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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. 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'}$2012-09-25
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    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.2012-09-25
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    @Erica Please stop vandalizing your questions. This behavior violates the norms of MSE. If it continues your account may be suspended.2012-10-13

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