Some universality results for dynamical systems
aa r X i v : . [ m a t h . D S ] D ec SOME UNIVERSALITY RESULTS FOR DYNAMICAL SYSTEMS
UDAYAN B. DARJI AND ´ETIENNE MATHERON
Abstract.
We prove some “universality” results for topological dynamicalsystems. In particular, we show that for any continuous self-map T of a perfectPolish space, one can find a dense, T -invariant set homeomorphic to the Bairespace N N ; that there exists a bounded linear operator U : ℓ → ℓ such thatany linear operator T from a separable Banach space into itself with k T k ≤ U ; and that given any σ -compact family F of continuousself-maps of a compact metric space, there is a continuous self-map U F of N N such that each T ∈ F is a factor of U F . Introduction
It is well known that the Cantor space 2 N , the Baire space N N and the Banachspace ℓ = ℓ ( N ) have some interesting universality properties: every compactmetric space is a continuous image of 2 N , every Polish space is a continuous imageof N N , and every separable Banach space is a quotient of ℓ .In this note, we will mainly be concerned with the following kind of generalizationof these classical facts: instead of just lifting the points of a given space X into oneof the above spaces, one would like to lift continuous self-maps of X .Throughout the paper, by a map we always mean a continuous mapping betweentopological spaces. For us, a dynamical system will be a pair ( X, T ), where X is atopological space and T is a self-map of X .A dynamical system is ( X, T ) is a factor of a dynamical system (
E, S ), and(
E, S ) is an extension of (
X, T ), if there is a map π from E onto X such that T π = πS , i.e. the following diagram commutes: E S / / π (cid:15) (cid:15) (cid:15) (cid:15) E π (cid:15) (cid:15) (cid:15) (cid:15) X T / / X When T and S are linear operators acting on Banach spaces X and E , we saythat T is a linear factor of S if the above diagram can be realized with a linearfactoring map π .Among other things, we intend to prove the following two “common extension”results. Mathematics Subject Classification.
Primary: 37B99,54H20. Secondary: 54C20,47A99.
Key words and phrases. universal, factor, ℓ , Cantor space, Baire space.The first author would like to acknowledge the hospitality and financial support of Universit´ed’Artois. • There exists a bounded linear operator U : ℓ → ℓ such that every linearoperator T from a separable Banach space into itself with k T k ≤ is alinear factor of U . (This is proved in Section 3.) • Given any σ -compact family F of self-maps of a compact metric space, thereis a map U F : N N → N N such that each T ∈ F is a factor of U F . (This isproved in Section 4.2.)Regarding the Baire space N N , we will also show that any Polish dynamical thedynamics of any continuous self-map T of a perfect Polish space X is in some sense“captured” by a self-map of N N ; namely, there is a dense, T -invariant set inside X which is homeomorphic to N N . This result is proved in Section 2.The reader may wonder why, in the second result quoted above, the domain ofthe map U F is N N and not the Cantor space 2 N . The reason is that, even for simplecompact families F , one cannot obtain a common extension defined on a compactmetrizable space; see Remark 4.6. We have not been able to characterize those σ -compact families F admitting a common extension defined on 2 N . However, wegive a simple sufficient condition which implies in particular that this holds true forthe family of all contractive self-maps of a compact metric space. This is provedin Section 4.3. We also show that for an arbitrary σ -compact family F , a rathernatural relaxation in the definition of an extension allows this; see Section 5.The kind of question we are considering here may be formulated in a very generalway. One has at hand a certain category C and, given a collection of objects F ⊆ C ,one wants to know if there exists an object U F ∈ C which is projectively universal for F , in the following sense: for any T ∈ F , there is an epimorphism π : U F → T from U F to T . Less ambitiously, one may just want to find a “small” family ofobjects U which is projectively universal for F , i.e. for any T ∈ F , there exists anepimorphsim from some U ∈ U to T .In this paper, we have been concerned with categories C whose objects are topo-logical dynamical systems and whose morphisms are factoring maps. For example,if C is the category of all compact metrizable dynamical systems then, as men-tioned above, there is no projectively universal object for C , but there is one forthe family of all contractive dynamical systems defined on a given compact metricspace. Moreover, it is well known that every self-map of a compact metrizablespace is a factor of some self-map of 2 N (see Section 4.1); in other words, the fam-ily of all dynamical systems defined on 2 N is projectively universal for the wholecategory. Likewise, if C is the category all linear dynamical systems on separableBanach spaces then (as annouced above) there is an object U living on ℓ which isprojectively universal for the family all ( X, T ) ∈ C with k T k ≤ N N happens to be projectively universal (see Section4.1).The second category we have in mind is that of dynamical systems defined on arc-like continua . An exotic space called the pseudo-arc plays a central role in thetheory of arc-like continua. As it turns out, the family of all dynamical systemsdefined on the pseudo-arc is projectively universal for this category (this is proved OME UNIVERSALITY RESULTS FOR DYNAMICAL SYSTEMS 3 in [9]). The pseudo-arc has a rich and long history; see e.g. [8] for a detailed surveyand [5] for an interesting model theoretic construction.A third interesting example is the category of all minimal G -flows , for a giventopological group G . Recall that a G -flow ( X, G ) is a compact topological space X endowed with a continuous action of G , and that a G -flow ( X, G ) is said to beminimal if every G -orbit { g · x ; g ∈ G } is dense in X . It is well known that thereexists a projectively universal object ( M ( G ) , G ) for this category, which is uniqueup to isomorphism. There is a vast literature on universal minimal flows (see [12]and the references therein, for example [3], [4], [6], [10], [11], [13]). In particular, theproblem of determining exactly when M ( G ) is metrizable has been well investigated;see [10]. Very much in the spirit of the present paper, it seems quite natural toask for which families F of minimal G -flows one can find a projectively universal G -flow defined on a metrizable compact space.2. Dense subsystems homeomorphic to N N In this section, our aim is to prove the following result.
