find the cluster points of this sequence Let $c$ be an irrational number and define $x_n = nc-[nc]$, where $[x]$ denotes the greatest integer in $x$.
a) Show that the terms of the sequence are distinct that is $x_n = x_m$ implies $n=m$.
I have done this part.
b) Prove that for each $\epsilon >0$ and $N \in \cal{N} $, there is an $n \ge N$ such that $ 0<x_n<\epsilon$. The book has hint and it reads ''Use the Archimedean principle to choose an $m \in \cal{N}$ such that $1/m < \epsilon$. For $1 \le k \le m$, let $I_k = [(k-1)/m,k/m)$ and note $I_k$s are disjoint, their union is $[0,1)$ and each has length $1/m$. Now consider $\{x_j: j=1,N+1,2N+1,...,mN+1\}$ and observe that by a), this set contains $m+1$ distinct numbers in $[0,1)$.''
Here's my reasoning. By definition we know that $0\le x_n <1$, since $c$ is irrational, $nc$ can not be integer so $x_n \ne 0 $. For $n \in \cal{N}$, by the hint, there is some $k$ such that $x_n \in I_k$, so $x_n < k/m < k \epsilon$.
If I'm done here, then the 'Now consider...' part of the hint haven't been used, and the conclusion seems to be true for all n, so what's wrong?
c) Let $x \in [0,1)$ Prove that for each $\epsilon > 0$ and $N \in \cal{N}$, there is an $n \ge N$ such that $|x-x_n| < \epsilon$. (Hint: Choose an $m \in \cal{N}$ such that $1/m < \epsilon$. Apply b) to choose an $n \ge N$ such that $0< x_n < 1/m $. Let $k$ be the unique integer between $1$ and $m$ such that $(k-1)/m \le x < k/m$. Now let $l$ be the largest positive integer such that $l x_n <k/m$.)
Again the hint is clear to me but seems confusing. By b) I already have $x_n <\epsilon \le x+\epsilon$, but I'm stuck in proving the other part of the inequality $ x-\epsilon < x_n$, and I couldn't see how the hint may be used.

For part b) : no, you’re not done yet, because you only have
$x_n < k\epsilon$ and you need $x_n < \epsilon$.
For part c) : formulas are not esoteric jargon. They usually mean something simple.
Thus, $0 < x_n < \epsilon$ (the goal of question b)) means “$x_n$ is near to 0”. Similarly,
$|x-x_n| < \varepsilon$ (the goal of question c)) means “$x_n$ is near to $x$”. This should get you
on the right tack.
