Seminar In Topological Dynamics

Bar Ilan University Department of Mathematics Seminar In Topological Dynamics EXAMPLES AND BASIC PROPERTIES Harari Ronen May, 2005 1

2 1. Basic functions 1.1. Examples. To set the stage, we begin with some standard examples of continuous maps (transformations) which will be used to illustrate different properties. Example 1.1 (Doubling map). Let X denote the unit interval with its endpoints identified, X = R/Z.Define a continuous map T : X X by T (x) = 2x (mod 1). ((mod 1) means drop the integer part), i.e: { 2x, 0 x < 1 T (x) = 2 ; 1 2x 1, 2 x 1. This is usually called the Doubling map, since it doubles distances on X. An equivalent formulation would be if we let X=K:={z C : z = 1} and then define T : X X by T (e 2πiθ ) = e 2πi2θ, where 0 < θ < 1. This is equivalent in the sense that there is a homeomorphism ρ : R/Z K given by ρ(θ + Z) = e 2πiθ which relates the two transformations. (R/Z R/Z K K). Example 1.2 (Rotation on the circle). Let X = R/Z and fix a number α [0, 1).We define a homeomorphism T : X X by T (x) = x + α (mod 1), i.e: T (x) = { x + α, 0 x + α 1; x + α 1, x + α > 1. (An equivalent formulation would be if we let X=K and then define T : X X by T (e 2πiθ ) = e 2πi(θ+α). This is equivalent in the sense that the homeomorphism ρ : R/Z K given by ρ(t + Z) = e 2πit relates the two transformations). Example 1.3 (Shift map). For k 2 let X k = n Z {1, 2,... k}denote the space of all sequences taking values {1, 2,... k} indexed by Z. In order to define a metric we first associate to two sequences x = (x n ) n Z and y = (y n ) n Z an integer N(x, y) :=min{n 0 : x N y N, or x N y N }.We define a metric on X k by: d(x, y) = { ( 1 2 )N(x,y), x y; 0, otherwise. First I ll show that it is a metric: (1) d(x, y) 0 since ( 1 2 )N(x,y) > 0, N(x, y) 0 by definition. (2) d(x, y) = d(y, x) easy. I ll prove the Triangle inequality : d(x, z) = ( 1 2 )N(x,z)

3 Define N 0 := N(x, z) := min{n 0 : x N z N or x N z N } Assume that x N0 z N0 so or x N0 y N0 or (x N0 = y N0 ) z N0 y N0 then or N(x, y) N 0 or N(y, z) N 0 so: d(x, z) = ( 1 2 )N0 ( 1 2 )N(x,y) + ( 1 2 )N(y,z) = d(x, y) + d(y, z). Definition 1.4. : Let X be a metric space, and Y is a subset of X. Y is sequentially compact if every sequence in Y contains a subsequence that converges in Y, or equivalently, every sequence in Y has at least one accumulation point in Y. Theorem 1.5. A compact space is sequentially compact. Remark 1.6. (without proof):let M be a metric space,a M.Then for every open set U such that x U,x is an accumulation point if (U {x}) A contains infinite elements. Proof. (of the theorem) Let X be a compact space. If X is not sequentially compact, then there is a sequence in X which does not have any accumulation points. Therefore there is a neighborhood U x for every point x X such that U x contains only a finite number of terms of the sequence. The collection {U x : x X} now forms an open covering of X, hence there must be a finite subcovering because of the compactness of X. However, this finite number of U x s contains at most a finite number of terms of the sequence, which is the contradiction. Theorem 1.7 (Bolzano-Weierstrass). Let (X, d) be a metric space. X is compact if and only if X is sequentially compact. Proof. I proved. Let U be an open cover of X, and δ be the Lebesgue number for the cover U. (Let M be a compact metric space, and U α an open covering of M. Then there exist δ > 0 such that for every x M there is α I such that B(x, δ) U α. Such δ is called the Lebesgue number for the covering). Lemma 1.8. (without proof) Let X be a metric space. If X is sequentially compact then the Lebesgue number is positive for every open covering of X. Then, by this Lemma, we know that δ > 0. For x X, there is member U x in U such that B(x, δ) U x. If no subcovering of U exists, then we are able to find a sequence (x n ) n N in X such that x n / n 1 i=1 U x i (x 1 can be chosen arbitrary). Because X is sequentially compact, there is a convergent subsequence of (x n ) n N in X. On one hand, d(x i, x j ) δ when i j by construction. On the other hand, points in the tail of the convergent subsequence can be chosen arbitrary close. Contradiction. Lemma 1.9. X k is a compact space. Proof. We shall actually show that X k is sequentially compact. Let x m = (x (m) n ) n Z (m = 1, 2,...) be a sequence in X k ; (the sequence looks like: x 1 = (x 1 n,..., x 1 0,..., x 1 n)) and so on... then we need to show that there exist a point x X k and a subsequence x (m l) x (l = 1, 2, 3,...). First observe that the zeroth terms x (m) 0 (m = 1, 2, 3,...) must take some value in {1,2,3,... k} infinitely often. Choose such an x 0 {1, 2, 3,..., k} with x (m) 0 = x 0, for infinitely many m. We continue inductively: For l > 0, choose x l {1, 2, 3,..., k} and x l {1, 2, 3,..., k} such

