How to proceed in my attempt at showing $\mathbb Z + \alpha \mathbb Z$ is dense in $\mathbb R$? I've been trying to prove that $S=\{s + \alpha t \mid s,t \in \mathbb Z\}$ where $\alpha$ is irrational is dense in $\mathbb R$ but now I seem to be completely stuck and I don't know if it's whether my proof idea is wrong in the first place or whether I just don't see how to proceed.
Here is what I have:
Let $y \in \mathbb R$ and $\varepsilon > 0$. The goal is to show that $S \cap B_\varepsilon (y)\neq \varnothing$. Say, ${1\over n}\le \varepsilon \le {1\over m}$. Then we consider $I = [y-{1\over n}, y + {1\over n}]$. We divide it into two halves. 
Pick $s_1, s_2, s_3 \in S$ and define $s_i'= s_i - \lfloor s_i \rfloor$. We shift $I$ by $-y$. Then one of the two halves of $I-y$ contains two of the $s_i'$. Hence $|s_i' - s_j'| < {1\over n}$.
This is where I'm stuck. I now have two irrational parts that are close enough to each other. Now somehow I need to use them to produce two points in $S$ that are near $y$. But I just don't see how. 

Please could someone help me construct the desired points or tell me
  my mistake and how to do this?

I also tried to find a proof of this using Google but since I don't know what this is called I couldn't find anything. 
 A: Hint: Show that the set $ \{ \{n\alpha\} : n \in \mathbb Z \} $ is dense in $ [0, 1] $, where $ \{ n\alpha \} $ denotes the fractional part of $ n\alpha $. Conclude that $ 0 $ is not an isolated point of the subgroup $ \mathbb Z + \mathbb Z \alpha $. Now, consider the integer span of an arbitrarily small number in this subgroup, and prove that you can approximate any real number "closely enough" using the elements in this span.
A: I'm going to walk through your proof with $\alpha = \pi$, OK? 
Let's take $y = 1.3$. And $\epsilon = 0.1$. So $n$ could be 11 and $m$ could be 9. (By the way: what would you do if $\epsilon = 2$? You'd have a tough time finding natural numbers $n$ and $m$ with the required properties! Also: notational convention isn't enough. You should explicitly say that $m, n \in \mathbb N$.)
Pick $s_1, s_2, s_3 \in S$ and define $s_i' = s_i - \lfloor s_i \rfloor$. 
OK. I'll pick $s = 0$ for all three, and $t = 0, 1, 2$, so 
$s_1 = 0, s_2 = \pi, s_3 = 2\pi$, and $s'_1 = 0, s'_2 = .1415..., s'_3 = .283...$
"Shift $I$ by $-y$". I suppose that by this you mean $I' = [-1/n, 1/2]$. But you should just say so. So in my case, 
$$I' = [-1/11, 1/11] \approx [-.091, .091]$$
Now you say that one of the two halves of $I'$ (I confess, I don't know what these halves are!) contains two of the $s'_i$. But in fact, $I'$ contains only one of the $s'_i$, namely $s'_1$.
So that's the first concrete error in your attempted proof. 
