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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!

EDIT

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 $\Bbb R$.

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    $\begingroup$ 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
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    $\begingroup$ 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

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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 $\sin(x + \pi) = - \sin(x) = \sin(-x)$.

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    $\begingroup$ 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
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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 $[-1,1].$

Proof:

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.

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    $\begingroup$ 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

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