Non-isomorphic projective planes on $\omega$ Let $X\neq \emptyset$. A set $L\subseteq {\cal P}(X)$ is said to be a projective plane on $X$ if the following conditions are met:


*

*if $x\neq y\in X$ there is a unique $l\in L$ such that $x, y \in l$,

*if $l\neq m \in L$, then we have $|l\cap m | = 1$, 

*there are four distinct elements of $X$ such that no member of $L$ contains more than $2$ of the four.


What is the maximal cardinality of a collection ${\cal C}$ of projective geometries on $\omega$ such that no two distinct members of ${\cal C}$ are isomorphic? (The notion of ismorphism of projective planes is defined below.)

Note. If $L, M$ are projective planes on $X$, we say they are isomophic if there is a bijection $\varphi: X\to X$ such that $l\in L$ if and only if $\varphi(l) \in M$.
 A: You ask for the number of isomorphism classes of projective planes on $\omega$.  I claim that it is exactly $2^{\aleph_0}$.
It is at most $2^{\aleph_0}$.
Indeed, a projective plane on $\omega$ can be encoded by the set of $\{x,y,z\}\subset\omega$ of cardinal $3$ which are aligned (i.e., such that $z$ lies on the unique line containing $x,y$), and the set of $3$-element subsets of $\omega$ is countable, so the set of its subsets has cardinal $2^{\aleph_0}$.
It is at least $2^{\aleph_0}$.
For this, it is enough to show that there are at least $2^{\aleph_0}$ pappian projective plane, i.e., planes of the form $\mathbb{P}^2(K)$ for some countable field $K$ : note that $K$ is determined, up to isomorphism by the incidence structure (choose $0,1,\infty$ on some line and construct addition and multiplication by the usual projective constructions).  So it is enough to show that there are at least $2^{\aleph_0}$ isomorphism classes of countable fields.  But this is easy: take any subset $S$ of the set of prime numbers and consider the field $K_S := \mathbb{Q}(\sqrt{p} : p\in S)$ (the set $S$ can be recovered as the set of primes $p$ having a square root in $K_S$).
