Suppose $A\subseteq\mathbb{R}$ and $x\in \mathbb{R}$. I want a measure $\mu(x,A)$ that determines "how well" $x$ can be approximated by $A$?
For example, if $A=\mathbb{Q}$, we use the irrationality measure:
Definition
Let $x$ be a real number, and let $R$ be the set of positive real numbers $\mu$ for which
$$0<\left|x-\frac{p}{q}\right|<\frac{1}{q^{\mu}}$$
has at most finitely many solutions $p/q$ for $p,q\in\mathbb{Z}$. Then the irrationality measure or irrationality exponent is defined as the "threshold" where $x$ is no longer approximable by the rational numbers.
$$\mu(x)=\inf\limits_{\mu\in R}\mu$$
In general notation, $\mu(x,\mathbb{Q})=\mu(x)$
However, if $x\in A$ (for this case, $x\in\mathbb{Q}$), I want $\mu(x,A)$ to equal infinity and other cases of $x$ to give the same result before as if $x\in A$, we don't have to approximate it with set $A$ (we could just take $x$).
I believe this can be done by changing the inequality $0<\left|x-\frac{p}{q}\right|<\frac{1}{q^{\mu}}$ to
$$\left|x-\frac{p}{q}\right|\le \frac{1}{q^{\mu}}$$
Problem
I would like to find a similar $\mu$ for $A$ in general, which is similar to the irrationality measure.
For instance, consider $A=\left\{\sqrt{a}:a\in\mathbb{Q}\right\}$
If $x$ is a real number, I would like to find a function $g:\mathbb{R}^2\to\mathbb{R}$ where if $R$ is a set of positive real numbers $\mu$ then
$$\left|x-\frac{\sqrt{p}}{\sqrt{q}}\right|\le g(p,q)^{\mu}$$
has at most finitely many solutions $\sqrt{p}/\sqrt{q}$ for $p,q\in\mathbb{Z}$ such that
$$\mu(x,A)=\inf\limits_{x\in R}\mu$$
Question 1: In this case, should $g(p,q)=1/\sqrt{q}$ or should it be something else?
Question 2: What about for other sets such as:
$$A=\left\{\sqrt{m}-\sqrt{n}:m,n\in\mathbb{N}\right\}$$
$$A=\left\{\frac{e^\left(\sqrt{a}-\sqrt{b}\right)}{\ln(c)}:a,b\in\mathbb{Z},c\in\mathbb{N}+1\right\}$$
$$A=\text{Irrationals}$$
I expect for most $x\in \mathbb{R}$, $\mu(x,A)$ should range between one and two?