Chapter 2 Dynamics of Continuous Maps

CHAPTER 2 Dyn a mi c s of Continuous Maps

It is sometimes said that the theory of dynamical systems is the study of the behaviours of orbits of maps. Among many ways of the study of dynamical systems, we select the path of general topology. We use expansivity and pseudo orbit tracing property, related to Anosov and Axiom A systems, as tools of study. This chapter provides these tools which are sufficiently powerful through out this book. 52.1 Self-covering maps Let X and Y be metric spaces. A continuous surjection f : X + Y is called

a covering map if for y E Y there exists an open neighborhood V, of y in Y

such that

f-'(v,)

=

U U ~(i # 'i +-

~i

n vil = 0)

i

where each of Ui is open in X and flui : Ui + V, is a homeomorphism. A covering map f : X + Y is especially called a self-covering map if X = Y. We say that a continuous surjection f : X + Y is a local homeomorphism if for x E X there is an open neighborhood Ux of x in X such that f(Ux) is is a homeomorphism. It is clear that every open in Y and flu= : Ux -+ f(Ux) covering map is a local homeomorphism. Conversely, if X is compact, then a local homeomorphism f : X + Y is a covering map. This is stated precisely as follows: T h e o r e m 2.1.1 (Eilenberg [El). Let X be compact and f : X --+ Y a continuous surjection. Iff is a local homeomorphism, then them exist two positive numbers X and 8 such that each subset D of Y with diameter less than X determines a decomposition of the set f -l (D) with the following properties: (1) f-'(D)=D1UDzU.*.UDk, (2) f maps each Di homeomorphically onto D, ( 3 ) if i # j then no point of Di is closer than 28 to a point of D j , (4) for every 11 > 0 them is 0 < E < X such that if the diameter of D is less than E then each diameter of Dj is less than 7 . If, in addition, Y is connected, then there is a constant k > 0 such that f : X + Y is a 12-to-one map. The numbers (Ale) in Theorem 2.1.1 are called Eilenberg's constants for the map f : X + Y.For the proof we need the following lemma. 31

CHAPTER 2

32

Lemma 2.1.2. Let X be a compact metric space and a a finite open cover of X . Then there exists 6 = 6(a)> 0 such that for any subset A if the diameter of A is less than 6 then A c B for some B E a . Such a number 6 is called a Lebesgue number of a. Proof. If this is false, for any n > 0 there exists a subset A, such that B for all B E a. diam(A,) = sup{d(a,b) : a , b E A,} 5 l / n and A, Take x , E A,, for n 2 1. Since X is compact, there is a subsequence {x,,} such that x,, 4 x as i -+ 00. Then x E B for some B E a. Since X \ B is compact, we have a = d ( x , X \ B ) = inf{d(x,b) : b E X \ B} > 0. Take sufficiently large ni satisfying ni > 2 / a and d(x,, ,x ) < a/2. Then, for y E Ani we have d ( y , ~ ) I d ( ~ i x n i ) + d ( z n i , l~/ n ) Ii + a / 2 < a

<

and thus y E B. Since y is arbitrary in A,,, we have Ani dicting. 0

c B, thus contra-

Proof of Theorem 2.1.1. We first prove that there exists 8 > 0 such that if x # y and f ( x ) = f (y) then d ( x , y) > 48. If this is false, then for any n > 0 there exist X n , Y n E X with 2 , # Yn and f(x,) = f(y,) such that d(x,, y , ) 5 l / n . Without loss of generality we suppose that x, -+ x and yn 4 y as n + 00. Then x = y. Since f:X + Y is a local homeomorphism, there exists an open neighborhood Ux of x such that flu. : Ux4 f(Ux) is a homeomorphism. Thus x,, yn 6 U, for sufficiently large n and so f (2,) = f (y,), thus contradicting. Let k > 0 and define a set Yk by

Then we have Y&n Y&l= 0 for integers k # k'. Next, we prove that there is K > 0 such that Y = Y1 U U YK and each of Yk is open. Since is compact, by the above fact it follows that Yk = 0 for integers k larger than a certain K > 0. Thus Y = Yl U U YK.To show openness of Y& (1 5 k 5 K ) take y E Yk. Then f-'(y) = { ~ 1 , . - * , x k } . We can choose open neighborhoods Uxi of x i such that flu=,is a homeomorphism and Uxi n Uxj = 0 if i # j. Clearly

x

-.. ...

We claim that f ( U z i )contains a neighborhood U(y) of y such that U(y) C Yk. Indeed, if this is false, for n > 0 there is yn E f(Uxi) such that

$2.1 Self-covering maps

33

Iff-l(y") = {z?,..., z g , z g + l , - - - } a n d z l zi (asn + m ) f o r l 5 i 5 k+1, then we have that f ( ~ = ) y and d ( z i , z j ) > 48 for 1 5 i 5 k 1. Therefore, y E Yk+1 U * * * U YK. This is a contradiction. We are now ready to prove Theorem 2.1.1. Let 8 > 0 and K > 0 be as above. Since X is compact and f is a local homeomorphism, we can choose a finite open cover a = { U1, ,Urn) satisfying (a) f ( U i ) is open in Y, (b) f:cl(Ui) --+ f(cl(Ui)) is injective, (c) diam(cl(Ui)) < 8 for 1 5 i 5 m. Here cl(E) denotes the closure of E. Let y E Y, then y E y k for some k (1 5 k 5 K ) and so define Q(y) = y k fl ( n { f ( U i ) : y E f ( U i ) } ) . Then Q(y) is an open neighborhood of y. Since y is arbitrary in Y,we can take y1,---,ye such that U = { Q ( y l ) , - - . ,Q(ye)} is a finite open cover of Y. Let X > 0 be a Lebesgue number of U and let D be a subset satisfying diam(D) < A. Then D C Q(y) for some y E {yl,... ,ye}. If y E y k for ,zk} and zi E Uni for some Uni E a. Notice some 12, then f-'(y) = {z1,-*that f-l(y) n Uni = { z i } for 1 5 i 5 k. We now define Di = Uni n f - ' ( D ) for 1 5 i 5 k. Then we have f ( D i ) = D for 1 5 i 5 k. Indeed, since Q(y) = Yk r l f(Uni)), if z E D then z = f(z) for some z E Uni. Thus z E Uni n f-'(z) c Uni n f - ' ( D > = Di and therefore z = f(z) E f ( D i ) , i.e. D c f(Di).Since /(Di)= f(Uni n f - ' ( D ) ) c f(Uni) n D = D,we have D = f ( D i ) for 1 5 i 5 k. (1)was proved. Since Di c Uni and D c f(Uni),fpi maps each Di homeomorphically onto D . Thus (2) was proved. (3) is computed as follows. If i # j , then d(Di,Dj) 2 d(Uni,Unj) 2 d ( z i , z j ) - diam(U,,,) - diam(Unj) > 48 - 8 - 8 = 28. Here d(A,B) is defined by d ( A , B )= inf{d(a,b) : a E A,b E B}. To show (4) let 1 > 0. For 1 5 i 5 m there is 0 < ~i < X such that if d(f(z),f(z)) < ~i (f(z), f(z) E f(cl(Ui))) then d(z,z ) < 7. This follows from the fact that f:cl(Ui) --f f(cl(Ui)) is injective. Put E = min{si : 1 5 i 5 m } . If diam(D) < E , then we have diam(Di) < 1 for 1 5 i 5 k. Since Y = Yk and each yk is open and closed, when Y is connected, we have Y = y k for some k, i.e. f is k-to-one. 0 --f

+

.- .

(n;=,

u,"==,

In the remainder of this section we describe well known theorems that will be used in the sequel. Let X be a topological space. A path in X is a continuous map from the unit interval [0,1] to X. If any two points in X are joined by a path, then X is said to be path connected. In general, a connected space need not be path connected. An arc in X is an injective continuous map from [0,1] to X. We say that X is arcwise connected if any two points in X are joined by an arc. It is clear that if X is arcwise connected then it is path connected.

Theorem 2.1.3. A Hausdorfl space X i s path connected if and only if X is arcwise connected.

CHAPTER 2

34

Proof. Let f : [0,1] + X be any path with f(0) # f(1). We now construct an arc h in X joining f(0) and f (1). Choose a closed subinterval I1 = [a1,bl] C [0,1] whose length 0 5 l ( I 1 ) = bl - al < 1 is maximal, satisfying the condition that f ( a 1 ) = f ( b 1 ) . The existence of such a subinterval is ensured by the fact that X is Hausdorff. Next, among all subintervals of [0,1] which are disjoint from 11 , choose an interval I2 = [az,bz] of maximal length satisfying f ( a z ) = f ( b 2 ) . Continue this process inductively, we can construct disjoint subintervals of maximal lengths e ( 1 1 ) 2 C(I2) 2 . 2 0 satisfying the condition that f is constant on the boundary of each I,.

--

arc

path

Figure 1 Let a : [0,1] + X be the unique map which is constant on each closed interval Ij and which coincides with f outside of these subintervals. Then a ( t ) coincides with f ( t ) for t E 8 I j . It is easily checked that a is continuous, and that for each point x E a([O,l]), a-'(x) c [0,1] is either one point or a closed interval. From now on we show that these conditions imply that A = a([O,11) is the image of some arc joining a(0) and a(1). To do so choose a countable dense subset { t l , t z , . - . } c [0,1]. Define a total ordering of A by a ( s ) < a ( t ) if a ( s ) # a ( t ) for s < t. Then an order preserving homeomorphism h: [0,1] -+ A can be constructed inductively as follows. Set h(0) = a(0) and h(1) = a(1). For each dyadic fraction 0 < m/2k < 1 with m odd, suppose that h((m - 1)/2k) and h ( ( m + 1 ) / 2 k ) have already been defined. Then there exists the smallest index j such that

and define h(m/2'") = a ( t j ) . It is not difficult to check that the h constructed in this manner on dyadic fractions extends uniquely to an order preserving map from [0,1] to A, which is a homeomorphism. 0 By definition a topological space X is locally connected if each point in X has an arbitrary small connected neighborhood. There exists a corresponding concept of locally path connected, or locally arcwise connected.

$2.2 Expansivity

35

Theorem 2.1.4. If a compact metric space X is locally connected, then X is locally path connected. Moreover every connected component of X is path connected. Proof. Let E > 0. Since X is locally connected, there exists a sequence of numbers 6, > 0 such that any two points with d ( z , z ' ) < 6, are contained in a connected set of diameter less than ~ / 2 , . We show that any two points z(0) and z(1) with d ( z ( O ) , z ( l ) )< 60 can be jointed by a path of diameter at most 4 ~ To . do this choose a sequence of denominators 1 = ko < k1 < kz < , eack of which divides the next. For each fraction of the form i / k n between 0 and 1 we choose an intermediate point z ( i / k , ) satisfying two conditions. (1) Any two consecutive fractions i / k n and (i l ) / k n correspond to points s ( i / k n ) and z ( ( i l)/kn) which have distance less than 6,. Thus any two such points are contained in a connected set C ( i ,k,) which has diameter less than ~ 1 2 The ~ . second condition is the following: (2) Each point of the form ~ ( j / k , + ~where ), j / k , + l lies between i / k n and (i l)/kn, belongs to the set C ( i ,k,), and thus

+

+

+

By induction on n the construction of such denominators 1, and intermediate points x ( i / k n ) is straightforward. Thus we may suppose that Z ( T ) has been defined for a dense set of rational numbers r in [0,1]. Next we prove that the correspondence T H Z ( T ) is uniformly continuous. Let r and T' be any two rational numbers such that z ( r ) and z ( r ' ) are defined. Suppose that IT - T'I < l/kn. Then we can choose i l k , as close as possible to both r and T ' . Then we have

and so

d(z(r),z(r')< ) 4E/k,.

This proves uniform continuity. Thus there is a unique continuous extension t H z ( t ) which is defined for all t E [0,1]. We have constructed the required path of diameter 5 4~ from z(0) to z(1). Therefore X is locally path connected. The rest of the argument is straightforward. 0

Theorem 2.1.5. Let X and Y be Hausdorff spaces and f : X --f Y a continuous surjection. If X is arcwise connected, then Y is arcwise connected. Proof. Take arbitrary points go,y1 in Y. Since f is surjective, there exist in X such that f(z0) = go, f ( z 1 )= 91. Since X is arcwise connected, we can find a path u joining 20 and 2 1 . Then f o u : I Y is a path joining yo and y1. This shows that Y is arcwise connected. 0

z 0 , q

--f

CHAPTER 2

36

52.2 Expansivity Let X be a metric space with metric d. A homeomorphism f : X --t X is ezpansive if there is a constant e > 0 such that x # y (2, y E X) implies

for some integer n. Such a constant e is called an ezpansive constant for f. This property (although not e ) is independent of the choice of metrics for X when X is compact, but not so for noncompact spaces. As one of notions that are weaker than expansivity we capture the notion which is called sensitive dependence on initial conditions. This is defined by the property that if there is 6 > 0 such that for each E X and each neighborhood U of z there exist y E U and n E Zsuch that d ( f " ( z ) , fn(y)) > 6. From definition it follows that X has no isolated points. A homeomorphism f:X + X is topologically transitive if there exists zo E X such that the orbit Of(z0) = {fn(zo): n E Z}is dense in X.

Remark 2.2.1. Let f : X + X be a homeomorphism of a compact metric space. Then f is topologically transitive if and only if for U,V nonempty open sets there is n E Zsuch that f n ( U )n V # 8. Suppose Of(z0) is dense in X and U,V are nonempty open sets. Then fn(zo) E U and f"(z0) E V for some n and m. Thus f k ( U )n V # 0 where k=n-m. To see the converse let {Ui : i > 0) be a countable base for X. For z E X we have Of(z) is not dense in X a Of(z) n Un= 0 for some n f"(z) E X \ U,,for each m E Z

n P-"(x\ un> u n ~-Yx\u,). 00

azE

for some n

m=-m 0

0

0

0

U=;,f-"(Un) is dense in X by the assumption, nz=-,(X \ f-" (Un)) is nowhere dense and thus lJ~..p=, f-"(X \ Un)is a set of first category. Since X is a compact metric space, {z E X : Of(z) is dense in X} Since

nz=-,

is dense in X.

Remark 2.2.2 ([B-S]). Let f : X -, X be a homeomorphism of a compact metric space. Suppose X is an infinite set. I f f is topologically transitive and

$2.2 Expansivity

37

Per(f) = {z E X : fn(z) = z for some n > 0) is dense in X, then f has sensitive dependence on initial conditions. This is easily checked as follows. We first show that Per(f) # X. For n > 0 let F,, = {z E X : fn(z) = z}. Then F,, is closed in X and F, = Per(f). Suppose Per(f) = X. Then there exists n > 0 such that int(F,) # 8. Since f is topologically transitive, it follows that F, = X, and so F,, is a finite set, thus contradicting. Notice that there is 6 > 0 such that for z E X there exists p E Per(f) such that d ( z , O f ( p ) )> 6. Indeed, if this is false, then for n > 0 there is z, E X such that d(z,,,Oj(q)) I l l n for all q E Per(f). Since X is compact, without loss of generality suppose 2, --+ z as n --+ 00. Then d(z, O f ( q ) )= 0 and thus 2 E O f ( q )for all q E Per(f). This can not happen. Let X = 614. For z E X let U ( z ) be an arbitrary neighborhood of z in X. Then U = U ( z )n Ux(z)is a neighborhood of z where Ux(z)= {y E X : d(z, y) < A}. Take p E Per(f) n U with f n ( p ) = p for some n > 0 and choose qz E Per(f) satisfying d(z, O j ( q z ) )> 6. Since v = f-j(Ux(fj(qz))) is a nonempty open set and f is topologically transitive, there is k E Z such that f k ( V )n U # 8. Notice that Ikl is chosen sufficiently large. Let j' > 0 be the integer such that lkl = nj' r where 0 <_ r < n and put j = j' 1. Then 0 < nj - lkl 5 n. Take y E f k ( V )n U. Since k = -nj' - r when k < 0, we have

urZ1

n;=-,,

+

since -n 5 -nj

+

+ k 5 n. Since f n j ( p ) = p, when ( 1 ) holds,

If (2) holds, then we have d ( f - " j ( y ) , f - " j ( p ) ) > 612. In any case we have d ( f f n j ( s ) , f * " j ( y ) ) > X or d ( f * " j ( z ) , f * " j ( p ) ) > A, i.e. f has sensitive dependence on initial conditions. Let f : X -+ X be a homeomorphism of a compact metric space. A finite open cover a of X is a generator (weak generator) for f if for every bisequence {A,} of members of a, f-,,(cl(A,,)) is at most one point f-,(A,,) is at most one point). Here cl(E) denotes the closure of a subset E. These concepts are due to Keynes and Robertson [K-R].

