Consider the sequence $ x_n = \{ \alpha n^2\}$, where $\{ \}$ means fractional part: $\{ x \} = x -\lfloor x\rfloor$. Prove that for all $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ the set of partial limits $x_n$ is the interval $[ 0 , 1]$
I can prove this statement for $x_n = \{ \alpha n\}$, but I don't know what do with $\{ \alpha n^2\}$.
Ok. I will show my proof for {}. Firstly, I'm not an English speaker and my english isn't good, for this reason I'm sorry for my mistakes. Let's move on to the proof.
- $\forall k, n k \neq n \Rightarrow x_n \neq x_k$: prove the opposite: $ an \lfloor an \rfloor = ak \lfloor ak\rfloor \Rightarrow a = \frac{\lfloor an \rfloor- \lfloor ak \rfloor}{n-k} \Rightarrow a \in \mathbb{Q}$ -- contradiction
- $x_n$ can be represented as points on a circle, adjacent two points lie at the same distance along the arc.
- We prove that the described set is everywhere dense : See consider any two points on the circle, let the distance between them $=l$. since no points match, there are two points on the circle with a distance less than $l$. let these points correspond to numbers $n_1, n_2$. then from passing $n_1 - n_2 $ points from the second we get a new point (number =$ 2n_2 - n_1$) in the sequence. the distance between them is less $l$. continuing this way we will get to the segment between the points fixed at the beginning. we have proved that there is at least one point between any two points. This means that the set is everywhere dense