# For $\alpha$ irrational, prove that $\{m\alpha + n \mid m,n \in \mathbb{Z}\}$ is dense in $\mathbb{R}$

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!!

• 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 $\mathbb{Z} g_\varepsilon$. Oct 16, 2023 at 20:37
• 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 $0\neq \vert g_1-g_2-m \vert \leq \varepsilon$. Oct 16, 2023 at 20:47
• 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? Oct 16, 2023 at 21:42
• @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? Oct 17, 2023 at 5:49
• I'm not sure that made any sense... Oct 17, 2023 at 5:52

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.

• 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 $\{i\alpha\} \ne \{j\alpha\}$? Oct 17, 2023 at 16:31
• 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? Oct 17, 2023 at 16:49
• 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. Oct 17, 2023 at 16:53
• @IzaakvanDongen I've edited my proof. Oct 17, 2023 at 16:58
• 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? Oct 17, 2023 at 17:03