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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.

student
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  • A general related result is Dirichlet's approximation theorem. Anyway, you should make the observation that $S$ is an additive group. In other words, if $x_1\in S$ and $x_2\in S$, then so are $x_1\pm x_2$. Also, by induction, if $z\in S$ then $nz\in S$ for all integers $n$. So if you can prove that to any $\epsilon>0$ there is an element $z_\epsilon\in S\cap (0,\epsilon)$. Integer multiples of $z_\epsilon$ then "cover the entire line with elements of S" with gaps shorter than $\epsilon$. The density follows from this. – Jyrki Lahtonen Sep 21 '16 at 12:41
  • (cont'd) In other words, you can forget about $y$. Doing the case $y=0$ is enough as long as you won't allow $z_\epsilon=0$. See also here, and here (Mariano's answer is better here because it works for any irrational - in spite of the title of that thread), possibly even earlier on our site ... (+1 to you and both the answers) – Jyrki Lahtonen Sep 21 '16 at 12:43

2 Answers2

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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.

Ege Erdil
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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.

John Hughes
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