Theorem 2.1.
Let X be a perfect Polish space. For any self-map T : X → X ,there is a dense G δ set Y ⊆ X homeomorphic to N N such that T ( Y ) ⊆ Y . This will follow from Lemmas 2.5 and 2.6 below.We start with a lemma where no assumption is made on the Polish space X .Recall that a space E is 0 -dimensional if it has a basis consisting of clopen sets. Lemma 2.2.
Let X be a Polish space, T : X → X be continuous and Z ⊆ X be -dimensional with T ( Z ) ⊆ Z . Then, there is Y such that (1) Z ⊆ Y ⊆ X and T ( Y ) ⊆ Y ; (2) Y is -dimensional and Polish (i.e. a G δ subset of X ).Proof. We need the following simple and well known fact, which easily follows frombasic definitions.
Fact 2.3.
Let E be Polish and let F ⊆ E be -dimensional. Then, for any ε > ,there is a sequence of pairwise disjoint open set { U n } in E such that diam ( U n ) < ε and F ⊆ S n U n . By Fact 2.3 we may choose a sequence { U n } of pairwise disjoint open subsetsof X such that the diameter of each U n is less than 1 and Z ⊆ S n U n . Now let V n := T − ( U n ). Then the { V n } are pairwise disjoint open sets. Moreover, as T ( Z ) ⊆ Z , we have that Z ⊆ S ∞ n =1 V n . Now we apply Fact 2.3 with E := V m and F := V m ∩ Z , for each m ∈ N . This produces a collection of pairwise disjointopen sets { U n } in X such that, for each n , the diameter of U n is less than 1 / U n ⊆ V m for some m . Hence the collection { U n } satisfy the following properties: • Z ⊆ S ∞ n =1 U n , • diam( U n ) < / n , • T ( U n ) ⊆ U m for some m .Continuing in this fashion, we get for each j ∈ N a collection of pairwise disjointopen sets { U jn ; n ∈ N } such that the following properties hold: • Z ⊆ S ∞ n =1 U jn , • diam( U jn ) < / j for each n , DARJI AND MATHERON • T ( U jn ) ⊆ U j − m for some m .Now we simply let Y := T ∞ j =1 S ∞ n =1 U jn . This Y has the required properties. (cid:3) Remark 2.4.
In the above lemma, one cannot make Y dense in X . For example,take X to be the disjoint union of R and a countable discrete set { a n ; n ∈ N } . Thetopology is the usual one. Let the map T on X defined as follows: it is the identityon R and it maps a n to the n th rational in R (under some fixed enumeration). Nowtake Z to be the set of irrationals on the line. There is no way to extend this Z toa dense, T -invariant and still 0-dimensional set Y . Lemma 2.5.
