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Consider the set $\{m+n\alpha: m,n \in \Bbb Z\}$ where $\alpha$ is an irrational number.

To prove that the above set is a dense set (any point either lies in that set or is a limit point of that set), I am not understanding (understanding intuitively, but formally how does this comes from the definition of dense set) why is it enough to prove that for any given $n$, there exist an integer $m$ such that the fractional part of $m\alpha$ is less than $\frac{1}{n}$.

user96000
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  • A duplicate of http://math.stackexchange.com/questions/136665/how-to-show-this-is-a-dense-set and several others. – lhf Oct 17 '13 at 02:34
  • @lhf: I am not asking the solution of the first problem. I want to know why it is enough to prove some other thing to prove the first problem. – user96000 Oct 17 '13 at 02:58

1 Answers1

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The fractional part of $m\alpha$ is $m\alpha-\lfloor m\alpha\rfloor$, which is non-negative. Suppose that

$$0\le m\alpha-\lfloor m\alpha\rfloor<\frac1n\;.$$

Let $(a,b)$ be any interval such that $b-a\ge\frac1n$; two consecutive integer multiples of $m\alpha-\lfloor m\alpha\rfloor$ differ by less than $\frac1n$, so there is an integer $k$ such that $$a<k\left(m\alpha-\lfloor m\alpha\rfloor\right)<b\;.$$

But $k\left(m\alpha-\lfloor m\alpha\rfloor\right)=(km)\alpha-k\lfloor m\alpha\rfloor$, where $km,k\lfloor m\alpha\rfloor\in\Bbb Z$, so we’ve found an element of the set in the interval $(a,b)$.

For any non-empty open interval $(a,b)$ there is an $n\in\Bbb Z^+$ such that $\frac1n<b-a$, so the set has non-empty intersection with every non-empty open interval and is therefore dense in $\Bbb R$.

Brian M. Scott
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