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Classification of Maps on Intervals That Exhibit Dense Periodic Points 
 

*Suja N Thomas,**P.B Vinod Kumar 

Rajagiri School of Engineering and Technology, Kochi APJ Abdul Kalam Technological University, Kerala, India. 

*sujanthomas72@gmail.com 

Muthoot Institute of Technology and Science, Kochi APJ Abdul Kalam Technological University, Kerala, India. 

**vinodkumarpb@mgits.ac.in 

 

Article History: 

Received: 14-10-2024 

Revised: 28-11-2024 

Accepted: 10-12-2024 

Abstract:  

In this paper, continuous maps on interval with dense set of periodic points such that 

the periods being set of natural numbers are studied.These maps named as Peri- 

odically Rich maps could be divided into two classes on Interval, such as the maps 

with period four points having increasing periodic orbit and the maps with two fixed 

points such that image of a subset of Interval does not belong to the same subset. 

Notion of bigness of Periodically Rich maps is also discussed. 

Keywords: Set of periods, Transitive maps, Dense Periodic points. 

 

1. INTRODUCTION 

Let X be a Hausdorff topological space and f : X → X be a continuous map. x ∈ X such that f 

(x) = x is called a fixed point of f . A point x ∈ X is called a periodic point if there exists n ∈ Z+ 

such that fn(x) = x. The smallest n for which fn(x) = x is called the period of f . The set of 

periodic points of period n is denoted by Pern(f ) and the set of periods of f is denoted by P (f ), 

i.e., 

Pern(f ) = {x ∈ X|fn(x) = x and fm(x) /= x, ∀m < n} P (f ) = {n ∈ Z+|Pern(f ) /= ∅}. 

Per(f ) = 
S

n∈Z+ 
Pern(f ) is the set of periodic points of f . The forward orbit of a point x ∈ X 

under the mapping f is denoted by Of (x),i.e., Of (x) = {x, f (x), . . . , fn(x), . . . }. If x ∈ Pern(f ) 

then {x, f (x), . . . , fn−1(x)} is called periodic orbit of x, in this case clearly |Of (x)| = n. The 

periodic orbit of a point x in I = [0, 1] is said to be increasing if x < f (x) < · · · < fn−1(x). A map f is 

said to be transitive, if for every pair of non-empty open subsets U and V in X, there exists a 

positive integer n such that fn(U ) ∩ V /= φ. 

In general, the set of periodic points Per(f ) can be empty, a finite set, a countable set or an 

uncountable set . Per(f ) can be even rich so that Per(f ) = X, which is one of the condition for f 

to be chaotic in the sense of Devaney [5]. If f is transitive on I then 



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Per(f ) = I [12]. A point x ∈ I is recurrent if for each neighborhood U of x there exists 

n ∈ Z+ such that fn(x) ∈ U ; the set of recurrent points is denoted by Rec(f ). It is proved in [3] that 

for map f on I, Per(f ) = Rec(f ). For maps on I, Per1(f )∪Per2(f ) is closed 

[4]. Considering the set of periods, P (f ) can also be empty, a finite or an infinite subset of Z+ or the 

whole Z+. Sharkovskii has proved that for maps on R or I, if n ∈ P (f ) then 

m ∈ P (f ) for every m which follows n in the Sharkovskii ordering 3 ≺ 5 ≺ 7 ≺ 9 ≺ 

... ≺ 2.3 ≺ 2.5 ≺ 2.7 ≺ ...22.3 ≺ 22.5 ≺ ...23.3 ≺ 23.5 ≺ 23.7... ≺ 23 ≺ 22 ≺ 2 ≺ 1. 

[10]  

At this juncture a natural question arises. Is there any relation between Per (f ) and P (f )? The 

question prompts to think about the ’bigness’ of set of periodic points. More specifically, the 

‘bigness’ of set of periodic points is treated in two ways, (1) Perf = X and (2) P (f ) = Z+. 

To understand, some examples are listed below. 
 