Let X be a perfect Polish space and T : X → X be continuous.Then, there is a Polish X ⊆ X dense in X and nowhere locally compact such that T ( X ) ⊆ X .Proof. We will find a meager, F σ set Z ⊆ X dense in X such that T − ( Z ) ⊆ Z . Itwill then suffice to let X := X \ Z . As X is Polish and Z an F σ meager set, wehave that X is Polish and dense in X . Moreover, X is nowhere locally compact.Indeed, assume that for some nonempty open set U in X , the set X ∩ U hascompact closure in X . Then, ( X ∩ U ) ∩ X is compact and hence is closed in X ,and it has empty interior in X since X \ X is dense in X . In particular, X ∩ U is nowhere dense in X , a contradiction because X is dense in X .To complete the proof, let us construct the desired Z . Let B , B , . . . be anenumeration of some countable basis of open sets for X . Let S = { ( n, m ) ∈ N × N : T m ( B n ) is meager } . Let U = S ( n,m ) ∈ S B n . Let A = S { T m ( B n ); ( n, m ) ∈ S } . Then, A is meager in X . Let D be a countable dense subset of X \ A . Clearly, each of T − m ( D ), m ≥ F σ . We claim, moreover, that T − m ( D ) is meager. This is clear for m = 0 as D is countable. To obtain a contradiction, assume that for some m ≥ T − m ( D ) isnot meager. As T − m ( D ) is F σ , it must contain a nonempty open set. Let n ∈ N be such that B n ⊆ T − m ( D ). Then, we have that T m ( B n ) is meager since it is asubset of the countable set D . Hence, T m ( B n ) ⊆ A , contradicting that D ∩ A = ∅ .Let Z = S m ≥ T − m ( D ). Then, Z is an F σ , dense, and meager subset of X .Moreover, T − ( Z ) ⊆ Z , concluding the proof of the lemma. (cid:3) Lemma 2.6.
Let X be a nowhere locally compact Polish space and let T : X → X be continuous. Then there is a dense G δ set Y ⊆ X such that T ( Y ) ⊆ Y and Y ishomeomorphic to N N . Moreover, one may require that Y contains any prescribedcountable set C ⊆ X .Proof. Let C be any countable subset of X . As X is separable, we may expand C tobe a countable set which is dense in X . Let Z = S ∞ n =0 T n ( C ). Clearly, T ( Z ) ⊆ Z .Moreover, Z is 0-dimensional as it is countable. Let Y be given by Lemma 2.2.Only that Y is homeomorphic to N N needs an argument. We already know that Y is 0-dimensional and Polish. Finally, Y is nowhere locally compact as X is and Y is dense in X . Hence, Y is homeomorphic to N N . (cid:3) Remark 2.7.
It follows in particular that if T : G → G is a self-map of a non-locally compact Polish group G then, for any given countable set C ⊆ G , there is Y ⊆ G homeomorphic to N N such that C ⊆ Y and T ( Y ) ⊆ Y . OME UNIVERSALITY RESULTS FOR DYNAMICAL SYSTEMS 5
Proof of Theorem 2.1.
This is now immediate by applying Lemma 2.5 and thenLemma 2.6. (cid:3) A universal linear operator
In this section, we prove the following result, which says essentially that anybounded linear operator on a separable Banach space is a linear factor of a single“universal” operator acting on ℓ . Theorem 3.1.
There exists a bounded linear operator U : ℓ → ℓ with k U k = 1 such that any bounded linear operator T : X → X on a separable Banach space X is a linear factor of ρ · U for every ρ ≥ k T k .Proof. We need the following fact. This is probably well known but we include anoutline of the proof for completeness.
Fact 3.2.
There exists a 1-1 map µ : N → N such that, for any other 1-1 map σ : N → N , one can find a set A ⊆ N and a bijection π A : N ։ A such that µ ( A ) = A and σ = π − A µπ A .Proof of Fact 3.2. For any 1-1 map σ : N → N , let G σ be the digraph on N inducedby σ ; that is : −→ ij is a directed edge of G σ iff σ ( i ) = j . Then, each component C of G σ has one of the following forms: • C is a cycle of length n for some n ∈ N , • there is i ∈ C such that C = { i , i , i , . . . } where all i k ’s are distinct andthe only directed edges are of the form −−−−→ i k i k +1 , or • C = { . . . , i − , i , i , . . . } where all i k ’s are distinct and the only directededges are of the form −−−−→ i k i k +1 .Now, consider any µ whose digraph contains infinitely components of each typedescribed above. (cid:3) To prove the theorem, we first observe that it suffices to construct U so thatevery bounded linear operator with norm at most 1 (on a separable Banach space)is a linear factor of U .Let µ be as in Fact 3.2. Let { e n } be the standard basis of ℓ , i.e., e n is 0 inevery coordinate except the n th coordinate where it takes the value 1. We define U : ℓ → ℓ simply by U ( e n ) = e µ ( n ) . It is clear that U is a bounded linear operatorof norm 1. Let T : X → X be a bounded linear operator acting on a separableBanach space X , with k T k ≤
1. Since X is separable and k T k ≤