4 that x (m) l = x l,..., x (m) 0 = x 0,..., x (m) l = x l,say, for infinitely many m. Finally, we define x = (x l ) l Z. For each l 0 we choose m l := m such that x (m) l = x l,..., x (m) 0 = x 0,..., x (m) l = x l ; then, d(x (ml), x) 1 and so d(x (ml), x) 0 as 2 l l +. 2. The Shift Map Definition 2.1. We can define a map σ : X k X k by (σx) n = x n+1, n Z,i.e. σ : (..., x 2, x 1, x 0, x 1, x 2,...) (..., x 1, x 0, x 1, x 2,...) Since this map shifts sequences by one place it is called the shift map. Lemma 2.2. The map σ : X k X k is a homeomorphism. Proof.. To show continuity we observe that if x y and d(x, y) = ( 1 2 )N then we know that x i = y i for N i N. Thus we have that (σx) i = x i+1 = y i+1 = (σy) i for i = (N + 1),..., N 1. This means that d(σx, σy) ( 1 2 )N 1 = 2d(x, y) and we see that σ is continuous. Clearly, σ : X k X k is invertible (since the inverse transformation σ 1 : X k X k simply shifts sequences back one place). Finally, the inverse map σ 1 : X k X k is continuous by the same sort of argument as above. Notation and terminology: Let S be the set: S = {1,..., k} If k N, then a k block or a a block of length k or a word of length k over S is an element of the set S k := S S (k times). Thus, if b S k then b = b 0 b k with b i S for i = 0,..., k. The b i are called the entries of b. The set of all finite words over S will be denoted by S ; so S := {S k : k Z + }. Let Ω := S Z. If x = (x n ) n Z is an element of Ω then we shall often write x = x 2 x 1 x 0 x 1 x 2. If x Ω and j Z then the block x j x j+n is often denoted by x[j; j + n]. An non-empty block b is contained(occurs at) in x Ω (j Z whenever b = x j x j+ b 1 = x[j; j + b 1] where the length b of b is the unique k Z + such that b S k. In this case we say that x contains b (at place j)or that b is a finite sub-block of x. If j = 0, then we also write x = ḃ Definition 2.3. If b is a finite block and j Z, then the cylinder based on b at place j is the set of all elements in Ω in which b occurs(contained in) at place j, that is, the set: C j [b] := {x Ω : x[j; j + b 1] = b} If b = then C j [b] =.