(nr=-,

CHAPTER 2

38

If a,P are open covers of X , then their joint a V /3 is given by a V P = { A n B : A E a , B E P}. An open cover p is a refinement of an open cover a (written a 5 p ) if every member of p is a subset of some member of a. It is clear that a 5 a V p and p 5 a V p. If f : X --t X is a continuous surjection, then f-'(a) = { f - ' ( A ) : A E a} is an open cover of X. It is easily checked that

Theorem 2.2.3. Let f: X + X be a homeomorphism of a compact metric space. Then the following are equivalent. (1) f is expansive, (2) f has a generator, ( 3 ) f has a weak generator. Proof. (2) + (3) is clear (3) + (2) : Let /3 = { B 1 , . ,Bd}be a weak generator for f and let 6 > 0 be a Lebesgue number for p. Let a be a finite open cover consisting of sets Ai with diam(cl(A;)) 5 6. If {Ai,} is a bisequence in a , for every n there is j , such that cl(Aj,) c Bj,, and so f-"(cl( Aj,)) c f-"(Bjn). Therefore a is a generator. (1) + (2) : Let 6 > 0 be an expansive constant for f and let a be a 00 finite cover consisting of open balls of radius 612. Suppose z , y E f-,(cl( A,)) where A, E a. Then d(fn(x), fn(y)) 5 6 for every n, and by the assumption x = y. (3) =+ (1) : Suppose a is a weak generator. Let 6 > 0 be a Lebesgue number for a. If d(f"(z), fn(y)) < 6 for n E Z,then for n there is A, E Q such that f "(x), f "(y) E A, and so x, y E f-"(A,) which is at most one point.

-

n,"=-,

Or=-,

n,=-,

n= :-,

0

Theorem 2.2.4. Let f : X X be a homeomorphism of a compact metric space and let k # 0 be an integer. Then f is expansive if and only if f k as expansive. .--)

vf!,'

Proof. This follows from the facts that if a is a generator for f then f-i (a)= a V f - l ( a ) V . V flkl-'(a) is a generator for f k , and that if a is a generator for f k then a is also a generator for f . 0

-.

The following theorem is easily checked. Thus we leave the proof to the readers.

92.2 Expansivity

39

Theorem 2.2.5. (1) Iff:X + X is expansive and Y is a closed subset of X with f (Y) = Y, then fly :Y -+ Y is expansive. (2) If f i : Xi Xi (i = 1,2) are expansive, then the homeomorphism f1 x fa : XI x X2 -+ X1 x X2 defined by --f

fl

x

fz(x1,22)= (f1(x1),f2(x2))

(219x2) E

Xl x xz

is expansive. Every finite direct product of expansive homeomorphisms is expansive. ( 3 ) If X is compact and f:X + X is expansive, then h o f o h-' : Y + Y is expansive where h: X + Y is a homeomorphism.

Let k be a fixed natural number and let Yk = (0, l , . . . ,k - 1). Put the discrete topology on Yk and define the product space Yf = n r m Y k equipped with the product topology. Then a homeomorphism u : Yf + Yf is defined by ( o ( x ) ) i = xi+l,i E Z, for x = ( x i ) . u is called the shift map. Theorem 2.2.6. Let k 2 2. Then the shift map (I :Yf

-+

Yf is expansive.

Proof.F o r O I i 5 k - l d e f i n e A i = { ( z , ) : z o = i } . Thena={Ao,...,Ak -1) is an open and closed cover of Yf. If x E u-"(Ai,) where Ain E a, then x = (... ,i-1,io,i1,-..) and thus a is a generator. 0

n= :-,

Theorem 2.2.7. Let f :X --t X be an expansive homeomorphism of a compact metric space. Then there exist k > 0 and a closed subset E of Yf such that Z is u-invariant (u(E) = C ) , and moreover there is a continuous surjection T : E + X satisfying ?r o u = f o T , i.e. the diagmm

E

A

.1x-x

E commutes.

f

Proof.Let 26 > 0 be an expansive constant for f. Choose a finite cover a! = {Ao,... ,&-I} by closed sets such that diam(Ai) < 6 for 0 5 i 5 k - 1, and such that Ai intersect only in their boundaries. Let 8 denote the union of fn(8) is a set of first category and the boundaries of Ai. Then 8, = so X\&, is dense in X. For x E X\8, we can assign uniquely a member of

u=:-,

y ,by x

G (a,)

whenever f "(x)E Aa, . Let A denote the collection of all sequences arising in this way. Then A C Yf and CP 0 o((xn)) = f

0

CP((~:,)>, (xn) E A.

CHAPTER 2

40

If cp is uniformly continuous, then cp : A -t X can be uniquely extended to a continuous map 7r : C + X, satisfying 7r 0 a ( ( z n ) )= f 0 r((xn)) for (2,) E C, where C = cl(A). It only remains to see uniform continuity of cp. Let E > 0. Since a is a generator for f , there is N > 0 such that each member of V ,! f"(a) has a diameter less than E . If this is false, then there is E > 0 such that for every j > 0 there are xj, y j E X with d(zj, yj) > E and Aj,i E a (-j 5 i 5 j) with zj,yj E f-i(Aj,i). Suppose x j -t z and yj + y. Then x # y . Since a is finite, there is A0 E {Aj,o : j E 9 ) such that z j , y j E A0 for infinitely many j. Thus, z , y E cl(A0). Similarly, for each n infinitely many elements of {Aj,, : j E Z} coincide and we have A, E a with z , y E f-"(cl(A,)). Therefore, z , y E O0 fwn(cl(An)). This is a contradiction. If ( a , ) , (b,) E A and a , = b, for In1 5 N , then cp((a,)) and cp((b,)) are in the same member of Vf, f"(a) and thus d(cp((a,,)),(p((b,)))< E . Therefore cp is uniformly continuous. 0

ni=-j

n,=-,

Let X be a metric space. A continuous surjection f :X -t X is positively expansive if there is a constant e > 0 such that if x # y then d( f ,(z),f "(y)) > e for some non-negative integer n ( e is called an expansive constant for f ) . For compact spaces, this is independent of the compatible metrics used, although not the expansive constants.

Theorem 2.2.8. Suppose X is compact. Let k positively expansive if and only if so is f k.

> 0 be an integer. Then f is

Proof. Let f :X -t X be positively expansive and e > 0 an expansive constant. Since f is uniformly continuous, there is 6 > 0 such that for 1 5 i 5 k d ( z , y )< 6

* d(f'(z),f'(y))< el

which implies that 6 is a n expansive constant for f k . Similarly, the converse is proved. 0 The following theorem is easily checked. Thus we leave the proof to the readers.

Theorem 2.2.9. (1) If f :X -t X is positively expansive and Y is a closed subset of X with f ( Y )= Y , then f p : Y -+ Y is positively expansive. (2) If fi : Xi -t Xi (i = 1,2) are positively expansive, then the continuous sujection f1 x f 2 : X1 x X2 -+ X1 x Xz defined by

fl

x

f2(51,22)

= (f1(51),f2(22))

(x1,22) E x1 x

x 2

i s positively expansive. Every finite direct product of positively expansive maps is positively expansive. ( 3 ) If X is compact and f : X -t X is positively expansive, then h o f o h-' : Y -t Y is positively expansive where h:X -t Y is a homeomorphism.

$2.2 Expansivity

41

Theorem 2.2.10 (Reddy [Rl]). If X is compact and f : X 4 X is a positively expansive map, then there exist a compatible metric D and constants 5 > 0 , X > 1 such that for x , y E X

D(x, Y) I 6 -*- D(f (21, f(Y)) 1 XD(x, Y).

For the proof we need the following result due to Frink [F'r]. Lemma 2.2.11. Let Vn be a decmasing sequence of open symmetric neighborhoods of the diagonal set A = { (z,x) : x € X } of X x X such that 00

v O = x x x ~,

v , = A

and Vn+lo Vn+lo Vn+l C Vn for all n 2 1. Then there is a metric D compatible with the topology of X such that for n 1 1 1 Vn c { ( z , ~ ) :D(s,Y)< -} c Vn-12n Here a subset A of X x X is said to be symmetric if A satisfies the property t h a t ( z , y ) € AH ( y , x ) E A , a n d A o B i s d e f i n e d b y A o B = { ( x , y ) : 3 z € X s.t. (x,z) E A , ( z , y ) E B}.

Proof. Define a function ( :X x X

4

R by

...

and for (z,y) E X x X denote by c(x, y) a finite sequence x = 20, x l , , z,, x,+1 = y. We write C(z,y) the collection of all such sequences c(x, y). Let us Put ~ ( xY), = i n f { C ((Si,xi+l) : ~ ( 2Y> , E c(z, Y)). i

It is easily checked that for any points x, y, z in X (i) (ii) (iii)

CHAPTER 2

42

If we hold that for n 2 1

n,"==,

then z = y whenever D ( z ,y) = 0, since (2, y) E V, = A. From (ii) and (iii) together with this fact, it follows that D is a metric compatible with the topology of To obtain (iv) it sufficies to show that for each finite sequence z = 2 0 , z~,"', zn+1 = Y

x.

For, if D(z,y) < 1/2", by (v) we have

and thus (z, y) E V,-l. The remainder of the proof is only to show that (v) is true. If R. = 0 then <(zo,z5 ~ )2 t ( z o , z l ) . Suppose (v) holds for 0 5 k 5 n - 1, i.e. t ( Z 0 , z k ) 5 2 Cfzl t ( t i , t i + l ) . Letting a = CyL: ( ( z i , z i + l ) , we have either

For the case (vi) denote as k 2. 1 the greatest integer among {k'} such that

Then we have

<

( ( ~ i , x i + ~ )a / 2

and by hypothesis

$2.2 Expansivity Choose m > 0 such that 1/2" 5 a (Zk+l,Zn) E Vm-1, from which ( 2 0 m - 2 5 0). Thus we have

43

< 1/2m-'.

Then (zO,zk)9 (zk,zk+l), E Vm-2 (we consider Vm-2 = l4j if

, ~ ~ )

For the case (vii) we have

and by hypothesis

) a Since [ ( z o , ~ 5

< 1/2"-',

we have (z0,zn) E

Vm-2.

Therefore

n-1

In any case we have (v). 0 Proof of Theorem 2.2.10. To show the existence of a compatible metric D that is our requirement, let e > 0 be an expansive constant for f . Write WO= X x X and for n 2 1 let

Wn = {(z,~) : d(fj(z),fj(~)) < e, 0 I j

nz=o

I n - 1).

Then it is easily checked that { Wn} is a decreasing sequence of open symmetric neighborhoods of diagonal set A such that Wn = A, and letting g = f x f :xxx+xxx,

Since W1 is open, there is 6 > 0 such that

CHAPTER 2

44

Thus we have Wl+N 0 Wl+N 0 WI+N C Now define a new sequence { Vn} by

V O = WO,

w1.

vn = Wl+(n-l)N,

2 1.

The sequence Vn are open symmetric neighborhoods of A. If we establish the relation Vn+l o Vn+l o Vn+l C Vn for n 2 0, by Lemma 2.2.11 there is a metric D for X such that

Vn

c {(z,~) : D(+,y) < 1/2n} c Vn-1,

n 2 1.

To use Lemma 2.2.11 we first prove that the relation is true for n 2 2. Take 2 2. Then there are a , b E X such that ( z , a ) , ( a , b ) , ( b , y E) Vn+l, and so for 0 5 j 5 (n- l ) N (z,y) E Vn+l o Vn+l o Vn+l for n

gj(z,y) = ( f W ,fj(.>)

0

( f Wf W 0 (fW7 fj(Y>>

E gj(Vn+l) 0 gj(Vn+l) ogj(Vn+l) c Wl+nN-j 0 Wl+nN-j 0 Wl+nN-j c vz 0 vz 0 VZ c Vl. Therefore, d(fj(z),fj(y)) < e for 0 5 j 5 (n - l)N and thus (z,y) E w l + ( n - l ) N = %. To obtain the conclusion, suppose 0 < D(z,y) < 1/32. Then there is n 2 -1 such that (x,y) E Vn+l - Vn+z and thus

1 15 ~ ( z , y 5) min{-321 ' -1. 2"+' 2n+3

This shows n 2 3. Since (z,y) E WlfnN - Wl+(n+l)N,there is j such that

n~

< j I (n + I)N,

Let ( z , w) = (f3N(z),f3"(y)). 0

d(fJ(z), fj(y)) 2 e.

Then we have

5 (n - 3)N < j

-3N

5 (n- 2 ) N ,

d( f F 3 N ( z ) , fj-3N(w>) = d(fj(z), f j ( Y ) ) 2 e.

Thus (z,w) @ Wl+(n-Z)N = Vn-1 and

D ( f 3 N ( 4 f3N(y)) , 2

1

> 2D(z,Y).

Put y = (1/2)& and define a metric D' by

c

3N-1

D'(Z,Y) =

i=O

Then D' has the desired property. 0

7 i D ( f i ( 4 , fa(?/>>.

52.2 Expansivity

45

Theorem 2.2.12. If a homeomorphism of a compact metric space is positively expansive, then the space is a set consisting of finite points. Proof. Since f is positively expansive, there are a compatible metric D, constants 6 > 0, 0 < X < 1 such that D ( x , y ) 6 implies D ( f - l ( x ) , f - l ( y ) ) 5 X D ( x ). Thus @- = { f - i : i 2 0) is uniformly equi-continuous. P u t @+ = : i 2 0). By a metric p defined by p ( f , g ) = max{D(f(z), g(x)) : x E X}, @- is totally bounded. Define a map G : $- -+ @+ by G(f -*) = f for i 2 0 Then G is pisometric. Therefore @+ is totally bounded. Since X is compact, @+ is uniformly equi-continuous and so there is v > 0 such that D ( x ,y ) < v implies D ( f i ( x ) f, i ( y ) ) 5 e for all i 2 0 ( e is an expansive constant). This shows x = y and therefore z is an isolated point. 0

<

.

Theorem 2.2.13. The closed interval I = [0,1] does not admit positively expansive maps. Proof. Suppose f : I -t I is a positively expansive map. Let 0 < E < 1 and denote as I z ( c ) the closed subinterval [z, x + c ] where x E [0, 1 -E]. By positive expansivity o f f there exists 0 < E < 1 such that fir.(.) : I z ( c ) + f ( I z ( E ) )is injective for all z E [ O , l - E ] . The point a E I o ( c / 2 ) which gives the greatest . a = 0 or a = ~ / 2 . value of f p , ( c 1 2 ) is one of the end points of I o ( E / ~ )Thus, Indeed, if a is an interior point of I o ( E / ~ then ) , we have that for some u > 0

so that there exist x1,xz E I o ( E / ~with ) z1 # 2 2 such that f ( x 1 ) = f ( x 2 ) . This cannot happen. If a = ~ / gives 2 the greatest values o f f on I o ( E / ~ then ) , the point E must be a point of the greatest value of f on I , l 2 ( ~ / 2 ) Continue . this argument, then we have that f: I -t I is a homeomorphism, which contradicts Theorem 2.2.12. 0

Theorem 2.2.14. Let f : X t X be a positively expansive map of a compact connected metric space. If f is an open map, then f has fixed points in X . Proof. Notice that f is a local homeomorphism. Since X is compact, f is a covering map. Positive expansivity ensures that there exists a compatible metric D and constants 6 > 0,X > 1 such that D ( z , y ) < 6 implies D(f (x),f ( y ) ) 2 XD(z,y). If V is an open set of diameter less than 6 and if q E f - ' ( V ) , then there is a unique open set containing q of diameter less than (1/X)6, which is mapped homeomorphically onto V. This enables us to lift chains of small open sets. Let {K : 1 5 i 5 n} be a fixed open cover of X so that the diameter of each is less than 6. Take xo E X with f ( x 0 ) # 20. Since X is connected, there is a chain of open sets chosen from among {X}from f ( x 0 ) to xo. Lift this chain to a chain of open sets from xo to x1 such that f (z1)= 20. Further, lift this chain to a chain of open sets from x1 to 2 2 such that f ( x 2 ) = 2 1 , and continue this inductively. Since the original chain had length at most n6, the

CHAPTER 2

46

length of the first lifted chain is at most (l/A)nS and in general the length of the t t h lifted chain is at most (l/A)"n6. If j < t then

D ( x j ,2 k )

I

1 '

1 - - - + (X)'}nS

{(X)J+l+

1 . nS

< ( -A) J

A-1'

Thus { z j } is a Cauchy sequence, which therefore converges to y E X. Since f ( x j ) = xj-1, we have f ( y ) = f(1imxj) = lim f ( x j ) = limzj-1 = y. 0 For f : X + X a homeomorphism of a compact metric space, define a local stable set W:(x,d) and a local unstabZe set W,"(x,d)by

W,"(z,d)= {Y E x : d(fn(z),fn(y)) I E,n 2 01, W:(x,d) = {y E X :d(f-"(x), f-"(y)) I E,n 10). The symbols

Wf(x, d ) ( 0 = s,u ) will be changed by W z ( z )later.

