[proofplan]
Assume that a first-order theory $T$ axiomatizes exactly the finite $L$-structures. We enlarge $T$ by adding, for each $n \in \mathbb{N}$, a sentence asserting that the universe has at least $n$ distinct elements. Every finite fragment of the enlarged theory has a finite model by the hypothesis that finite $L$-structures of arbitrarily large size exist. Compactness then gives a model of the entire enlarged theory, which is forced by $T$ to be finite and by the added sentences to be infinite, a contradiction.
[/proofplan]
[step:Introduce sentences forcing arbitrarily large finite lower bounds]
For each $n \in \mathbb{N}$, define the first-order $L$-sentence $\sigma_n$ by
\begin{align*}
\sigma_n := \exists x_1 \cdots \exists x_n \bigwedge_{1 \leq i < j \leq n} x_i \neq x_j .
\end{align*}
Thus an $L$-structure $M$ satisfies $\sigma_n$ exactly when $|M| \geq n$. Define the $L$-theory
\begin{align*}
\Gamma := T \cup \{\sigma_n : n \in \mathbb{N}\}.
\end{align*}
[/step]
[step:Show that every finite fragment of the enlarged theory has a model]
Let $\Gamma_0 \subset \Gamma$ be finite. Since $\Gamma_0$ contains only finitely many of the sentences $\sigma_n$, define
\begin{align*}
N := \max\bigl(\{1\} \cup \{n \in \mathbb{N} : \sigma_n \in \Gamma_0\}\bigr).
\end{align*}
By hypothesis, there exists a finite $L$-structure $M$ with $|M| \geq N$. Since $M$ is finite and $T$ is assumed to axiomatize exactly the finite $L$-structures, we have $M \models T$. Also, for every $\sigma_n \in \Gamma_0$, we have $n \leq N \leq |M|$, so $M \models \sigma_n$. Therefore $M \models \Gamma_0$.
[guided]
We must verify finite satisfiability of $\Gamma$, because compactness applies exactly to theories whose finite subsets have models. Let $\Gamma_0 \subset \Gamma$ be a finite subset. The only new sentences beyond $T$ are the size lower bounds $\sigma_n$, and a finite set $\Gamma_0$ can mention only finitely many of them. We collect the largest requested lower bound by setting
\begin{align*}
N := \max\bigl(\{1\} \cup \{n \in \mathbb{N} : \sigma_n \in \Gamma_0\}\bigr).
\end{align*}
The set inside the maximum is finite and nonempty, so $N$ is well-defined. The hypothesis on $L$ supplies a finite $L$-structure $M$ with $|M| \geq N$. Since $T$ is assumed to define exactly the finite $L$-structures, this finite structure satisfies every sentence of $T$, hence satisfies every sentence of $\Gamma_0 \cap T$. For each sentence $\sigma_n \in \Gamma_0$, the inequality $n \leq N \leq |M|$ gives $n$ distinct elements of the universe of $M$, so $M \models \sigma_n$. Hence $M$ satisfies every sentence in $\Gamma_0$.
[/guided]
[/step]
[step:Apply compactness to obtain a model of all size lower bounds]
Since every finite subset of $\Gamma$ has a model, the First-Order [Compactness Theorem](/theorems/2748) (citing a result not yet in the wiki: First-Order Compactness Theorem) implies that there exists an $L$-structure $M_\infty$ such that $M_\infty \models \Gamma$.
[guided]
The previous step proves the exact hypothesis needed for compactness: every finite subset $\Gamma_0 \subset \Gamma$ is satisfiable. Applying the First-Order Compactness Theorem (citing a result not yet in the wiki: First-Order Compactness Theorem), we obtain an $L$-structure $M_\infty$ satisfying the whole theory $\Gamma$. This means simultaneously that $M_\infty \models T$ and that $M_\infty \models \sigma_n$ for every $n \in \mathbb{N}$.
[/guided]
[/step]
[step:Derive the contradiction between finiteness and all lower bounds]
Because $M_\infty \models T$ and $T$ axiomatizes exactly the finite $L$-structures, $M_\infty$ is finite. Let $m := |M_\infty| \in \mathbb{N}$. Since $M_\infty \models \sigma_{m+1}$, the structure $M_\infty$ has at least $m+1$ distinct elements, contradicting $|M_\infty| = m$. Therefore no such first-order $L$-theory $T$ exists.
[/step]