1, one can finda countable dense subset D of B X , the unit ball in X , so that T ( D ) ⊆ D . Thismay be done by starting with any countable set dense in the unit ball and takingits union with its forward orbit. We let { z n } be a enumeration of D so that eachelement of D appears infinitely often in { z n } . Now we define σ : N → N . For any i ∈ N , we have T ( z i ) = z j for some j . Moreover, since every element of D appearsinfinitely often in the sequence { z n } , we may choose a 1-1 map σ : N → N such that T ( z i ) = z σ ( i ) for all i ∈ N . Now, let A ⊆ N and π A be as in Fact 3.2. Then there isa uniquely defined bounded linear operator π : ℓ → X (with k π k ≤
1) such that π ( e i ) = (cid:26) z π − A ( i ) if i ∈ A ,0 if i / ∈ A .The operator π is onto because π ( B ℓ ) contains all of D , which is dense in B X ;see e.g. [2, Lemma 2.24]. DARJI AND MATHERON
Finally, let us show show that T is a factor of U with witness π , i.e. . πU = T π .It is enough to check that πU and T π are equal on { e i } as all maps are linear andcontinuous and { e i } spans ℓ . Let i / ∈ A . Then, by Fact 3.2 µ ( i ) / ∈ A . Hence, wehave that neither i nor µ ( i ) is in A , so that π ( U ( e i )) = π ( e µ ( i ) ) = 0 and T ( π ( e i )) = T (0) = 0 . Now suppose i ∈ A . Then, T ( π ( e i )) = T ( z π − A ( i ) ) = z σ ( π − A ( i )) = z π − A ( µ ( i )) = π ( e µ ( i ) ) = π ( U ( e i )) . (cid:3) Remark 3.3.
One really has to consider all multiples of U : it is not possibleto find a single bounded linear operator U so that every bounded linear operator T : X → X is a linear factor of it. Indeed, if an operator T is a linear factor of someoperator U , then there is a constant C such that k T n k ≤ C k U n k for all n ∈ N ; inparticular, λId cannot be a factor of U if | λ | > k U k . To see this, let π witness thefact that T : X → X is a factor of U : Y → Y . By the Open Mapping Theorem,there a constant k such that B X ⊆ kπ ( B Y ). Let A ≥ k π k ≤ A .Then, C = kA has the property that k T n k ≤ C k U n k for all n ∈ N .In spite of this last remark, one can find a single Fr´echet space operator whichis universal for separable Banach space operators. (Interestingly, it is not known ifthere exist quotient universal Fr´echet spaces.)
Corollary 3.4.
Let E := ( ℓ ) N , the product of countably many copies of ℓ , en-dowed with the product topology. There exists a continuous linear operator U E : E → E such that any bounded operator T acting on a separable Banach space is alinear factor of U E .Proof. Let U : ℓ → ℓ be the operator given by Theorem 3.1. Then define U E : E → E by U E ( x , x , x , . . . ) := ( U ( x ) , · U ( x ) , · U ( x ) , . . . ) . This is indeed a continuous linear operator, and it is readily checked that U E hasthe required property. Indeed, if T : X → X is a bounded linear operator ( X aseparable Banach space) then one can find n ∈ N such that T is a linear factor of n · U , with factoring map π : ℓ → X . Then the operator e π : E → X defined by e π ( x , x , . . . ) := π ( x n ) shows that T is a linear factor of U E . (cid:3) Common extensions of compact dynamical systems
In this section, we prove some “common extension” results for self-maps of com-pact metric spaces . In what follows, we denote by D the collection of all dynamicalsystems ( X, T ) where X is a compact metric space; or, equivalently, the collectionof all self-maps of compact metric spaces.4.1. The extension property.
A map π : E → X between topological spaces hasthe extension property if, for every map φ : E → X , there is a map S = S φ : E → E such that φ = πS . Note that such a map π is necessarily onto (consider constantmaps φ ). The following theorem can be found in [14]. Theorem 4.1.
For any compact metric space X , there exists a map π : 2 N → X with the extension property. OME UNIVERSALITY RESULTS FOR DYNAMICAL SYSTEMS 7
This is a strengthening of the very well known result saying that every compactmetric space is a continuous image of 2 N . In fact, from this lemma one gets im-mediately the following result from [1]. (It seems that the two results were provedindependently at about the same time.) Corollary 4.2.
Every ( X, T ) ∈ D is a factor of some dynamical system of theform (2 N , S ) .Proof. Given π : E → X with the extension property, just consider the map φ := T π : 2 N → X . (cid:3) At this point, we digress a little bit by showing that a result of the same kindholds true for
Polish dynamical systems, the “universal” space being (as shouldbe expected) the Baire space N N . This may be quite well known, but we couldn’tlocate a reference. Proposition 4.3.
