Does there exist an uncountable $P \subset \mathcal{P}(\mathbb{N}) $ with the property that for any distinct $x,y \in P$, $|x \cap y|$ is prime? 


A more general, but likely harder, question: is it possible to characterize the set $\mathcal{A}$ of all subsets $S \subset \mathbb{N}$ with the following property: there is some uncountable $P \subset \mathcal{P}(\mathbb{N})$ such that for all distinct $x,y \in P$, $|x \cap y| \in S$. 


It's easy to show no $S$ finite belongs to $\mathcal{A}$. It's also not hard to show every $\text{mod} \ m$ equivalence class, belongs to $\mathcal{A}$.


One reasonable sounding idea is that this is related to the asymptotic density of $S$ in $\mathbb{N}$. 



Cardinality of families of subsets of $\mathbb{N}$ whose intersections are finite
I guess this is a variant of Noah's construction:


Let $S$ be an infinite subset of $\mathbf{N}=\{0,1,2,\dots\}$. Define a leafless rooted tree $V\_S$, starting from a root at level 0, such that given a vertex $v$ of level $n\ge 0$, $v$ has exactly 1 successor if $n\notin S$ and exactly 2 successors if $n\in S$.


Let $\mathcal{W}\subset\mathcal{P}(V\_S)$ be the set geodesic rays not containing the root, based at a vertex of level 1; since $S$ is infinite, $\mathcal{W}$ has cardinal $2^{\aleph\_0}$. Then for any distinct $U,V\in\mathcal{W}$, $U\cap V$ is either empty, or is the geodesic segment from the some vertex of level 1 to the branching point $v$ at which $U,V$ fork; this has cardinal $n$, the level of $v$; also in case $0\notin S$, $U\cap V$ cannot be empty. Conversely every such geodesic segment has the form $U\cap V$ for suitable $U,V$, and so is the empty set if $0\in S$. So the set of $|U\cap V|$ when $U,V$ range over distinct elements of $\mathcal{W}$, is exactly $S$.