Example 1. A map for which Per(f ) = X and P (f ) /= Z+ 

Let f : I —→ I be defined as 

            2    x + 1/2,  0 ≤ x ≤ 1/4 

f1(x) =        - 2x + 3/2, 1/4 < x ≤ 3/4 

   
2x-  3/2, 3/4 < x ≤ 1 

 

 
FIGURE 1. Graph of f1 

 



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1 

1 

1 

1 

Fixed point of f1 is at x = 1/2. 

f1([0, 1/2]) = [1/2, 1] and f1([1/2, 1]) = [0, 1/2] 

f 2([0, 1/2]) = f1(f1([0, 1/2])) = f1([1/2, 1]) = [0, 1/2] 

f 2([1/2, 1]) = f1(f1([1/2, 1])) = f1([0, 1/2]) = [1/2, 1] 

f 3([0, 1/2]) = f1(f 2([0, 1/2])) = f1([0, 1/2]) = [1/2, 1] 
1 1 

f 3([1/2, 1]) = f1(f 2([1/2, 1])) = f1([1/2, 1]) = [0, 1/2] 
1 1 

So, f 3(x) /= x for all x ∈ [0, 1] except for x = 1/2. 

Similarly, if n is odd, fn(x) /= x, ∀x ∈ [0, 1] except for x = 1/2. 

If n is even, fn([0, 1/2]) = [0, 1/2] and fn([1/2, 1]) = [1/2, 1]. i.e., fn has fixed points 
1 1 1 

in [0, 1] for even n. So there are points with even periods in [0, 1]. Therefore 

P (f1) = Z+\{3, 5, 7, . . . }. 

 

On the other hand,if x is any irrational number in [0, 1] then Of1 (x) = I. So f1 is transitive on I [11]. 

Hence Per(f1) = I [12]. 

Example 2. Consider a map with P (f ) = Z+ and Per(f ) /= X. Let f2 : I → I be defined as 

 

 

FIGURE 2. Graph of f2 

    3x, 0 ≤ x ≤ 1/3 

f2(x) =     1 ,  1/3 < x ≤ 2/3 

    3- 3x, 2/3 < x ≤ 1 

Of2 (9/28) = {9/28, 27/28, 3/28}. 

 



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2 

4 

 

i.e., 9/28 is a periodic point of period 3. Hence by proposition 1, P (f2) = Z+. 

Now consider open sets U and V in (1/3, 2/3). fn(U ) = fn(V ) = {0}, ∀ n ∈ Z+. So, 
2 2 
 

there does not exists n ∈ Z+ such that fn(U ) ∩ V /= ∅. Therefore Per(f2) /= I. 

Example 3. Tent map defined on I. 

f3 (x) =
        2x, 0 ≤ x ≤ 1/2 

                2(1 — x), 1/2 < x ≤ 1 

 

P (f3) = Z+ and Per(f3) = I [9]. 

Example 4. Consider f4 defined on I by f4(x) = x2. 

Fixed points are 0, 1 and fn(x) → 0, ∀x ∈ (0, 1). So no periodic points other than 

fixed points. 

Here neither Per(f4) /= I nor P (f4) /= Z+. 

 

 

FIGURE 3. Graph of f3 

 
FIGURE 4. Graph of f4 

 

Example 5. Let Sı = {z ∈ C |ı z ı= 1} and f5 : Sı —→ Sı be defined as f5(z) = z2. 

   

Per(f5) = Sı and | Pern(f5) |= 2n — 1, n ≥ 1 [6] 

Example 6. Let f6 : R —→ R be defined as f (x) = x2−7 . 
  6                   2 

 

 



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FIGURE 5. Graph of f6 

 

Per3(f6) /= ∅ and therefore P (f6) = Z+ [10]. 
 

Also Per(f6) ∩ Q = ∅ and Per(f6) = R. 

 

Example 7. Let f7 : C —→ C be defined as f7(z) = z2 — z. P (f7) = Z+\{2} [2]. Here Per(f7) /= 

C. 

From the examples given above, it is observed that there is no implication between 
 

P (f ) = Z+ and Per(f ) = X. 

 
Definition 1. Let f be a continuous map defined on a topological space X. We say that f is 

Periodically Rich if f has both a dense set of periodic points and the set of periods is equal to Z+. 

 

i.e., f is Periodically Rich if (1) Per(f ) = X and (2) P (f ) = Z+. 

Let PR(X) be the set of continuous maps on X which are Periodically Rich. If PR(X) /= 

∅, we say that X admits Periodically Rich maps. For example Interval I admits Periodi- cally Rich 

maps since tent map satisfies both the conditions of definition. 

 

2. PERIODICALLY RICH MAPS ON INTERVAL 

 

The following proposition is a corollary to Sharkovskii’s result. 

 

Proposition 1. For continuous maps f : I → I if Per3(f ) /= ∅ then P (f ) = Z+. 

 

Proposition 2. Let f : I —→ I be a continuous map. If f is transitive and Per3(f ) /= ∅ 

then f ∈ PR(I). 

