**Background**
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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.
* $P\subset 2^\omega$ is a rational perfect set if it's nonempty, perfect (closed without isolated points) and $P\cap \mathbb Q$ is dense in $P$.


It's possible to show that being a perfect tree is absolute for transitive models of ZFC, therefore if $M$ is such a model model, then $\mathbb M^M=\mathbb M\cap M$.


Let $M$ be a ctm and $G$ be $\mathbb M^M$-generic over $M$. The Miller real is defined as:


$$f=\bigcup\bigcap G$$


It's possible to show that the domain of $f$ is $\omega$, that $\bigcap G=\{f|n: n \in \omega\}$ and that $G=\{p \in \mathbb M^M: \forall n \in \omega\,(f|n \in p)\}$.


**Question**
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I have read in [this article](http://www.matmor.unam.mx/~michael/reprints_files/MAD_families_and_the_rationals.pdf) (theorem 4, first implication) that if $F\subset 2^\omega$ and $F \in M$ is such that $(F$ is closed in $2^\omega$ and $F$ does not contain any perfect rational subset$)^M$, then if $f$ is a miller real, it follows that $f \notin \operatorname{cl} F$. But I don't know why.


I managed to prove that under these hypothesis, $(\operatorname{cl} (F\cap \mathbb Q)$ is countable$)^M$ and therefore $\operatorname{cl}^M (F\cap \mathbb Q)=\operatorname{cl} (F\cap \mathbb Q)$, but I don't know if this helps.



Miller real is not in the closure of sets under some conditions
For any closed set $F\in V$, one can show $D\_F=\{p\in \mathbb{P}: \exists \text{open }U \ [p]\subset U\& U\cap F=\emptyset\}$ is dense. Just a comment though, I believe when people are talking about Miller forcing, another form (essentially the same) is more common, i.e subtrees of $\omega^{<\omega}$ such that each node has an extension that splits infinitely (has infinitely many immediate extensions). You are welcome to provide more details if you'd like :-). 


