$\epsilon$-dense subsets on $\mathbb R/\mathbb Z$.

Question

Let $\langle M, d\rangle$ be a metric space. We say that $A \subset M$ is $\epsilon$-dense if every open ball of radius $\epsilon$ contains a point of $A$.

Now let $T=\mathbb R/\mathbb Z$, the quotient group packed with the quotient topology. It's easy to show that $\delta(x+\mathbb Z, y+\mathbb Z)=\min\{|x-y+a|: a \in \mathbb Z\}$ defines a metric on $T$. Consider the metric space $T^2$ with the metric $\delta_2(\langle a, b\rangle, \langle c, d\rangle)=\max\{\delta(a, b), \delta(c, d)$}. It's easy to see that the open ball of radius $\epsilon$ on $T^2$ is the product of two balls of radius $\epsilon$ and the same center on $T$.

By the Kronecker's Theorem, it's easy to show that if $\xi$ is irrational then the set $\{\langle x+\mathbb Z, \xi x+\mathbb Z\rangle: x \in \mathbb R\}$ is dense, and I have done it. However, what I want is to show that given $\epsilon>0$ there exists $l>0$ such that whenever $a \geq l$ and $I$ is an open interval of lenght $a$ then the set $\{\langle x+\mathbb Z, \xi x+\mathbb Z\rangle: x \in I\}$ is $\epsilon$-dense.

Can someone help me?

Answer 1

HINT: The pigeonhole argument sketched here shows that there is an $m\in\Bbb Z^+$ such that $m\xi\bmod 1<\epsilon$. Choose $p\in\Bbb Z^+$ large enough so that $p(m\xi\bmod 1)\ge 1$. Let $x\in\Bbb R$, and let $y=x\xi\bmod 1$; then

$$\{(x+km)\xi\bmod 1:k=0,\ldots,p\}=\{(y+km\xi)\bmod 1:k=0,\ldots,p\}$$

is $\epsilon$-dense in $[0,1)$. Use this to show that it suffices to take

$$\ell>\frac1{\xi}+pm\xi\;.$$

(If you were using closed intervals $I$, you could make that an equality.)

Answer 2

It might take a few lemmas to tighten this argument up, but I think the result pretty much follows from:

1. What you've already proven about $\mathbb R$
2. $\mathbb R=\bigcup_n(-n,n)$
3. The metric $\delta_2$ is translation-invariant, so $(-n,n)$ represents all intervals.

You shouldn't have to recapitulate the usage of Kronecker's Theorem, although that will work as well (Brian's answer).