5 Definition 2.4. A continuous flow is a pair < X, σ > where X is a topological Hausdorff space and σ : R X X is a mapping such that: (1) σ is continuous. (2) σ(0, x) = x for all x X. (3) σ(t, σ(s, x)) = σ(t + s, x) for all x X and t, s R. A mapping σ : R X X satisfying conditions (2), (3) is called an action of R on X. If condition (1) is fulfilled then it is called a continuous action. Definition 2.5. A subshift is a subflow of (X, σ) defined by a closed non-empty invariant subset. Definition 2.6. For every subset C of S we define: H (C) := {x Ω : no c C occurs in x} H(C)is closed and invariant under σ. So if H (C) then it defines a subshift. 3. Transitivity In this section we shall introduce some basic properties of continuous maps T : X X on compact metric spaces X. Definition 3.1. We say that a homeomorphism T : X X of a compact metric space X is transitive if there exists a point x X such that its orbit {T n x : n Z} = {..., T 2 x, T 1 x, x, T x, T 2 x,..., } is dense in X. We call such a point a transitive point. We say that a continuous map T : X X of a compact metric space X is (forward)transitive if there exists a point x X such that its orbit {T n x : n Z + } = {x, T x, T 2 x,..., } is dense in X. We call such a point x X a (forward) transitive point. We can check each of the examples in section 1.1 for this property. Example 3.2. We shall show that this example is forward transitive when k = 2, other cases being similar. Consider the sequence 1, 2, 11, 12, 21, 22, 111, 112, 121, 122, 221,..., 222, 111,..., 1111,... We can write down x n {1, 2}, n 0, as the nth term in the sequence 1211122122111112121122221...2221111... Finally, consider the point x [0, 1] given by the series x = + (x n 1) n=0 2. We n+1 claim that the point x is a (forward) transitive point. Observe that T x = 2( + (x n 1) n=0 2 )(mod 1)= x n+1 0 + + (x n+1 1) n=0 2 (mod1) = + (x n+1 1) n+1 n=0 2. n+1 Similarly, T k x = + x n+k 1 n=0 2 (mod1). n+1 To show that the set {T n x : n 0} is dense it suffices to show that for each interval of the form [ p, p+1 ], which 0 p 2 l 1, we can find N 0 with T N x [ p, p+1 ]. 2 l 2 l 2 l 2 l Given p we can write it in binary form as i 0,..., i N+1, with i 0,..., i n 1 {0, 1}. But for some N we can find x N = i 0, x N+1 = i 1,..., x N+n 1 = i n 1.This means that T N x [ p, p+1 ], as required. 2 l 2 l Proposition 3.3. Let M be a metric space. The following are equivalent: (1) M is compact.

6 (2) Every infinite set A M has an accumulation point. (3) Every sequence in M has a sub-sequence which converges. Proof. (1) (2) : assume M is compact.let A M an infinite set. We need to prove that A has an accumulation point.asuume that A does not have an accumulation point,which means that every point x M is not an accumulation point of A.If x A it means that there is an environment x U such that U A = {x}. We shall take this environment and mark it as U x. The family of the sets U x is a cover of M.I claim that this cover does not have a finite sub-cover. Why? because every x A has only one element in the cover that contains him which is U x.therefore every sub-family which doesn t include each of the U x for x A,will not be a cover. Therefore there is no finite sub-cover - contradiction to M being compact. (2) (3) : Assume every A M infinite has an accumulation point. Let {x n } be a sequence in M. We shall prove that {x n } has a sub-sequence which converges.there are two cases: (1) If the same element in M appears infinite number of times in the sequence {x n } so we have a sub-sequence which converges. (2) If all the elements appear a finite number of times so we have an infinite different elements. Let A be the set of values that appear in the sequence {x n } and in this case A is infinite (obvious). So A has an accumulation point, x. We show that {x n } has a sub-sequence converges to x. we shall take the ball B(x, 1).The point x is an accumulation point of A so B(x, 1) A.But A is the set of values of the sequence {x n }, which means x n1 B(x, 1).In B(x, 1 2 ) there are an infinite elements from A so there are elements that appear after the n 1 place.we shall take n 2 > n 1 such that x n2 B(x, 1 2 ) and so on... We shall look at B(x, 1 k ) which has infinite elements from A so there is an element that appear after the n k 1 place. We shall take n k > n k 1 such that x nk B(x, 1 k ) and it is the next element of the sequence. We claim that lim x n k = x k because 0 < d(x nk, x) < 1 k and that s why x n k x. (3) (1): Remark (without proof): If every sequence {x n } in M,has a sub-sequence which converges,then for every open cover{u α } α I of M there exists ɛ-lebesgue number of the cover. Assume that every sequence {x n } has a sub-sequence that converges. We need to prove that M is compact. Let U α be a cover of M. We need to show that it has a finite subcover.let ɛ be the Lebesgue number of the cover. We claim that there are finite elements x 1,..., x n such that {B(x n, ɛ)} is a cover of M.Assume that there is not such a finite set. We take some x 1 M.B(x 1, ɛ) is not covering M.We shall take x 2 / B(x 1, ɛ).b(x 1, ɛ) and B(x 2, ɛ) are not covering M. Let take x 3 / B(x 1, ɛ) B(x 2, ɛ). And so on... after taking x 1,..., x k, {B(x i, ɛ)} k i=1 is not a cover of M and we choose x k+1 / k i=1 B(x i, ɛ).in the sequence {x k } k=1 we

