Density of the rationals in the reals

Question

While studying measure theory I have encountered the following set, $$U_\varepsilon=\bigcup_{n\in \mathbb{N}}(q_n-\varepsilon /2^n,q_n+\varepsilon/2^n),$$ where $(q_n)_{n\in \mathbb{N}}$ is an enumeration of the rationals in $[0,1]$. We have $$m^*(U_\varepsilon)\le \sum_{n\in \mathbb{N}}2\varepsilon/2^n=2\varepsilon,$$ and thus for $\varepsilon$ small enough, $U_\varepsilon$ does not contain every irrational in $[0,1]$.

If we are allowed to take any function $f$ so that $f(n)\to 0$ and look at $$U_{f,\varepsilon}:=\bigcup_{n\in \mathbb{N}}(q_n-\varepsilon f(n),q_n+\varepsilon f(n)),$$where $(q_n)_{n\in \mathbb{N}}$ is an enumeration of the rationals in $[0,1]$,

1. Does there exist such a function $f$ so that $[0,1]\subset U_{f,\varepsilon}$ for every $\varepsilon >0$?

2. What are the asymptotics of $f$ ensuring $[0,1]\subset U_{f,\varepsilon}$ for every $\varepsilon >0$?

3. What is the dependency in the enumeration?

And more generally,

4. What else do we know about this concept?

Answer 1

It seems the following.

3.For any monotonic function $f$ such that $f(1)\le 1/2$ and $f(n)\to 0$ and any irrational $\alpha\in [0,1]$ there exists an enumeration $(q_n)_{n\in \mathbb{N}}$ of the rationals in $[0,1]$ such that $U_{f,1}\not\ni\alpha$. For this purpose, while enumerating the rationals of $[0,1]$ it suffices to choose $q_n\not\in (\alpha-f(n), \alpha+f(n))$. Since $f(n)\to 0$, this restriction allows us to enumerate all rationals of $[0,1]$.

4.Let $\alpha\in [0,1]$ be any irrational number and $(q_n)_{n\in \mathbb{N}}$ be any enumeration of the rationals in $[0,1]$. For each natural $n$ put $f(n)=\min \{|q_m-\alpha|:m\le n\}$. Then $U_{f,1}\not\ni\alpha$, but $f(n)\to 0$, because rationals are dense in $[0,1]$.

1.Let $(q_n)_{n\in \mathbb{N}}$ be any enumeration of the rationals in $[0,1]$. For each natural $n$ let $g(n)$ denotes the number of the last occurrence of a rational with a denominator $n$. Clearly, $g(n)\to\infty$. If the function $f$ is monotonically non-increasing and $f(g(n))>1/n$ then $[0,1]\subset U_{f,\varepsilon}$ for every $\varepsilon >0,$ because if $n\ge 1/\varepsilon$ then $U_{f,\varepsilon}\supset \{q_m:m\le g(n)\}+[-1/n,1/n]\supset [0,1].$