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.

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You can get $\epsilon$ with \epsilon and the rest of the Greek alphabet similarly. Also, to get the prime on the subscript of $x_j$, put it in braces: x_{j'} gives $x_{j'}$ – Ross Millikan Sep 25 '12 at 19:49
Why did you erase your question a few minutes ago? It's nice to have an archive of old questions, in case someone in the future has a similar question. – Snowball Sep 25 '12 at 21:35
@Erica Please stop vandalizing your questions. This behavior violates the norms of MSE. If it continues your account may be suspended. – Bill Dubuque Oct 13 '12 at 0:32

2 Answers

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 $a$.

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Why will they be close to $a$? – Berci Sep 25 '12 at 19:54
They step through the interval in steps smaller than $\epsilon$, so one of them will land close to $a$. Then around again and you will again land close to $a$, so there are points close to $a$ at arbitrarily high index, which is a cluster point. – Ross Millikan Sep 25 '12 at 20:10
I totally understand what you are saying! How would I write it rigorously, though? – Erica McDonald Sep 25 '12 at 20:48

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 $kx_m=x_{km}$.

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