First-Order Logic: If $\mathcal{N}\models\phi$ for every $\mathcal{N}$, then $\mathcal{M} \cong \mathcal{N}$.

Question

Given that the alphabet $\mathcal{A}$ is finite and that $\mathcal{M}$ is a finite $\mathcal{L}_A$ structure, prove that there is an $\mathcal{L}_A$-sentence $\phi$ such that for every $\mathcal{L}_A$-structure $\mathcal{N}$, if $\mathcal{N}\models\phi$, then $\mathcal{M} \cong \mathcal{N}$.

Answer 1

If the universe of $\mathcal{M}$ is finite then the interpretations of any constant, function or relation in the language is a finite object. Since there are finitely many of them, one can come up with a sentence with as many variables as there are elements in $\mathcal{M}$, then take the conjunction of the sentence that says that they are all unique and nothing else exists, in conjunction with formulas that spell out the interpretation of everything, element by element. Can you see how?