-1

-2 A A A

2 1 2

-3

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

Figure 3.7: Q3 , Q—¦2 , and Q—¦3 .

3 3

80 CHAPTER 3. CONJUGACY

Proposition 3.7.1 If

√

5+2 5 .

c<’ = ’2.368 . . . (3.9)

4

then Λ is totally disconnected, that is, it contains no interval.

In fact, the proposition is true for all c < ’2 but, following Devaney [?] we will

only present the simpler proof when we assume (3.9). For this we use

Lemma 3.7.1 If (3.9) holds then there is a constant » > 1 such that

|Qc (x)| > » > 1, ∀x ∈ I\A1 . (3.10)

Proof of Lemma. We have |Qc (x)| = |2x| > » > 1 if |x| > 1 » for all x ∈ I\A1 .

2

So we need to arrange that A1 contains the interval [’ 1 , 1 ] in its interior. In

22

other words, we need to be sure that

1

Qc ( ) < ’p+ .

2

The equality

1

Qc ( ) = ’p+

2

translates to √

1 + 1 ’ 4c

1

+c=’ .

4 2

Solving the quadratic equation gives

√

5+2 5

c=’

4

1

as the lower root. Hence if (3.9) holds, Qc ( 2 ) < ’p+ .

Proof of Prop. 3.7.1. Suppose that there is an interval, J, contained in Λ.

Then J is contained either in I0 or I1 . In either event the map Qc is one to one

on J and maps it onto an interval. For any pair of points, x and y in J, the

mean value theorem implies that

|Qc (x) ’ Qc (y)| > »|x ’ y|.

Hence if d denotes the length of J, then Qc (J) is an interval of length at least

»d contained in Λ. By induction we conclude that Λ contains an interval of

length »n d which is ridiculous, since eventually »n d > 2p+ which is the length

of I. QED.

Now consider a point x ∈ Λ. Either it lies in I0 or it lies in I1 . Let us de¬ne

s0 (x) = 0 ∀x ∈ I0

81

3.7. SEQUENCE SPACE AND SYMBOLIC DYNAMICS.

and

s0 (x) = 1 ∀x ∈ I1 .

Since all points Q—¦n (x) are in Λ, we can de¬ne sn (x) to be 0 or 1 according to

c

whether Q—¦n (x) belongs to I0 or I1 . In other words, we de¬ne

c

±

if Q—¦n (x) ∈ Io

0 c

sn (x) := . (3.11)

Q—¦n (x) ∈ I1

1 if

c

higher iterates of Qc .

So let us introduce the sequence space, Σ, de¬ned as

Σ = {(s0 s1 s2 . . . ) | sj = 0 or 1}.

Notice that in contrast to the space X we introduced in Section 3.4, we are

not excluding any sequences. De¬ne the notion of distance or metric on Σ by

de¬ning the distance between two points

s = (s0 s1 s2 . . . )

and

t = (t0 t1 t2 . . . )

to be

∞

|si ’ ti |

def

d(s, t) = .

2i

i=0

It is immediate to check that d satis¬es all the requirements for a metric: It

is clear that d(s, t) ≥ 0 and d(s, t) = 0 implies that |si ’ ti | = 0 for all i, and

hence that s = t. The de¬nition is clearly symmetric in s and t. And the usual

triangle inequality

|si ’ ui | ¤ |si ’ ti | + |ti ’ ui |

for each i implies the triangle inequality

d(s, u) ¤ d(s, t) + d(t, u).

Notice that if si = ti for i = 0, 1, . . . , n then

∞ ∞

|sj ’ tj | 1 1

¤

d(s, t) = = n.

j j

2 2 2

j=n+1 j=n+1

Conversely, if si = ti for some i ¤ n then

1 1

d(s, t) ≥ ≥ n.

2j 2

So if

1

d(s, t) <

2n

82 CHAPTER 3. CONJUGACY

then si = ti for all i ¤ n.

Getting back to Λ, de¬ne the map

ι:Λ’Σ

by

ι(x) = (s0 (x)s1 (x)s2 (x)s3 (x) . . . ) (3.12)

where the si (x) are de¬ned by (3.11).

The point ι(x) is called the itinerary of the point x. For example, the ¬xed