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Let $\alpha$ be an irrational number.

Is it true that the set $\{\frac{m}{\alpha} + n | m,n \in \mathbb{Z} \}$ dense in $\mathbb{R}$?

If it is, how do we prove it?

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  • $\begingroup$ It is a group, so it is enough to find a sequence converging to $0$ $\endgroup$ – Yanko Oct 7 '17 at 21:49
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Yes it is

Lemma: Let $H$ be a subgroup of $\mathbb{R}$, then $H$ is dense if and only if there exists a convergence sequence $h_n\rightarrow \xi$ for some $\xi\in \mathbb{R}$ and $h_n\in H$ with the property that $h_n$ is not eventually constant.

Proof: The first direction is obvious, For the other direction, let $h_n\rightarrow \xi$. Let $\varepsilon>0$, as $h_n$ is a Cauchy sequence there exists $h\in H$ for which $|h|<\varepsilon$, as $H$ is a group you have that $kh\in H$ for every $k\in\mathbb{Z}$, it follows that $\{kh:k\in\mathbb{Z}\}$ is $\varepsilon$-dense in $\mathbb{R}$, as $\varepsilon$ is arbitrary, we have that $H$ is dense in $\mathbb{R}$.

Now, the set you mentioned is a group, hence it is enough to find a convergence sequence which is not a eventually constant.

For every $m$, let $n_{m} = \lfloor m/\alpha\rfloor$, and consider the set $\frac{m}{\alpha}-n_m$ this is a set of infinitely many different numbers in $[0,1]$ hence there is a converging sub sequence $\frac{m_k}{\alpha}-n_{m_k}$, using the Lemma this completes the proof.

Note: another way to do this is to use the fact that an irrational rotation of the circle has a dense orbit (with rotation $1/\alpha$)

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  • $\begingroup$ $\dfrac{1}{\alpha}\in\mathbb R\setminus\mathbb Q\iff \alpha\in\mathbb R\setminus\mathbb Q$. It is enough to prove that the fractional parts $\{n\alpha\}=n\alpha-\lfloor{n\alpha}\rfloor$ form a dense set in $[0,1]$ when $n$ runs $\mathbb Z$ and this can be done using the Pigeonhole Principle without mention of groups. $\endgroup$ – Piquito Oct 7 '17 at 23:17
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    $\begingroup$ yes, thanks for clarifying the note. $\endgroup$ – Yanko Oct 7 '17 at 23:19
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(This is almost the same as yanko's argument, but from a different perspective)

Lemma. If $H$ is a subgroup of $\Bbb R$, then $H=c\Bbb Z$ for some $c$, or $H$ is dense in $\Bbb R$.

Proof. Let $H$ be a non-cyclic subgroup of $\Bbb R$. In particular, $H$ contains non-zero elements and, as $x\in H$ implies $-x\in H$, it contains some positive elements. Let $s=\inf(H\cap \left]0,\infty\right[)$, which is a non-negative real.

Let $U\subseteq \Bbb R$ be a non-empty open set. Then it contains some open interval $\left]a,a+\delta\right[$ with $\delta>0$.

If $s<\delta$, pick $h\in H$ with $h>0$ and $s\le h<\delta$. For $n\in\Bbb N$ large enough, we have $-nh<a$. Fix such $n$. Then for $m\in\Bbb N$ large enough, we have $(m-n)h>a$. For the minimal such $m$, we have $(m-n-1)h\le a$, hence $a<(m-n)h\le a+h<a+\delta$, i.e., $h\in H\cap U$, contradiction.

Thus $s\ge \delta>0$. Assume $s\notin H$. Then we can pick $h\in H$ with $s< h<2s$ and $h'\in H$ with $s<h'<h$. Then $h-h'\in H$ contradicting $0<h'-h<s$. Therefore $s\in H$ and $s\Bbb Z\subseteq H$. By assumption, $H\ne s\Bbb Z$. So pick $h''\in H\setminus s\Bbb Z$. Wlog. $h''>0$. For $n\in \Bbb N$ large enough, we have $ns>h''$. Then $(n-1)s<h''<ns$ and so $0<h''-(n-1)s<s$, contradicting the definition of $s$ as infimum. $\square$

Now note that all non-zero elements of $c\Bbb Z$ are rational if $c$ is rational, and all non-zero elements of $c\Bbb Z$ are irrational if $c$ is irrational. As the group $H:=\{\frac m\alpha+n\min m,n\in\Bbb Z\,\}$ contains the non-zero reational $1$ and the non-zero irrational $\frac 1\alpha$, hence $H$ is not cyclic, and by the lemma, $H$ is dense.

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