7 have that for every k 1 < k 2, d(x k1, x k2 ) ɛ.if the sequence has a sub-sequence that converges, say to x, there were two different elements of the sub-sequence in B(x, ɛ 2 ) and then the distance between them was less than ɛ.contradiction. We got elements {x 1,..., x n } M such that M = n k=1 B(x k, ɛ).for all {1 k n} there is (and we take it)α k I such that B(x k, ɛ) U αk and then M = n k=1 U α k, and therefore M is compact. Example 3.4. There are two different cases, depending on whether or not α is irrational. First assume that α is irrational, then the map T : X X (as in example 1.2)can be shown to be transitive (and even forward transitive) where x = 0,say. It suffices to show that the orbit {T n 0} n Z + is dense.since this is an infinite set in R/Z we can choose x R/Z and a subsequence n i + with T ni 0 = n i α(mod1) x.for any sufficiently small ɛ > 0 we can choose n i > n j with T ni 0 x < ɛ 2 and T nj 0 x < ɛ 2 and thus T ni 0 T nj 0 = T ni nj 0 < ɛ.moreover, T ni 0 T nj 0, since if not this would contradict α being irrational. Thus the points T (ni nj)k 0, k 1 form an ɛ-dense subset of R/Z. Since ɛ can be chosen arbitrary small this completes the proof of transitivity. Now assume that α = p q with p, q Z having no common divisors and q 0. For any x X the orbit {T n x : x Z} would be a finite set {x, x + 1 q,..., x + q 1 q (mod1)}.in particular T is not transitive. Example 3.5. We shall show that this example is forward transitive.the sequence {x n } n N = {1, 2,..., k, 1, 1,..., 1, k, 2, 1,..., 2, k,..., z 0, z 1,..., z N 1...} }{{} All strings appear in X k (in which all finite strings appear once) is a forward transitive point. To see this choose any point z X k (z = (z n ) n Z ) and for any ɛ > 0 choose N > 0 sufficiently large that ( 1 2 )N < ɛ. If we choose r such that x r = z 0,..., x r+n 1 = z N 1 then we see that (σ r x) 0 = x r = z 0,..., (σ r x) N 1 = x r+n 1 = z r+n 1 and so d(σ r x, z) ( 1 2 )N < ɛ. 4. Other characterizations of transitivity The following result gives equivalent conditions for a homeomorphism of a compact metric space to be transitive. Theorem 4.1. The following are equivalent. (1) T : X X is transitive. (2) If U is an open set with T U = U then either U is dense or U =. (3) If U, V are non-empty open sets then for some n Z we have that T n U V. (4) The set {x X : the orbit {T n x} n Z is dense in X} is dense in G δ set. Definition 4.2. A set which can be represented as the intersection of a countable collection of open sets is called a G δ set.

8 Proof. 1. 2. Assume x X has a dense orbit. Assume that T U = U. We can choose n Z such that T n x U.Moreover, for any m Z we have that T m x T m n U = U. Since the orbit of x is dense (i.e. m Z T m x X is dense)we see that U is dense. 2. 3. The T -invariant union n Z T n U is dense in X by assumption 2. Thus n Z T n U V and so n Z with T n U V. 3 4 Consider a dense set {x n } n Z and consider the balls of radius 1 k,k 1, denoted by B(x n, 1 k ). We can identify {x X : {T m x} m Z is dense in X }= + n=0 + k=1 + m= T m B(x n, 1 k ) (i.e. n 0, k 1, m Z with T m x B(x n, 1 k )). 4. 1. This is immediate. 5. Transitivity for subshifts of finite type In section 1.1 we defined the shift transformation σ : X k X k on X k = n Z {1,..., k}.for any closed σ invariant subset X X k (i.e. σ(x) = X) we consider the restriction σ X. We can use the same notation σ : X X. Definition 5.1. A subshift Λ is called a subshift of finite type (abbreviation: ssft) whenever Λ = H (C) for a finite set C of blocks. Definition 5.2. For n N and B S n, we define: (all n blocks are in B). H n (B) := {x Ω : x[i; i + n 1] B for all i Z} H n (B) then it is a subshift. Definition 5.3. Let Λ be a subshift. Than we define: Λ = H 2 (B) := {x Ω : x n x n+1 B for all n Z} for some set B of 2 blocks over S. Consider a finite directed graph Γ with s vertices labeled faithfully, by the symbols 0, 1,..., s 1 of S and let for b 1, b 2 S the pair (b 1, b 2 ) be a directed edge (from b 1 to b 2 ) in Γ iff the 2 block b 1 b 2 belongs to B. Thus the following graph represents the set: {00, 20, 23, 32, 12, 15, 51, 46} of 2 blocks over the symbol set {0, 1, 2, 3, 4, 5, 6}. Let M be the adjacency matrix of the graph Γ, associated with the ssft H 2 (B): M has entries m ij for i, j = 0,..., s 1 and { 1, if ij B,(i,j) an edge in Γ; m ij := 0, if ij/ B, that is (i,j) not an edge in Γ. Clearly, H 2 (B) is determined by M: If x Ω, then x H 2 (B) iff x n x n+1 B for all n Z, iff m xnx n+1 = 1 for all n Z. In the following table a description is given of essentially all possible ssft s of order 2 over the symbol set S := {0, 1}. For each such a subshift the transition matrix M, the corresponding graph Γ and a schematic description of (Λ, σ) are given. The arrows between points in Λ indicate the direction in which a point σ n x runs through its orbit if n increases.

