sequence accumulating everywhere in $[-1,+.1667em1]$

We want to show that if $k$ is an irrational number, then any real number of the interval  $[-1,\mathrm{\hspace{0.17em}1}]$  is an accumulation point of the sequence

$\sin 2k\pi ,\sin 4k\pi ,\sin 6k\pi ,\mathrm{\dots }$ (1)

In other words, the real numbers (1) come arbitrarily close to every number of the interval.

Proof.  Set on the perimeter of the unit circle, starting e.g. from the point  $(1,\mathrm{\hspace{0.17em}0})$,  anticlockwise the points

$P_0,P_1,P_2,\mathrm{\dots }$ (2)

with successive arc-distances $2k\pi$.  Since $k$ is irrational, $2k\pi$ and the length $2\pi$ of the perimeter are incommensurable.  Therefore no two of the points $P_i$ coincide, whence we have an infinite sequence (2) of distinct points.  We can see that these points form an everywhere dense set on the perimeter, i.e. that an arbitrarily short arc contains always points of (2).

Let then $\epsilon$ be an arbitrary positive number.  Choose an integer $n$ such that

and the perimeter of the unit circle, starting from the point  $(1,\mathrm{\hspace{0.17em}0})$,  into $n$ equal arcs.  Each of the points $P_1,P_2,\mathrm{\dots }$ falls into one of these arcs, because the arcs $2\pi /n$ and $P_0{P}_1$ are incommensurable.  Thus, among the $n+1$ first points $P_1,P_2,\mathrm{\dots },P_{n+1}$ there must be at least two ones belonging to a same arc.  Let $P_{\mu }$ and $P_{\nu }$ () belong to the same arc.  Then the length $l$ of the arc $P_{\mu }{P}_{\nu }$ is less than $\epsilon$.  Starting from $P_{\mu }$ one comes to $P_{\nu }$ by moving on the perimeter $\nu -\mu$ times in succession arcs with length $2k\pi$ (when one has possibly to go around the perimeter several times).  Repeating that procedure, starting from the point $P_{\nu }$, one comes to the point  $P_{\nu +(\nu -\mu )}=P_{2\nu -\mu }$, and furthermore to $P_{3\nu -2\mu }$, to $P_{4\nu -3\mu }$, and so on.

The points

$P_{\mu },P_{\nu },P_{2\nu -\mu },P_{3\nu -2\mu },\mathrm{\dots }$ (3)

form on the perimeter a sequence of equidistant points, a subsequence of (2).  Since the arc-distance of successive points of (3) equals to  , whence it is evident that any arc with length at least $\epsilon$ contains at least one of the points (3).  Consequently, the points (2) are everywhere dense on the perimeter of the unit circle.  Thus the same concerns their projections (http://planetmath.org/ProjectionOfPoint) on the $y$-axis, i.e. the sines (1) on the interval  $[-1,\mathrm{\hspace{0.17em}1}]$.

Title	sequence accumulating everywhere in $[-1,+.1667em1]$
Canonical name	SequenceAccumulatingEverywhereIn11
Date of creation	2013-03-22 19:13:39
Last modified on	2013-03-22 19:13:39
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	7
Author	pahio (2872)
Entry type	Example
Classification	msc 54A05
Related topic	LissajousCurves