Measuring how "close" $\alpha\in[0,1]\setminus\mathbb{Q}$ is to being rational

Question

Let $\mathbb{N}_+$ denote the set of positive integers and let $\mathbb{N}_0 = \mathbb{N}_+\cup\{0\}$. Fix $\alpha\in[0,1]\setminus \mathbb{Q}$. For $n\in\mathbb{N}_+$ we let the approximation radius of $n$ be $$\text{rad}_\alpha(n) = \min\Big\{\Big|\alpha-\frac{x}{n}\Big|:x\in\mathbb{N}_0\text{ and } x \leq n\Big\}.$$

Inductively define the strictly increasing approximation radius sequence $(\text{appr}_\alpha)_{n\in\mathbb{N}_+}$ by

$\text{appr}_\alpha(1) = 1$ and $\text{appr}_\alpha(n+1) = \min\{x\in\mathbb{N}_+: x>n \text{ and }\text{rad}_\alpha(x) < \text{rad}_\alpha(n)\}$ for all $n\in\mathbb{N}_+$.

For $f,g: \mathbb{N}_+\to \mathbb{N}_+$ we say that $f\leq^*g$ if there is $N\in\mathbb{N}_+$ such that for all $x\in\mathbb{N}_+$ with $x\geq N$ we have $f(x) \leq g(x)$.

Question. Given any function $f: \mathbb{N}_+\to \mathbb{N}_+$, is there $\alpha\in[0,1]\setminus\mathbb{Q}$ such that $f\leq^*\text{appr}_\alpha$?

Answer 1

I think the answer is yes.

We will assume that $\text{appr}_{\alpha}$ is defined for $\alpha\in\mathbb{Q}$ also but $\text{appr}_{\alpha}(n) = \infty$ for all large $n$. Consider another function $\text{Appr}_{\alpha}(n)\colon \mathbb{N}\to\mathbb{N}_+^2$ which will return the pair $(x, \text{appr}_{\alpha}(n))$, where $x$ is the optimal numerator from the definition of $\text{rad}_{\alpha}$. For positive integers $n,x, N$, define the set $$ A_{n,x,N} = \left\{\alpha\mid \text{Appr}_{\alpha}(n) = (x,N)\right\}. $$

Lemma. Let $n, x, N, N'$ be positive integers and $I = [x/N,r]$ be a segment such that $I\subset A_{n,x,N}$. Then there exists $M > N'$ such that for some $y$ and $r' > y/M$ we have $[y/M, r']\subset (I\cap A_{n + 1,y,M})$.

First let us derive the statement from the lemma. Let $I_0 = [1/2, 3/5]\subset A_{2, 1,2}$ be some starting segment. We have $n_0 = 2, x_0 = 1, N_0 = 2$. On each step apply lemma to $n = n_k, x = x_k, x = N_k$ and $N' = f(n_k + 1)$ and define $n_{k + 1} = n_k + 1, x_{k + 1} = y, N_{k + 1} = M$ then we have the nested sequence of the segments $I_0, I_1, \ldots$ such that $I_k = [x_k/N_k, r_k]$ and $I_k\subset A_{n_k,x_k,N_k}$. There exists $\alpha\in \cap I_k$. By definition we have $\text{appr}_{\alpha}(k) = N_k \ge f(k)$.

Proof of the lemma. There exists $r_1$ such that for all $\alpha\in[x/N, r_1]$ we have $\text{appr}_{\alpha}(n + 1) > N'$ (this is all points such that $x/N$ is closer to them then any other fraction with denominator $\le N'$). Take an arbitrary rational number $y/M\in (x/N, r_1)$. Again we can choose $r'$ such that $y/M$ is the closest rational number with denominator $\le M$ for all point in $[y/M, r_1]$. The lemma follows.