Find all $\alpha\in\mathbb{R}$ such that $\forall n\,\exists m\;|\alpha-m/n| < 1/(3n)$

[elim]

Sol. Let $A$ be the set we are looking for.
Then clearly $x\in A\iff x+1\in A$ and
$A=\small\displaystyle{\bigcap_{n\in\mathbb{N}^+}\bigcup_{m\in\mathbb{Z}} \big(\frac{m}{n}-\frac{1}{3n},\frac{m}{n}+\frac{1}{3n}\big)}$
So if $E = A\cap ( -1/2,1/2],$ then $A={\small\displaystyle{\bigcup_{m\in\mathbb{Z}}}} \{x+m\mid x\in E\}$
Let $S_n = ( -1/2,1/2]\cap {\small\displaystyle{\bigcup_{m\in\mathbb{Z}} \big(\frac{m}{n}-\frac{1}{3n},\frac{m}{n}+\frac{1}{3n}\big)}}$
We claim that $\displaystyle{\bigcap_{k=1}^n S_k =\big (-\frac{1}{3n},\frac{1}{3n}\big).}$ This is true for $n=1$
and by induction it's true for all $n$ since
$\small\displaystyle{\frac{1}{n+1}-\frac{1}{3(n+1)} -\frac{1}{3n}=\frac{n-1}{3n(n+1)}\ge 0}$ makes
$\small\displaystyle{\big(-\frac{1}{3n},\frac{1}{3n}\big){\large\cap}\bigcup_{m\in\mathbb{N}^+}(\frac{m}{n+​1}-\frac{1}{3(n+1)},\frac{m}{n+1}+\frac{1}{3(n+1)})=\varnothing}$
Consequently $E ={\small\displaystyle{\bigcap_{n\in\mathbb{N}^+}}} S_n = \{0\}.$ and $A = \mathbb{Z}.\quad\square$