 

Proof. If f is transitive then Per(f ) = I [12] and Per3(f ) /= ∅ implies P (f ) = Z+ by proposition 



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1. Hence f ∈ PR(I).   

Proposition 3. Let f : I —→ I be a transitive map. If there exists x ∈ Per4(f ) with increasing 

orbit then f ∈ PR(I). 

 

Since f is transitive, Per(f ) = I [12]. We use the following lemmas from the literature 

[1] to show that Per3(f ) /= φ. 

 

Lemma 1. Let f : I → I be a continuous map and let J, K c I be closed intervals with 

f (J) m K, then there exists a closed interval H c J with f (H) = K. 

 

Lemma 2. Let f : I → I be a continuous map. Suppose H and K are closed intervals with H c K 

c I and f (H) = K, then f has a fixed point in H. 

 

Proof of Proposition 3: Let x ∈ Per4(f ) with increasing orbit, i.e., 

x < f (x) < f 2(x) < f 3(x). 

f ([x, f (x)]) m [f (x), f 2(x)] 

f ([f (x), f 2(x)]) m [f 2(x), f 3(x)] 

f ([f 2(x), f 3(x)]) m [x, f (x)] 

 
So by lemma 2, there exists 

F1 c [x, f (x)] such that f (F1) = [f (x), f 2(x)] 

F2 c [f (x), f 2(x)] such that f (F2) = [f 2(x), f 3(x)] 

F3 c [f 2(x), f 3(x)] such that f (F3) = [x, f (x)]. 

Now, f 2(F3) = f (F1); f 2(F1) = f (F2); f 2(F2) = [x, f (x)].Therefore [x, f (x)] = f 3(F1). So by 

lemma 2, f 3 has a fixed point in F1, say y. Now y /∈ Per2(f ), since f 2(y) ∈ [f (x), f 2(x)]. 

Therefore y ∈ Per3(f ). 

So Per3(f ) /= ∅. Now, by Proposition 2, f ∈ PR(I). 

Proposition 4. Let f be a continuous map such that 

(i) f (x) = x for some x ∈ (0, 1) 

(ii) f (1) = 1 



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(iii) For every J c I, f (J) /⊆ J 

(iv) f is transitive. Then f ∈ PR(I). 

Before proving proposition, we prove a lemma. 

Lemma 3. Let f : I → I be a map and there exists a < b < c in (0, 1) such that 

(i) f (a) = a 

(ii) f (c) ≤ a 

(iii) f (b) ≥ c 

then P (f ) = Z+. 

Proof. Let a < b < c and f (a) = a. Let f (b) = br, f (c) = cr. For given conditions, cr ≤ a and c ≤ br. 

So c ∈ [cr, br]. By Intermediate value theorem, there exists x ∈ [b, c] such that f (x) = c. Now since x ∈ 

(a, br) , f (a) = a,and f (b) = br by Intermediate value theorem, there exists y ∈ (a, b) with f (y) = x. 

Therefore cr = f (c) < y < x = f (y) < c = f (x). By standard result [8], P (f ) = Z+ 

Proof of Proposition 4: By condition (i), there exists a ∈ (0, 1) such that f (a) = a and 

 

 

FIGURE 6. Case 1 of Proposition 4 

by condition (ii), f (1) = 1. Let us take one of the two possibilities f (0) /= 1 or f (0) = 1. 

Case 1: If f (0) /= 1. 

Let m = sup{f (x)|x ∈ (0, a]}. Clearly m > a. There are two possibilities either 

m /= 1 or m = 1. 

Consider m /= 1. Let b = inf{x ∈ (a, m]|f (x) > m}, then for all x ∈ [a, b], f (x) ≤ 

m, andf (b) > m. Let ar = sup{x ∈ [a, b]|f (x) = x}, ar ∈ [a, b). Now f (x) > x for 

all x ∈ (ar, b] and f ([ar, b]) = [ar, m]. Let n = min{f (x)|x ∈ [b, 1]} and c = inf{x ∈ 

[b, 1]|f (x) = n}. 

 

Now ar > f (c) = n. 



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Claim: There exists d ∈ (ar, c) such that f (d) > c. 

If c < m, take d = b. If c ≥ m, there exists y ∈ (b, c) such that f (y) > c, take d = y. Claim is 

proved. 

So we have, ar < d < c with f (ar) = ar, f (c) ≤ f (ar) and c < f (d). By lemma 3, 

P (f ) = Z+. 

Now consider the other case m = 1. 

