Is subset relation axiomatizable? We know that the ZFC axioms define the elementary class associated with them. And we can extend the signature to a binary relation symbol P and add a defining axiom that says P is the subset relation. So, the subset relation is at least a psuedo-elementary class. My question is, is it also an elementary class?
 A: I believe it is not an elementary class. I sketch a proof, below, but have not checked in detail an important piece. I might come back and verify the details if I have time.
In the language $L$ containing only $\subseteq$, let $T$ be the theory asserting that $\subseteq$:


*

*is a lattice,

*has a lower bound but no upper bound,

*has relative complements,

*is atomic (in that everything has an atom below it),

*has infinitely many atoms.


Clearly, the $L$-reduct of ZFC contains $T$. I will show that


*

*$T$ is complete, and

*$T$ has a model not isomorphic to the reduct of any model of ZFC.


Completeness
Note that, in $T$, finite (including 0-ary) unions and finite (excluding 0-ary) intersections, as well relative complement are defined, so we might as well include them in $L$ and expand $T$ accordingly. We can similarly expand $L$ and $T$ with predicates $|A| = n$ for every natural number $n$, defined as saying $A$ extends exactly $n$ atoms. A standard back-and-forth argument should verify that this theory is complete and eliminates quantifiers (in the expanded $L$). I didn't check in detail that the back-and-forth works as I expect it does.
Model of $T$ that isn't a reduct
Fix an infinite set $X$, and let $M$ be the $L$-structure consisting of the finite subsets of $X$, with $\subseteq$ interpreted as the usual subset relation. Then $M \models T$. $M$ cannot be the reduct of a model of ZFC, however, since any model of ZFC contains an infinite set (and therefore a set extending infinitely many atoms).
