I'm studying the model theory notes of https://faculty.math.illinois.edu/~anush/Courses/2016F.Math571.MT/Henson--van%20den%20Dries.pdf for model theory. I'm finding it difficult with the 6.11 exercise:
Let $L$ be the first order language with two binary function symbols ∩ and ∪, a unary function symbol c, and two constant symbols 0 and 1. For each set S let P(S) denote the $L$-structure based on the power set of S. That is, the underlying set of P(S) is the collection of all subsets of S, we interpret ∩,∪,c as intersection, union, and complement, respectively, and we interpret 0,1 as ∅,S, respectively. Let K be the class of all $L$- structures that are isomorphic to P(S) for some set S. Show that K is not axiomatizable.
I think the solution involves arguing that K is axiomatizable and then finding a contradiction. If anyone can give a comprehensive answer I would appreciate it.
Suppose for a contradiction that the class $K$ is axiomatisable by some first-order theory $T$. Consider the structure $P(\mathbb{N})$, which is in $K$. This structure is infinite (it has cardinality $2^{\aleph_0}$). By the downward Löwenheim-Skolem theorem there is then a countable elementary substructure $A \preceq P(\mathbb{N})$. Being an elementary substructure means in particular that $A$ satisfies the same sentences as $P(\mathbb{N})$. So because $P(\mathbb{N}) \models T$ we also have $A \models T$. That would mean that $A$ is in the class $K$. By definition of $K$ we then have that $A$ is of the form $P(S)$ for some infinite set $S$. Because $|S| \geq \aleph_0$ we have: $$ |A| = |P(S)| = 2^{|S|} \geq 2^{\aleph_0}. $$ But this contradicts that $A$ is countable. So we find our contradiction and conclude that the class $K$ cannot be first-order axiomatisable.