Open discrete cones covering cover sphere?

Question

This seems to be true, but I can't prove it. Define a cone to be $K\subset \mathbb{R}^n$ such that if $x\in K$ iff $\lambda x\in K$ for all $\lambda>0$. A discrete open cone $K'$ is the intersection of a open cone $K$ with $\mathbb{Z}^n$. Suppose $\{K_i'\}$ is a collection of discrete open cones covering $\mathbb{Z}^n\backslash\{0\}$. Does it follow that $\{K_i\}$ cover $\mathbb{R}^n\backslash\{0\}$? Notice that it would be enough if they covered the sphere. Essentialy, what I'm trying to prove is that if we have an open cover of $\mathbb{Z}^n\backslash\{0\}$ by cones, then it admits finite subcover. Is that true? If there is another way to prove it would be fine.
Edit: so the approach above didn't work, I'm wondering whether this is true or not: does every cover of $\mathbb{Z}^n$ by discrete open cones admits finite subcover? For $\mathbb{R}^n$ this seems to be true by compacity of the sphere, but for the lattice I'm not sure...

Answer 1

It's to long for a comment. We only show, that $\mathbb Q^n\setminus\{0\}$ is covered by $\{K_i\}$.

Let $u\in \mathbb Q^n\setminus\{0\}$ be arbitrarily chosen. There is a positive integer $k$ such that $ku\in\mathbb Z^n\setminus\{0\}$. Since $\mathbb Z^n\setminus\{0\}$ is covered by the collection of discrete cones $K'_i$, there is a discrete cone $K'_i$ with $ku\in K'_i$. From $K'_i=K_i\cap \mathbb Z^n$ follows $ku\in K_i$. And since $K_i$ is a cone, we get $u=\frac 1k ku\in K_i$. Hence the collection $\{K_i\}$ covers $\mathbb Q^n\setminus\{0\}$.

Even if you assume, that the $K_i$ are open, we don't necessarily have a cover of $\mathbb R^n$.

For $K_1=\{(x,y)\in\mathbb R^2\mid y<x\sqrt 2\}$ and $K_2=\{(x,y)\in\mathbb R^2\mid y>x\sqrt 2\}$ and $K'_i=K_i\cap\mathbb Z^2$, we have, that $K'_i$ are discrete cones and $\bigcup_{i\in \{1,2\}}K'_i=\mathbb Z^2$, but $\bigcup_{i\in \{1,2\}}K_i\neq \mathbb R^2$.