On Continuity of a Function Concerning Dynamics on Circle

Dynamics is the study of the motion of a body, or more generally evolution of a system with time (see e.g. [2], [3]). In dynamics on circle, motion takes place through the points of a circle. This movement of points is tried to be represented by a function defined on the circle. Such functions are called circle maps. The study of dynamics on circle also includes knowing about the continuity of the concerned functions. A circle in the plane IRxIR is the boundary of a disc with centre at a point of the plane and some radius. Corresponding to discs in IRxIR, there are spheres in IRxIRxIR, in fact, for every natural number n, there are spheres in IRn+1 (=IRxIRx…….IR(n+1) times). Thus the boundary of a sphere with centre at a point of IRn+1 and some radius can be considered as a ‗circle‘ in IRn+1. Since any two circles in IRxIR are homeomorphic, circle in the plane means the circle at the origin of the plane with unit radius for the study involving dynamics on circle. Such a circle is denoted by S1. The corresponding circle in IRn+1 is denoted by Sn. But here we shall consider only S1. A circle map is a function whose domain and codomain both are S1. The continuity of a function at a point assures that the points which are close to the point of continuity (i.e. the points which are in a certain neighborhood of the points of continuity) are mapped, by the function, as close to each other as we want. Dynamics on the circle and dynamics of circle maps have been studied by many see e.g. [3], [4] and [5]. Every real number can be written as the sum of its integral part and fractional part. Let x  IR. Using Archimedean property of real numbers (It states that given a real number x there exists an integer n, n ≥ x) and well-ordering property of natural numbers (It states that every non-empty subset of natural numbers has the least element), there exists smallest integer nx such that x ≤ nx. This integer nx is called the integral value of x and is denoted by [x]. Since [x] ≤ x < [x] + 1, 0 ≤ x – [x] < 1. x – [x] is called the fractional part of x, and is denoted by rx. By definition S1 = {(cosx, sinx) : x  [0, 2)}. It can be seen that S1 = {(cos2y, sin2y) : y  [0, 1)}. Let x  IR. Then x = [x] + rx where rx  [0, 1). Therefore S1 = {(cos2x, sin2x) : x  IR}. For x  IR, eix = cosx + isinx is a periodic function with periodicity 2. There is (the covering map)  : IR  C = the set of complex numbers, defined as (x) = e2ix = cos2x + isin2x. As a point of IRxIR, (x) = (cos2x, sin2x). Thus  is also a function from IR to IRxIR. It can be seen (e.g. see [1]) that (x+n) = (x) for every integer n. Thus  is a periodic function with periodicity 1. Since S1 = {(cos2x, sin2x) : x  IR}, the function , maps IR onto the circle S1  IRxIR, or C. For every integer n,  : [n, n+1)  S1 is one-one and onto. In particular,  : [0, 1)  S1 is one-one and onto. If we identify 0 and 1 of [0, 1], then S1 can be identified with [0, 1]. A circle map is a continuous function f : S1  S1. For example, for a fixed , 0 <  < 2, if we define f : S1  S1 as f(cosx, sinx) = (cos(x+), sin(x+)), f is a circle map. Also, for a fixed , 0 <  < 2, if we define f* : S1  S1 as f*(cosx, sinx) = (cos(x+2π), sin(x+2π)), f* is a circle map. If we define, for x  IR, F : IR  IR, F(x) = x + /2, then we note that that  o F = f o . Let x  IR. (F(x)) = π(x + /2) = (cos2π(x + /2), sin2π(x + /2)) = (cos(2πx+), sin(2πx+)). (x) = (cos2x, sin2x). f((x)) = f(cos2x, sin2x) = (cos(2πx+), sin(2πx+)). Such a function F is called a lift of the circle map. If we define F* : IR  IR, F*(x) = x + , for x  IR, then it can be seen that  o F* = f* o . As F*(x) = x + , π(F*(x)) = π(x+) = (cos2π(x+), sin2π(x+)). (x) = (cos2x, sin2x). f*((x)) = (cos2π(x+), sin2π(x+)). Thus

F* is called a lift of the circle map f*. Thus the

continuous. Some results about the function F are obtained in [1] without assuming the continuity. For each x  IR, x = [x] + rx with 0 ≤ rx < 1. This gives a function r : IR  [0, 1) taking x to rx. The function r is called the fractional part function. The fractional part function apart from being used in proving some properties of subsets of IR (as seen below) is helpful in some part of the study of dynamics of circle maps, in particular, and functions in dynamical systems in general. In this article we shall study some subsets of IR where the fractional part function is continuous. The continuity of a real-valued function of a real variable is defined using –. Suppose the domain Y of a real-valued function of a real variable is a proper subset of IR. If Y is an interval, the same – definition of continuity works except at the end point(s), where we talk of left-hand/right-hand continuity depending upon the end point. When the domain Y is not necessarily an interval, we need an exact definition of continuity, to be precise in our considerations. An easy way for that is to consider subspace topology. But that definition is theoretical and not easy to apply. We have an equivalent definition of continuity of a real-valued function of a real variable defined on Y, in terms of –, which is easy to use.