L e m m a 2.2.15. f : X -+ X is an expansive homeomorphism with expansive constant e if and only if there exists e > 0 such that for all y > 0 there is N > 0 such that for all x E X and all n 2 N

c W,(fp"(.)ld), f -"(WeU(x,d)) c W;(f -n(x), 4. f"(W,.(x,d))

Proof. If this is false, then we can find sequences x,, yn E X , m n > 0 such that yn E W:(xn,d), limn-,oomn = 00 and d(fmn(x,), fmn(yn)) > y. Since Yn E WeJ(x,,d),we have

d ( f i o fmn(x,), f i o f m n ( y n ) ) I e

If fmn(x,) -+ x and fmn(y,) + y as n i E Z.On the other hand, we have

-+ 00,

for all i 2 -m,. then d(fi(x), fi(y)) 5 e for all

d(x,y) = n+oo lim d(frnn(xn),frnn(y*))L 7 , thus contradicting. Conversely if d(fn(x), f n ( y ) ) 5 e for all n E Z,for any y y E Wy9(x,d). Since y is arbtrary, we have x = y. 0 bY

> 0 we have

For x E X the stable set and unstable set of a homeomorphism f are defined

Wa(x,d ) = {y E X : lim d( f n ( z ) , f"(y)) = O}, n-oo

W " ( x , d ) = {y E X : lim d(f"(x), f"(y)) = 0). n+-w

$2.2 Expansivity Lemma 2.2.16. If f:X+ X is an expansive homeomorphism, for E number less than an expansive constant

Proof, If

47

>0 a

y E Un,of-nW:(fn(z),d), then there is n 2 0 with fn(y) W:(fn(x), d). Thus, for y > 0 there is N 2 0 such that for all m 2 N

E

and so d(frn+"(y), frn+"(z))5 7 for m 2 N. Therefore, y E Wa(z,d). Conversely, if t E Wa(z,d) and E > 0, then we can find N 2 0 such that d(fn(x),fn(y)) 5 c for n >_ N. Thus d(fj o f N ( z ) , f jo fN(y)) 5 e for all j 2 0. Therefore, f N ( y ) E W:(fN(x),d) and so

For expansive homeomorphisms of a compact metric space we have the following theorem (compare to Theorem 2.2.10). Theorem 2.2.17 (Reddy [R2]). If f:X+ X is an expansive homeomorphism, then there exist a compatible metric D and constants E > 0 , c > 0 and 0 < X < 1 such that

for all n 2 0 .

We say that the metric D as in Theorems 2.2.10 and 2.2.17 is a hyperbolic metric for the map f: X + X.

Proof,Define WO= X

x X and for n 2 1

where e is an expansive constant. Then {Wn} is a decreasing sequence of open symmetric neighborhoods of the diagonal set A. Since f is expansive, if

CHAPTER 2

48

(2, y) 4 A, there is n > 0 with (5, y) 4 W,. This implies there are E > 0 and N 2 1 such that

n,, - W, = A. Thus

We now define a new sequence of open symmetric neighborhoods of A by

vo = wo, vn = Wl+(*-l)N, Obviously,

n 2 1.

V, = A. To use Lemma 2.2.11 we must prove that V,+l

o Vn+l o V,+l

c V,

for n 2 0.

It is clear that V1 o V1 o V1 c VOand by the choice of E and N , V2 o Vz o VZ C V1. Let k 2 2 and take (z,y) E vk+1Ovk+1Ovk+1. Then we prove that (z,y)E v k . Let z , w E X be points such that (z,w), (w,z),(z,y) E Vk+1. If ( p , q ) is any of these three points, then we have

from which

d(fi(z),fi(y)) < e,

lil < 1

+ (k - l)N.

Therefore, (2,y) E Vk. Let D be a compatible metric satisfying

vn c {(.,y)

:D(z,v)<

1

;?.}

c K-1,

n 2 1.

Let A c X x X and denote by zA the subset {y E X : (z,y)E A}. Then we have for n 2 1

$2.2 Expansivity

49

Indeed, the first equality is computed as

Similarly the second equality is computed. Let N be as above. By induction we have for n

21

f N ( W , " (4 z , n zWn) = W,"(fN(.), d) nf N ( 4 K + N , f - N ( ~ , " ( ~n , dZW,,) ) = ~ , " ( f - ~ ( ~ )n, fd- )N ( ~ ) ~ , , + N . FYom now on we prove that D is a hyperbolic metric. Since there is q > 0 such that D(z,y) < q implies d ( z , y ) < e, we have W,(z,D) c W,"(z,d)for z E X. Put v = min{q, 1/4} and take y E W,"(z,D).We first show that

Since (z,y) E

Let

V,, and (z,y)4 Vn+l 3 N1/2m+a(A,D)for some n, we have

L = 3N. It follows from induction that

Define X = (1/2)*. To obtain the conclusion we show that if y E W,"(z,D) then D(f"(z), f"b)) I 8XrnD(z,Y), m L 0. Put m = kL+j fork 2 0 and 0 5 j for some n 2 1, we have

< L. Since (f"(z),

fkWE W,"(f"(4,d)

f"(y))

n fkL(4W1+(n-1)N,

E Vn-Vn+l

CHAPTER 2

50

D(f kL+j Since m = bL

+ j, we have D ( f"(z),f "(Y))

1

I 4 ( j ) k D ( z , Y )= 4XkLD(z9Y)

< 8A"+jD(z, y) = 8XmD(z,y). By replacing f by f - ' , we can establish the required property for y E W,U(z,D).0 We denote as H" the half space of R",i.e. H" = { ( z 1 , z 2 , - *, *~ n E) R" : z, 2 0). Then H" 2 R"-'. A metric space M is called an n-dimensional topological manifold if for each p E M there is a neighborhood U ( p ) such that U ( p ) is homeomorphic t o an open subset V of H". Let qp denote the point in V corresponding to p . We write a M = { p E M : qp E Rn-'} and OM is called the boundary of M. If 8 M = 0, we can take R" instead of H". In this case M is called a topological manifold without boundary. We here say that a topological manifold M is closed if it has no boundary and is compact, connected. If U is an open subset of M and 1c, is a homeomorphism from U onto an open subset of R",then (U, $) is called a chart (or a coordinate neighborhood) of M . For p E U we can write 1c,(p)= ( ~ ' ( p ) , ,~ " ( p ) ) Then . zl, ,Z" is continuous functions on U. We say that (zl,-.,xn) is a local coordinate system of M at the chart (U,$), and that ( ~ ' ( p ) , . . ., z " ( p ) ) is a local coordinate of a point p in U. Let S = {(U,,$a) : a E A} be a family of charts of M. If {U, : a E A} is an open cover of M, then S is called an atlas (or a system of coordinate neighborhoods) of M . A continuous surjection f : X + X of a metric space is said to be ezpanding if f is positively expansive and it is an open map. If X is compact, then an expanding map f : X + X is a local homeomorphism, and so f is a selfcovering map by Theorem 2.1.1. In the case when X is a closed topological manifold, we have that every positively expansive map of X is expanding in our sense because it is a local homeomorphism by the following theorem (cf. PPI *

. ..

-

$2.2 Expansivity

51

Theorem 2.2.18 (Brouwer theorem). If U and V are homeomorphic subsets of R" and U is open in R", then V is open in R". The following theorem is a wider result which contains Theorem 2.2.13.

Theorem 2.2.19 (Hiraide [Hi7]). No compact connected topological manifold with boundary admits a positively expansive map. For the proof we need the following lemma.

Lemma 2.2.20. Let X be a compact connected locally connected metric space and f :X + X a positively expansive map. If a closed proper subset K satisfy the conditions :

then K = 0 . Let M be a compact connected topological manifold and 8M denote the boundary of M. If M admits a positively expansive map f , then f: M + M is locally injective. By applying Theorem 2.2.18 we have that f (M \ 8M)c M \ 8M)and flM\BM is an open map. Therefore BM = 0 by Lemma 2.2.20. Proof of Lemma 2.2.20. Since f: X + X is positively expansive, there exists a metric D satisfying all the properties of Theorem 2.2.10. For any x E X define U,(x)= {y E X : D(z,y) < E } and denote as Cc(x)the connected component of x in Uc(x).Then C,(x)is open in X. Since X is connected and locally connected, X is locally arcwise connected by Theorems 2.1.3 and 2.1.4. Let 6 > 0 be as in Theorem 2.2.10 and take E with 0 < E < 612. We first show that for x E X \ K

G(.) c x \ K

-

f (Cc(.))

3

CAL(f(2))

where X > 0 is as in Theorem 2.2.10. Suppose y E Cx,( f (x))\ f (C,(x)>. Since Cx,( f (x)) is arcwise connected, there is an arc w : [0,1] + Cx,(x)such that w ( 0 ) = f(x) and w(1) = y. Since C,(x)c X \ K and Cc(x) is open in X , f (Cc(x)) is open by (2). Since w ( 0 ) = f(x) E f(Cc(x)), there is 0 < to < 1 satisfying w([O,to))C f(Cc(x)) and w ( t 0 ) 6 /(C,(z)).Then we have

Since cl(Cc(s)) c U b p ( x ) ,by the choice of 6we have that f l C ~ ( ~ * ( + ) ) : cl(Cc(x)) homeomorphism. Since w([O,to]) c cl( f (C,(x))), there is an arc G : [O,to] + cl(Cc(z)) such that f o = w. Since f o G([O,to))= w([O,to))c f(G(z)),we have G([o,to))c G ( x ) and since 4 t 0 ) $ f(Cc(Z)), obviously G(to) 6 Cc(x). + cl( f (Cc(x))) is a

CHAPTER 2

52

On the other hand, since w ( t 0 ) E Cx,(f(z)), we have

AD(G(to),z)5 D(w(to),f(z))< and so

S j ( t 0 ) E Uc(z).Thus G(t0) E Cc(z) since z E S j ( [ O , t o ) ) C Cc(z) C Uc(z).This is a contradiction. Therefore, Cx,(f(z)) C f(Cc(z)).

Let us define

X(E) = {z E X \ K : Cc(z)c X

\ K}.

Since K # X, there is 0 < EO < 6/2 such that X(Q) # 0. To obtain the conclusion of the lemma suppose K # 8. Take z E X(EO). Then we have C X C O ( f ( 2 ) ) c f(Cc,(z)).Thus, f(X(E0)) c X(AE0). It is easily checked that X(AEO)c X(pe0) c X ( E Ofor ) 1 < p < A. We show that cl(X(k0)) C X(peo). To do this let { z i } be a sequence of X ( k 0 ) and suppose x i + z as i + 00. Since CEO(.)is open in X, we have zi 6 Cco(z) for sufficiently large i. Thus Cco(z)C CPco(z:i) c Cxco(zi)which implies z E X ( ~ E O Therefore ). cl(X(Ae0)) c X ( p ~ 0 ) . We have shown that f(cl(x(e0))) = Cl(f(X(E0))) c Cl(X(AE0)) c X(PE0) c X(E0).

n:=o

Thus Y = fi(cl(X(eo))) is a nonempty closed set and f ( Y ) = Y . Since Y # X and X is connected, Y is clearly not open in X. For A c X and a > 0 define Na(A) = U a E A C Q ( a Then ). N(P-l)co(Y) c X(EO).Indeed, let z E Y and z E C(P-l)co(z). Then Cco(z)c CPCO(z). Since e E Y C X ( ~ E Owe ) , have Cco(z)C CPco(z) C X \ K and so z E X(EO). Since f ( Y )= Y , we have

nm(p-l)co(~)) cn cY 00

Yc

00

m(Eo))

i=O

i=O

nzo

and thus Y = fi(N(p-l)co(Y)). Since C ( p - l ) c o ( zC) Cclco(z) C X \ K for C x ( , - l ) c 0 ( f ( ~and ) ) thus f(N(P--l)CO

L

E Y , we have f ( C ( P - l ) c o ( z3) )

( Y ) )3 N x ( P - l ) c o ( y ) 3 ~(jA-1)co(Y).

Therefore, Y = N(P-l)co(Y) and Y is open in X. This is a contradiction. 0 We can construct an example of positively expansive map which is not open (an example due to Rosenholtz ([R])).Consider the subset X in the plane defined by

Remark 2.2.21.

x = {% : I%[

3 1 = 1) u {% : 1% - -1 = -} 2 2

u {% : 1%

+ -132

1 = -}. 2

$2.2 Expansivity

53

Give X the arclength metric. Then a map is defined as follows : stretch each of the small circles onto the big circle, stretch each of the upper and under semicircles of the big circle first around a small circle, and then across the other semicircle and finally around the other smaller circle. More precisely we describe the map f by

2(2 - 3/2) 2(2 3/2) z6/2 - 3/2 z3 -2*/2 3/2

+

+

R e ( z ) 2 1, R e ( z ) 5 -1, 1/2 5 Re(z) 5 1, - 1/2 2 R e ( z ) 5 1/2, - 1 5 Re(e) 5 -1/2.

if if if if if

Then it is easily checked that f is positively expansive, but not open. 1

-11

I

Figure 2

If f : X + X is a homeomorphism of a compact metric space and x is a point in X , then a-limit set of x , denoted as a ( x ) , consists of those points y E X such that y = lim+,w f " j ( x ) for some strictly decreasing sequence of integers nj. The w-limit set of x , denoted as w ( x ) , is similarly defined for strictly increasing sequence. The set a ( x ) ( ~ ( x )is) closed, nonempty and f invariant, i.e. f ( a ( x ) )= a ( . ) for z E X . If, for some point x , a ( ~and ) W(X) each consists of a single point,then we say that x has converging semi-orbits under f . Theorem 2.2.22 (Reddy [R3]). If f : X -+ X is an expansive homeomorphism of a compact metric space, the set of points having converging semiorbits under f is a countable set. Proof. Since f is expansive, it follows that f has at most finitely many fixed points, say q l , . ,q k . Let A denote the set of points having converging semiorbits under f . Suppose A is uncountable.

.

54

CHAPTER 2

If x E A, both a(.) and ~ ( z are ) fixed points. Let A ( i , j ) be the set of points satisfying a(.) = qi and ~ ( z=) qj. Then A is the union of the finitely many sets A ( i , j ) , so that one of these, say B , is uncountable. For each N > 0 let B ( N ) be the collection of points x E B such that, for n 2 N,fn(z) is ) f-n(x) is e/a-close to a(.). Since B is the union of the e/a-close to ~ ( zand sets B ( N ) , one of them (say B ( M ) ) must be infinite. Since X is compact, there exist distinct points y and z of B ( M ) with d ( y , z ) < e/2 so small that fn(z) is e-close to fn(y) if 171.1 5 M, where e denotes an expansive constant for f . Thus, from the definition of B ( M ) , we have that d(fn(y),fn(z)) < e for all n E Z, thus contradicting the choice of e. Therefore, A is countable. 0 Theorem 2.2.23 (Jacobsen and Utz [J-U]). There ezists no expansive homeomorphism of a closed am. Proof. If f : [0,1] --f [0,1] is a homeomorphism, then f has either f (0) = 0 and f(1) = 1, or f(0) = 1 and f(1) = 0. In both cases f a induces a homeomorphism satisfying f 2 ( 0 ) = 0 and f a ( l ) = 1. P u t Fix(f2) = {z E [0,1] : f2(x ) = z}. Then Fix(f2) is closed. If Fix(f2) = [0,1], then all points of [0,1] have converging semi-orbits under f and so f is not expansive by Theorem 2.2.22.

Figure 3

If Fix(f2) # [0,1], then U = [0,1]\Fix(f2) is a nonempty open set. Thus U is the union of the countable intervals Ij where Ij’s are mutually disjoint. For x E U there is Ij such that z E Ij. Then the extremal points of Ij are fixed points of f a . Thus we have that for x E Ij, x > fa((.) > f4(z)> or z < f2(z) < f4(z)< . In any case {f2j(x) : j 2 0) converges to a fixed point of f Since Ij is uncountable, f is not expansive. 0 m e - ,

Let f:X 4 X be a homeomorphism of a compact metric space. If, for all x E X,the orbit Of(”) = { fn(x), n E Z} is dense in X ,then f is called a minimal homeomorphism. Obviously every minimal homeomorphism is topologically transitive.

$2.2 Expansivity

55

Theorem 2.2.24. A homeomorphism f : X + X is minimal if and only if E = 0 or X whenever f (E) = E and E is closed.

Proof. Suppose f is minimal. If E is closed and f ( E ) = E # 8, then E 3 Of(x) for z E E and cl(Of(x)) C E. Thus we have E = X. Conversely, for z E X,cl(Of(x)) is a closed f-invariant set and thus cl(Of(x)) = X. 0

Figure 4

A closed subset E which is f-invariant is called a minimal set with respect to f if f p is minimal.

Theorem 2.2.25. Every homeomorphism f:X

+X

has a minimal set.

Proof. Let 0 denote the collection of all closed nonempty f-invariant subsets of X. Clearly 0 # 0 because of X E 0.0 is a partially ordered set under the inclusion. Every linearly ordered set of 0 has the least element. Thus 0 has a minimum element by Zorn's lemma. This element is a minimal set with respect to f. 0

Theorem 2.2.26. There exists no expansive homeomorphism of the unit circle S'. Proof, Suppose f:S' + S' is expansive. By Theorems 2.2.22. and 2.2.23, f has no periodic points. Indeed, suppose f has distinct periodic points z and y. Then f k(x ) = z and f k ( ( y ) = y for some L > 0. For simplicity put g = f k . There exists a closed arc C in S1 such that g(C) = C , or g a ( C ) = C. If g(C) = C, then glc : C + C is expansive. This is impossible by Theorem 2.2.23. If f has only one fixed point, for any point 2 E S ' the semi-orbit {x, f(z), f'(x),...) converges to the fixed point, and also {x, f-'(x), f - 2 ( z ) , - - -converges } to the fixed point. Thus f:S' + S' is not expansive by Theorem 2.2.22. Let M be a minimal set of f in S'. Since f :M + M is expansive and minimal, we see that M is totally disconnected (see Theorem 2.2.44 below). Notice that M is an infinite set because f has no periodic points.

