Find integer $k$ such that $k \alpha_i \bmod{1}$ are simultaneously small for all $i$

Question

A classical result shows that if $\alpha$ is irrational, then $\{k \alpha \bmod{1}\}_{k \in \mathbb{Z}}$ is dense over $[0,1]$. Can we extend this result as follows?

Suppose $\alpha_1,\dots,\alpha_m$ are such that $(1,\alpha,\dots,\alpha_m)$ is linearly independent over $\mathbb{Q}$. For any $\varepsilon > 0$, there exists an integer $k$ such that $k \alpha_i \bmod{1} \leq \varepsilon$ for all $1 \leq i \leq m$.

Edit: it was originally asked assuming only that all $\alpha_i$ are irrational. As observed in the comments, for $m\ge 2$ and $\varepsilon<\frac12$, $\alpha_2=-\alpha_1$ yields an obvious counterexample.

Answer 1

(reflecting the comments)

The answer is yes, by Kronecker's theorem. Namely, the assumption that $(1,\alpha_1,\dots,\alpha_m)$ is linearly independent over $\mathbf{Q}$ means that $(\alpha_1,\dots,\alpha_m)$ generates a dense sub(semi)group of the torus $(\mathbf{R}/\mathbf{Z})^m$, and hence this subsemigroup meets $[0,\varepsilon]^m$ (infinitely many times).

In the original setting (assuming only $\alpha_i$ irrational for some $i$), it is still true that it meets $[-\varepsilon,\varepsilon]^m$ infinitely many times: indeed this assumption ensures that the closure of the sub(semi)group is non-discrete and accumulates at zero. (And using that every nonempty closed subsemigroup of a compact group is a subgroup.) But for $m\ge 2$, taking $\alpha_2=-\alpha_1$ irrational shows that it doesn't work with $[0,\varepsilon]^m$ instead.