I have succeeded in proving that $\sin(\Bbb Z)$ is dense in $[-1, 1]$ , however I would also like to prove that this result implies that $\sin(\Bbb N)$ is also dense in $[-1,1]$.

So far, I have tried to proceed by contradiction, noticing first that

$$\sin(\Bbb Z) = \sin(\Bbb N) \cup \sin(-\Bbb N) = \sin(\Bbb N) \cup -\sin(\Bbb N);$$

then there is at least a point in $[-1,1]$ such that one of its neighbourhoods contains infinitely many points of $\sin(-\Bbb N)$ but zero points of $\sin(\Bbb N)$.

However I haven't been able to go further since I do not understand which property of sine must be used in order to complete the proof (if the proof is actually possible given only that hypothesis).

Any help is highly appreciated!


I would like to clarify the reasoning I have done so far:

(1) I proved that any additive subgroup of $(\Bbb R, +)$ either has a positive least element or is dense in $\Bbb R$;

(2) I proved that the set $\{ n+m\alpha: n,m \in \Bbb Z\}$, where $\alpha$ is irrational, is dense in $\Bbb R$ by showing that it is an additive subgroup of $(\Bbb R, +)$ which doesn't have a least positive element;

(3) now it is easy to prove that $\sin(\Bbb Z)$ is dense in $[-1,1]$ by using the periodicity (let $\alpha = 2\pi$), the surjectivity in $[-1,1]$ and the sequential continuity of the sine function.

In light of recent comments, I think it would be easier to show that $ A = \{ n+m\alpha: n \in \Bbb N,m \in \Bbb Z\}$ is dense in $\Bbb R$.

  • 3
    $\begingroup$ I doubt that you can easily pass from $\mathbb{Z}$ to $\mathbb{N}$. Perhaps you can go through your proof and see if you can simply replace $\mathbb{Z}$ with $\mathbb{N}$ there? $\endgroup$
    – freakish
    Commented Feb 15, 2022 at 12:40
  • 3
    $\begingroup$ I don't see a trivial solution. For example if $f(\mathbb N)\subseteq [0,1]$ is dense, and $f(-x)=-f(x)$ then $f(\mathbb{Z})$ is dense in $[-1,1]$, but $f(\mathbb{N})$ is not. So the property you listed are not enough to prove this statement for $\sin x$. $\endgroup$
    – Kolja
    Commented Feb 15, 2022 at 12:46
  • $\begingroup$ @freakish The fact is that on my proof I need to use $\Bbb Z$ to show that a certain set is an additive subgroup of $(\Bbb R , +)$ $\endgroup$ Commented Feb 15, 2022 at 12:47
  • $\begingroup$ Is it enough to only use the fact that $\mathbb N$ is a monoid - there is an identity, and it is closed with respect to addition? $\endgroup$
    – Kolja
    Commented Feb 15, 2022 at 12:48
  • $\begingroup$ (+1), Interesting and thought-provoking question... $\endgroup$ Commented Feb 15, 2022 at 12:50

2 Answers 2


Just a remark that one can prove $\sin \mathbb{Z}$ dense in $[-1,1]$ implies that $\sin \mathbb{N}$ is dense in $[-1,1]$ directly without using diophantine approximation.

Assume that $\sin \mathbb{Z}$ is dense in $[-1,1]$.

By assumption, there exists an integer $n$ such that $\sin(n)$ is as close to $-1$ as you wish. Also, for any $\alpha \in [-1,1]$, there are arbitrarily large integers $m$ such that $\sin m$ is a close to $\alpha$ as you wish.

Now either infinitely many of those $m$ are positive, in which case $\alpha$ is a limit point of $\sin \mathbb{Z}$, or they are eventually all negative. But then $-m + 2n$ will be a positive integer for $m$ big enough, one can easily compute from the double angle formula that $\sin (-m + 2n)$ is very close to $- \sin(-m) = \sin(m)$ which is very close to $\alpha$.

The basic idea is that when $\sin(n)$ is close to $-1$ then that $2n$ is very close to an odd multiple of $\pi$, and $\sin(x + \pi) = - \sin(x) = \sin(-x)$.

  • 1
    $\begingroup$ In the phrase "$\alpha$ is a limit point of $\sin \mathbb Z$, perhaps $\mathbb Z$ should be $\mathbb N$? $\endgroup$
    – Lee Mosher
    Commented Feb 15, 2022 at 22:50
  • $\begingroup$ Very nice solution, by the way. $\endgroup$
    – Lee Mosher
    Commented Feb 15, 2022 at 22:51

We will use the theorem:

Given any irrational number, $\alpha>0,$ there are infinitely many positive integers, $p,q$ such that $$\left|\frac pq-\alpha\right|<\frac1{q^2}.$$

That can proven using continued fractions.

Claim: If $f$ is continuous and periodic with period $T>0,$ and $f(\mathbb Z)$ is dense in $[-1,1],$ then $f(\mathbb N)$ is dense in $[-1,1].$


If $T$ is rational, then $f(\mathbb Z)$ is finite and thus can’t be dense in any interval.

So $T$ must be irrational.

Since $f$ is continuous and periodic, it is uniformly continuous.

Now, given $\epsilon>0,$ and $\alpha\in [-1,1],$ let $m\in\mathbb Z$ be such that $|f(m)-\alpha|<\frac{\epsilon}2.$

Since $f$ is uniformly continuous, there is a $\delta>0$ such that if $|x-y|<\delta,$ $|f(x)-f(y)|<\frac{\epsilon}2.$

Find $p,q\in\mathbb Z^+$ such that $$\left|\frac pq-T\right|<\frac1{q^2}$$ and $p>-m$ and $\frac1q<\delta.$

Then $|p-Tq|<\delta,$ so $$\begin{align} |f(m+p)-\alpha|&\leq |f(m+p)-f(m)|+|f(m)-\alpha|\\ &=|f(m+(p-Tq))-f(m)|+|f(m)-\alpha|\\&<\frac\epsilon2+\frac\epsilon2=\epsilon \end{align} $$

But $m+p\in\mathbb N.$

You certainly don’t need the whole first claim, just the narrower:

If $\alpha>0$ is irrational, then there are two sequences of positive integers, $p_n,q_n$ such that $\lim_{n\to\infty}(p_n-\alpha q_n)=0.$

I just happen to know the stronger lemma well, so I often lead with it.

  • 1
    $\begingroup$ Thank you for your answer! Unfortunately I have already seen this approach (the one using Dirichlet Approximation Theorem) and I asked this question because I thought that there would be an easier way than assuming the theorem at the beginning of your answer. Nevertheless, thank you so much again! $\endgroup$ Commented Feb 15, 2022 at 15:36
  • $\begingroup$ Yeah, you don’t really need that lemma, just the weaker Lemma that, for $\alpha>0$ irrational, the set $p-\alpha q$ is dense at $0$ for $p,q$ positive integers. If it isn’t dense at $0,$ there is a non-zero infimum of $$\{|p-q\alpha|\mid p,q\in\mathbb Z^+\}.$$ $\endgroup$ Commented Feb 15, 2022 at 16:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .