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Let $c\in\mathbb{R}\setminus\mathbb{Q}$. Define $x_n=cn-[cn]$. I am trying to show that $0$ is a cluster point of $\{x_n\}$ as follows: for any natural number $N$ and $\epsilon>0$, we can find an $m\in\mathbb{N}$ such that $1/m<\epsilon$. We can then partition the interval $[0,1)$ into the subintervals

$I_k = [\frac{k-1}{m},\frac{k}{m}], 1\leq k\leq m.$

If we now take $x_j$ for $j=1,N+1,2N+1,...,mN+1$, we see that there exists $k$ such that $I_k$ contains two distinct numbers $x_{j_1},x_{j_2}$, where $j_1$ and $j_2$ are elements of $\{1,N+1,...,mN+1\}$. This is because there are only $m$ intervals $I_k$, while there are $m+1$ elements of the aforementioned set. This then implies that $|x_{j_1}-x_{j_2}|<1/m$.

I am now trying to use this observation to find an $x_n$ with $0<x_n<\epsilon$, hence showing that $0$ is a cluster point of the sequence, but I'm stuck since I only managed to find two elements of the sequence which differ by no more than $1/m<\epsilon$.

Aaron Andersen
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  • the fractional parts {nc} are actually uniformly distributed on $[0;1[$: I think that you can prove that $\int_0^1 g = \lim_{N\to+\infty} \sum_{n=0}^{N-1} g({nc})$ as in https://dept.math.lsa.umich.edu/~rauch/558/Kronecker.pdf (but with $\cos(2\pi n c)$ and $\sin(2\pi n c)$ for dense set of functions) see https://wiki.math.ntnu.no/_media/ma8106/2020v/stoneweierstrass.pdf for the Stone-Weierstrass theorem – reded Jan 23 '24 at 21:05
  • basically ${y}-{x}={y-x}$ when ${y}-{x}\geq 0$ – reded Jan 23 '24 at 21:37

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