This might be quite a trivial question but this has been bugging me, I don't know what the resolution is. Context: the irrationality measure $\mu=\mu(x)$ of some real number $x$ is defined to be that unique positive real number for which:
For all $\varepsilon\gt0$, the inequality: $$\left|x-\frac{p}{q}\right|\lt\frac{1}{q^{\mu-\varepsilon}}$$Has infinitely many solutions and the inequality: $$\tag{$\ast$}\left|x-\frac{p}{q}\right|\lt\frac{1}{q^{\mu+\varepsilon}}$$Has only finitely many solutions.
The denominators $q$ are taken positive, and all references to $(p,q)$ throughout this post are for rational numbers in lowest terms.
According to online sources, the above is equivalent to the statements:
For all $\varepsilon\gt0$, there exists a constant $C>0$ such that: $$\left|x-\frac{p}{q}\right|\ge\frac{C}{q^{\mu+\varepsilon}}$$Holds for all $(p,q)$. However, there does not exist any constant $D\gt0$ such that: $$\left|x-\frac{p}{q}\right|\ge\frac{D}{q^{\mu-\varepsilon}}$$Holds for all $(p,q)$.
In showing these two to be equivalent, one direction is completely trivial - if only finitely many solutions to such an inequality exist, we can probe the finite solution set and find a constant $C$, sufficiently small, to be able to reverse the inequality sign. However, I am stuck on how to show: $$\left|x-\frac{p}{q}\right|\ge\frac{C}{q^{\mu+\varepsilon}}$$Implies that only finitely many solutions to the reverse inequality $(\ast)$ exist.
The progress I was able to make:
Assume the existence of a constant $C>0$ such that $|x-p/q|\ge Cq^{-\varepsilon-\mu}$ holds for all $(p,q)$. Take any $\varepsilon'>0$. $Cq^{-\varepsilon-\mu}\ge q^{-\varepsilon'-\varepsilon-\mu}$ holds iff. $q^{\varepsilon'}\ge\frac{1}{C}$, and this can be arranged for large $q>Q$, for some $Q\in\Bbb N$. Thus the inequality $(\ast)$ could only be satisfied if $q\le Q$, so only finitely many solutions exist.
However, I don't know how to pass from $\varepsilon+\varepsilon'$ (for all $\varepsilon'\gt0$) to just $\varepsilon$. There is no clear continuity structure.
I would appreciate any pointers!
The key point is that both statements are for all $\epsilon >0$ - this doesn't mean $\epsilon$ has to be the same in both sides of the implication!
In detail, we are assuming, for all $\epsilon'>0$, there is a constant $C>0$ such that $\lvert x-p/q\rvert \geq C/q^{\mu+\epsilon'}$ for all $p,q$, and want to show that for all $\epsilon>0$ the inequality $\lvert x-p/q\rvert < 1/q^{\mu+\epsilon}$ has only finitely many solutions.
Use the assumption with $\epsilon'=\epsilon/2$. If the second part fails, then there are infinitely many $q$ such that $$\frac{C}{q^{\mu+\epsilon/2}}\leq \lvert x-p/q\rvert <\frac{1}{q^{\mu+\epsilon}}.$$ This implies $q^{\epsilon/2}\leq C$, which is a contradiction for sufficiently large $q$.