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latter case that we want to study.

To visualize the what is going on, draw the square whose vertices are at

(В±p+ , В±p+ ) and the graph of Qc over the interval [в€’p+ , p+ ]. The bottom of

the graph will protrude below the bottom of the square. Let A1 denote the

open interval on the x-axis (centered about the origin) which corresponds to

this protrusion. So

A1 = {x|Qc (x) < в€’p+ (c)}.

Every point of A1 escapes from the interval [в€’p+ , p+ ] after one iteration.

Let

A2 = Qв€’1 (A1 ).

c

Since every point of [в€’p+ , p+ ] has exactly two pre-images under Qc , we see that

A2 is the union of two open intervals. To п¬Ѓx notation, let

I = [в€’p+ , p+ ]

and write

I\A1 = I0 в€Є I1

where I0 is the closed interval to the left of A1 and I1 is the closed interval to

the right of A1 .Thus A2 is the union of two open intervals, one contained in I0

and the other contained in I1 . Notice that a point of A2 escapes from [в€’p+ , p+ ]

in exactly two iterations: one application of Qc moves it into A1 and another

application moves it out of [в€’p+ , p+ ].

Conversely, suppose that a point x escapes from [в€’p+ , p+ ] in exactly two

iterations. After one iteration it must lie in A1 , since these are exactly the

points that escape in one iteration. Hence it must lie in A2 .

In general, let

An+1 = Qв€’в—¦n (A1 ).

c

Then An+1 is the union of 2n open intervals and consists of those points which

escape from [в€’p+ , p+ ] in exactly n + 1 iterations. If the iterates of a point x

eventually escape from [в€’p+ , p+ ], there must be some n в‰Ґ 1 so that x в€€ An . In

other words,

An

nв‰Ґ1

is the set of points which eventually escape. The remaining points, lying in the

set

О› := I\ An m,

nв‰Ґ1

are the points whose iterates remain in [в€’p+ , p+ ] forever. The thrust of this

section is to study О› and the action of Qc on it.

Since О› is deп¬Ѓned as the complement of an open set, we see that О› is closed.

Let us show that О› is not empty. Indeed, the п¬Ѓxed points, pВ± certainly belong

to О› and hence so do all of their inverse images, Qв€’n (pВ± ). Next we will prove

c

77

3.7. SEQUENCE SPACE AND SYMBOLIC DYNAMICS.

3

2

1

0

-1

-2 A1

I0 I1

-3

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

Figure 3.5: Q3 .

78 CHAPTER 3. CONJUGACY

6

5

4

3

2

1

0

-1

-2 A2 A1 A2

-3

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

Figure 3.6: Q3 and Qв—¦2 .

3

79

3.7. SEQUENCE SPACE AND SYMBOLIC DYNAMICS.

3

2

1

0

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