proved that $T$ admits only infinite models. $\endgroup$
    – Maman
    Nov 9, 2021 at 10:50
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    $\begingroup$ How? You proved it admits models of all infinite sizes, not that it only admits infinite models. If we don’t require the theory is complete, there are many counterexamples... for instance the theory of linear orders has both infinite and finite models. $\endgroup$ Nov 9, 2021 at 13:40
  • $\begingroup$ @spaceisdarkgreen I see ! Indeed good remark ! So My mistake ... $\endgroup$
    – Maman
    Nov 9, 2021 at 15:00

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