A set of sentences that has a model iff that model is finite Is there a set of sentences $\Sigma$ such that for any structure $A$, $A$ is a model of $\Sigma$ iff the universe of $A$ is finite?
I'm a little unsure. The first implication seems reasonable, that there is a set of sentences that has no infinite model. But I can't think of a set of sentences that is modelled by every finite structure. Not sure how to go about the proof though...
 A: As you pointed out, there are set of sentences that imply that any model is finite, for instance $\exists x, \forall y, y=x$ only has models of cardinality $1$.
However, there can be no set of formulas $\Sigma$ such that $A\models \Sigma \iff A$ is finite.
Indeed, assume there were such a set (formulated in the language $L$), and consider constant symbols $\{c_n, n\in \mathbb{N}\}$ that are not in $L$, and such that $c_n\neq c_m$ whenever $n\neq m$. Then consider $L'$ the language $L$ with these constants, and $\Sigma'$ the theory consisting of $\Sigma$ and the formulas $c_n\neq c_m$ whenever $n\neq m$. 
Then any model of $\Sigma'$ is finite since its $L$-reduct models $\Sigma$ + assumption on $\Sigma$. And any model of $\Sigma'$ is infinite, due to the conditions on the $c_n$'s. So $\Sigma'$ cannot have any model : it is inconsistent.
However, $\Sigma'$ is finitely consistent, since any finite subset $S\subset \Sigma'$ is contained in a set of the form $\Sigma\cup \{c_n\neq c_m\mid n\neq m, n,m\in J\}$ where $J$ is finite, if you interpret $c_n$ by $n$, then $J\models \Sigma$ since $J$ is finite, and so $J\models S$.
Therefore the compactness theorem implies that $\Sigma'$ is consistent : a contradiction
