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correction pointed out by Greg Martin
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Martin Sleziak
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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.

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 $B$ and the asymptotic density of this set is equal to one.

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.

deleted 3 characters in body
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Martin Sleziak
  • 4.7k
  • 4
  • 35
  • 40

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})&=f(a_k) \end{align*}\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 $B$ and the asymptotic density of this set is equal to one.

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})&=f(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 $B$ and the asymptotic density of this set is equal to one.

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 $B$ and the asymptotic density of this set is equal to one.

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Martin Sleziak
  • 4.7k
  • 4
  • 35
  • 40

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})&=f(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 $B$ and the asymptotic density of this set is equal to one.