IR is used to denote the real numbers. IN is the set of natural numbers. Z is the set of integers. A circle in the plane IRxIR is the boundary of the disc with centre at the origin of the plane and unit radius, it is denoted by S1. S1 = {(cosx, sinx) : x  [0, 2)} = {(cos2y, sin2y) : y  [0, 1)} = {(cos2x, sin2x) : x  IR}. A circle map is a continuous function f : S1  S1. Lift of a circle map f : S1  S1 is a continuous function F : IR  IR such that (i)  o F = f o  (ii) there exists some k  Z such that F(x + 1) F(x) + k for every x  IR. Archimedean property of real numbers : Given a real number x there exists an integer n, n ≥ x. Well-ordering property of natural numbers : Every non-empty subset of natural numbers has the least element. Let x  IR. Using Archimedean property of real numbers and well-ordering property of natural numbers, there exists smallest integer nx such that x ≤ nx. This integer nx is called the integral value of x. [x] is used to denote nx. Since [x] ≤ x < [x] + 1, 0 ≤ x – [x] < 1. x – [x] is called the fractional part of x, and is denoted by rx. For each x  IR, x = [x] + rx with 0 ≤ rx < 1. This gives a function r : IR  [0, 1) taking x to rx. The function r is called the fractional part function. Let Y  IR. cl(Y) denotes the closure of Y. A collection {Yj : j  J} of subsets of IR is called separated or pair {(n, n + s) : n  Z}. Let Cs = {(n + s, n + 1) : n  Z}. Let Cs* = CsZ = {[n + s, n + 1) : n  Z}. Let Bs = {x  IR : r(x) ≤ s}. Let Bs* = {x  IR : r(x) ≥ s}. Let Es = {x  IR : r(x) = s} = BsBs*. For n  Z and 0 < s < 1, let Asn = [n, n + s] and Vsn = [n, n + s). We can define As and Asn, also for s = 0; A0 turns out to be Z and A0n = {n}. Let Y  IR. Let g : Y  IR. Let x  Y, g : Y  IR is continuous at x if g : (Y, *)  IR is continuous at x, where * is the induced topology of the usual topology of IR. g : Y  IR is continuous if g : Y  IR is continuous at every x  Y.

As seen in the introduction, for a fixed  with 0 <  < 2, we have two circle maps f and f*, defined as f(cosx, sinx) = (cos(x+), sin(x+)) and f*(cosx, sinx) = (cos(x+2π), sin(x+2π)). F : IR  IR, F(x) = x + /2, and F* : IR  IR, F*(x) = x +  are lifts of f and f* respectively. For  = /4, /2 and 3/4, we have the following graphs of f. Since  is taking a particular value, we can write f in place of f.

 = /4

x=0:0.01: pi; x1=cos(x); y1=cos(x+pi/4); x2=sin(x); y2=sin(x+pi/4); plot(x1,y1) grid on

 = /4

-1-0.8-0.6-0.4-0.200.20.40.60.81-1 -0.8 -0.6 -0.4 -0.2

0

0.2 0.4 0.6 0.8

1

x=0:0.01: pi; x1=cos(x); y1=cos(x+pi/2); x2=sin(x); y2=sin(x+pi/2); plot(x1,y1) grid on

 = /2

-1-0.8-0.6-0.4-0.200.20.40.60.81-1 -0.8 -0.6 -0.4 -0.2

0

0.2 0.4 0.6 0.8

1

 = 3/4

x=0:0.01: pi; x1=cos(x); y1=cos(x+3*pi/4); x2=sin(x); y2=sin(x+3*pi/4); plot(x1,y1) grid on

