Let $\mathbb{N}$ denote the set of non-negative integers. If $f:\mathbb{N}\to\mathbb{N}$ is a map, we set for $k\in \mathbb{N}$:


* $f^{(0)}(k) = k$, and
* $f^{(n+1)}(k) = f(f^{(n)}(k))$ for all $n\in\mathbb{N}$.


We say $r\in \mathbb{N}$ is a *rocket element of $f$* if $f^{(n)}(r) < f^{(n+1)}(r)$ for all $n\in \mathbb{N}$ and denote the set of all rocket elements of $f$ by $\text{Roc}(f)$. 


If $A\subseteq \mathbb{N}$ we define the *upper density $\mu^+(A)$* of $A$ by $$\mu^+(A) = \limsup\_{n\to\infty}\frac{|A\cap \{0,\ldots n\}|}{n+1}.$$


By $S\_\mathbb{N}$ we denote the set of all bijections $f:\mathbb{N}\to \mathbb{N}$.


**Question.** Is there $f\in S\_\mathbb{N}$ such that $\mu^+(\text{Roc}(f)) = 1$?


**Bonus question.** If not, what is the value of $\sup\{\text{Roc}(f): f\in S\_\mathbb{N}\}$?


(**Note.** Only the question needs to be answered for acceptance, but I am also keen to know about the "bonus question" if the answer cannot be directly inferred from the answer to the question.)



"Rocket elements" in bijections $f:\mathbb{N}\to \mathbb{N}$
Choose any infinite set $A\subseteq\mathbb N$ such that $\mu^+(A)=0$. Enumerate both $A$ and $B=\mathbb N\setminus A$ as 
\begin{align\*}
A&=\{a\_1<a\_2< \dots < a\_n < \dots\}\\
B&=\{b\_1<b\_2< \dots < b\_n < \dots\}
\end{align\*}
Then you can definite the function $f\colon\mathbb N\to\mathbb N$
\begin{align\*}
f(b\_k)&=b\_{k+1}\\
f(a\_1)&=b\_1\\
f(a\_{k+1})&=a\_k
\end{align\*}
for $k=1,2,\dots$. It is clear that $f$ is a bijection.


For this function $f$, the set of the rocket elements is either $B$ or $B\cup\{a\_1\}$ and the asymptotic density of this set is equal to one.