CHAPTER 2

56

We claim that isolated points are not in M. Indeed, if x E M is an isolated point, then Ua(z)n M = {z} for some 6 > 0 and so {z} is open in M. Since M is minimal, the closure of the orbit of x equals M and thus fn(x) E Ua(x) n M = {x} for some n E Z, a contradiction. Since S'\ M is open, S' \ M is expressed as countable disjoint union S'\ M = Uj Ij of open arcs I, in S1.Obviously diam(1j) + 0 as j -+ 00 and for fixed j,fn(lj)# Ij for all n # 0. Thus, diam(f"(1j)) t 0 as In1 + 00. Take

-

n

a, b E Ij and let fno(ab) be the arc with the longest length in { f "(ab) : n E Z}. Let E > 0. Then we can choose I, and a, b E I, such that the length between n

and fno(b) is less than E . Then, diam(f"(ab)) < E for all n E Z and therefore f : S' -+ S' is not expansive. This is a contradiction. 0 f"o(a)

The following theorem was proved in Hiraide [Hi51 and Lewowicz [L2].

Theorem 2.2.27. There exists no expansive homeomorphism of the 2-sphere

S2.

We omit the proof. If the theorem is of interest, the reader should see [Hi51 and [L2]. Let X be a compact metric space with metric d and Xzdenote the product topological space X z = {(?i) : x; E X,i E Z}. Then X z is compact. We define a compatible metric d for Xz by

A homeomorphism u : X z

-+

Xz,which is defined by

.((xi)) = (yi) and yi = xi+' for all i E

Z,

is called the shift map. For f : X -+ X a continuous surjection, we let

Xf= { ( z i ) : z;E X and f(xi) = xi+l,i E Z}. Then Xf is a closed subset of X z . Moreover we have .((xi)) = (f(xi)) for all (xi) E Xf,and so Xf is u-invariant, i.e. u(Xf) = Xf. The space Xf is called the inverse limit space constructed by f and we sometimes write Xf= lim(X, f ) . The restriction u = qx, : Xf+ Xfis called the shift map t determined by f . Let i E Z and denote as p i : Xf -+ X the projection defined by (xi) H xi. Then pi o u = f o pi holds, i.e. the diagram

Xf A Xf Pi

j,

1Pi

x - x f

commutes,

52.2 Expansivity

57

and PO o ui = f o po for all i 2 0. If, in particular, f is bijective, then it is clear that each pi is bijective. In the case where X is a torus and f: X + X is a toral endomorphism, we can show that the inverse limit space Xfhas a structure of compact connected finite dimensional abelian group, which is called the solenoidal group. The properties of solenoidal groups will be discussed in Chapter 7. It is easily checked that Xf is homeomorphic to the space

X; = {(xi)?

: zi

E X and f (zi+l) = z i , i 2 0)

equipped with the product topology, and the shift map (T is topologically conjugate to a homeomorphism u' defined by (~'((zi)?)= (f (zi))? for (zi)? E X;, i.e. there exists a homeomorphism h: Xf+ X; such that the diagram

Xf 0 Xf

1.

commutes.

A continuous surjection f:X -+ X is c-expansive if there is a constant e > 0 (called an ezpansive constant) such that for (xi), (yi) E Xfif d(zi,yi) 5 e for all i E Z then (zi) = (yi). This is independent of the compatible metrics used. The notion of c-expansivity is weaker than that of positive expansivity. For homeomorphisms c-expansivity implies expansivity. Theorem 2.2.28. Let k and only if so is f k .

>0

be an integer. Then f:X + X is c-expansive if

Proof. We notice that f is uniformly continuous. Let e constant for f . Then we have d(z,y)

< 6 -----r. @(z),fi(y))

> 0 be an expansive

< e, 1 5 i 5 k

for some 6 > 0. This 6 is an expansive constant for f k . In fact, for (zi),(yi) E Xf suppose d ( z k i , yki) < 6 for all i E Z. Then we have d(zi, yi) < e for i E Z. By expansivity, zi = yi for i E from which z k i = y k i for i E Z. The converse is proved by the same way. 0

z,

For the relation between continuous surjections and their inverse limit systems we have the following Theorem 2.2.29. Let X be a compact metric space. A continuous surjection f: X 3 X as c-ezpansive if and only tf u : Xf--+ Xf is eapnsive. Proof. If (zi) # (yi) then there is k E Z such that d(zk,Y&) > e by cexpansivity. Then i(uk((zi)),uk((yi))) 2 d(zk,Yk) > e, where d((zi),(yi)) = 00 d(zi, yi)/21il. Therefore (r is expansive.

xi=-,,

CHAPTER 2

58

Conversely, let e

> 0 be an expansive constant for u and let d(si, y i ) < e / 4

(i E Z) for (xi),( y i ) E Xf.Then we have (i(d'((xi)),a"((yi))) 5 e for n E Z, and hence ( S i ) = ( y i ) by expansivity of u. Therefore e / 4 is an expansive constant for f : X t X. 0 The following theorem is easily checked. Thus we leave the proof to the readers.

Theorem 2.2.30. (1) Iff:X --t X is c-expansive and Y is a closed subset of X with f ( Y )= Y , then fly : Y + Y is c-expansive. (2) If fi : Xi + Xi (i = 1,2) are c-expansive, then the continuous surjection f1 x fz : XI x X Z + X I x X2 defined by

is c-expansive. Every finite direct product of c-expansive maps is c-expansive. ( 3 ) If X i s compact and f:X + X is c-expansive, then h o f o h-l : Y + Y is c-expansive where h : X + Y is a homeomorphism.

Theorem 2.2.31. The closed interval does not admit c-expansive maps.

Proof. Let I denote the closed interval [0,1]. For f:I + I a continuous surjection, its inverse limit system (If,u) is defined by

Suppose that u2 : If + If is expansive. It is easily checked that (If,u2)is topologically conjugate to ( I f a , ~ ) .Since f is surjective, there exist a , b E I with 0 5 a < b 5 1 such that either

(1) f(a) = 0 and f ( b ) = 1

or

(2) f ( a ) = 1 and f ( b ) = 0.

If we have (2), then we see that f2(a') = 0 and f 2 ( b ' ) = 1 for some u', b' E I with 0 5 a' < b' 5 1. Thus, to obtain the conclusion it is enough to prove (1). Take two points u,v E I ( . < v) such that (3)

f ( u ) = u,

f ( v ) = 'u,

f(x) # x for u < x

< v.

This is ensured by the facts that f has at least two fixed points and the set of fixed points is not dense in I. (3) is divided into the following cases,

( 4 ) f(x) > x for u < x We first check the case ( 4 ) .


or

(5) f(z)< x for u < x

< v.

82.2 Expansivity

59

If fllU,+,1 : [u,v] -+ [u,v] is bijective, then the inverse limit system of ( [ u ,v], is a subsystem of (If,.). Thus the system is expansive. But it contradicts Theorem 2.2.23. Therefore we can find two points x1 and x2 such that u < z1< x2 < 21, f(x1) = f(x2).

fllU,,l)

Now we define

yo = f(xo) = max{f(x) : u

< x 5 x2}.

Then, u < xo 5 x2 5 yo. Since the graph of f on [u,v] is over the diagonal set of I x I, there exists a sequence x8 > xy > in [u,v] with

-

xo0 - x 0 , f(XP) = x;+

Since x i

-+

u as n + 00, for 0

< E < x2 - xo we have

u

for some N

> 0.

i 2 1.

< 2% < U+& /2

Take 6 > 0 so small that for z,y E I

< 26 * If"(.)

Ix - yI

Write c1 = 2% - 6 and c2 = x& IfN(c')

- f"(y)l < E / 2 ,

0 5 n 5 N.

+ 6. Then

- fN(x&)I = I f N ( c i ) - 501 < &/2, i = 1,2,

and thus f N ( c ' ) < xo + E < x2. By the definition of yo we have f N + l ( c i ) 5 yo for i = 1,2. If f N + ' ( c i ) = yo for some i, then we can find a sequence {cj :j

2 O} c [u,v]

such that ci = ci and f ( c i ) = cj-, for j 2 1. Notice that {ci : j 2 0) is strictly decreasing. Define a sequence { b j } in [u,v] by

b3. -- f N - j ( ci) , bN+j

Since Ici - 20"

OljlN, j

= Ci,

> 0.

< 26 for i = 1,2, we have

If"(c')

- f n ( X g ) I = 1bN-n

0

- XN-"~

< &,

0

5 n 5 N.

Sinceu+~/2>x&>x$+~ > . . . > u a n d x & ~[u,v],wehave 21

+

&

> ci = c; > cf >

*

>u

CHAPTER 2

60

and thus lbj - z ! 1 < E for j 2 N. Therefore, lbj < E for all j 2 0. Since fN+'(ci) = f(zo) = yo, we have f j ( b 0 ) = fj(z8) for j > 0. Therefore, for j > 0, j-th components of a"(bo, bl ,* * * )

and

Q"(z:, z : ,

* *

-)

are closer than E for all n E Z. Let z3 = max{ fN+'(c') : i = 1,2}. If z3 < yo, then there exist c3, c4 with

such that

f N+'(C3)

= fN+'(C4)

= c4.

By (4) we can find sequences {cj : j 2 O } , i = 3,4, in [u,v]such that ci = ci for j 2 1. They are strictly decreasing sequences. By the and f(c3) = similar argument we have that for j 2 0, j-th components of

g"(fN ( C3o ) , f N-1 ."(f N ( C4o ) , f N-1

3

(co),-**,c:,c;,---), 4

,c;,c:,4

(%I),***

are closer than E for all n E Z. Therefore (T : If + If is not expansive. The case ( 5 ) is obtained in the same argument. 0

For toral endomorphisms we obtain the following theorem from Lemmas

2.2.33 and 2.2.34 below.

Theorem 2.2.32. (1) A toml automorphism is expansive if and only i f it is an automorphism of type (0. (2) A toml endomorphism is positively ezpansive i f and only if it is an endomorphism of type (II). (3) A toml endomorphism which is not injective is c-expansive but not positively expansive if and only i f it is an endomorphism of type (III). Lemma 2.2.33. Let f be a linear map of the euclidean space B", where n 2 1, and let d be the euclidean metric for R". Then f is expansive under d i f and only i f f is hyperbolic, i.e. it ha8 no etgenvalues of modulus 1.

Proof. Suppose all eigenvalues o f f are off the unit circle. Then Rn splits into the direct sum B" = E" €3 E" of subspaces E" and E" such that f ( E d )= E" and f ( E " )= E", and such that there are c > 1,0 < X < 1 so that

$2.2 Expansivity

01

for n 2 0. Therefore, f is expansive under d. Conversely, if f has an eigenvalue of modulus 1, then Rn splits into the direct sum of subspaces as follows : (i) R" = EC@ E" @ E", (ii) f(E') = E"(0 = c , s , u ) , (iii) there are c > 0 and 0 < X < 1 such that (1) and (2) hold, and (iv) flEC is linearly conjugate to a linear map coresponding to Jordan form (in the real field)

where either

for 1 5 j 5 k. Here I is a 2 x 2 identity matrix and

0 < 8j

R(8j) =

< T.

By (iv) we can find a subspace F of Ec such that f p is an isometry under d. Therefore, f is not expansive. 0 Let

x and X be metric spaces with metrics 2 and d respectively, and let

X be a continuous surjection. Then T is called a locally isometric covering map if for each z E X there exists a neighborhood U ( z ) of x such that T - ~ ( v ( x ) )= u, ((Y# + u, n ual= 0) T :

-+

U 0

where U, is open and

TIU,

x

: U,

-+

U ( x ) is an isometry.

Lemma 2.2.34. Let f : -+ and g : X -+ X be continuous sujections and T : -+ X a locally isometric covering map. Suppose T o f = g o T and X i s compact, and suppose there exists 60 > 0 such that for each x E and 0 < 6 5 60 the open ball Ua(z)of x with radius 6 is connected and

x

x

T

is an isometry. Then

: Ua(.)

+

u6(.(.))

CHAPTER 2

62

( 1 ) f is ezpansive (under 2 ) i f and only if g is expansive, when f and g are homeomorphisms, (2) f i s positively expansive if and only if g is positively expansive, ( 3 ) f is c-expansive if g is c-expansive, and i f f is c-expansive and satisfies the following condition (C): for - E > 0 there ezists N > 0 such that for (pi),(q;) E Ef one has d(p0,qo) 5 E whenever a ( p ; , q i ) 5 e for all i with lil 5 N where e > 0 i s an expansive constant for f , then g is c-expansive.

Proof. From the assumptions it follows that for each x E X and 0

< 6 5 60

and if p , q E T - ' ( x ) and p # q, then v&(p) r l u6(q)= 0. Indeed, it is clear that T - ' ( u 6 ( z ) ) ZI U,,,-l(z,U6(p). Let a E ?r-'(u6(.)). Since qU6(,,) : us(.) 4 U6(.(a)) is bijective and x E u6(7r(a)), we have a E Ua(p) for some p E T-'(x), and so (i) holds. If b E Ua(p)n Ua(q) # 8, then p = q because ?TIUs(b) : U6(b) U6(a(b)) is bijective. To obtain the lemma, we first show that f is uniformly continuous. Let 0 < E 5 60 and take 0 < 6 5 60 such that for x , y E X

Now, if &,q)

< 6, then

, implies and so n o f ( q ) E U,(g o ~ ( p ) )which

Since 6 5 60,we have that f(Ua(p))is connected. Notice that U , ( f ( p ) ) is the connected component of f ( p ) in .-'(Uc(go7r(p))). Then f (Ua(p))C Uc(f ( p ) ) , and thus f is uniformly continuous. (1) : Suppose g is an expansive homeomorphism with expansive constant e, and let 7 = min{e,60}. For p,q E if f i ( p ) , f ' ( q ) ) 5 7 for i E Z,then we have 4g"(.(P)),gi(.(q>>) = ' ( P I , f i ( q ) ) I 79 i E Z,

x, a(

4f

and so ~ ( p=) r(q).Since z(p,q) 5 60, it follows that p = q.

52.2 Expansivity

63

Conversely, suppose f is expansive. Choose 0 < 6 < 60 such that if Z ( p , q ) < 6 then a ( f ( p ) , f ( q ) )< 60 and Z(f-'(p),f-l(q)) < 60,and put 7 = min{e,S} where e is an expansive constant. For x, y E X with d(z, y) 5 7 we take p, q E such that ~ ( p =) x, ?r(q) = y and a ( p , q ) = d(z,y). If d(gi(x),gi(y)) I: 7 for i E Z, then

since a(f(p), f ( q ) ) < 60. Inductively, we have a ( f i ( p ) ,f i ( q ) ) 5 7 for all i 1 0. In the same way, a(fi(p),fi(q))5 7 for all i I: 0. By expansivity we have p = q, and therefore x = y. (2) : This is proved in the same manner as (1). (3) : Suppose g is c-expansive (with expansive constant e > 0). Let 7 = min{e,60} and ( p i ) , ( q i ) E Since A o f = g o A , it is clear that ( r ( P i ) ) , ( r ( q i ) ) E X,. If a ( p i , q i ) I: 7 for i E Z,then d(n(pi), r ( q i ) ) 5 7 for i E Z,and so pi)) = ( r ( q i ) ) , which implies (pi) = (qi). Conversely, suppose f is c-expansive and satisfies the condition (C), and suppose d(xi, yi) _< 7 (i E Z)for ( z i ) ,(yi) E X, where 7 = min{e, 6 ) and 0 < 6 5 60 is chosen such that a ( f ( p ) ,f ( q ) ) I: 60 whenever &I, q ) < 6. For each n 2 0 we take ,p ! E A - ' ( z - n ) , q!!,, E ~-'(y-,) such that && ,!I), 5 7, and define ( p ; ) , (q;) E by

xf.

xf

) (zi)and ( A ( $ ) ) = (yi), and &?,qy) 5 7 for all i 2 -n. Then ( ~ ( p ? ) = By the condition (C) it follows that d (p t,q t) + 0 as n --+ 00. Thus zo = yo. Similarly, we have zi = yi for i E Z. 0

Remark 2.2.35. -

The condition (C) in Lemma 2.2.34 (3)is always satisfied if

X is compact. However, it remains a question of whether it is true for general case.

Let X be a compact metric space. Topological dimension of the space X is said to be less than n if for all 7 > 0 there exists a cover Q of X by open sets with diameter < 7 such that each point belongs to at most n 1 sets of a. (If the topological dimension is of interest, the reader should see Hurewicz and Wallman [H-W]).

+

Theorem 2.2.36. Let X be a compact metric space. X is 0-dimensional if and only if it is totally disconnected (i.e. the connected component of each point is a singre point). Before starting with the proof we prepare the following

CHAPTER 2

64

Lemma 2.2.37. Let 0 be the collection of open closed subsets of X. Then 0 is a base of X if and only if X is totally disconnected. Proof. Take x , y E X with x # y. Then we can find an open set U, such that x E U, and y # U,. Suppose 0 is a base of X. Then x E B c U, for some B E 0.Since X \ B is open closed and y E X \ B , the connected component of x , ~ ( x )is, a subset of B . However, since y is arbitrary, we have

and therefore X is totally disconnected. For fixed x E X let G be an open neighborhood of x . Then it suffices to find an open closed subset B satisfying x E B c G . If X is totally disconnected, then each x E X is expressed as { x } = U ( x ) where { U ( x ) } is the collection of open closed subsets containing x. Indeed, let L = n U ( x ) . To see {x} = L it suffices to prove that L is the connected component of x. If this is false, then L is expressed as L = A U B where A and B are closed and A n B = 0. Thus we have that x E A or x E B . We suppose x E A. Choose open subsets A1 and B1 such that A c A1, B C B1 and Al n B1 = 0. Then L c A1 U B 1 . Since L = U ( x ) ,there is U ( x ) such that L c U ( x ) c A1 U B1. Then it is easily checked that A: = U ( x )n A1 and Bi = U ( x )n B1 are open closed in X. Since x E A, A: is an open closed subset containing x and L n A: = 0. This contradicts the definition of L. Therefore, for y E X we can find an open closed subset H , satisfying y E H , and x # Hy.Since X \ G is compact, a finite cover { H,, , ,H,, } of X \ G exists. Notice that each HVi satisfies x @ ITvi. H = lJy HVi is open closed and x # H. Therefore x E B C G where B = X \ H. 0

n

n

--

Proof of Theorem 2.2.36. Suppose X is totally disconnected. By Lemma 2.2.37 the collection of all open closed subsets is a base of X. Let I' be an open cover of X and let x E X. Then there is U, E I' such that x E U, and then there is an open closed V, satisfying z E V, C U,. Since {Vz : x E X} covers X, choose a finite cover { W1, * ,Wk} and define

.

