A ray with initial point $0$ need to pass arbitrarily near to a integer point. How can I prove that for a ray $\overrightarrow{0x}=\{\lambda x| \lambda \in \mathbb{R}_{\geq 0}\}$ in $\mathbb{R}^n$ we have $d(\mathbb{Z}^n \setminus \{0\},\overrightarrow{0x})=0$?
 A: For a rational $x$ it is simple since: 
$x=\frac{q}{p}$ for $q,p\in \mathbb{Z} $
and just choose $\lambda=p$ 
For a irrational $x$ use the fact that:
for any $x \in \mathbb{R}$ there exists a sequence ${\{a_n\}\in \mathbb{Q}}$ , $a_n \rightarrow x$
A: We can prove a stronger result, namely that the (Euclidean) distance between $\{ kx : k \in \mathbb{N}_{>0} \}$ and the set of lattice points $L$  is zero, for any vector $x \in \mathbb{R}^n$.
Let $p(v,k)$ be the $k$-th coordinate of $v$, for any $v \in \mathbb{R}^n$ and $k \in [1..n]$.
Let $f(v) = ( p(v,k) - \lfloor p(v,k) \rfloor : k \in [1..n] )$ [Keep only the fractional part of each coordinate.]
Then $\{ f(kx) : k \in [1..m^n+1] \}$ is a set of points in a unit hypercube.
Thus by pigeonhole principle, $f(ax),f(bx)$ are in the same hypercube of width $\frac{1}{m}$ for some distinct $a,b \in [1..m^n+1]$.
Also $(a-b)x = f(ax)-f(bx)+c$ where $c$ has integer coordinates, and $|a-b| \in [1..m^n]$.
Therefore $d( \{ kx : k \in [1..m^n] \} , L ) \to 0$ as $m \to \infty$.
I've left out some details but I'm sure you can fill them in.