For any Polish space X , there exists a map π : N N → X withthe extension property. Consequently, any self-map T : X → X is a factor of aself-map of N N .Proof. We show that the “usual” construction of a continuous surjection from N N onto X produces a map with the extension property.Denoting by N < N the set of all finite sequences of integers, let ( V t ) t ∈ N < N be afamily of non-empty open subsets of X satisfying the following properties: • V ∅ = X and diam( V t ) < −| s | for all t = ∅ ; • V ti ⊆ V s for each t and all i ∈ N ; • S i ∈ N V ti = V t for all t .Let π : N N → X be the map defined by { π ( α ) } := T k ≥ V α | k . We show thatthis map π has the extension property.Let us fix a map φ : N N → X . Since φ is continuous, one can find, for each k ∈ N ,a family S k ⊆ N < N such that S { [ s ]; s ∈ S k } = N N and each set φ ([ s ]), s ∈ S k iscontained in some open set V t ( s ) with | t ( s ) | = k . Moreover, we may do this in sucha way that the following properties hold true: • if k ≥
2, every s ∈ S k is an extension of some e s ∈ S k − ; • if s ∈ S k is an extension of e s ∈ S e k , (with e k ≤ k ), then t ( s ) is an extensionof t ( e s ); • S k is an antichain, i.e. no s ∈ S k is a strict extension of an s ′ ∈ S k .(To get the third property, just remove from S k the unnecessary s , i.e. thoseextending some s ′ ∈ S k with s ′ = s .)Note that for each fixed k ∈ N , the family { [ s ]; s ∈ S k } is actually a partitionof N N because S k is an antichain; hence, for any α ∈ N N , we may denote by s k ( α )the unique s ∈ S k such that s ⊆ α . Now, define S : N N → N N as follows: if α ∈ N N then S ( α ) | k := t ( s k ( α )) for all k ∈ N . It is readily checked that S is continuousand φ = πS . (cid:3) Let us now come back to compact metrizable dynamical systems. From Corollary4.2, one can easily prove that there is a “universal extension” for all dynamicalsystems (
X, T ) ∈ D . However, this map is defined on a huge compact nonmetrizable space. In what follows, for any compact metric spaces X and Y , we endow C ( X, Y )with the topology of uniform convergence, which turns it into a Polish space.
DARJI AND MATHERON
Proposition 4.4.
There exists a compact -dimensional space E and a map U : E → E such that every ( X, T ) ∈ D is a factor of ( E , U ) .Proof. By Corollary 4.2, it is enough to show that there exists a 0-dimensionalcompact dynamical system ( E , U ) which is an extension of every Cantor dynamicalsystem (2 N , S ).Consider the space E := (2 N ) C (2 N , N ) endowed with the product topology. This isa 0-dimensional compact space. For any z = ( z S ) S ∈C (2 N , N ) ∈ E , we define U ( z ) ∈ E as follows: U ( z ) S = S ( z S ) for every S ∈ C (2 N , N ) . Then, U is continuous because each z U ( z ) S is continuous. Moreover, each S ∈ C (2 N , N ) is indeed a factor of U by the very definition of E and U : just considerthe projection map π S : E → N from E onto the S -coordinate. (cid:3) Remark 4.5.
If one just wants to lift those dynamical systems (
X, T ) ∈ D whichare transitive , i.e. have dense orbits, then one can take ( E , U ) := ( β Z + , σ ), where β Z + is the Stone- ˇCech compactification of Z + and σ : β Z + → β Z + is the uniquecontinuous extension of the map n n . Indeed, given a point x ∈ X with dense T -orbit, the map n T n ( x ) can be extended to a continuous map π T : β Z + → X showing that ( X, T ) is a factor of ( β Z + , σ ). Remark 4.6.
If ( E , U ) is a compact dynamical system which is universal for D inthe sense of Proposition 4.4, then the space E cannot be metrizable. Proof.
For any g ∈ T , let us denote by R g : T → T the associated rotation of T .We show that if ( E , U ) is a compact dynamical system that is “only” a commonextension of all rotations R g , then E already cannot be metrizable.Let us fix a point e ∈ E . Since E is assumed to be compact and metrizable,one can find an increasing sequence of integers ( i k ) such that the sequence U i k ( e )is convergent. Now, given g ∈ T , let π g : E → T be a map witnessing that U is an extension of R g . Setting ξ g := π g ( e ), we then have g i k ξ g = π g U i k ( e ) forall k ; and since π g is continuous, it follows that for every g ∈ T , the sequence g i k is convergent. But it is well known that this is not possible. (One can provethis as follows: if g i k → φ ( g ) pointwise, then R T φ ( g ) g − i k dg → (cid:3) Lifting to the Baire space.
