Cluster points of multiples of the fractional part of an irrational number. Would anyone like to help me complete this proof? I need some help understanding where to go next. The book is giving me hints and I am trying to follow along, but I am getting confused about how to finish.
Let $c$ be irrational with $0<c<1$. Let $x_n=nc-[ nc] =nc \mod 1$, with $[nc]$ meaning $\operatorname{floor}(nc)$. Determine the cluster points of the sequence $x_n$.
Let $\varepsilon>0$
Ok, so first I prove that $x_n=x_m$ implies $n=m$, which is easy, since $c$ is irrational. So every $x_n$ is unique.
Secondly,  I can use the Archimedian property to pick $m$ such that $\frac1m < \varepsilon$ . Then I can divide up the interval $[0,1)$ into $m$ pieces like this: for $1 \leq k \leq m$ I can let $I_k=\left[\frac{k-1}m,\frac km\right)$.
Now I can take $\{{x_j : j=1, N+1, 2N+1,\ldots,mN+1}\}$ , which has $m+1$ distinct values, and thus by the pigeonhole principle, there must be $x_j$ and $x_{j'}$ that are both in the same $I_k$ and hence $|x_j-x_{j'}|<\varepsilon$.
So here I am not sure where to go now. Would anyone care to help me out? I am trying to find the cluster points.
 A: The cluster points are all points of $[0,1]$.  Given a particular point $a$, an $\epsilon \gt 0$ and a number $N$, you need to show that there is an $n \gt N$ that has $|x_n-a|\lt \epsilon$.  Once you find $x_j$ and $x_{j'}$ with $|x_j - x_{j'}| \lt \epsilon$, any time you add $j-j'$ to the subscript, it steps by that amount.  If you keep doing these steps, one of them will land within $\epsilon$ of $a$.
A: Step 1. Show that $\inf_{n\in\mathbb N}x_n=0$. 
Proof. Assume that $\inf_{n\in\mathbb N}x_n=a>0$. 
Case I. There is an $n$, such that $x_n=a$, but as $[1/a]a<1<[1/a]a+a$, then 
$$
0<(1+[1/a])a-1<a.
$$
Thus for $k=1+[1/a]$
$$
x_{kn}=(1+[1/a])a-1<a,
$$
which contradicts the assumption that $\inf_{n\in\mathbb N}x_n=a$.
Case II. For every $\varepsilon>0$, there exists an $n$, such that $a<x_n<a+\varepsilon$. In that case we can find $m>n$, such that
$$
a<x_m<x_n<2a,
$$
and if $b=x_n-x_m$, then $0<b<a$, and if $\ell=[x_m/b]$ and $j=m-n$, we have that
$$
x_{m+\ell j}=x_m-\ell b<b<a,
$$
which is again a contradiction. That $\inf_{n\in\mathbb N}x_n=0$.
Step 2. For every $0\le c<d\le 1$ there exists an $n$, such that $x_n\in[c,d]$.
Proof. if $c=0$, it is straightforward from Step 1. Otherwise, let $\varepsilon=\min\{c,d-c\}$. Then let $x_n<\varepsilon$. Clearly, some multiple of $x_n$ lies in  $[c,d]$, i.e., 
$kx_m\in[c,d]$. But $kx_m=x_{km}$.
