From this description it is clear that PER(S) consists of points with even-

tually repeating binary expansions, these are the rational numbers. They are

dense. We can see that S is transitive as follows: We are given intervals I and J.

70 CHAPTER 3. CONJUGACY

Let y = .b1 b2 b3 . . . be a point of J, and let z = .a1 a2 a3 . . . be a point of I which

is at a distance greater than 2’n from the boundary of I. We can always ¬nd

such a point if n is su¬ciently large. Indeed, if we choose n so that the length

1

of I is greater than 2( n’1) , the midpoint of I has this property. In particular,

any point whose binary expansion agrees with z up to the n’th position lies in

I. Take x to be the point whose ¬rst n terms in the binary expansion are those

of z, followed by the binary expansion of y, so

x = 0.a1 a2 a3 . . . an b1 b2 b3 b4 . . . .

The point x lies in I and S n (x) = y. Not only is S transitive, we can hit any

point of J by applying S n (with n ¬xed, depending only on I) to a suitable

point of I. This is much more than is demanded by transitivity. Thus S is

chaotic on [0, 1).

Of course, once we know that S is chaotic on the open interval [0, 1), we

know that it is chaotic on the closed interval [0, 1] since the addition on one

extra point (which gets mapped to 0 by S) does not change the de¬nitions.

Now consider the map t ’ e2πit of [0, 1] onto the unit circle, S 1 . Another

way of writing this map is to describe a point on the unit circle by eiθ where

θ is an angular variable, that is θ and θ + 2π are identi¬ed. Then the map is

t ’ 2πt. This map, h, is surjective and continuous and is one to one except at

the end points: 0 and 1 are mapped into the same point on S 1 . Clearly

h—¦S =D—¦h

where

D(θ) = 2θ.

Or, if we write z = eiθ , then in terms of z, the map D sends

z ’ z2.

So D is called the doubling map or the squaring map. We have proved that it

is chaotic. We can use the fact that D is chaotic to give an alternative proof of

the fact that Q’2 is chaotic. Indeed, consider the map h : S 1 ’ [’2, 2]

h(θ) = 2 cos θ.

It is clearly surjective and continuous. We claim that

h —¦ D = Q’2 —¦ h.

Indeed,

h(D(θ)) = 2 cos 2θ = 2(2 cos2 θ ’ 1) = (2 cos θ)2 ’ 2 = Q’2 (h(θ)).

This gives an alternative proof that Q’2 (and hence L4 and T ) are chaotic.

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3.5. SENSITIVITY TO INITIAL CONDITIONS

3.5 Sensitivity to initial conditions

In this section we prove that if f is chaotic, then f is sensitive to initial conditions

in the sense of the following

Proposition 3.5.1 (Sensitivity.) Let f : X ’ X be a chaotic transforma-

tion. Then there is a d > 0 such that for any x ∈ X and any open set J

containing x there is a point y ∈ J and an integer, n with

|f —¦n (x) ’ f —¦n (y)| > d. (3.5)

In other words, we can ¬nd points arbitrarily close to x which move a distance

at least d away. This for any x ∈ X. We begin with a lemma.

Lemma 3.5.1 There is a c > 0 with the property that for any x ∈ X there is a

periodic point p such that

|x ’ f —¦k (p)| > c, ∀k.

Proof of lemma. Choose two periodic points, r and s with distinct orbits, so

that |f —¦k (r) ’ f —¦l (s)| > 0 for all k and l. Choose c so that 2c < min |f —¦k (r) ’

f —¦l (s)|. Then for all k and l we have

2c < |f —¦k (r) ’ f —¦l (s)|

= |f —¦k (r) ’ x + x ’ f —¦l (s)|

¤ |f —¦k (r) ’ x| + |f —¦l (s) ’ x|.

If x is within distance c to any of the points f —¦ l(s) then it must be at a greater

distance than c from all of the points f —¦k (r) and vice versa. So one of the two,

r or s will work as the p for x.

Proof of proposition with d = c/4. Let x be any point of X and J any open

set containing x. Since the periodic points of f are dense, we can ¬nd a periodic

point q of f in

U = J © Bd (x),

where Bd (x) denotes the open interval of length d centered at x,

Bd (x) = (x ’ d, x + d).

Let n be the period of q. Let p be a periodic point whose orbit is of distance

greater than 4d from x, and set

Wi = Bd (f —¦i (p)) © X.

Since f —¦i (p) ∈ Wi , i.e. p ∈ f ’i (Wi ) = (f —¦i )’1 (Wi ) for all i, we see that the

open set

V = f ’1 (W1 ) © f ’2 (W2 ) © · · · © f ’n (Wn )

72 CHAPTER 3. CONJUGACY

is not empty.

Now we use the transitivity property of f applied to the open sets U and

V . By assumption, we can ¬nd a z ∈ U and a positive integer k such that

f k (z) ∈ V . Let j be the smallest integer so that k < nj. In other words,

1 ¤ nj ’ k ¤ n.

So

f nj (z) = f nj’k (f k (z)) ∈ f nj’k (V ).

But

f nj’k (V ) = f nj’k f ’1 (W1 ) © f ’2 (W2 ) © · · · © f ’n (Wn )

‚ f nj’k (f ’(nj’k) Wnj’k )

= Wnj’k .

In other words,

|f nj (z) ’ f nj’k (p)| < d.

On the other hand, f nj (q) = q, since n is the period of q. Thus

|f nj (q) ’ f nj (z)| = |q ’ f nj (z)|

= |x ’ f nj’k (p) + f nj’k (p) ’ f nj (z) + q ’ x|

≥ |x ’ f nj’k (p)| ’ |f nj’k (p) ’ f nj (z)| ’ |q ’ x|

≥ 4d ’ d ’ d = 2d.

But this last inequality implies that either

|f nj (x) ’ f nj (z)| > d

or

|f nj (x) ’ f nj (q)| > d

for if x were within distance d from both of these points, they would have to be

within distance 2d from each other, contradicting the preceding inequality. So

one of the two, z or q will serve as the y in the proposition with m = nj.

3.6 Conjugacy for monotone maps

We begin this section by showing that if f and g are continuous strictly mono-

tone maps of the unit interval I = [0, 1] onto itself, and if their graphs are both

strictly below (or both strictly above) the line y = x in the interior of I, then

they are conjugate by a homeomorphism. Here is the precise statement:

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3.6. CONJUGACY FOR MONOTONE MAPS