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



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> $\text{appr}\_\alpha(1) = 1$ 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 $n\in\mathbb{N}\_+$.
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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 $f\leq^\*\text{appr}\_\alpha$?



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