Let x ∈ [0, a] such that f (x) = m = 1. Let n = min{f (x)|x ∈ [a, 1]} and let 

 

FIGURE 7. when m = 1, n ≥ x 

f (c) = n, c ∈ [a, 1]. Either n /= 0 or n = 0. If n /= 0 and n ≥ x,then there exists z ∈ (x, a) 

such that f (z) ≥ x. So x < z < a with f (x) ≥ f (a) and x ≤ f (z). By lemma 3, P (f ) = Z+. 

If n /= 0 and n < x, then there exists z ∈ (n, a) such that f (z) > n. Here z ∈ (n, x) ∪ (x, 

a). If z ∈ (n, x) then z < x < c with f (z) > z and f (x) > x. De- fine g(y) = f (y) — y. By 

Intermediate value theorem, there exists s ∈ (z, x) such that g(s) = 0 and g(z) < 0, g(x) > 0. 

Therefore s ∈ (z, x) is fixed for f . So s < x < c with fixed s, f (c) ≤ f (s) and c ≤ f (x). By 

lemma 3, P (f ) = Z+. 

 

FIGURE 8. when m = 1, n < x 



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Case 2: If f (0) = 1 

Let c ∈ (0, 1) such that f (c) = 0. If such c doesn’t exist, p = inf{f (x)|x ∈ I} > 0. Hence [p, 0] 

is invariant under f . 

So there exists z ∈ (c, 1) such that f (z) = c. Therefore P (f ) = Z+. 
 

By condition (iv), Per(f ) = I. Therefore f ∈ PR(I). Hence the proof.   

We summarize the above propositions 3 and 4 as given below 

 

Proposition 5. Let f be transitive map on I. Then f ∈ PR(I) if f belongs to anyone of the following 

two classes. 

Class 1. 

There exists x ∈ Per4(f ) with increasing periodic orbit. 

Class 2. 

(i) f (x) = x for some x ∈ (0, 1) 

(ii) f (1) = 1 

(iii) For every J c I, f (J) /⊆ J. 

Now, we shall give an example to show that transitivity of f is not a necessary condition for f to be in 

PR(X). 

Example 8. Let f : R → R be defined as 

 3x,  0 ≤ x ≤ 1/3 

 -3x + 2,  1/3 ≤ x < 2/3 

f (x) =  3x — 2,  2/3 < x ≤ 1 

 f (x — 1) + 1,  x ≥ 1 

   f   (x + 1) — 1,  x ≤ 0 

 



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FIGURE 9. Graph of f 

 
f|[n,n+1] has one fixed point in (n, n + 1) 

 fi|[n,n+1] has 3i — 2 fixed points in (n, n + 1). 

If x and y are fixed points of f i in (n, n + 1), then 

 

|fi(x) — fi(y)| < (1/3)i−1. 
 

Therefore Per(f ) = R. 

Hence f ∈ PR(R). Here f is not transitive. 

Note: Let C(I) be the space of all continuous function defined on I and let 

(i) T (I) = {f ∈ C(I)|f is transitive} 

 
FIGURE 10.  The relationship between Per(I),P (I),T (I),and PR(I). 
 

(ii) Per(I) = {f ∈ C(I)|Per(f ) = I} 

(iii) P (I) = {f ∈ C(I)|P (f ) = Z+}. 

Then we have, T (I) c Per(I) and PR(I) = Per(I) ∩ P (I). Also T (I) ∩ P (I) /= ∅ (see figure 10). 



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2 

4 

2ϴ 

It is proved that T (I) ∼= T (I) ∼= l2 [7]. 

Now, we shall prove the following proposition. 
 

Proposition 6. P (I) = C(I). 
Proof. Let c be a fixed point of f , i.e., f (c) = c. Given an ϵ > 0, we will construct a 

g ∈ P (I) such that P (g) = Z+.  

Case 1: c /= 1. 

Since f is continuous at c, there exists a δ > 0 (choose δ < є ) such that 

 

|f (x) — c| < 
ϵ 

, for all x ∈ [c, c + δ]. 
2 

Let θ = δ , A be [c, c + 2θ] and B be [c + 2θ, c + 3θ]. 

Define g1 : A → A as 

 

 

 

FIGURE 11. Graph of g1 

 

g1 (x) = 
            2x-c,           x ∈ [c, c + θ] 

 

    -2x + 3c + 4θ.  x ∈ [c + θ, c + 2θ]. 