9 Lemma 5.4. For every n N and i, j S the following are equivalent: (1) m (n) ij > 0. (2) x Λ : x 0 = i and x n = j. Proof. (1) = (2): By (1) there is a path of length n in Γ form vertex i to vertex j. This path can be extended to a two-sided infinite path in Γ, representing a point x Λ with the desired property. Definition 5.5. Let A be a k k matrix with entries 0 or 1. We call the matrix irreducible if 1 i, j, k, N > 0 such that A N (i, j) > 0. Example 5.6. When k = 3 the matrix A= 0 1 1 0 1 1 is irreducible (n = 4 1 0 0 1 1 0 for example. However, the matrix A = 1 1 0 is not irreducible. (These 0 0 1 properties are readily checked). Definition 5.7. Given a k k matrix A with entries 0 or 1.We define {X A = (x n ) n Z {1,..., k} : A(x n, x n+1 ) = 1, n Z}. n= We define the subshift of finite type σ : X A X A to be the restriction σ XA. The following gives necessary and sufficient conditions for σ : X A X A to be transitive. Theorem 5.8. A subshift of finite type σ : X A X A is transitive if and only if A is irreducible. Proof. Assume that σ is transitive. Consider the sets [i] 0 := {(x n ) n Z X A : x 0 = i} for i = 1,..., k. These sets are open. Given 1 i, j k we know that there exists N > 0 such that σ N [j] 0 [i] 0.Choose (x n ) n Z σ N [j] 0 [i] 0 ;then we know that x 0 = i and x N = j. Notice that k k A N (i, j) = A(i, r 1 )A(r 1, r 2 ) A(r N 2, r N 1 )A(r N 1, j). r 1=1 r N 1 =1 But since A(i, x 1 ) = A(x 1, x 2 ) = = A(x N 1, j) = 1 we see that A N (i, j) 1. (Another way to look at this is by paths: We can write: m (n) ij = s 1 s 1 p 1=0 p 2=0 s 1 p n 1=0 m ip1 m p1p 2 m pn 1j Then m (n) ij is equal to the total number of paths in the directed graph Γ that begin in vertex i, end in vertex j and have length (=number of consecutive edges) n).

10 Conversely, assume that for 1 i, j k we have that A N (i, j) 1.Given U, V open sets we can choose (i n ) n Z U and (j n ) n Z V such that for M > 0 sufficiently large U [i M, i M 1,..., i M ] M M := {(x n ) n Z X A : x k = i k, M k M}, V [j M, j M 1,..., j M ] M M := {(x n ) n Z X A : x k = j k, M k M}. By hypothesis we can find N > 0 such that A N (i M, j M ) 1. This means that we can find a string x 1,..., x N 1 such that A(i M, x 1) = A(x 1, x 2) =... = A(x N 1, j M ) = 1 and then define i n, if n M; x n = x n M, if M+1 n M + n 1; j n (2M+N), if M+N n. then we have that x U σ N V i.e U σ N V.

