Approximation of irrationals in $[0,1]$ Enumerate the rationals in $[0,1]$ as follows:
$\{0,1, 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5,... \}$, and let $F_n$ be the set consisting of the first $n$ elements of this enumeration. Is there a sequence $\{\epsilon_n\}$ of strictly positive numbers (tending to zero) such that the following holds?

For all $x\in [0,1]$,
$d(x, F_n)<\epsilon_n$ infinitely often $\implies$ $x$ is rational

(Here $d(x, F_n)=\min |x-y|: y\in F_n\}$)
 A: No. $x := \sum_{j=1}^\infty \frac{1}{10^{n_j}}$ for some $n_1 < n_2 < \dots$ (based on $(\epsilon_n)_n$) works. 
If you need details: let $n_1 = 1$ and take $n_2$ large enough so that $\frac{1}{10^{n_2}} < \frac{\epsilon_{k_1}}{10}$, where $\frac{1}{10^{n_1}}$ is the $k_1$'st rational (according to OP's enumeration). Then take $n_3$ large enough so that $\frac{1}{10^{n_3}} < \frac{\epsilon_{k_2}}{10}$, where $\frac{1}{10^{n_1}}+\frac{1}{10^{n_2}}$ is the $k_2$'nd rational. Then take $n_4$ large enough so that $\frac{1}{10^{n_4}} < \frac{\epsilon_{k_3}}{10}$, where $\frac{1}{10^{n_1}}+\frac{1}{10^{n_2}}+\frac{1}{10^{n_3}}$ is the $k_3$'rd rational. Etc. Then, for each $j \ge 1$, $|x-(\frac{1}{10^{n_1}}+\dots+\frac{1}{10^{n_j}})| \le \sum_{t=0}^\infty \frac{1}{10^{n_{j+1}+t}} = \frac{10}{9}\frac{1}{10^{n_{j+1}}} \le \frac{10}{9}\frac{\epsilon_j}{10} < \epsilon_j$.
Of course we may also choose the $n_j$'s so that $\frac{n}{n_j} \to 0$ as $j \to \infty$ so that $x$ is irrational (see Hardy and Wright irrational sums)
