5

How to prove that the set $\{ m+n\alpha : m\in\mathbb{N},n\in\mathbb{Z}\}$, ($\alpha$ is an irrational number) is dense in $\mathbb{R}$?

Using the fact every additive subgroup of $\mathbb{R}$ is either discrete or dense in $\mathbb{R}$, we can prove that the set $\{ m+n\alpha : m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$. But how to prove the denseness of the above set?

Martin Sleziak
  • 53,687
Parvesh Kumar
  • 175

1 Answers1

6

In this answer I use the pigeonhole principle to give a short, easy proof that if $\alpha$ is irrational, then $\{n\alpha\bmod 1:n\in\Bbb Z\}$ is dense in $[0,1)$, where $x\bmod 1=x-\lfloor x\rfloor$ is the fractional part of $x$. It’s easy to check that the proof still works if we replace $\Bbb Z$ either by $\Bbb Z^+$ or by $\Bbb Z^-$.

Now let $(a,b)$ be an open interval of real numbers; we can certainly find an interval $(c,d)\subseteq(a,b)$ such that $(c,d)\cap\Bbb Z=\varnothing$. Let $k=\lfloor c\rfloor$; then $0<c-k<d-k<1$, so there is an $n\in\Bbb Z^+$ such that $n\alpha\bmod 1\in(c-k,d-k)$. If $\ell=\lfloor n\alpha\rfloor$, $n\alpha=\ell+(n\alpha\bmod 1)$, so

$$c-k+\ell<n\alpha<d-k+\ell\;.$$

Now let $m=k-\ell$ to find that $c<m+n\alpha<d$ and hence that $m+n\alpha\in(a,b)$.

This doesn’t quite work, because $m$ may be negative. To get around that problem, we pick infinitely many pairwise disjoint subintervals of $(c,d)$. Let $\langle c_k:k\in\Bbb N\rangle$ be a strictly increasing sequence lying entirely within the interval $(c,d)$, and for $k\in\Bbb N$ let $I_k=(c_k,c_{k+1})$. If $\alpha>0$, the argument that I applied to $(c,d)$ applies equally well to each of the intervals $I_k$, so for each $k\in\Bbb N$ there is an $n_k\in\Bbb Z^-$ such that $\lfloor c_k\rfloor+(n_k\alpha\bmod 1)\in I_k$; if $\alpha<0$, for each $k\in\Bbb N$ there is an $n_k\in\Bbb Z^+$ such that $\lfloor c_k\rfloor+(n_k\alpha\bmod 1)\in I_k$. In either case the numbers $n_k\alpha$ are negative, and because the intervals $I_k$ are pairwise disjoint, the numbers $n_k$ are distinct, and the set $\{n_k\alpha:k\in\Bbb N\}$ is unbounded. There must therefore be a $k\in\Bbb N$ such that $\lfloor c_k\rfloor-\lfloor n_k\alpha\rfloor>0$, and if we set $m=\lfloor c_k\rfloor-\lfloor n_k\alpha\rfloor$, then $m\in\Bbb N$, and we have $m+n_k\alpha\in I_k\subseteq(a,b)$.

Thus, $\{m+n\alpha:m\in\Bbb N\text{ and }n\in\Bbb Z\}$ meets every non-empty open interval $(a,b)$ and is therefore dense in $\Bbb R$.

Community
  • 1
Brian M. Scott
  • 616,228
  • Sir you nicely give the answer of above problem. Can you give me answer of my this question. http://math.stackexchange.com/questions/1436617/limit-points-of-particular-sets-of-real-numbers – neelkanth Sep 16 '15 at 11:03
  • (1) and (2) are now solved...please can you tell me about (3) and (4)? – neelkanth Sep 16 '15 at 11:05
  • @neela: I’m going to be away from the computer for a few hours, but I’ll take a look when I get back. – Brian M. Scott Sep 16 '15 at 11:14
  • @Brain M. Scott Thanking you. Whenever you get the time please loot at these parts also. – neelkanth Sep 16 '15 at 11:55
  • I'm sorry, what is $a_k$? I am just trying to understand the last part of your argument. – student Sep 21 '16 at 15:40
  • Do you mean $c_k$? – student Sep 21 '16 at 15:57
  • 1
    It all seems magic to me and I really can't figure out how you came up with this clever argument of dividing $(c,d)$ into intervals that have as endpoints the elements of a strictly increasing sequence $c_k$. – student Sep 21 '16 at 16:03
  • This works because $c_k$ is bounded. Right? – student Sep 21 '16 at 16:11
  • 2
    @student: Yes, $a_k$ was a typo for $c_k$. And yes, it works because the $c_k$ are bounded, since they lie in the interval $(c,d)$, while the $n_k\alpha$ are unbounded. – Brian M. Scott Sep 21 '16 at 16:48
  • Thank you very much for your help!! – student Sep 21 '16 at 16:59
  • @student: You’re welcome! – Brian M. Scott Sep 21 '16 at 17:00
  • @BrianM.Scott instead of the argument above could we say that since $\mathbb{R}$ is a $T_{1}$ space the interval $(c, d)$ must contain infinitely many points of ${n\alpha \mod 1: n\in\mathbb{Z}^{\pm}}$? – Karthik Kannan Feb 19 '21 at 08:03
  • @KarthikKannan: $(c,d)$ need not contain any of those points: those points are all in $(0,1)$, and $(c,d)$ could be anywhere on $\Bbb R$. – Brian M. Scott Feb 19 '21 at 21:12
  • @BrianM.Scott sorry I meant $(c-k, d-k)$ where $k:= \lfloor c\rfloor$ as defined above. I was wondering whether we could use the argument that $\mathbb{R}$ is $T_{1}$ to avoid constructing the sequence $(c_{k})$. – Karthik Kannan Feb 20 '21 at 03:28
  • 1
    @KarthikKannan: You mean argue that since $\Bbb R$ is $T_1$, any dense subset of $(c-k,d-k)$ must be infinite, so there is an infinite $N\subseteq\Bbb Z$ such that $0>n\alpha\in(c-k,d-k)$ for each $n\in N$? Yes, that would work. I don’t know whether I simply didn’t notice it at the time, or whether I didn’t want to have to prove it (since there’s no general topology tag) and saw that I could avoid doing so by using what I’d already proved at the top. – Brian M. Scott Feb 20 '21 at 06:07
  • @BrianM.Scott thank you very much for the clarification. – Karthik Kannan Feb 20 '21 at 08:17
  • @KarthikKannan: You’re very welcome. – Brian M. Scott Feb 20 '21 at 08:17