 
(see figure 11 ) 

g1 on [c, c + 2θ] is topologically conjugate to the tent map on [0, 1] defined by 

T (x) =
 2x,         ≤ x ≤ 1/2 

 2(1 — x), 1/
2 ≤ x ≤ 1 

via the homeomorphism h : [c, c + 2θ] → [0, 1] defined as h(x) = x−c . 
 

Therefore Per(g1) = [c, c + 2θ]. 



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2 
θ 

 

 

7 

x∈A x∈A 

Let g2 : B → [0, 1] be defined as 

g  (x) = c + (x — c — 2θ)

  
f (c + 3θ) — c

  

. 

i.e., the graph of g2 is a line segment joining (c + 2θ, c) and (c + 3θ, f (c + 3θ)) (see figure 12 ). 

 

FIGURE 12. Graph of g2  

FIGURE 13. Graph of g at c /= 1 

Now, define g : [0, 1] → [0, 1] as 

 f (x),  x ∈ [0, 1]\(A ∪ B)  

                                    g(x) =     g1(x), x ∈ A 

 g2(x),  x ∈ B 

(see figure 13 ). 

 

 0,   x ∈ [0, 1]\(A ∪ B) 

                |f (x) — g(x)| =        f (x) — g1(x)|,  x ∈ A 

    (x) — g2(x)|,  x ∈ B 

 

Hence 

 f — g  = max{max |f (x) — g1(x)|, max |f (x) — g2(x)|} 
x∈A x∈B 

≤ max

  

max |f (x) — c| + max |c — g1(x)|

  

, 

   

max |f (x) — c| + max |c — g2(x)|

  

 
 
 
Let x0 = c + 4 θ. 

x∈B 

< max{ 
ϵ 

+ 
ϵ 

,  ,
ϵ 

+ 
                     2 4 2 

x∈B 

ϵ 
} = max{

3ϵ 

2 4 

 

, ϵ} = ϵ. 



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Therefore x0 ∈ Per3(g). Hence P (g) = Z+. 

Case 2: If c = 1. 

Let A be [1 — 2θ, 1] and B be [1 — 3θ, 1 — 2θ]. 

Let g1 : A → A be defined as 

g (x) =            
2x-1, x ∈ [1 — θ, 1] 

 3 — 2x — 4θ, x ∈ [1 — 2θ, 1 — θ] 

Let g2 : B → [0, 1] be defined as 

 

𝑔2(𝑥)  =  𝑓 (1 —  3𝜃)  +
(𝑥 —  1 +  3𝜃)(1 —  𝑓 (1 —  3𝜃))

𝜃
 

Let g : [0, 1] → [0, 1] be such that 

                     f (x),  x ∈ [0, 1]\(A ∪ B)  
 
                                    g(x) =    g2(x) ,  x ∈ B 

                       g 1    ( x ) ,      x ∈ A 

(see figure 14) 
 

 
FIGURE 14. Graph of g at c = 1 

Then as in the earlier case, given ϵ > 0, 

 f — g  < ϵ. 

(1 — 4/7θ) ∈ Per3(g) since 



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    Hence the proposition.   

Corollary 1. Let f ∈ C(I). Then 

(i) Given ϵ > 0, there exists g ∈ P (I) such that |f — g| < ϵ and g/J ∈ PR(J) for some interval J 

c I. 

(ii) {g : I → I| given ϵ > 0, |f — g| < ϵ and g/J ∈ PR(J)}  = |Fix(f )|. 

Proof. (i)Take 
            

J =
         [c,c+ 2θ] if c /= 1 

                 [ 1-2θ, 1]  if c = 1. 

g/J ∼= T/[0,1]. Therefore g ∈ PR(J). 

 

(ii) The cardinality of set of all g’s in Corollary (i) is the number of fixed points of f . Because at 

each fixed points of f we can construct g as above.   

Note: If h is a homeomorphism on I, then 

h ∈/ PR(I) as Per(h) = Per1(h) ∪ Per2(h). Here P (h) = {1, 2}. 

 

3. CONCLUSION 

 

Studies of Periodically Rich maps on various topological spaces would contribute to literature of the 

chaotic dynamical systems.The set of periodic points is a critical con- cept in studying chaotic maps 

as it captures the recurrence and regularity in the otherwise seemingly erratic dynamics.The set of 

periods reflects the variety and richness of periodic behavior in chaotic systems, often being dense in 

natural numbers, indicating that the sys- tem exhibits periodic behavior at every scale. 

Acknowledgment: 

We gratefully acknowledge the administrative support, facilities and opportunities pro- vided by the 

management of Rajagiri School of Engineering & Technology, Kochi and IHRD, Kerala. 

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