F1 = W i , F2 =W2\Wl,

...,

Fk=Wk\(W1U..*UWk-1).

Then { F1,. ,Fk} is a refinement of r, and a point of X belongs to only one set of {Fl, * * ,Fk}. Therefore X is zero dimensional. Suppose X is zero dimensional. Take arbitrarily a, b E X with a # b. We set U = X \ { a } a n d V = X \ { b } . Then{U,V}coversX. Let a = { F 1 , . . - , F k } be a closed cover of X such that Fi n Fj = 0 for i # j and a is a refinement of { U,V}. Without loss of generality we suppose a E 4 . Since X = F1 U F and F1 n F = 0 where F = F2 U U Fk, The connected component of a, c(a), is a subset of F1 and so c(a) c X \ {b}. Since b is arbitrary, we have that c(a) c n { X \ {b} : b E X , a # b } = { a } , i.e. X is totally disconnected.

.. -

---

52.2 Expansivity

65

We here construct the Cantor set which is of particular importance. To do as follows. First, denote the closed interval [0,1] by C1. Next, delete from C1 the open interval (1/3,2/3) which is its middle third, and denote the remaining closed set by C2. Clearly, C2 = [O, 1/31 U[2/3,11.

so we proceed the argument

Next, delete from Cz the open intervals (1/9,2/9) and (7/9,8/9) which are the middle thirds of its two pieces, and denote the remaining closed set by C3. It is easy to see that

Figure 5

If we continue this process, at each stage deleting the open middle third of

each closed interval remaining from the previous stage, we obtain a sequence of closed sets C,,,each of which contains all its successors. The Cantor set C is defined by m

n= 1

and it is closed. C consists of those points in the closed interval [0,1] which ultimately remain after the removal of all the open intervals (1/3,2/3),

(1/9,2/9),

(7/9,8/9),

* * * *

Clearly C contains the end points of the closed intervals which make up each set Cn ; 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, . a

*

*

The set of these end points is clearly countable. However, the cardinal number of C itself is the cardinal number of the continuum.

CHAPTER 2

66

To prove this it suffices to exhibit an injection f of [0,1) into C. We construct such a map as follows. Let z E [0, 1) and let z = .blb2 be its binary expansion. Each b,, is either 0 or 1. Let a, = 2b,, and regard

---

f(z)= 3-lal-k 3-2a2

+---

as the ternary expansion of a real number f(z)in [0,1). The reader will easily convince himself that j(z) is in the Cantor set C. In fact, since a1 is 0 or 2, f(z)is not in [1/3,2/3). Since a2 is 0 or 2, f(z)is not in [1/9,2/9) or [7/9,8/9) ; etc. Also, it is easy to see that f: [0, 1) C is surjective. This shows that C is the set of the numbers z E [0,1] with --f

For T 2 1 we call the set C n [3+i,3+(i subinterval with mnk T if

+ l)] (0 5 i 5 3r - 1) a Cantor

We denote as I ( i , r ) the i-th Cantor subinterval with rank Obviously 2'

c = UI(i,T),

T

from the left.

I(i,r)=I(2i- l,T+l)U1(2i,T+l).

i=l

Remark 2.2.38. Let X be a compact totally disconnected metric space having no isolated points. Then X is homeomorphic to the Cantor set C. This is checked as follows. Let d and d' be metrics for X and C respectively. Since X is totally disconnected, we can choose open closed subsets X1,X2,*** ,X2k1(IZ1> 0) such that diam(Xi) < diam(X)/2 2k1

X = UXi, i=l

(1 5 i 5 2'')

xinxj=~ (i+j)

Since each of Xi has no isolated points, as above we can choose open closed subsets X ~ J ,* * ,Xi,)ka such that

-

diam(X;,j)

< diam(X)/22 (1 5 j 5 2'l),

2ka

x i = u xi,j, j=1

Xi,j nXi,ji = 0 ( j

+j').

52.2 Expansivity

07

Continue inductively this manner. Then there is a sequence {kj}gl of positive integers and open closed subsets Xil,...,i,( 1 5 it 5 2", 1 5 I 5 m ) satisfying the conditions

< diam(X)/2"

(1 5 it 5 2",1 5 I 5 m ) .

(i>

di=(Xi, ,...,i,)

(ii)

Xil,...,i,-l,j n xil,...,im-l,jl = 0 ( j # j ' ) ,

xil,...,im-l

(iii)

2hm

=

U Xil,...

,i,-l,j-

j=1

If (i~,... , i n , . - - )E n g l { l , 2 , . - - ,2kj}, then we have {z} = for some z E X. Define k ( m ) = Cj"=, kj for rn 2 1 and

nj"=,Xi, ,...,

ij

n= :,

I(p(rn),k(rn)),and so we define Then there is y E C such that {y} = h ( z ) = y. Then h: X + C is a one-to-one continuous surjection Indeed, for E > 0 take mo 1 1 such that 3k1+*'*+kmo > l / and ~ choose 6 such that

< 6 < min{d(Xil ,...,imo-l,j,Xil,...,imo-l,j~) : 1 5 in 5 2kn, 1 5 n 5 mo - 1, 1 5 j # j ' 5 2km-0). If d ( z , y ) < 6(z,y E X), then h ( z ) and h ( y ) belong to the Cantor subinterval with the rank Cj"=ol kj. Thus we have d'(h(z),h ( y ) ) < E and therefore h is 0

continuous. That h is bijective is easily checked.

Theorem 2.2.39. Let f : X + X be an expanding map of a compact metric space X. Then the topological dimension of X is finite. Proof. To see dim(X) < 00 we consider the inverse limit system ( X f , a ) of (X, f ) . Since f is expanding, by definition f is a local homeomorphism and positively expansive. Then, by using Theorems 2.1.1 and 2.2.10, we see that there is a base of neighborhoods for Xf each of whose members is the direct product of a neighborhood of X with Cantor set. Since f is positively expansive, d : Xf + Xf is expansive. If Xf is finite dimensional, obviously the dimension of X is finite. That dim(Xf) < 00 follows from the following result. 0

Theorem 2.2.40 (Mafib [Mall). If f : X + X is an expansive homeomorphism of a compact metric space X, then the topological dimension of X is finite . For the proof we need the following lemmas. Let e constant and fix 0 < E < e/2.

> 0 be an expansive

CHAPTER 2

68

Lemma 2.2.41. There is a 6 > 0 such that ifd(z,y) E

for some n

< 6 (z,y E X) and

5 sup{d(fj(z),fj(y)) : o 5 j 5 n } 5 2.5

2 0 , then d(fn(z), f"(y)) 2 6.

Proof. If this is false, then there are sequences Z,,yn E X, m, > Cn > 0 such

that

lim d(xn,yn) = 0,

n+oo

lim

n+oo

d(fmm(xn),fmm(Yn))

2 E, :0 I m 5 m n } 5 2

= 0,

d(f'n(xn),r'"(yn))

sup{d(fm(zn), f"(Yn))

~ .

Since X is compact, we suppose that fC* (xn)+ x and f '"(y,) + y as n + 00. Then d(z,y) 2 E and d(fn(x), f"(y)) 5 28 for all n E Z.Therefore Lemma is proved. 0

Lemma 2.2.42. For all p E

> 0 there exists N

< SUP{d(f"(4, f"(Y))

= N ( p ) > 0 such that : In1

I NI

whenever d(z,y) 2 p. Proof. If this is false, then there are sequences s,,y, E X with d ( z n , yn) 2 p such that suP{d(P(zn), fj(yn>> : IjI I n ) I E If x, + x and y, + y as n + 00, then d ( z , y ) 2 p and d(fn(x),fn(y)) 5 E for all n E Z. 0 Proof of Theorem 2.2.40. By using Lemmas 2.2.41 and 2.2.42 we derive the conclusion. Let 6 be as in Lemma 2.2.41 and choose a cover { Ui : 1 5 i 5 1) of X by open sets with diameter < 6. We show that dim(X) I L2. For each n 2 0, choose 6, > 0 such that d(x,y) 5 6, implies d(fj(z),fj(y)) < E for all Ijl 5 n. We define

-

utj = fn(ui)n f-"(uj)

and x y for x,y E Ucj if there exists a finite sequence x = 20, X I , . *, x p = y such that d(z,,x,+~)< 6, for all 0 5 T 5 p - 1 and 2, E Ucj for all 0 5 T I p. Denote as nk Ui,; , 1 5 12 5 k ( i , j , n )

the 6,-components of Uej (i.e. the equevalence class of Ucj under the relation x y). It is easily checked that each U,$ is open and they cover X. We claim that N

2.2 Expansivity

69

Indeed, if this is false, then we could find p > 0 and a large number n, say N(p) given in Lemma 2.2.42, such that in some U$! there are z,y with d(z,y) > p. Let z = zo,q,. ,z p = y be a sequence in U;, such that d ( Z r , z r + l ) 5 6, for d O 5 T 5 p - I.Let US put n

> 2N(p),

.-

sr = sup{d(fm(zr), ~ " ( Z O ) ): Iml I n } . Then sr

> E by Lemma 2.2.42. 81

Choose T such that

S,I

< E,

From the choice of 6, it follows that

Isr+1

- 8.1

€7

1 IT

I E if T' < T and a, > E . d(f-,(zo),

since f-,(z~),f-"(z,) E U;.Since f"(z,)) < 6 and SO 8,

I

f-"(zr))

< P*

Then

8,

5 2~ and therefore

I6

fn(zg), fn(z,)E

U,,we have d( fn(zO),

= su~{d(f"(zo), f"(zr)) : I m l I n } I 2 ~ .

For simplicity put zb = f-"(zo) and z',= f - " ( ~ , ) . Then sup{d(fm(zb),

f"(zi)) : o 5 m 5 2n} 5 2 ~ .

Since d(z;,z',) < 6, by Lemma 2.2.41 we have

d(f2n(zb), f2"(.5))

= d(f"(zo),

fn(z?-)) 2 6,

thus contradicting d(fn(zO), f"(z,)) < 6. Therefore (1) is proved. It only remains to show that for each n, any point of X belongs to at most L2 sets of the cover

{U*$ :1 I i , j I e,1 5 k 5 k ( i , j , n ) } . Suppose that

n{uc:;.m: 1 5 m 1 s } # 0 .

If ( i m , j m ) = ( i ~ , j ~then ) , we have UzT :'m = Uz!& since they are 6,components of Ui",,jn. and have nonempty intersection. This implies that to different values of m correspond different values of the couple (im,jm). Therefore, s I L2. 0

Remark 2.2.43. Theorem 2.2.40 is true for a c-expansive continuous surjection. This is checked in the way as follows. Let 3 = { Fi : 1 5 i 5 L } be a finite closed cover of X such that each Fi has sufficiently small diameter. For n > 0 we define FCj = fn(F;)n f-"(Fj) for 1 5 i,j I 1. Then 3" = {FC,} is a finite closed cover of X. Use this cover 3" in the proof of Theorem 2.2.40. Then we can show that the dimension of X is finite.

CHAPTER 2

70

Theorem 2.2.44 (Maii6 [Mall). If f:X + X is an ezpansiue homeomorphism of a compact metric space, then every minimal set for f is zerodimensional. For the proof we must investigate the structure of orbits for expansive homeomorphisms. We first prepare the following

Lemma 2.2.45. Let X be a compact connected metric space and A a proper closed subset of X . If x E A and C is the connected component of x in A, then C n BA # 0 where BA denotes the boundary of A. Proof, Let C be the collection of all open and closed subsets K under the relative topology of A such that C C K C A. Since X is compact, we have C = n { K : K E C } . To obtain the conclusion suppose C n 8A = 0. Then U{X\K : K E C} 3 BA. Since OA is compact and X\K is open and closed, BA is covered by the finite union of X\Ki. Since Ki = K is open and closed, we have K E C and K n t3A = 0. Notice that K is expressed as K = U n A for some open subset U of X because K is open in A. Since A = BA U int(A), we have K = U n A = U n (BA U int(A)) = U n int(A). Thus K is open in X. Since K is closed in X , we have X = K and so BA = 0. Therefore, A = X which is a contradiction. 0

ni

Let 7

>

0. For z E X let C , ( x ) be the connected component of z in : d ( z , y ) 5 7},and write BB,(x) = {y E X : d ( z , y ) = 7). We prepare some lemmas that are used in proving Theorem 2.2.44.

B,(z)= {y E X

Lemma 2.2.46. If, for some x E X and some m Wa(z,d ) # 0, then X contains a periodic point.

> O,fm(Wa(x,d)) n

Proof.Take y E W " ( x , d ) nfm(Ws(x,d)) and put z = f-m(y). Then fm(z) E W"(z,d)= W " ( z , d ) . Thus, lim,,,d(f" o fm(z), fn(z)) = 0. Since X is compact, a subsequence { fnj(z)} converges to w . Therefore, d ( w ,fm(w)) =

limn-,a, d(fjn o fm(z), f i n ( % ) ) = 0.

Define Ci(z) and X:(x) as the connected components of x in W,"(z,d)n and W:(z, d) fl &(Z) respectively. Let e > 0 be an expansive constant for f . We fix 0 < E < e / 2 . B6(Z)

Lemma 2.2.47. If dim(X) > 0, then there exists 0 < 7 < 0 < 6 < 7 there is a E X such that

Xi(.)

E

such that for

n a & ( U ) # 0 or C;(a) n BBa(a) # 0.

Proof. Let C , ( x ) denote the connected component of x in B,(x). Since dim(X) > 0 , X is not totally disconnected. Thus there exist x E X and 7 > 0 such that W x c )n BB,(x:)# 0 by Lemma 2.2.45, and so &(Z) n a&(z) # 0 for 0 < 6 < 7.

$2.2 Expansivity

Suppose there is 0 < 6

< -y

71

such that

x!(Y> n aB6(y> = 0 for all y E X. Then we prove that for some a E X

X:(a) n 8Ba(a)# 0.

Figure 6

To find such an a E X we construct a collection of compact connected sets A,, n 2 0, and a sequence of points xn E A, such that for a certain subsequence {m,} the following conditions hold :

If we establish the above conditions, then the conclusion is obtained a8 follows. Take a subsequence {zni}of {z,} such that tni + a as i --t 00. We fix {zni} and define

It is clear that the point a is contained in A. We show that A is connected. Indeed, suppose A is not connected. Then there exist nonempty closed sets 4, FZsuch that F1 nFZ = 0 and A = 4 LJFz. Since { z , ~ , ~is} a subsequence of the sequence {xni}, we have + a it9 j + 00. The point a belongs to F1 or F2. Without loss of generality we

CHAPTER 2

72

suppose the former, i.e. a E F1. For b E F2 let {b,i,j) be a sequence such that bni,j E Anivj and b,i,j -+ b as j -+ 00. Since F1 n F2 = 0,there exist neighborhoods U ( F i ) of Fi such that U(F1)n U(F2) = 0. But each hni,jis connected. Thus we can find yni,j E Ani,j satisfying Yni,j 6 U(F1)u U(F2).

If yni,j 4 y as j + 00, then we have y E A, but y By (iv) we have

6 F1 UF2, thus contradicting.

Since z, = fmn(z,+l)by (i), from (iv)

from which

“nij

b,.BJ.

Figure 7

By induction on n we have f’(An)CBe(f’(zn)),

If n

-+ 00,

then fj(A)

05j
c B,(fj(a)) for 0 5 j < 00,

and thus

A c W,d(a,d).

Since zni,j+ a, there is yni,jE A,i,j n a B a ( ~ , , satisfying ~,~) 5 = !im d(Yni,j, zniqj)= d(y, a ) . 3 --)‘=a

52.2 Expansivity

73

Therefore, A n aBa(a) # 8. Since A C W i ( a , d ) and A is connected, we have E $ ( a )n a&(.) # 0. The remainder of the proof is only to construct the sequence {A,} of closed connected subsets satisfying (i), (ii), (iii) and (iv). Let x and 0 < 6 < E be as above. Put A0 = E:a(Z), then A0 n & 3 6 ( 2 ) # 8. Notice that E:(y) n 0&(y) = 8 for all y E X. To obtain the conclusion suppose Ao, 111,. -.,A,-1 are constructed. Then we have An-1

P WZ(zn-1, d).