As observed above, if one wants to lift too manymaps at the same time then one cannot do it with a compact metrizable space. Onthe other hand, our next result says that for “small families” of maps, one can finda common extension defined on a
Polish space.In what follows, for any compact metric spaces X and Y , we endow C ( X, Y )with the topology of uniform convergence, which turns it into a Polish space.
Theorem 4.7.
Let X be a compact metric space. Given any σ -compact family F ⊆ C ( X, X ) , one can find a map U F : N N → N N such that every T ∈ F is a factorof U F . This will follow from the next three lemmas, the first of which is a refinement ofTheorem 4.1.Let us say that a map π : E → X between compact metric spaces has the strongextension property if for every given map Φ : 2 N → C ( E, X ), there exists a map F : 2 N → C ( X, X ) such that for all p ∈ N we have that Φ( p ) = πF ( p ). OME UNIVERSALITY RESULTS FOR DYNAMICAL SYSTEMS 9
Lemma 4.8.
Let X be a compact metric space. Then, there is a map π : 2 N → X which has the strong extension property.Proof. We follow the proof of Theorem 4.1 given in [7, p. 110]. The constructionof the map π is the same as in that proof. We only need to verify that π has theadditional stronger property.We recall the construction of π . As in [7], one can easily construct inductively asequence { J i } of finite sets of cardinality at least 2 and collections { V s } and { W s } of nonempty open subsets of X such that the following conditions hold for all k ≥ s ∈ Q ki =1 J i : • the diameters of V s and W s are less than 2 − k ; • n V t ; t ∈ Q ki =1 J i o is a cover of X ; • { W sj ; j ∈ J k +1 } is a cover of V s ; • V sn = V s ∩ W sn .We let Λ := Q ∞ i =1 J i , which is homeomorphic to 2 N . For α ∈ Λ, we define π ( α )in the obvious way: { π ( α ) } = ∩ ∞ k =1 V α | k . Then, π is a well-defined map from Λonto X .Now let Φ : 2 N → C (Λ , X ) be a map. With each p ∈ N and Φ( p ) we willassociate a map F ( p ) : Λ → Λ as in [7]. However, we must do this in a continuousfashion.First, for each k ≥
1, we let 2 − k > ε k > ε k is aLebesgue number for the covering { W sj ; j ∈ J k +1 } of V s for all s ∈ Q ki =1 J i .For any fixed p ∈ N , the uniform continuity of Φ( p ) allows us to choose anincreasing sequence of positive integers { m k } such that for each k ≥ s ∈ Q m k i =1 J i , the diameter of the set Φ( p )([ s ]) is less than ε k /
4. (Here [ s ] denotes thoseelements of Λ which begin with s .) Moreover, as the map Φ is continuous, Φ(2 N )is a uniformly equicontinuous subset of C (Λ , X ); so we may in fact choose { m k } independent of p . Hence, we have an increasing sequence of positive integers { m k } such that for all p ∈ N and all k ≥ s ∈ Q m k i =1 J i , the diameter of the setΦ( p )([ s ]) is less than ε k / { l k } such that if p, p ′ ∈ N satisfy p | l k = p ′ | l k , then d (Φ( p ) , Φ( p ′ )) < ε k /
4. (Here, d is the “sup” metric on C (Λ , X ).) Note that this implies in particular that for all q ∈ { , } l k and s ∈ Q m k i =1 J i , the diameter of the set E s,q := (cid:8) Φ( p )([ s ]); p | l k = q (cid:9) is less than ε k . This, in turn, implies (by the choice of ε k ) that E s,q is containedin an open set V t for some t ∈ Q ki =1 J i ; and, moreover, that if E s,q was alreadyknown to be contained in V e t for some e t ∈ Q k − i =1 J i , then t may be chosen to be anextension of e t . Let us summarize what we have:(i) For all k ≥
1, for all q ∈ { , } l k and for all s ∈ Q m k i =1 J i , thereexists t = t ( q, s ) ∈ Q ki =1 J i such that for all p ∈ N extending q , wehave that Φ( p )([ s ]) ⊂ V t .(ii) Everything can be done recursively in such a way that if k ′ ≥ k , q ′ ∈ { , } l k ′ is an extension of q and s ′ ∈ Q m k ′ i =1 J i is an extensionof s , then t ′ = t ′ ( q ′ , s ′ ) is an extension of t .Let us now define the desired F : 2 N → C (Λ , Λ). Fix p ∈ N . Then, F ( p ) : Λ → Λis defined as follows. For each α ∈ Λ, we define F ( p )( α ) so that for all k ≥ F ( p )( α ) | k = t ( p | l k , α | m k ). By the above, F ( p ) is a well-defined function, and it is continuous. Moreover, F itself is continuous: if p, p ′ ∈ N with p | l k = p ′ | l k , then d ( F ( p ) , F ( p ′ )) < ε k < − k . That Φ( p ) = πF ( p ) for all p ∈ N follows in the samemanner as in [7]. (cid:3) Corollary 4.9.
