Let $r\in [0,1]\setminus\mathbb{Q}$, let $\mathbb{N}$ denote the set of non-negative integers and let $\mathbb{N}_+=\mathbb{N}\setminus\{0\}$. For $n\in\mathbb{N}_+$ let $$\alpha_r(n)=\min\{|r-\frac{m}{n}|: m\in\mathbb{N}\}$$ be the best approximation of $r$ that can be obtained using $n$ as the denominator.
We inductively define $\text{appr}_r:\mathbb{N} \to \mathbb{N}$, the approximation sequence of $r$:
- $\text{appr}_r(0) = 1$, and
- $\text{appr}_r(n+1) = \min\{m\in\mathbb{N}: m \geq \text{appr}_r(n) \text{ and } \alpha_r(m) < \alpha_r(\text{appr}_r(n))\}$ for all $n\in \mathbb{N}$.
Let $\mathbb{N}^\mathbb{N}$ denote the set of all functions $f:\mathbb{N}\to\mathbb{N}$. Given $f, g:\mathbb{N}\to\mathbb{N}$ we say $f\leq^* g$ iff there is $N\in\mathbb{N}$ such that $f(k) \leq g(k)$ for all $k\geq N$.
We say ${\cal D}\subseteq \mathbb{N}^\mathbb{N}$ is dominating if for all $f\in \mathbb{N}^\mathbb{N}$ there is $d\in {\cal D}$ such that $f\leq^* d$, and we say ${\cal B}\subseteq \mathbb{N}^\mathbb{N}$ is unbounded if for all $f\in\mathbb{N}^\mathbb{N}$ there is $b\in {\cal B}$ such that $b\not\leq^*f$. (A diagonalization argument shows that every unbounded (and therefore also every dominating) family must be uncountable, and there are interesting set-theoretical considerations in this context, see here.)
Question. Is $$\{\text{appr}_r: r\in[0,1]\setminus\mathbb{Q}\}\subseteq \mathbb{N}^\mathbb{N}$$ dominating? If not, is it unbounded?
