I've been having a hard time deciphering Kunen's suggestion in the following exercise:
Overall, I'd like to get a little more direction in solving this exercise. Specifically, I would like some help determining precisely who the poset $\mathbb{P}$ is. From what is said there, it seems to me that the set $\mathbb{P}$ is the product $[2^{<\omega}]^{<\omega} \times [E]^{ <\omega}$, but, if this is correct, I haven't been able to find the adequate partial order that makes everything work well.
On the other hand I imagine that, once it is revealed who $\mathbb{P}$ is, it will be more or less easy to say which are the dense subsets that one needs to apply $\mathsf{MA}(|E|)$; however, if it's not too much trouble, a little guidance on this matter would also be very helpful.
The way I would define the poset that is described is as follows:
A condition $p \in \mathbb{P}$ is a triple $(t,n,e)$, where $n \in \omega$, $t$ is a function from $2^{\leq n}$ to non-empty rational open boxes of $\mathbb{R}\times \mathbb{R}$ such that $s \subseteq s' \in 2^{\leq n}$ implies $t(s') \subseteq t(s)$ (the hint also says that incompatible $s$ should map to disjoint boxes, but this will be irrelevant by a density argument) and $e$ is a finite subset of $E$. Whenever $(x,y) \in e$ it gives you a promise that the Cantor set defined by the generic tree contains $(x,y)$. Thus we should additionally require that for every $(x,y) \in e$ there is some $s \in 2^n$ such that $$(x,y) \in t(s).$$
Now the order is the "natural" extension relation: $$(t',n',e') \leq (t,n,e) $$ iff $t \subseteq t'$, $n \leq n'$ and $e \subseteq e'$.
Can you make it from here?