 = 3/4

Remark.3.1. IR = {V1n : n  Z} = {[n, n + 1) : n  Z}. Proof. Let x  IR. x = [x] + r(x) with 0 ≤ r(x) < 1. Therefore x  [m, m + 1), where m = [x]. Thus IR = {[n, n + 1) : n  Z}. Remark.3.2. (i) For every s, 0 ≤ s < 1, As = {Asn : n  Z}. (ii) IR = {[n, n + s) : n  Z, 0 < s < 1}. (iii) IR = {Asn : n  Z, 0 < s < 1}. Proof. (i) It follows by definitions of As and Asn. (ii) Let x  IR. x = [x] + r(x) with 0 ≤ r(x) < 1. Let t be such that 0 ≤ r(x) < t < 1. Therefore x  [m, m + t), where m = [x]. Thus IR = {[n, n + s) : n  Z, 0 < s < 1}. (iii) By (ii), IR = {[n, n + s) : n  Z, 0 < s < 1}  {Asn : n  Z, 0 < s < 1}. So (iii) follows. Lemma.3.3. For every s, 0 < s < 1, As = Bs. Proof. Let x  As. Then x  [n, n + s] for some n  Z. Since n ≤ x ≤ n + s < n + 1, n = [x]. Since x = [x] + r(x) and n + r(x) ≤ n + s, therefore r(x) ≤ s. So x  Bs. Conversely, let x  Bs. [x] ≤ x = [x] + r(x) ≤ [x] + s as r(x) ≤ s. Therefore x  [[x], [x] + s]. So x  As. Remark.3.4. For every s, 0 < s < 1, Es = {n + s : n  Z}. Proof. Let x  Es. Since r(x) = s, x = [x] + s. Therefore x  {n + s : n  Z}. Conversely, let x  {n + s : n  Z}. Then x = m + s for some m  Z. So x – m = s. Since [x – m] = – m + [x] and 0 < s < 1, therefore m = [x]. This implies that r(x) = s. Therefore x  Es. Lemma.3.5. Let H  IR. Let h : H  IR. Let x  H. If there exists a * > 0 such that for y  (x  *, x + Proof. (i) Let  > 0. Let  = min{, *}. Let y – x < . Then y  (x  *, x + *) as  ≤ *. Therefore by the given condition (i), h(x) – h(y) = x – y <  ≤ . (ii) Let  > 0. Let  = min{/, *}. Let y – x < . Then y  (x  *, x + *) as  ≤ *. Therefore by the given condition (ii), h(x) – h(y) ≤ x – y <  ≤ .

