TITLE: Distance of scaled point on the unit sphere to an integer
QUESTION [6 upvotes]: Let $S:=\{ x\in \mathbb{R}^d:||x||_2=1\}$ be the d-dimensional unit sphere, where $||x||_2$ is the euclidean norm.
Let $\epsilon>0$ and $s\in S$ be an arbitrary point on the sphere.
Is it correct that there exists an $\alpha>0$ and a $k\in \mathbb{Z}^d\setminus\{0\}$ such that the
distance between $\alpha s-k$ is less than $\epsilon$?
In other words can i scale every point on the unit sphere such that the distance to an non-zero integer is arbitrarily small

REPLY [2 votes]: The statement is true for any vector $v\in\mathbb R^d$, not necessarily on the unit sphere. For any $x\in\mathbb R^d$, let $x\text{ mod }\mathbb Z^d$ denote the unique element in $y\in [0, 1)^d$ such that $x-y\in\mathbb Z^d$.
Because $[0, 1)^d$ is bounded, the sequence $\{nv \text{ mod } \mathbb Z^d\mid n\in\mathbb N\}$ has a convergent subsequence $\{n_iv\text{ mod } \mathbb Z^d \mid i\in\mathbb N\}$. In particular, for large enough $i>j$, there is $\|n_i v \text{ mod } \mathbb Z^d - n_j v \text{ mod } \mathbb Z^d \|= \|(n_i-n_j) v \text{ mod } \mathbb Z^d\|<\epsilon$. That is $(n_i-n_j)v$ is very close to an integral point, and we may take $\alpha = n_i - n_j$ (in particular $\alpha$ can be chosen as an integer).
Note that $\alpha$ can be taken to be as large as needed, and when $\alpha$ is large, $\|\alpha v\|$ is not close to $0$, so the $k$ can always be taken to be nonzero.
Also the same argument applies to any lattice with compact quotient, not necessarily $\mathbb Z^d$.