Let X be a compact metric space and let M be a compact subsetof C ( X, X ) . Then, there is a compact set N ⊆ C (2 N , N ) such that every T ∈ M isa factor of some S ∈ N .Proof. Since M is compact and metrizable, there is a map G : 2 N → C ( X, X ) suchthat M = G (2 N ). Now, choose a map π : 2 N → X having the strong extensionproperty, and consider the map Φ : 2 N → C (2 N , X ) defined by Φ( p ) = G ( p ) π .Let F : 2 N → C (2 N , N ) be a map such that Φ( p ) = πF ( p ) for all p ∈ N . Then G ( p ) π = Φ( p ) = πF ( p ), so that G ( p ) is a factor of F ( p ) for every p ∈ N (because π is onto). Hence, we may take N := F (2 N ). (cid:3) Lemma 4.10.
Let N be a compact subset of C (2 N , N ) . Then, there exists map U N : N N → N N such that every S ∈ N is factor of U N .Proof. Recall first that the Polish space C (2 N , N ) is homeomorphic to N N . Indeed,it is easy to check that C (2 N , N ) is 0-dimensional (because 2 N is), and it is also easyto convince oneself that every equicontinuous family of self-maps of 2 N has emptyinterior in C (2 N , N ), so that C (2 N , N ) is nowhere locally compact.As N is a compact subset of C (2 N , N ) ≃ N N , we may enlarge it if necessary andassume that N is homeomorphic to 2 N . Then Z := C ( N , N ) is homeomorphic to N N . Now we define U N : Z → Z as in the proof of Proposition 4.4 (changing thenotation slightly): if z ∈ Z then U N ( z )( S ) := S ( z ( S )) for every S ∈ N . Then U N is easily seen to be continuous (recall that Z = C ( N , N ) is endowed with thetopology of uniform convergence), and the evaluation map π S at S shows that each S ∈ N is a factor of U N . (To see that π S : Z → N is onto, consider constant maps z ∈ Z .) (cid:3) Lemma 4.11.
Let Z := 2 N or N N , and let U , U , . . . be maps from Z to Z . Then,there is a map U : Z → Z such that each U n is a factor of U .Proof. The only property needed on the space Z is that Z is homeomorphic to Z N . Define U : Z N → Z N by U ( z , z , . . . ) := ( U ( z ) , U ( z ) , . . . ). Then, T iscontinuous. Moreover, if π n : Z N → Z is the projection of Z N onto the n -thcoordinate, we have that π n U = U n π n for all n ≥
1. Identifying Z N with Z , thisshows that U has the required property. (cid:3) Proof of Theorem 4.7.
Let F be a σ -compact of C ( X, X ), and write F = S n M n where M n are compact sets. By Corollary 4.9, we may choose compact sets N n ⊆C (2 N , N ) such that every T ∈ M n is a factor of some S ∈ N n . By Lemma 4.10,we may choose for each n a map U n : N N → N N such that every S ∈ N n is afactor of U n . Now, by Lemma 4.11, there is a map U : N N → N N such that eachmap U n is factor of U . Since factors of factors are again factors, we may thus take U F := U . (cid:3) Lifting to the Cantor space.
The appearance of the Baire space N N inTheorem 4.7 may look rather surprising since we are dealing with self-maps of compact metric spaces. This suggests the following question. OME UNIVERSALITY RESULTS FOR DYNAMICAL SYSTEMS 11
Question 4.12.
Let X be a compact metric space. For which families F ⊆ C ( X, X )is it possible to find a common extension U defined on the Cantor space 2 N ?Let us denote by I the family of all subsets F of C ( X, X ) having the aboveproperty. It follows from Lemma 4.11 that I is closed under countable unions; andclearly, I is hereditary for inclusion. Hence, I is a σ -ideal of subsets of C ( X, X ).Since σ -ideals of sets are quite classical objects of study, it might be interesting toinvestigate the structural properties of this particular example.Having said that, we are quite far from being able to answer Question 4.12in a satisfactory way. However, in this section we give a rather special sufficientcondition for belonging to I .In what follows, we fix a compact metric space ( X, d ), and we also denote by d the associated “sup” metric on C ( X, X ).For each i ∈ Z + , let us denote by e i : C ( X, X ) → C ( X, X ) the map defined by e i ( S ) := S i = S ◦ · · · ◦ S | {z } i times . We shall say that a family
N ⊆ C ( X, X ) has uniformly controlled powers if thesequence { e i | N ; i ≥ } ⊆ C ( N , C ( X, X )) is pointwise equicontinuous; in otherwords, if whenever a sequence ( S k ) ⊆ N converges to some S ∈ N , it holds that(1) S ik ( x ) k →∞ −−−−→ S i ( x ) uniformly with respect to x and i. Example 4.13.
