We can define Miller Forcing as the poset of nonempty perfect rational trees. That is, we define:
$p\subset 2^{<\omega}$ is a perfect tree iff it is closed downwards (for all $s, n$, if $s \in p$ and $n \in \omega$, then $s|n \in p$) and every branch splits (for all $s \in p$ there exists $t \in p$ such that $s\subset t$ and $t^\frown(0), t^\frown(1) \in p$.
$\mathbb Q\subset 2^{\omega}$ is the set of binary sequences that are eventually $0$ ($\mathbb Q=\{s^\frown \textbf{0}:s \in 2^{<\omega}\}$, where $\textbf{0}$ is the constant zero function.
Given a perfect tree $p$, $[p]=\{x \in 2^\omega: \forall n \in \omega\,(x|n \in p)\}$. It's possible to show that $[p]$ is a perfect set, and that every perfect set is of this kind.
A perfect tree $P$ is a rational perfect tree iff $\mathbb Q\cap [p]$ is dense in $[p]$.
$\mathbb M$ is the set of all perfect rational trees ordered by inclusion $p\leq q \leftrightarrow p\subset q$
If $p$ is a perfect tree, we define $\operatorname{Br}(p)=\{s \in p: s^\frown (0), s^\frown (1) \in p\}$ (the set of the branching points of $p$) and $\operatorname{Br}_n(p)=\{s \in \operatorname{Br}(p): |\{t \in \operatorname{Br}(t): t\subset s\}|=n\}$ (the set of the $n$-ths branching points of $p$. There are $2^{n-1}$ of these points).
We define for $p, q \in \mathbb M$, $p\leq_n q$ iff $p\leq q$ and $\operatorname{Br}_n(p)=\operatorname{Br}_n(q)$.
A fusion sequence is a sequence of elements of $\mathbb M$, $(p_n)_{n \in \omega}$ such that for every $n$, $p_{n+1}\leq_{n+1}p_n.$
Let $(p_n)_{n \in \omega}$ be a fusion sequence. It's easy to show that $p=\bigcap_{n \in \omega}p_n$ is a perfect tree. Is it true that $p$ is also a rational perfect tree? i.e., is is true that $[p]\cap \mathbb Q$ is dense in $[p]$?