Lemma.4.1. Let {Yj : j  J} be a collection of subsets of IR such that {Yj : j  J} is closed. Then {Yj : j  J} are pair wise separated iff {Yj : j  J} are pair wise disjoint and each Yj is closed. Proof. If {Yj : j  J} are pair wise disjoint and each Yj is closed then clearly {Yj : j  J} are pair wise separated. Now suppose that {Yj : j  J} are pair wise separated. Let k  J. Let x  cl(Yk). Suppose x  Yk. For j  J, j ≠ k, since cl(Yk)Yj = , x  Yj. Thus x  {Yj : j  J}. Therefore x  B = IR – {Yj : j  J}. Since B is open and x  cl(Yk), BYk ≠ . But BYk =  as B  IR – Yk. Corollary.4.2. Let {Bj : j  J} and {Yj : j  J} be such that IR – {Yj : j  J} = {Bj : j  J}. If {Bj : j  J} is open then {Yj : j  J} are pair wise disjoint and each Yj is closed iff {Yj : j  J} are pair wise separated. Proof. By the given condition {Yj : j  J} is closed. Therefore the result follows by Lemma.4.1. Lemma.4.3. Let IR = {Yj : j  J} be such that YjYk =  for every j, k  J, j ≠ k. If for each j  J, Yj = HjKj with HjKj = , then IR – {Hj : j  J} = {Kj : j  J}. Proof. Let j, k  J, k ≠ j. Since YjYk = , YjHk = . Thus Yj  IR – Hk. So Yj – Hk = Yj(IR – Hk) = Yj. This implies that {Yj – Hk : k  J, k ≠ j} = Yj. Therefore, {Yj – Hk : k  J} = Yj – Hj. Now IR – {Hj : j  J} = ({Yj : j  J}) – ({Hk : k  J}) = {(Yj – ({Hk : k  J}) : j  J} = {(Yj – Hk : k  J) : j  J} = {Yj – Hj : j  J} = {Kj : j  J}, as Yj – Hj = Kj. Lemma.4.4. For every s, 0 < s < 1, IR – As = Cs = {(n + s, n + 1) ; n  Z}. Proof. By Remark.3.1, IR = {[n, n + 1) : n  Z}. Let J = Z. Let, for n  Z, Yn = [n, n + 1). Then YnYk =  for every n, k  Z, n ≠ k. Let Hn = [n, n + s] and Kn = (n + s, n + 1). Then Yn = HnKn and HnKn = . Therefore by Lemma.4.3, IR – As = Cs = {(n + s, n + 1) : n  Z}. Remark.4.5. For every s, 0 < s < 1, Bs* = {[n + s, n + 1) : n  Z}. Proof. Bs* = {x : r(x) > s}{x : r(x) = s}. Therefore, by definition of Bs, Bs* = (IR  Bs)Es = (IR – As)Es as Bs = As by Lemma.3.3. Since, by Lemma.4.4, IR – As = Cs Remark.4.6. Let 0 < s < 1. For every n  Z, n is a limit point of Bs*, so Bs* is not closed. Proof. By Remark.4.5, Bs* = {[n + s, n + 1) : n  Z}. Since n  Z iff n – 1  Z, Bs* = {[n –1 + s, n) : n  Z}. For n  Z, n is a limit point of Bs*, so Bs* is not closed. Remark.4.7. IR – Aso = Cs* = {[n + s, n + 1) ; n  Z}, where Aso = {(n, n + s) : n  Z}. Proof. IR = {[n, n + 1) : n  Z}. Let J = Z. Let for n  Z, Yn = [n, n + 1). Then YnYk =  for every n, k  Z, n ≠ k. Let Hn = (n, n + s) and Kn = [n + s, n + 1). Then Yn = HnKn and HnKn = . Therefore by Lemma.4.3, IR – Aso = Cs*= {[n + s, n + 1) ; n  Z}. Remark.4.8. The Lemma.4.4 can also be proved using the function r i.e using Lemma.3.3. We give the proof below. Lemma.4.9. (i) IR – As = Cs = {(n + s, n + 1) ; n  Z}. (ii) As is closed. Proof. (i) Since by Lemma.3.3, As = Bs, we prove that IR – Bs = {(n + s, n + 1) : n  Z}. IR – Bs = {x  IR ; r(x) > s}. Let x  IR – B. So r(x) > s. x = [x] + r(x). We claim x  ([x] + s, [x] +1). Since r(x) > s, [x] + r(x) > [x] + s. Since r(x) < 1, [x] + r(x) < [x] + 1. Therefore [x] + s < x < [x] + 1. Hence x  ([x] + s, [x] + 1). For the converse, suppose that x  (n + s, n + 1) for some n  Z. Since n + s < x < n + 1, n = [x]. So r(x) = x – [x] > s. Therefore, x  IR – Bs. (ii) Since {(n + s, n + 1) ; n  Z} is open, by (i), IR – As is open. Therefore, As is closed. The following is the direct proof that As is closed. We need the following Remark for that. Remark.4.10. Let a, x, b  IR such that a < x < b. Let  = min{x – a, b – x} then a ≤ x –  and x +  ≤ b. So (x  , x + )  (a, b). Proof. Since  = min{x – a, b – x},  ≤ x – a and  ≤ b – x. Therefore a ≤ x –  and x +  ≤ b. Therefore (x  , x + )  (a, b). Proof of As is closed: Suppose x  As. So x  [[x], [x] + s]. Therefore r(x) > s as x = [x] + r(x). Thus [x] + s < x < [x] + 1. Let  = min{r(x)  s, 1  r(x)}. Since x – ([x] + s) = r(x)  s, and [x] + 1 – x = 1 – r(x), by Remark.4.10, (x  , x + )  ([x] + s, [x] + 1). Therefore (x  , x + )  [[x], [x] + s] = . Let n  Z, n ≠ [x]. Either n < [x] or [x] + 1 ≤ n. Let n < [x]. By Remark.4.10, s ≤ r(x)  . Therefore, n + s < x  . Suppose [x] + 1 ≤ n. Using Remark.4.10, x +  ≤ [x] +1 ≤ n. Thus (x  , x + )  [n, n + s] =  for every n