Indeed, if A,-1 c W;(x,-l,d), then E;(zn-l) 3 An-l since E:(zn-l) connected component of z,-~ in &(&-1) n W,"(z,-,,d). Thus E;(Zn-l)

n BB6(%-1)

thus contradicting. Therefore, An-l\W,"(Zn-l,d) some m > 0 we have

n oB6(%-1) # 8,

# 8, from which choose a point y. Then, for

s u ~ { d ( f - ~ ( z )f-"(zn-l)) , (v)

3 An-1

is the

: E E &-I}

2 d ( f - m ( y ) , f-m(%-l)) > E , : E E An-l,O 5 j < m } < E . ~~p(d(f-'(z),f-'(x,-1))

For this m we write m,-l = m and z, = f-mn-l(x,-l). Denote as An the connected component of x, in B6(3&,)n f-"'n-I(A,-l). Then (ii) holds, and A, n aB6(%,) # 8 implies (iii). By (v) we have (iv). 0

Lemma 2.2.48. For 0 < 6 < E there exists N > 0 such that for all x E X and y E W,"(z,d)with d(y,x) = 6, there is 0 < n 5 N such that 4f-n(y),

f-"(z)) > E -

Proof. If this is false, then there exist sequences x, E X,31, E W,"(Zn,d ) such that d(x,, y,) = 6 and d(f-j(x,), f-j(y,)) 5 E for 0 5 j 5 n. If x, + x and yn + y as n + 00, then x # y and d(fn(z),fn(y)) 5 E for all n E Z, thus contradicting. 0 Lemma 2.2.49. There exists 60 > 0 such that

w:(z,~)n B 6 ( 4 = w;J~,d)n &(z) for all

x E X and 0 < 6 < 60.

Proof.If this is false, for n > 0 then there exist x, E X and 0 such that W;(Zn,d) n B6,(%) # W&(%,d) n B6,(%)-

< 6, < 1/n

CHAPTER 2

74

Thus we can find yn E W&(z,,d)and m, d(Zn,Yn)

+

0 as n

-+

00,

d(fmn(zn),fmn(yn))

Since yn E W&(z,,d),for all m with m

4f" frnn(z*), f" 0

> 0 satisfying > ~ , >n 0.

> -m, 0

frnn(Yn))

from which

I

z

d(frn(z),frn(y))5 2&, m E where fmn(zn) + z and fmn(yn) + y as n --t 00. Therefore, z = y, but d ( z , y ) 2 E . This is a contradiction. 0

Figure 8

Lemma 2.2.50. Let Z = inf{d(z,y) : d(f-l(z),f-'(y)) > &,z,y E x} and let 6, > 0 be as an Lemma 2.2.49. For 0 < 6 < min{60,Z/2} there exists N = N ( 6 ) > 0 such that if z E X and A c Wi(z, d ) i s a compact connected set containing x and if A n Ba(z) # 0, then there exist 0 < m < N , a,P E f-"(a) and compact connected sets Aa,Ap such that

(4 (b) (c)

(4

P E Ap, ha n OB6(a) # 0, a E Aa,

a E W&(P,d),

n aB6(P)# 0,

inf{d(z,w) : z E Ba(a),w E B @ ) } > 6, ha c W2"c(a, d ) n B6(a), c W;c(P,d ) n B6(P)*

Proof. Let N = N ( 6 ) > 0 be as in Lemma 2.2.48. Since A f l d B & ( z )# 0 and z E A C W:(z, d ) by assumption, for y E W : ( x , d ) with d(z, y) = 6 we can find 0 5 m 5 N - 1 such that (el

(f)

s~p{d(f-~~+'~(z),f-(~+')(z)): z E A}

> - d(f-("+')(y), f-(m+"(z)) > E , sup{d(f-j(z),f-j(z)) : z E A,O I j 5 m } 5 E.

f 2.2 Expansivity

75

By (f) and the fact that A C W:(z,d)

(4

f-"(A)

Since f-"(z) E f-"(A),

we have

(h)

c W,"(f-"(z),

f-"(z) E W;(w,d)

4.

for w E f-"(A).

Figure 9 From (g) and (h) it follows that for all w E f-"(A)

f-v) c W&(w,4.

(i)

By (e), d(f-l o f-"(z), f-" (z))2 z.

f-'

o

f-"(x)) > E for some z E A, and so ~ I ( f - ~ ( z ) ,

Figure 10 Since z , x E A, we have diam(f-"(A)) 2 Z. Since 36 < Z, we can find a,@ E /-"(A) such that d ( a , @ )> 36. Obviously a 6 B6(p) and p pl &(a). Therefore, (c) holds. Let A,,Ap be the connected components of a and p in f-"(A) n &(a), f-"(A) n B,s(@)respectively. From (g) it follows that (a) holds. Since

CHAPTER 2

76

A,

Ap

C f-"(A)

and Ap C f-m(A),

by (i) we have A, C W,d,(a,d)and

c W.&(p,d).Thus (d) holds. It remains to check (b).

This is followed by replacing X , A and C with f--(A), f-m(A) n &(a) and A, respectively (in fact, since fqm(A) nBa(a) is a closed ball with radius 6 of a,we have i3(f-"(A) n &(a)) = {z E f-"(A) : d ( z , a ) = a} ). 0

Proof of Theorem 2.2.44. It is enough t o verify that if f: X -+ X is minimal and expansive then dim(X) = 0. Suppose dim(X) > 0 and 7 > 0 are as in Lemma 2.2.47. For 0 < 6 < E choose N = N ( 6 ) > 0 as in Lemma 2.2.48 and let 60 > 0 be as in Lemma 2.2.49. Take Z > 0 as in Lemma 2.2.50, and for 0 < 6 < min(bo,F/2,7} define

If 71 = 0, then X is a set consisting of finite points. Indeed, for all n 2 0 there exist zn,yn E X and 0 5 in, j, 5 N ( 6 ) such that

We can suppose that n 2 0. Thus

2,

-+

z,y,

4

y as n

-+

00

and j , = j,i, = i for all

Notice that i # j. Suppose i - j > 0, then we have fj-;(z) = y E W&(z, d ) C Wa(z,d) and f + - i ( ~ a ( ~ , d ) ) n ~ ~ (d )3# 0 , since f j - i ( z ) E fj-;(Wa(z,d)). By Lemma 2.2.46, X contains a periodic point zo and thus the orbit Of(z0) equals X (since f : X + X is minimal). It remains to check the case 71 > 0. For this case we shall derive a contradiction by showing the existence of a nonempty open set U and a point p such that f n ( p ) # U for all n 2 0. If we have U and p as above, then B n U = 0 and f ( B ) c B where B is the closure of the semi-orbit {p, f ( p ) , .}. Since A = f n ( B )is nonempty and f ( A ) = A , we have X = A , thus contradicting. We first construct a collection of compact connected sets A,, a sequence of points 2, E A, and a nonempty openset U c X that satisfy the following

nz=o

-.

52.2 Expansivity Since 6 define

77

< 7,there is u E X with X$(u)fl 6 & ( U ) # 0 by Lemma 2.2.47. We 20

= U,

A0

= E;(Q)

and show the existence of U # 0 with diam(U) < 71/2 such that U n A. # 0. Suppose U n A0 # 8 for all open sets U. Then A,, is dense in X and so A0 = X (since A0 is closed). Since diam(A0) 5 26 < e and f is expansive, X consists of one point set, thus contradicting dim(X) > 0.

/

Figure 11

If mo = 0, then (a), (b) and (d) hold for Ao. To construct A1 satisfying (c) we use Lemma 2.2.50. In fact, since A0 n 6&(xO) # 0 and Ao C W:(zo,d), by Lemma 2.2.50 we can find 0 < ml < N ( 6 ) , a , P E f - m l ( A ~ ) and compact connected sets A,, Ap satisfying (a), (b), (c) and (d) of Lemma 2.2.50. Then

Indeed, if y E Un{U,”o fj(Ap)}, then y E fj(Ap) for some 0 5 j 5 m l . Thus there is a E A p such that fj(a) = y. By assumption of (e), for z E U n fj(A,) there is w E A, such that f j ( w ) = x. Since x, y E U, (f)

d(fj(w>,P(4)= d(x,y)

and since w E A, and

t E Ap

d(w, z )

>6

< 71/2

(by (c) and (d) of Lemma 2.2.50).

By Lemma 2.2.50 (d), w E W i ( a ,d) and by Lemma 2.2.50 (a), a E W&(P,d ) . Thus, w E W & ( a , d ) . However d(fi(w),fj(z)) = d ( z , y ) 1 71 by definition, thus contradicting (f).

CHAPTER 2

78

Since p E f - m l ( A ~ ) n Ba(/3),we denote as A1 the connected component of /3 in f-ml(Ao) n B @ ) and put z1 = 0. Then we have (a') A1 n 8B6(zl) # 0 (by Lemma 2.2.50) , (by Lemma 2.2.50 (d)) , (b') A1 c Wc(zl,4

(4

Ai C f-"'(Ao),

(d')

{Uf j ( A 1 ) l n u = 0

ml

j=O

(by (el)

Replace A0 by A1 and repeat the above technique. Then there exist a compact connected set A2 and m2 > 0 satisfying (a'), (b'), (c') and (d'). Continuing inductively this technique, we have {A,},{z,} and U satisfying (a), (b), (c) and (d). By (c) we have

(Al) 3 ... 3 fml+***+mm(A,) 3 ..* and thus fml+".+mn (A,) # 0, from which take a point p. Obviously f-ml(p) E A, By (d) we have f-ml+j(p) 4 U for 0 5 j 5 ml and thus

n,

A,, 3

fml

f - ' ( p ) $2u, 0 5 i 5 m1.

Since f - ( m i + m a ) (PI E

A29

by

(4

f - ' ( p ) $ U, o 5 i 5 ml and consequently f - ' ( p ) $ U for all i 2 0. 0

+ m2

Let f: X --t X be a continuous surjection of a compact metric space. The map f is said to be minimal if, for all z E X ,the orbit O f ( z )= {fn(z) :n 2 0) is dense in X. A closed subset E is said to be a minimal set if f ( E ) = E and f p : E --t E is minimal. Remark 2.2.51. It remains a question of whether Theorem 2.2.44 is true for a c-expansive continuous surjection. We remark that the question can not be shown by the technique of the proof of Theorem 2.2.44, nor by the way through the inverse limit as in Theorem 2.2.39 (because the map is not a local homeomorphism). 82.3 Pseudo orbit tracing property

A sequence of points {zi : a < i < b} of a metric space X is called a Spseudo orbit off if d(f(zi), .;+I) < 6 for i E ( a , b - 1). Given E > O a S-pseudo orbit { z i } is said to be &-traced by a point z E X if d ( f i ( z ) , z i )< E for every i E (a,b). Here the symbols a and b are taken as -00 5 a < b 5 00 if f is bijective and as 0 5 a < b 5 00 if f is not bijective. We say that f has the pseudo orbit tracing property (abbrev. POTP) if for every E > 0 there is S > 0 such that every 6-pseudo orbit of f can be &-tracedby some point of X. For compact spaces this property is independent of the compatible metrics used.

$2.3 Pseudo orbit tracing property

79

Remark 2.3.1. In the case where X is compact, a homeomorphism f : X -+ X has P O T P if for every E > 0 there is 6 > 0 such that every (one sided) 6-pseudo orbit {zi :i 2 0) o f f can be E-traced by some point of X. Indeed, let {zi : i E Z} be a &pseudo orbit of f . For each n 2 0 define a (one sided) &pseudo orbit { z r : i 2 0) by zr = zi-n for all i 2 0, and let zn be a tracing point for {z? : i 2 0). Then it is easily checked that an accumulation point of { f n(zn)} is a tracing point for {zi : i E Z}. Theorem 2.3.2. Let X be a compact metric space and denote as id the identity map of X. Then id : X -+ X has P O T P if and only if X is totally disconnected. Proof. If X is totally disconnected, for E > 0 there exists a finite open cover {Uo,...,Un} of X such that Ui n Uj = 0 for i # j and diam(Ui) < e for a. Choose 6 with 0 < 6 < min{d(Ui,Uj) : i # j} and fix a &pseudo orbit { z k : k E Z} for id. Then there is Ui such that {zk} c Ui and so we can find in Ui an €-tracing point for {zk}. Therefore id : X --f X has POTP. Conversely, suppose dim(X) # 0. Then there is a closed connected subset F such that diam(F) > 0. Since F is compact, diam(F) = d(z0, yo) = €0 for some z0,yo E F. Let ~1 = E 0 / 3 . Since F is connected, for any 6 > 0 we can construct a &pseudo orbit from xo to yo in F which is not €1-traced in X. This is a contradiction. 0

Theorem 2.3.3. Let f : X + X be a continuous map of a compact metric space and let It > 0 be an integer. Then f has P OT P if and only if so does

fk.

Proof. We first notice the following (l),(2) and (3). (1) Let E > 0. Then there is E > el > 0 such that each €1-finite pseudo orbit {zi : 0 5 i 5 12) satisfies d( f i ( ~ o ) zi) , < ~ / 2 , 0 5 i 5 It and d(z, y)

< €1

implies max{d(fi(z), fi(y)) : 0

I i I k} < ~ / 2 .

(2) Let €1 be as in (1). Then there is 61 > 0 such that each &-pseudo orbit f k is El-traced by some point. (3) Let €1 and 61 be as in (1) and (2). Then there is 6 > 0 such that each &finite pseudo orbit { z i : 0 5 i I 12) is 61-traced by zo E X. With these properties we show that each &pseudo orbit {yi : i 2 0) for f is E-traced by some point. Write

for

zi = y k i , For fixed i, {yki+j : 0 5 j

I 12)

i 2 0.

is a &finite pseudo orbit for f . By (3) we have

d(f'(yki), Yki+j)

< 61, 0 I j 5 12.

ao

CHAPTER 2

I f j = k then d ( f k ( y k i ) , y k i + k ) = d ( f k ( z i ) , z i + l ) < 61. Thus { Z i } is a 61pseudo orbit for f k . By (2) there is y E X such that d ( f k i ( y ) , z i ) < c1 for i 2 0. Thus, d ( f k i ( y ) , y k i ) < E I for i 2 0. On the other hand, since (yki+j : 0 5 j 5 k} is an El-pseudo orbit for f, by (1) we have

Since i is arbitrary, we have d( fn(y), yn) 6-pseudo orbit {yi}. 0

< E for

n >_ 0 and y &-tracesthe

Theorem 2.3.4. Let X be a compact metric space. If f :X omorphism with POTP, then so is f - l .

4

X i s a home-

Proof. For every E > 0 let 6 > 0 be a number such that each 6-pseudo orbit { X i } is &-tracedby a point y E X. Choose 6' > 0 such that d ( z , y) < 6' implies d ( f ( z ) ,f(y)) < 6 and put g = f - ' . It is enough to see that each 6'-pseudo orbit { p i } for g is &-tracedby some point. Since d ( g ( y i ) , y i + l ) < 6' for i E Z, we have d ( y j , f ( y i + l ) ) < 6 for i E Z. Letting zi = y-i for i E Z , { z i } is a 6-pseudo orbit for f and thus there is y E X with d ( f i ( y ) , z j ) < E for i E Z. This implies d ( f - ; ( y ) , z - i ) < E for i E Z and therefore d ( g i ( y ) , y i ) < E for i E Z. 0

Theorem 2.3.5. Let X and Y be metric spaces and X x Y the product topological space with metric

where dl and dz are metrics for X and Y respectively. Let f : X g : Y + Y be continuous maps and let f x g be the map defined by

+

X and

Then f x g has POTP if and only if both f and g have POTP. Proof. Suppose f x g has POTP. For every E > 0 let {zi} and {yi} be 6-pseudo orbit for f and g respectively. Then {(z;,yi)} is a 6-pseudo orbit for f x g . Thus there is (2, y) E X x Y with d ( ( f x g ) i ( z , y), ( z i ,yi)) < E for i 2 0. Then, d l ( f ( z ) , z i ) < E and d 2 ( g i ( y ) , y i ) < E for i 2 0. Also the converse is proved. 0

$2.3 Pseudo orbit tracing property

81

Theorem 2.3.6. Let f : X -, X be a continuous map of a compact metric space X and let h : X -+ Y be a homeomorphism. Then g = h o f o h-' has POTP if and only i f so does f . Proof. For every E > 0 there is 61 > 0 such that d(z,y) < ~1 implies d'(h(z), h ( y ) )< E where d' is a metric for Y . If f has POTP, then there is b1 > 0 such that each 61-pseudo orbit { x i } for f is sl-traced by some point. Let 6 > 0 be a number such that d'(z,y) < 6 implies d(h-l(z), h-l(y)) < 6,. Then it is enough to see that each 6-pseudo orbit {yi} for g is &-traced by some point. To do this put xi = h-'(yi) for i 2 0. Since d'(g(yi),yi+l)< 6 for i 2 0, we have d(f(zi),zi+l) = d(h-' 0 g(yi),h-l (yi+l))< 61 for i 2. 0. Thus { z i }is a 61-pseudo orbit for f and so d ( f i ( y ) , z i ) < e l ( i 2. 0 ) for some y E X. Therefore, d'(h o f i ( y ) , h(zi))= d'(gi o h ( y ) ,y i ) < E for i 2. 0. 0

Theorem 2.3.7. Let X be a compact metric space. A continuous surjection f:X+ X has POTP if and only if for every E > 0 there is 6 > 0 such that for ( z i ) E Xz with d(zi,zi+1)< 6 (i E Z) there exists a point ( y i ) E X f SO that d ( y i , z i ) < E for i E Z. Proof. This is easily checked as in Remark 2.3.1. 0 By Theorem 2.3.7 we can give the definitions of &pseudo orbit and &-tracing point for continuous surjections as follows. If ( z i ) E Xz has the property that d(f(zi),zi+l)< 6 for i E Z, then ( z i ) is called a &pseudo orbit of f . If (yi) E Xf satisfies d ( y i , z i ) < 8 (i E Z), then ( y i ) is called an s-tracing point for (zi).