11 6. Appendix 6.1. Minimality and the Birkhoff recurrence theorem. In this section I want to present a simple but important recurrence result, called the Birkhoff recurrence theorem. Our starting point is to define the following property. Definition 6.1. A homeomorphism T : X X is minimal if for every point x X the orbit {T n x : n Z} is dense in X. The following is obvious from the definitions. Proposition 6.2. A minimal homeomorphism is necessarily transitive. We can now consider each of the examples from section 1.1 and ask which of these are minimal. Since example 1 is not a homeomorphism we begin with example 2. Example 6.3. Lemma 6.4. When α is irrational then T (x) = x + α is minimal. Proof. It suffices to show that for every x R/Z and every neighbourhood (y ɛ, y + ɛ) (y R/Z, ɛ > 0) we can find n 1 such that T n x (y ɛ, y + ɛ). We already know that T is transitive (i.e. there exists at least one transitive point x 0 R/Z with dense orbit). Fix y R/Z and use the transitivity to choose a subsequence n i with T ni x 0 (y x + x 0 ) as i +. Thus T ni x = n i α + x (mode 1) = n i α + x 0 + (x x 0 ) (mod 1) = T ni (x 0 ) + x x 0 (mod 1) y + (x 0 x) + (x x 0 ) = y (mod 1) Example 6.5. The shift map is not minimal since it contains a fixed point (e.g. x = (..., 1, 1, 1,...)). The following theorem gives equivalent definitions. Theorem 6.6. Let T : X X be a homeomorphism of a compact metric space. The following properties are equivalent. (1) T is minimal. (2) If T E = E is a closed T invariant set, then either E = or E = X. (3) If U is an open set then X = n Z T n U. Proof. (1) (2) Assume that T E = E and choose x E.Hypothesis (1) gives that X = cl({t n x} n Z ) E X. (2) (3) Given a non-empty open set U let E = X ( n Z T n U). By construction T E = E and E X(since U ) and so by hypothesis (2) we have that E =. Thus X = n Z T n U.

12 (3) (1)Fix x X and an open neighbourhood U X. Since X T n U for some n Z (by hypothesis (3)) we have that T n x U. This shows that the orbit {T n x} n Z is dense in X. Using property (2) we get the following surprising result that every homeomorphism contains a minimum homeomorphism. Theorem 6.7. Let T : X X be a homeomorphism of a compact metric space X. There exists a non-empty closed set Y X with T Y = Y and T : Y Y is minimal. Proof. This follows from an application of Zorn s Lemma. Let ε denote the family of all closed T -invariant subsets of X with the partial ordering by inclusion, i.e. Z 1 Z 2 iff Z 1 Z 2. Every totally ordered subset (or chain ) {Z α } has a least element Z = α Z α (which is non-empty by compactness of X). Thus by Zorn s lemma, there exists a minimal element Y X(i.e. Y, Y ε with Y Y implies that Y = Y ). By property (2) of theorem 6.6 this can be reinterpreted as saying that T : Y Y is minimal. As a corollary we get the following simple but elegant result. Corollary 6.8. (BIRKHOF F RECURRENCE T HEOREM) Let T : X X be a homeomorphism of a compact metric space X. We can find x X such that T ni x x for a sub-sequence of the integers n i +. Proof. By theorem 6.7 we can choose a T variant subset Y X such that T : Y Y is minimal. For any x Y X we have the required property. Example 6.9. Consider the case X = R/Z and T : X X defined by T x = x + α (mode 1), where α is an irrational number. Let ɛ > 0; then we can find n > 0 (by Birkhoff s theorem) such that αn (mode 1) ɛ, i.e. there exists p N such that ɛ αn p ɛ. Rewriting this, we have that for any irrational α, p, n N such that α p n ɛ n. This is a (marginal) improvement on the most obvious estimate.

13 References [1] P.Billingsley, Ergoduc Theory and Information, Addison-Wesley, New York, 1965. [2] R. Devaney, An Introduction to the Modern Theory of Dynamical Systems, Addison-Wesley, New York 1989. [3] J. de Vries, Elements of Topological Dynamics, Kluwer Academic Publisher, p.163-175, 49-50. [4] A. Katok and B. Hasselblatt, Introduction to Modern Theory of Dynamical Systems, C.U.P., Cambridge, 1995. [5] W.Szlenk, An Introduction to the Theory of Smooth Dynamical Systems, Wiley, New York, 1984. [6] P. Walters, An Introduction to Ergodic Theory, Graduate Texts in Mathematics, 79, Springer, Berlin, 1982.