sequence accumulating everywhere in
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 , anticlockwise the points
with successive arc-distances . Since is irrational, and the length of the perimeter are incommensurable. Therefore no two of the points 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 be an arbitrary positive number. Choose an integer such that
and the perimeter of the unit circle, starting from the point , into equal arcs. Each of the points falls into one of these arcs, because the arcs and are incommensurable. Thus, among the first points there must be at least two ones belonging to a same arc. Let and () belong to the same arc. Then the length of the arc is less than . Starting from one comes to by moving on the perimeter times in succession arcs with length (when one has possibly to go around the perimeter several times). Repeating that procedure, starting from the point , one comes to the point , and furthermore to , to , and so on.
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 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 -axis, i.e. the sines (1) on the interval .
|Title||sequence accumulating everywhere in|
|Date of creation||2013-03-22 19:13:39|
|Last modified on||2013-03-22 19:13:39|
|Last modified by||pahio (2872)|