Theorem 2.3.8. Let X be a compact metric space. If a continuous sujection f : X + X has POTP then u :Xf + Xf obeys POTP. Proof. Let a = diam(X) and E > 0. Choose N > 0 with c ~ / 2 ~<-E,~ and let 7 > 0 be a number such that

By POTP o f f there is 6' > 0 such that any 6'-pseudo orbit o f f is ytraced. Choose 6 > 0 with 0 < 2N6 < 6'. Let k > 0 and suppose {(zy): 0 5 n 5 k} is a finite 6-pseudo orbit of u in Xf.Then we have

where i((zi),(pi)) = CZ-, d(zi,y;)/21il, and hence {"IN : 0 5 n 5 k} is a 6'-pseudo orbit of f , from which we can find y E X such that d ( f "(y), z 2 N ) 5 7 (0

5 n 5 k)*

CHAPTER 2

82

+

3 ~ / 8 €14

+ €14 < E

becauseof d ( f n ( y i ) , x ? ) 5 ~ / 8 ( l i 5 l N) by the fact that d ( f n ( y - N ) , z E N ) 5 7 . 0

Theorem 2.3.9. If u : Xf + Xf has POTP and f:X homeomorphism, then f has POTP.

--+

X is a local

Proof. For 8 > 0 let 6 > 0 be a number such that every &pseudo orbit of u is E-traced by some point of Xf.Let a = diam(X) and choose N such that 0 <~ r / 2 < ~ 6. - ~Then we can find 7 > 0 such that if d ( z , y ) 5 7 then there are {zi : -N 5 i 5 N} and (9; : - N 5 i 5 N} such that 20 = Z, f ( z i )= zi+l and yo = y, f ( y i ) = yi+l and d(zi,yi) < 618 for lil 5 N. Let { z i :0 5 i < XI} be a 7-pseudo orbit of f. Then for i 2 0 we can find ( z i ) E Xf such that zi = zi and { ( z i , ) } is a &pseudo orbit of u. Hence there is (2,) E Xf such that i ( u i ( z n ) (zk)) , 5 E for 0 5 i < 00, and then E

2 J(ui(zn),(zf)) 2 d ( f i ( z o > , zi) = d ( f i ( z o ) , zi).o

Theorem 2.3.10. Let X be a compact metric space. A positively expansive map f: X + X has POTP i f and only if f is an open map ( i e . expanding). Proof. I f f is expanding, then f is a local homeomorphism and X has a hyperbolic metric (Theorem 2.2.10). Thus, by using the technique of Theorem 1.2.1 (2) we see that the map has POTP. Conversely, suppose f has POTP. Let U be an open set of X and for x E U choose E > 0 such that the €-neighborhood, Uc(z),is contained in U. Then there is 0 < 6 5 e such that every 6-pseudo orbit of f is E-traced, where e is an expansive constant for f. If z E Ua(f(z)), then a sequence { z , z , f ( z ) , f 2 ( z ) , . . . ,f i ( z ) , * - *is} a &pseudo orbit o f f , and so it is €-traced by some point y E X. Then d ( f ' ( f ( y ) ) , f ' ( z ) )< 6 5 e for all i 2 0, which we have f ( U , ( z ) ) 3 U a ( f ( z ) ) . Therefore implies f(y) = z. Since y E Uc(z), f ( U ) is open in X. 0

$2.3 Pseudo orbit tracing property

83

Theorem 2.3.11. Let X be a compact metric space. If X is connected and f :X --t X is an expanding map, then the set of periodic points of f is dense in X . Proof. Let D be a hyperbolic metric for f and let (60,8) be Eilenberg's constants for (X,f ) . Then there are 61 > 0 and X > 1 such that D(z,y) 61 (z,y E X) implies D(f (z),f (3)) 2 XD(z, y). Notice that 8 is chosen such n f-l(y) # 0 where that 8 < el. Then D(f(z),y) < 8 implies U6,lx(z) Ua(z)= { z E X : D ( e , z ) < a}. It is enough to see that for each v > 0, U v ( z ) contains a periodic point of f. Fix 0 < v < ~ 1 / 2 .Since f has POTP by Theorem 2.3.10, there is 6 > 0 such that every &pseudo orbit of f is v-traced by some point of X. Let { U1,..,Un} be a finite open cover of X such that the diameter of each Ui is less than 8. Then we can find T > 0 such that (n 1)8/Xr-' < 5. Take and fix zo E X. Since X is connected, there is a finite sequence

<

.

+

{Yo = ZO,Ylr...,Y&,Y&+l = f'(.O)} such that yj, yj+l E Uij for 0 5 j 5 k and k 5 n. Find :y E X with properties that f(y:) = yk and D(y:,f'-'(zo)) < 8/X. Since D(yk-1, f (yk)) < 8, there is y:-l E X such that f (&) = yk-1 and D(y:_,, ):y < 8/A. Continuing this fashion, we obtain a finite sequence 1

{YA = ~l,Yl,..*,Y:,Y:+l

= fr-YzO)}

such that f ( y i ) = yj-1 and D(y&l,yj) < 8/X for 0 5 j 5 12. Since d(yk, f'-'(zo)) < 8/X, we have y i E X such that f(yi) = yk and D ( y i , f r - 2 ( z ~ )<) 8/X2. Since D(y:-,, f(yi)) < 8/X, we have also yi-l E X with f ( ~ i -=~ ) and D(&, ,y;) < 8/X2. In this manner, a finite sequence 2 2 { d = za,YI,..*,Y&,Y&+1 = f'-2(zo)}

is constructed and by induction on 1 a finite sequence

e e { d = Ze, Y:, .-.,Y&, ~ k + l= fr-e(zo)) exists. Then we have D(zr-l,f(zo)) I~ ( Y ~ - ' , Y ~ - ~ ) 5 n8/Xr-' < 6

+ . . a

+~($-',f(zo))

and hence (20, ~'-1,. ..,z1,20, %,.-I,. ..} is a periodic 5-pseudo orbit of f. Since f has POTP, there is a v-tracing point p E U,(zo)for this 6-pseudo orbit. Clearly f r ( p ) is also a v-tracing point. Thus d( f'(p), fi+'(p)) < 2v < 61 for all i 2 0. Since f is positively expansive (with expansive constant & I ) , we have p = f'(p). 0

CHAPTER 2

a4

Theorem 2.3.12. Let X be a compact metric space and denote the product space by Xz = where each Xi i s a copy of X . A metric d for Xz i s defined by

nraX;

Then the shift map u : Xz

+ Xz

has POTP.

Proof. For every E > 0 choose 6 with 0 &pseudo orbit for u. Then, for i E Z

for all k E Z, and so d(zb+, ,zL+')

< 26 <

E.

< 21'16 (i, k E Z).

Let {x' : i E Z} be a

Define a point z by

z = (... ,x;1,x;,z;,...). for i, k E Z. When k 2 0, we have

Then x E Xz and (ui(z))k = x;" k--1

and when k < 0 the inequality d(z:,x;++") < 2Ik1+l6 is also calculated. Therefore, d(zi,ui(z)) 26 < E for i E Z, i.e. the point x is an &-tracing point. 0

<

Let X be a compact metric space with metric d. A product space Xz = { (zi) : xi E X, i E Z} h a metrics

for z = ( z i ) , = ~ (yi) E X z . The reader can check that these metric generate the same topology of Xz. Theorem 2.3.13. Under the notations of Theorem 2.3.12, if topological dimension of X is non-zero, then u : Xz + Xz is not expansive. Proof. This follows from facts that dim(Xz) < 00 whenever u is expansive (In fact, since dim(X) > 0, we have dim(Xz) = &dim(Xi) = 00, [H-W]).0

52.3 Pseudo orbit tracing property

85

Theorem 2.3.14. Under the notations and the assumptions of Lemma 2.2.34, g : X -+ X has POTP if and only if f : -+ has POTP.

x x

f:x x

Proof. That -, is uniformly continuous follows from the proof of Lemma 2.2.34. We first show that if -+ has POTP then so does g : x + x. Let 60 > 0 be as in Lemma 2.2.34. For E > 0 there is 0 < 6 < 6o such that Let { y i } be any each 6-pseudo orbit { p i } for f is E-traced by some point of 6-pseudo orbit of g . Fix k > 0 and choose p - k E such that ?T(p-k) = y-k and fix p 4 . Inductively we choose pi E such that r ( p i ) = yi for i > - -12. Since f ( P i ) E we have a 0 f ( p i ) = g 0 x ( p i ) = g ( y i ) - Since yj+l E U a ( g ( y i ) ) , there is a unique pi+l E U h ( f ( p i ) ) such that r ( p i + l ) = yi+l. By the choice of { P i : i 2 - k } , { p i } is a &pseudo orbit for f and thus there is an €-tracing Let r ( y k ) = z k . Then, for point y k of { p i } , i.e. z ( f i ( y k ) , p i ) 5 E for i 2 4.

f:x x

x

x,

x

x.

i20

d(gi(zk)7y i ) = d(gi 0 r ( y k ) ,Y i ) = d ( r 0 f i ( v k > r, ( p i ) )

-

= d ( r i ( Y k ) , p i )I E.

If k

+

00, then { Y i

{ z k J } such that

: i E Z} is a 6-pseudo orbit and there is a subsequence z as j -+ 00. Thus d ( g n ( z k j ) , y n ) 5 E for n 2 -k, and for all n E Z. Therefore g has POTP.

zkj -+

so d ( g n ( z ) , y , ) 5 E To show the converse, let 60 > 0 be as above. Then, for E > 0 we can find 0 < 6 < 60/2 such that each 6-pseudo orbit for g is E-traced by some point of X . Since f : -+ is uniformly continuous, we may suppose that &I, q ) < E implies (P),f (4)< bop. Let { p i } be a 6-pseudo orbit for f and let yi = % ( p i ) for i. Obviously { y i } is a 6-pseudo orbit for g . Indeed, since f ( p i ) , p i + l ) < 6 for all i,

af

x x

z(

d ( g ( Y i ) , y i + l ) = 4.0

f ( p i ) , r ( p i + l ) )= a ( f ( p i ) , p i + l ) < 6.

Thus there is z E X such that d ( g i ( z ) , y i ) < E for all i. Since d(z,yo) there is q E U6(pO) such that r ( q ) = z. Note that

d(.

0

< E,

f i ( q ) ,.(pi>> = d(gi 0 r(q),y i ) = d(gi(z>7y i ) < E

for all i. Since r is locally isometric, we have z ( q , p o ) < E when i = 0. If -d ( f - ' ( q ) , p i - l ) < E for i 2 1, then we have z ( f i ( q ) , p i ) < E . Indeed, since d ( f o f i - ' ( q ) 7 f ( ~ i - 1 ) )< 60/2 and a ( f ( p i - l ) i p i ) < 6,we haveZ(fi(q),pi) < 60 and -

d ( f ' ( q ) , p i ) = d ( r 0 f ' ( q ) , r ( p i ) ) = d ( g i ( z ) ,y i ) < E Thus the point q E-traces the one sided 6-pseudo orbit { p i : i 2 0). Similarly we can show that a ( f i ( q ) , p i ) < E for i I 0. Therefore the conclusion is obtained. 0

CHAPTER 2

86

Theorem 2.3.15. Let f be a linear map of the euclidean space R" and let d be the euclidean metric for W". Then f has POTP under d if and only iff is hyperbolic, i.e. it has no eigenvalues of modulus 1. Proof. If f is hyperbolic, then R" splits into the direct sum of f-invariant subspaces, W" = Es@E", such that f: Es + E d is contracting and f : EU -+ E" is expanding. See the proof of Lemma 2.2.33. Thus, in the same manner as the proof of Theorem 1.2.1 (2) we have that both flEa and fpUhave POTP. Since f is linearly conjugate to f l E a x flEY, by Theorem 2.3.5 it follows that f has POTP. Conversely, suppose f has an eigenvalue with absolute value one. Then there is an f -invariant subspace F such that f 1~ : F --f F is an isometry under d. Let E > 0. Since f is a linear map, by using the Jordan form of f (in the real field) we can choose K = K ( E )> 0 such that llp"(x)11 -+ 00 as i -+ 00 whenever 1 1 ~ 1 1< E and I(fi(x)ll > K for some i 2 0. Since f p is an isometry, for any 6 > 0 we can construct in F an &pseudo orbit which comes from the origin and goes to a set of points x with 11x11 = 2K, which implies that f does not have POTP. 0 Theorem 2.3.10. A toral endomorphism has POTP if and only if it i s hyperbolic. Proof. This is obtained from Theorems 2.3.14 and 2.3.15. 0 Let k be a natural number and Yk = { l , . . .,k}. A metric d for Yf is defined by d ( x ,y ) = 2-"' if m is the largest natural number with xn = yn for all In\< m, and d ( x , y ) = 1 if xo # yo. Such a metric is uniformly equivalent to the metric d' defined as above. If S is a closed subset of Yf with u(S) = S, then u p : S -+ S is called a subshift. We sometimes write u p as u : S -+ S. A subshift u : S -+ S is said to Yk be of finite type if there exist some natural number N and a subset C C with the property that z = (zi)E S if and only if each block ( x i , - - -, z ; + N ) in z of length N 1is one of the prescribed blocks. The smallest such natural number N is called the order of the subshift of finite type.

n,"

+

Theorem 2.3.17. Every subshift of finite type is topologically conjugate to one of order 1. Proof. We just take a new symbolic space Y z consisting of the allowable blocks of length N, and then the conjugacy cp : Yf + Y z is given by cp(.

.. , X - 1 , 2 0 , 5 1 , " ' )

=

(...,

(q,( 7 ),(?) XN-2

The shift 5 :Y z -+ Y z is of order 1 and

'p o

XN-1

XN

o = 5 o cp holds. 0

,... ) .

$2.3 Pseudo orbit tracing property

87

Let A be a k x 12 matrix of 0's and 1'8, called a transition matrix. We define the compact set Then XA is invariant under the shift map ( ~ ( X A = ) EA). Such a shift map u p A is called a Markov subshift. Every subshift of finite type with order 1 is a Markov subshift.

Theorem 2.3.18 (Walters [W3]). Let S be a closed subset of Yf and let u : S + S be the subshift. Then u : S + S has POTP if and only if u is a subshift of finite type. Proof. If u : S + S is a Markov subshift, obviously u is of order 1. Let > 0 and take m > 0 such that 2-" < E . Then xi = yi for lil 5 m when d ( z , y ) < 2-('"+l). Let { x i : i E Z} be a 2-("+l)-pseudo orbit for up, i.e.

E

{xi}

cS

d ( u ( z i ) , x i + + '<) 2-(m+')

and

for all i E Z. Then xf = z:+' for all i E Z. Now put x, = 30" for n E define z = (x,) E Y f . Then we have (xn,%,+I)

~ t +=' )(z:O",z;L),

= (x:O",

nE

Z and

Z,

and thus x E S. Since (uiz)j = xi+j = xf for ljl 5 m and i, we have

d(ui(x),x') 5 2-"

<&

for i E Z. Therefore x is the &-tracingpoint of { x i } . If u : S + S has POTP, then we show that u is a Markov subshift. Since u : Yf + Yt is expansive ( under d ), so is u p . Let E > 0 be an expansive constant for up and let 0 < 5 < &/2. Then there is N > 0 such that

5 N -1 I d ( z , y ) < 6. Denote by B the set of all blocks xi-^, - * ,x i , * * - , xi+^) in z with length 2N + 1 for all x E S and write xi

= yi, lil

S(B)={x~Y~:(zi-~,...,zi,...,xi+~ ~E ) €ZB} ,. Obviously, S ( B ) 3 B and u ( S ( B ) )= S ( B ) . It is easy to see that S ( B ) is closed and u ~ ( Bis) of finite. Since ( y i - ~ , ,y i + ~ E ) B for each y E S ( B ) and i E Z, for i E Zthere is xi E S such that yi+j = xf for Ijl 5 N. Thus, d ( u i ( y ) , z i ) < 6 for i E Z,i.e. { x i } is a &pseudo orbit for up. Since u p is POTP, ther is z E S such that d(ui(z),zi) < &/2 for all i E Z. Then we have

--

d(u'(z), u i ( y ) ) 5 d(u"z), xi)

+ d ( x i ,ui(y)) <

&

for all i E Z. Since z,y E S and u p is expansive, we have z = y and so y E S. Therefore, S ( B ) = B. 0

aa

CHAPTER 2

Remark 2.3.19. The family defined by

{fa : s

fd(x)

=

{ s(2 sx

E [0,1]) of tent maps

fd

: [0,2] + [0,2)

x ) (1 5 x 5 2)

has the following properties : (1) fd has POTP for almost all parameters (with the exception of Lebesgue measure zero of [O, 2]), (2) the set of parameters for which f d does not have POTP is locally uncountable. This is a result due to Coven-Kan-York [C-K-Y]. A set is said to be locally uncountable if its intersection with any open set is uncountable. We introduce the family of functions pn,n 2 1, on [&,2] by pn(s) = f,"(l).A point sE 21 is said to be a periodic point with period n if f,"(l)= 1 (p,(s) = 1) but f!(l) # 1 for 1 5 k < n. Show that periodic parameters are dense in 21 by using the mean value theorem. Then (1)will be obtained. (2) will be shown by applying the kneading theory and Theorem 2 of [D-G-PI. See [C-K-Y] for the details.