point of As. Hence As is closed. As mentioned in the introduction, we need to have a working definition of the continuity of a real-valued function of a real variable is defined using , when the domain Y of the function is a proper subset of IR. If Y is an interval, the same  definition of continuity works except at the end point(s). When the domain Y is not necessarily an interval, we have the following definition which works even when Y is an interval. But first we have the theoretical definition. Then we have an equivalent definition of continuity of a real-valued function of a real variable in terms of , which we use later. Remark.4.11. Let Y  IR. g : Y  IR. Let x  Y. (i) g is continuous at x iff, for given  > 0, there exists  > 0 such that for y  (x  , x + )Y, g(x) – g(y) < . (ii) If there exists  > 0 such that for y  (x  , x + )Y, g(x) – g(y) ≤ x – y, then g is continuous at x. Proof. (i) It follows as (x  , x + )Y is open in the induced topology on Y. (ii) For given  > 0, if we take * = min{, }, then g is continuous at x. Remark.4.12. Let H  Y  IR. Let g : Y  IR. Let x  H. (i) If g : H  IR is not continuous at x, then g : Y  IR is not continuous at x. (ii) The converse of (i) is not true. That is, if g : Y  IR is not continuous at x, then g : H  IR may be continuous at x. Proof. (i) g : H  IR is not continuous at x, so by Remark.4.11, there exists some  > 0 such that whatever  > 0 we take there exists y  (x  , x + )H such that g(x) – g(y) ≥ . Since H  Y, y  (x  , x + )Y. Therefore, in view of Remark.4.11, g : Y  IR is not at x. (ii) Take H = Z and Y = IR. Every g :

Remark.4.13. Let x, y  IR. If [x] = [y], then x – y = r(x) – r(y). r(x) < r(y) iff x < y. Proof. x = [x] + r(x) and y = [y] + r(y). Therefore x – y = r(x) – r(y) as [x] = [y]. Now it follows that r(x) < r(y) iff x < y. Proposition.4.14. Let As = {[n, n + s] ; n  Z}. r : As  IR is continuous. Proof. Let x  As. First we find  > 0 such that for y  (x  , x + )As, [y] = [x]. x  [m, m + s] for some m  Z. Case(i) x  (m, m + s). Let  = min{x – m, m + s – x}. By Remark.4.10, (x  , x + )  (m, m + s). Let y  (x  , x + )As. Then y  (m, m + s). Therefore [y] = [x] as s < 1. Case(ii) x = m + s. Take  = ½(min{s, (1- s)}). By Remark.4.10, (x  , x + )  (m, m+1). Let y  (x  , x + )As. Then y  (m, m+1). Therefore [y] = m (1  s)}). Now by Remark.4.10,(x  , x + )  (m  1, m + s). Let y  (x  , x + )As. Then y  [m, m + s) as (m  1, m + s)As  [m, m + s). Therefore [y] = [x]. Thus in every case, for y  (x  , x + )As, [y] = [x]. By Remark.4.13, x – y = r(x) – r(y). Now, by Remark.4.11 (ii), r is continuous at x. Remark.4.15. By Proposition.4.14, for every s with 0 < s < 1, r is continuous on As. The collection {As : s  IR, 0 < s < 1} is a totally ordered subset of the p.o. set (IR, ) because for s, t  IR, 0 < s, t < 1, As  At, or At  As depending upon s ≤ t, or t ≤ s. Let x  IR. x = [x] + r(x). Since 0 ≤ r(x) < 1, x  At for every t such that r(x) < t. Therefore {As : s  IR, 0 < s < 1} = IR. By definition Aso = {(n, n + s) : n  Z}. It can be seen that {Aso : s  IR, 0 < s < 1} = IR – Z. Remark.4.16. We have seen above that, for s = 0, As = Z, i.e. A0 = Z. It can be seen (below) that r is continuous also on A0. Remark.4.17. (i) r is continuous on IR – Z. (ii) r is continuous on Z. (iii) r is continuous on H, if H  IR – Z or Z. Proof. (i) Let x  IR – Z. Since [x] < x < [x] +1, by Remark.4.10 there exists a  > 0 such that (x  , x + )  ([x], [x] + 1). Therefore, for y  (x  , x + ), [y] = [x]. By Remark.4.13, x – y = r(x) – r(y). Therefore r is continuous at x. Therefore r is continuous on IR – Z. (ii) Let m  Z. Let  > 0. For  < 1, (m  , m + )Z = {m}, therefore r(y) – r(m) = 0 for every y  (m  , m + )Z. (iii) Restriction of a continuous function is continuous.

In the above considerations there are many subsets of IR on which the fractional part function r is continuous. There may be other subsets of IR on which r is continuous. But we do not have any more information about such subsets of IR. It is worth investigating to know other subsets of IR on which r is continuous, or to have some information about such subsets.

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