I've found some duplicate questions on the website about this exact proof but I can't really understand them. The only definition that I know for denseness is that if $M \subset \mathbb{R}$ then $M$ is dense in $\mathbb{R}$ if for any interval $(a,b)$ there's at least one element from $M$ in $(a,b)$. None of the proofs used this fact or at least they didn't use it in a way that was obvious to me. Please help!!

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    $\begingroup$ Can you elaborate which part you have trouble understanding? The proof goes essentially like this: $1.)$ $G:=\{ m\alpha +n \ \vert \ m,n \in \mathbb{Z}$ is a subgroup of $\mathbb{R}$. It is enough to show that $\inf \{ \vert g \vert \ \vert \ g\in G \setminus \{0\} \} =0$. Indeed, this means that we can find for every $\varepsilon>0$ some $g_\varepsilon\in G\setminus \{0\}$ such that $\vert g_\varepsilon \vert\leq \varepsilon$. However, then $\mathbb{Z} g_\varepsilon \subseteq G$ and every point in $\mathbb{R}$ has distance at most $\varepsilon$ to $\mathbb{Z} g_\varepsilon$. $\endgroup$ Oct 16, 2023 at 20:37
  • $\begingroup$ As the differences of elements in $G$ are again in $G$ it is enough to find any two distinct elements in $G$ which are close. In order to do so one would like to use the pigeonhole principle. Let $H=\mathbb{Z}\alpha$. Consider the quotient map $P:H \rightarrow \mathbb{R}/\mathbb{Z}$ (note it is injective). By the pigeonhole principle we get that the image of $H$ must accumulate at some point. However, this means, that for every $\varepsilon>0$ there exists $g_1, g_2\in H$ (with $P(g_1)\neq P(g_2)$) and some $m\in \mathbb{Z} \subseteq G$ such that $0\neq \vert g_1-g_2-m \vert \leq \varepsilon$. $\endgroup$ Oct 16, 2023 at 20:47
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    $\begingroup$ The proofs at the question you link do skip over some justification! Severin mentions the intended argument. I'll spell it out a bit more: if you show that for any $\epsilon > 0$, you can find $m_\epsilon, n_\epsilon$ with $0 < m_\epsilon \alpha + n_\epsilon < \epsilon$, it follows that your set is dense in $\Bbb R$. To show it intersects the interval $(a, b)$, take $\epsilon = b - a$. The set of integer multiples of $m_\epsilon \alpha + n_\epsilon$ is contained in your set. But it must intersect $(a, b)$, since adjacent multiples have distance less than $b - a$. So it's dense! Does that help? $\endgroup$ Oct 16, 2023 at 21:42
  • $\begingroup$ @IzaakvanDongen So I can take the interval $(a,b)$ and split it into $k$ subintervals then take some $x_1, \dots, x_{k+1}$ and from there'll know that at least 2 of them are in the same interval hence less than $b-a$ apart. I hope I'm thinking in the right direction. What should I be taking for the $x_i$ values? $\endgroup$
    – Marin
    Oct 17, 2023 at 5:49
  • $\begingroup$ I'm not sure that made any sense... $\endgroup$
    – Marin
    Oct 17, 2023 at 5:52

1 Answer 1


This is what I've come up with so far:

Let $S = \{ m\alpha + n \mid m,n\in \mathbb{Z}\}$ for a fixed irrational number $\alpha$ and $\{x\} = x - \lfloor x\rfloor, x \in \mathbb{R}$. To prove that $S$ is dense in $\mathbb{R}$ we need to find $m_\epsilon$ and $n_\epsilon$ s.t. $0 \lt m_\epsilon\alpha + n_\epsilon \lt \epsilon, \forall \epsilon\gt0$.

Let $k \gt \frac{1}{\epsilon}$ and consider the sequence of $k+1$ values: $\{0\alpha\}, \{1\alpha\}, \{2\alpha\}, \dots, \{k\alpha\}$. Each element of this sequence is in the interval $[0, 1]$. If we split the interval $[0,1]$ into $k$ subintervals, each of length $\frac{1}{k}$ then by the pigeonhole principle we know that at least two of the values in the sequence are in the same subinterval $[\frac{t}{k}, \frac{t+1}{k}]$, for $0 \le t \lt k-1$. Let's call those values $\{i\alpha\}$ and $\{j\alpha\}$ where $\{i\alpha\}\lt\{j\alpha\}$. Then, $\{j\alpha\}-\{i\alpha\} = (j\alpha - \lfloor j\alpha\rfloor) - (i\alpha - \lfloor i\alpha\rfloor) = (j-i)\alpha - (\lfloor i\alpha\rfloor - \lfloor j\alpha\rfloor)\in S$, since it's of the form of $S$'s elements. Therefore, $\mid \{j\alpha\} - \{i\alpha\}\mid \lt \frac{1}{k} \lt \epsilon \implies S$ is dense in $\mathbb{R}$.

I need help with the final step I'm not exactly sure it makes much sense.

  • $\begingroup$ Nice work. I would write it as follows: "... Let's call those values $\{i\alpha\}$ and $\{j\alpha\}$ where $\{i\alpha\} < \{j\alpha\}$. Then $(j - i)\alpha + (\lfloor i\alpha \rfloor - \lfloor j\alpha \rfloor) = \{j\alpha\} - \{i\alpha\}$ is of the desired form, and since $\{i\alpha\}$ and $\{j\alpha\}$ are in the same interval, we have $0 < \{j\alpha\} - \{i\alpha\} \le \tfrac 1k$ (to justify this further, use the fact that $\{j\alpha\} \le \tfrac{t + 1}k$ and $\{i\alpha\} \ge \tfrac tk$)." PS: do you see why we can assume $\{i\alpha\} \ne \{j\alpha\}$? $\endgroup$ Oct 17, 2023 at 16:31
  • $\begingroup$ So I don't need the $x \in [\frac{t}{k}, \frac{t+1}{k}]$ part? Also $\{i\alpha\} \neq \{j\alpha\}$ since $i \neq j$, right? $\endgroup$
    – Marin
    Oct 17, 2023 at 16:49
  • $\begingroup$ Indeed not. Two real numbers being in the interval $[c, d]$ automatically implies their difference is at most $d - c$, for the reason I gave in parentheses. In your answer now, it's not true that $\lvert x - \{(j - i)\alpha\}\rvert < \tfrac 1k$ in general - $x$ could be quite big while $ \{(j - i)\alpha\}$ is quite small. $\endgroup$ Oct 17, 2023 at 16:53
  • $\begingroup$ @IzaakvanDongen I've edited my proof. $\endgroup$
    – Marin
    Oct 17, 2023 at 16:58
  • $\begingroup$ You should assume $\{i\alpha\} < \{j\alpha\}$, rather than $i < j$ (otherwise the difference might accidentally be negative). Aside from that, I think you could do with a sentence justifying why $\{i\alpha\} \ne \{j\alpha\}$. What do you mean by "the final step"? Proving that this version of density implies your version of density? $\endgroup$ Oct 17, 2023 at 17:03

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