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A basic result in analysis is that if $\varepsilon>0$ and $r$ is irrational, then one can find an integer $n$ so that $$\{nr\}<\varepsilon,$$ where $\{x\}$ is the fractional part of $x$. My question is roughly how much this choice of integer $n$ can be specialized. To be precise:

Question 1. Let $n_k$ be a sequence of distinct integers. Under what circumstances can we find a $k$ so that $$\{n_kr\}<\varepsilon?$$ A similar question can be asked for continuous periodic functions with irrational periods. For example:

Question 2. Let $n_k$ be as before. Does there exist a $k$ so that $$|\sin(n_k)-x|<\varepsilon$$ for any $x\in [-1,1]$?

This would be helpful in addressing this unanswered question. In this particular case, the integers are dense in $\mathbb{R}/2\pi\mathbb{Z}$, and so another way to phrase the question is to ask whether every infinite subset of the integers is also dense in $\mathbb{R}/2\pi\mathbb{Z}$.

Certainly if the sequence $n_k$ only misses a finite number of integers, then the answer to both questions is yes. But what if the sequence is all even integers, or all nonnegative integers, or something else entirely?

Does anyone know of any results of this type, or sources for where I could look to read further about this? Alternatively, are there any known cases where this type of result fails?

  • I feel like you aren't going to get very far---heuristically, if I have a collection of $1/\varepsilon$ integers, only one of them is going to satisfy the given inequality. Which means that I potentially have a lot of choices when constructing a sequence to avoid satisfying the inequality. – Xander Henderson Feb 20 '25 at 01:16
  • $nr$ modulo $1$ is uniformly distributed which means that if you fix $\epsilon$ roughly an $\epsilon$ fraction of the integers (this of course is not quite rigorous but hopefully it's clear what it means heuristically) satisfy the required inequality hence without narrowing down the sequence it's hard to say more – Conrad Feb 20 '25 at 02:42

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