[a,

[a,

52.4 Topological Anosov maps (TA-maps)

Let f:X+ X be a continuous surjection of a compact metric space. Let

E

> 0. We set for x E X

as before (henceforth, as the metric d is fixed, the local stable set W:(t, d ) is written by W , " ( x ) )and for x = ( x i ) E Xf

W,u(x)= {yo E M : 3(&) E Xfs.t. d(Xi,Yi) 5 E , i 5 0). Lemma 2.4.1. Let f:X + X be a c-expansive continuous suljection with expansive constant e . For 7 > 0 there exists ny > 0 such that for x = ( x i ) E Xf and x E X (1) f"(W,.(m)) c W , " ( f " ( x ) )for n 2 ny, (2) if y = ( y i ) E Xf and d ( y i ,x i ) 5 e for i 5 0 ( i.e. yo E W,"(X) ), then d ( y - , , x - , ) 5 7 for n 2 ny ( i.e. W:(X) c fn(Wy(cr-nx)) for n 2 ny ).

Proof. If (1) is false, then there are x n , y n E X and m, 2 n such that yn E W,b(xn) and d ( f m n ( x n )fmn(y")) , 2 7 . Fix i > 0. If fmm-i(xn) + xLi and fmm-i(yn) + yLi as n + 00, then d ( x L i , y L i ) 2 e since d ( f j ( x n ) , fj(y*)) 5 e for 0 5 j < m,. We have d(xb,yh) 1 7 when i = 0. Thus 96 E W t ( x ' )

$2.4 Topological Anosov maps (TA-maps)

89

where x' = (2:) because fi(zLi) = zb and fi(yLi) = yb for i 2 0. Since y* E W:(xn),we have

d(fmn+j (2, 1, fmn+j(y")) 5 e ( j 2 -m,) and hence g; E Wl(zb). Since Wc(x')n W ; ( x & )3 yk, we have z& = y& by c-expansivity, thus contradicting. If (2) is false, then we can find ( x : ) , ( y l ) E Xf and m, 2 n such that - 7 and d(z?,yl) 5 e for i 5 0. From this we obtain a d(z:m,, ~-m, > contradiction. 0

For a continuous surjection f:X + X we define the stable and unstable sets

W " ( z )= {y E X : lim d(f"(z), fn(y)) = 0) n+oo

W u ( ( x i ) )= {yo E X : 3(yi) E Xfs.t. lim d(x-i,y-;) = 0) i-rw

for z E X and (xi) E Xf. Let f : X --t X and 9: Y --f Y be continuous surjections of compact metric spaces. Suppose h o f = g o h holds for some homeomorphism h: X 4 Y . Then the following are easily checked : (1) if x E X and W s ( z )is the stable set of f , then h ( W " ( z ) )is the stable set of g at h(x), (2) if (zi)is a point in Xf and Wu((zi)) is the unstable set of f , then ( h ( x i ) ) belongs to Ygand h ( W u ( ( z i ) ) )is the unstable set of g at (h(Xi)).

Remark 2.4.2.

Lemma 2.4.3. Let f: X t X be a c-ezpansive continuous suvjection with expansive constant e and let e > E > 0 and z E X . Then WYX) =

u f-i(w:(fi(x)))

i20

and for x = ( z i ) E Xf W y x )=

u

fi(w,u(o-i(x))).

i20

Proof. This follows from Lemma 2.4.1. Now we shall give Anosov property for continuous surjections of metric X spaces, in a general setting. We say that a continuous surjection f : X is a topological Anosov map (abbreviated TA-map ) if f is c-expansive and

90

CHAPTER 2

has POTP. If, in particular, f is a homeomorphism, then f is said to be a topological Anosov homeomorphism if it is expansive and has POTP. A TAmap f : X --+ X is special if f satisfies the property that W'((x;)) = W " ( ( y i ) ) for every ( x i ) , ( y i )E Xf with xo = yo. The notion of special TA-maps is a generalization of that of special Anosov differentiable maps. Let X and Y be metric spaces. We say that two continuous maps f:X X and g:Y --+ Y are topologically conjugate if there exists a homeomorphism h : Y --+ X such that f o h = h o g . In this case any orbit of g is mapped by h to a homeomorphic orbit of f . If h is a continuous surjection, then f is said to be topologically semi-conjugate to g , in other words, f is called a factor of 9. Let f : X -+ X and g : Y --+ Y be continuous surjection of compact metric spaces. If f and g are topologically conjugate, then (1) f is a TA-homeomorphism if and only if so is g, (2) f is a TA-map if and only if so is g , (3) f is a special TA-map if and only if so is g . Indeed, (1) and (2) follow from Theorem 2.3.6 together with Theorems 2.2.5 (3) and 2.2.30 (3). By Remark 2.4.2 (2) we obtain (3).

Remark 2.4.4.

In the remainder of this chapter we discuss the relation between TA-maps and the topological stability of homeomorphisms (or self-covering maps) on closed topological manifolds. Let X be a compact metric space with metric d. A homeomorphism f : X --+ X is said to be topologically stable in the class of homeomorphims if for E > 0 there is 6 > 0 such that for a homeomorphism g with d( f (x),g ( x ) ) < 6 for all x there is a continuous map h so that h o g = f o h and d ( h ( z ) , x )< E for all x. A self-covering map f is said to be topologically stable in the class of selfcovering maps i f f satisfies the conditions as above in the class of self-covering maps.

Theorem 2.4.5 (Walters [W3]). If a homeomorphism f:X -+ X of a compact metric space is topological Anosov, then f i s topologically stable in the class of homeomorphisms. Proof. Let e > 0 be an expansive constant for f and fix 0 < E < e / 3 . Let 0 < 6 < e / 3 be a number with the property of POTP. By expansivity it is checked that there is a unique x E X which &-traces a given &pseudo orbit {xi}. Indeed, let y E X be another €-tracing point of {xi}. Then we have

for all i E Z and thus x = y . Let g : X --+ X be a homeomorphism with d(g(x), f(x)) < 6 for all 2 E X. Let x E X. Since d( f o g"(x),g"+'(x)) < 6 for n, { g " ( x ) } is a &pseudo orbit of f. Thus there is a unique point h ( z ) E X whose f-orbit &-traces { g " ( z ) } .

$2.4 Topological Anosov maps (TA-maps)

91

This defines a map h : X + X with d(f" o h(x),gn(x)) < E for n and x E X. Letting n = 0, we have d(h(x), x ) < E for x E X. Since d(f" o h o g(x),g"+'(x)) < E for n E Z and

d(f"

0

f 0 h(x),g"+'(x)) = d(f"+'

0

h(x),g"+'(x))


for n E Z, we have h o g(x) = f o h(x) for x E X. Finally, we show that h is continuous. Let X > 0. Then we can choose N > 0 such that d(fn(x),fn(y)) < e for In1 5 N implies d(x, y) < A. This is checked as follows. Suppose this is false. Let a be an open cover of X with diameter < e. Then there is E > 0 such that for each j > 0 there are x,, y, E X with d ( q , yj) > E and Aj,i E a (-j 5 i 5 j) with xj, yj E f-j(Aj,i). Since X is compact, we can suppose that x j -+ x and yj -+ y. Thus x # y. Consider the sets Aj,o for j. Then infinitely many sets of them coincide since a is finite. Thus xj, yj E A0 for infinitely many j and so x, y E 20.Similarly, for infinitely many Aj," they coincide and we have A,, E a with 2, y E f-"(&). Therefore x, y E 00 thus contracting the fact that f is expansive. Choose 7 > 0 such that d(x,y) < 1implies d(g"(x),gn(y)) < e/3 for In1 5 N. If d(x, y) < 7 then

ni=-j

nn=-=f-=(a,,),

d(f" 0 h(x), f" 0 W Y ) ) = d(h 0 9"(x), h 0 9"(Y)> 5 d(h 0 9"(z), 9"(x>>+ d(g"(x), 9YY)) + 49"(Y), h 0 9YY)) 5 E + e/3 + E < e, In1 5 N . Therefore, d(x, y) 0

-m

< 7 implies d(h(x), h(y)) < X and continuity of h is proved.

-1

0

m-1

rn

Figure 12

Remark 2.4.6. Let h:X -+ X be as above and suppose that X is a closed topological manifold. Then h : X + X is surjective. For the proof the readers may refer to Remarks 6.7.10 and 10.6.4 which will be mentioned in Chapters 6 and 10 respectively. However, in general h is not surjective. We give an example due to Walters.

92

CHAPTER 2

Consider the shift u : %z -t Yt where Y2 = {O,l}. Then and has POTP. Let m > 0 and define g: yz" -t y2" by

xi zi+l

x-,

Q

is expansive

lil > m, -m 5 i < m, i=m

Then d ( g ( x ) , u ( x ) )5 1/2m where d is the metric for yz" given as before. If m is sufficiently large then we have h o u = g o h for a continuous map with d(h(z),z)< 1/2,(x E Y,").Since g2m+'(z)= x for all z E Yt,we have h ( z ) = h o gam+l(x) = a2,+l o h ( x ) for all x E y,". Therefore h is not surjective.

Theorem 2.4.7. Let M be a closed topological manifold and f: M + M a self-covering map, but not injective. If f is a TA-map and has topological stability in the class of self-covering maps, then f is positively expansive. For the proof we need some properties of TA-maps on closed toplogical manifolds. Thus we shall postpone the proof until Theorem 6.7.11 in $6.7 of Chapter 6. Let f: X + X be a homeomorphism of a compact metric space. A point p E X is nonwandaring for f if, for any neighborhood U of p and any integer no > 0, there exists n with In1 > no such that fn(U) n U # 8. The set n(f) of nonwandering points is closed and f-invariant (f(fl(f) = n(f>).The limit sets w ( q ) and a(q),for q E X,are contained in n(f). Such a set fl( f ) is called the nonwandering set. Let g: Y + Y be a homeomorphism of a compact metric space and h:X + Y a continuous surjection. If g o h = h o f, then we can easily prove that h ( n ( f ) )c f l ( g ) . If, in particular, h is bijective, then h ( f l ( f ) )= f l ( g ) .

Theorem 2.4.8. Let f: M -t M be a homeomorphism of a closed topological manifold. Iff is topologically stable in the class of homeomorphisms, then the set of all periodic points of f, Per( f), is dense in n(f ). P m f . Since f is topologically stable, for E > 0 there exists 6 > 0 satisfying the definition of topological stability. We may suppose that 6 < E . Take and fix x E fl( f ) , and choose a coordinate neighborhood U of x contained in u6/2(x) (where u6/2(x) = {y E M : d ( x , y ) < 6/2}). Notice that U is chosen such that it is connected. Since x E n(f),there is 12 > 0 such that f k ( U )n U # 0 and f ' ( U ) n U = 0 for 0 < i < k. Thus f'(x') E U for some x' E U. Since U is a connected coordinate neighborhood, we can construct a homeomorphism 9: M + M satisfying 9 0 f"x') = x', g ( U ) = u, g ( y ) = y for all y $ U .

52.4 Topological Anosov maps (TA-maps)

93

Clearly, since d ( g ( z ) ,z)< 6 for z E M, we define cp = go f . Then, d(cp(z),f (z)) < 6 for E M. Therefore we can find a continuous surjection h: M t M such that h o cp = f 0 h, d ( h ( z ) , z )< E (zE M ) .

It is easy to see that cpk(z’)= 2’. Since h o cpk = f k o h, we have h(z’) = f k ( h ( z ’ ) )and thus h(z’) E Per(f). On the other hand, since d(z,h(z’)) 5 d(z,z‘) + d(z’,h(z‘)) < 2 ~ we , have h(z’) E UzL(z).Since E is arbitrary, it follows that Per( f ) is dense in GI( f ). 0 Theorem 2.4.9(Walters [W3]). Let M be a closed topological manifold with dimension 2 2. I f f : M --+ M is topologically stable in the class of homeomorphisms, then f has POTP. For the proof we need the following two lemmas.

Lemma 2.4.10. Let f : M --t M be a homeomorphism of a closed topological manifold. Let k 2 0 be an integer and let 7 , q be positive numbers. Then for , z k } with d ( f ( z i ) , z i + l< ) ~ ( 50 i 5 k - 1 ) there any set of points {zo,z1,--* exists a set of points {zh, z: , - ,z;} such that

Proof. This is proved by induction on 12. For k = 0 this is true. Suppose the lemma is true for k - 1 and we prove it for k. Let T > 0 and > 0 be given. Choose X > 0 such that d ( z , y) < X implies d( f (z),f ( 9 ) ) < 7 . We may suppose q < min{.r, A}. Let {zo,z1,. ,Z k } be a .r-pseudo orbit of f . By assumption we can choose {zb, z;, , such that d(zi,3:) < X(0 5 i 5 k - l ) , d ( f ( z { ) , z { + l<) 27 (0 5 i I k - 2) and z{# x$ if i # j ( i I k - l , j I k - 1). Since we have

-.

we can choose xk E M so that xk d ( f ( z k - i ) , z Z ) < 27. 0

--

# xi if

j

5 k - l , d ( x k , x k ) < q and

Lemma 2.4.11. Let M be a closed topological manifold with dimension 2 2. Then for E > 0 there is 6 > 0 such that if a finite collection 3 = {(pi,qi) E M x M : 1 5 i 5 T } satisfies the following conditions :

CHAPTER 2

94

then there exists a homeomorphism f : M -+ M such that

Proof. Let n = dimM. From the dimension theory we can take a finite open cover V of M which can be represented as the union of n 1 families V1,. ,Vn+l, such that each element of V is a connected coordinate neighborhood and for 1 L n 1 all elements of Ve are disjoint. Moreover, given E > 0 we can choose the open cover V such that the diameters of all elements of V are less than E. Let 6 > 0 be a Lebesgue number of V and assume that a finite collection F = { ( p i , q i ) E M x M : 1 5 i I T } satisfies the conditions as in the lemma. Let 3 e = {(pi,q i ) E F : 3 6 E Ve s.t. pi, qi E Vj}. Then F1 U U Fn+l= F. Since dim M = n 2 2, for each 1 5 L n 1 there is a homeomorphism f e : M + M such that (1) f & i ) = qi for (pitqi) E Fe, ( 2 ) f e ( p i ) = pi for (pi,qi) 9 Fe, and ( 3 ) f e is the identity outside elements of Ve. Let f = f i o ...o f n + l . Then f ( p i ) = qi for all ( p i , q i ) E F,and d(f(z),z) < (n 1 ) for ~ 3: E M . 0

+

-.

< < +

< +

---

+

Proof of Theorem 2.4.9. Let E > 0 be given and let 6 correspond to E as in the definition of topological Stability. We may assume that S has the property as in Lemma 2.4.10 (by taking 6 sufficiently small if necessary). Suppose {ZO, 21, ,Z k } is any pseudo orbit such that d( f (xi),zi+l)< 6 / 2 for 0 5 i < k - 1. By Lemma 2.4.10 there exists {zh,zi,*-, z i } such that d(zi,3::) < E(O < i k ) , d ( f ( z : ) , z : + , ) < 6(0 5 i k - 1),z: # 3: if i # j ( i , j < k) and f ( z : )# f ( z $ )if i # j ( i , j 5 k - 1). By Lemma 2.4.11 there is a homeomorphism cp : M + M with d(cp(z),z) < S(z E M ) and cp o f ( 3 : : ) = Z:+~(O 5 i 5 k - 1). Let g = 'p o f. Then d(g(z), f ( 3 : ) ) < S(z E M) and g(z:) = z:+,(O i 5 k - 1). By topological stability there is a continuous surjection h: M --+ M such that d ( h ( z ) ,z) < E and h o g(z) = f o h ( z ) for all z E M. Then

--<

<

<

d ( f i 0 h ( Z & ) , Z i ) = d(h ogi(z&),zi)= d ( h ( z : ) , z i )

< d ( h ( z i ) ,3:) + d(z:,Zi) <€+€=BE

(O
<

Thus, for each pseudo orbit { 3 : 0 , 3 : 1 , . . - ,zk} with d ( f ( z ; ) , z i + l )< 6 / 2 ( 0 i 5 k - 1) there exists y E M such that d( f i ( y ) , z i ) < 2 4 0 5 i 5 k - 1). From the proof of Theorem 2.3.3 it follows that f has POTP. 0

$2.4 Topological Anosov maps (TA-maps)

95

Remark 2.4.12. It is known that Theorem 2.4.9 is also true for the case where dim(M) = 1. In fact, we can prove that if a homeomorphism f of the unit circle S' is topologically stable in the class of homeomorphisms, then Per( f) is a finite set and each periodic orbit, O f ( p ) ,is sink or source, i.e. there is a neighborhood U of O f ( p )in S' such that f ( z ) + O f ( p ) (i + XI) for z E U or f ( z ) --* O f ( p ) (i + -m) for z E U. From this fact we obtain that the map has POTP. It remains a problem of whether or not if a homeomorphism f:M + M of a closed topological manifold is topologically stable in the class of homeomorphisms then the restriction fp(f): a(f)+ n(f) is expansive. This problem is a topological version of the stability conjecture. The conjecture was solved by MaiiC [Ma3], in C' differentiable dynamics.