Obviously, every finite family
N ⊆ C ( X, X ) has uniformly con-trolled powers. Example 4.14.
Let N be a compact subset of C ( X, X ) , and assume the eachmap S ∈ N has the following properties: S is -Lipschitz with a unique fixedpoint α ( S ) in X , and S i ( x ) → α ( S ) for every x ∈ X . (This holds for example if d ( S ( x ) , S ( y )) < d ( x, y ) whenever x = y .) Then N has uniformly controlled powers.Proof. Observe first that the map S α ( S ) is continuous on N (because it has aclosed graph and acts between compact spaces). Moreover, S i ( x ) → α ( S ) uniformly with respect to S and x as i → ∞ . (This follows from Dini’s theorem applied tothe continuous functions Φ i : N × X → R defined by Φ i ( S, x ) := d ( S i ( x ) , α ( S )).The sequence (Φ i ) is nonincreasing because each S ∈ N is 1-Lipschitz.) So there isa sequence ( ε i ) tending to 0 such that d ( S i ( x ) , α ( S )) ≤ ε i for all x , i and S ∈ N . Now, let ( S k ) be a sequence in N converging (uniformly) to some map S . Set α k := α ( S k ) and α := α ( S ). For every x and i , we have on the one hand d ( S ik ( x ) , S i ( x )) ≤ d ( S ik , S i ) , and on the other hand d ( S ik ( x ) , S i ( x )) ≤ d ( S ik ( x ) , α k ) + d ( α k , α ) + d ( α, S i ( x )) ≤ ε i + d ( α k , α ) . Since α k → α and S ik → S i for each fixed i ∈ Z + , this gives (1). (cid:3) Proposition 4.15.
Let
F ⊆ C ( X, X ) . Assume that F can be covered by count-ably many compact sets N having uniformly controlled powers. Then F admits acommon extension U F defined on N . Proof.
By Lemma 4.11, it is enough to prove the result for each one of the compactfamilies N . Moreover, since every self-map of a compact metric space is a factor ofsome self-map of 2 N , it is enough to show that N admits a common extension U defined on some compact metric space E .For any i ∈ Z + and a ∈ X , denote by e i,a : N → X the map defined by e i,a ( S ) := S i ( a ). Since N has uniformly controlled powers, the family of all suchmaps e i,a is pointwise equicontinuous on N ; so its uniform closure E is a compactsubset of Z = C ( N , X ), by Ascoli’s theorem. Note also that E contains all constantmaps because { e i,a } already does (take i = 0 and let a vary).Now, consider the map U N : Z → Z defined in the proof of Lemma 4.10, U N ( z )( S ) = S ( z ( S )) . The family { e i,a } is invariant under U N , because U N ( e i,a ) = e i +1 ,a . Hence, E is also invariant under U N . Moreover, since E contains all constant maps, eachevaluation map π S : E → X is onto. So the map U := U N | E has the requiredproperty. (cid:3) Corollary 4.16.
The family F of all d -contractive self-maps of X has a commonextension defined on N .Proof. For each c ∈ (0 , N c the family of all c -Lipschitz self-mapsof X . Then N c is compact by Ascoli’s theorem, and it has uniformly controlledpowers by Example 4.14. Since F = S n ∈ N N − n , the resut follows. (cid:3) Remark 4.17.
In general, the family of all 1-Lipschitz self-maps of X , or even thefamily of all isometries of X , does not have a common extension to 2 N . Considerfor example the rotations of X := T (see the proof of Remark 4.6).We conclude this section with a kind of abstract nonsense characterizing thebelonging to a subclass of the σ -ideal I . Fact 4.18.
Let N be a compact subset of C ( X, X ) , and let U N : C ( N , X ) →C ( N , X ) be the (usual) map defined by U N ( z )( S ) = S ( z ( S )) . The following areequivalent. (i) There is a map U : 2 N → N and a continuous map S q S from N into C (2 N , X ) such that U is an extension of each S ∈ N , with factoring map q S . (ii) There is a compact set E ⊆ C ( N , X ) invariant under U N such that, foreach S ∈ N , the evalution map z z ( S ) maps E onto X .Proof. Since every compact metrizable dynamical system has an extension to 2 N ,the implication (ii) = ⇒ (i) follows easily from the proof of Proposition 4.15.Conversely, assume that (i) holds true, and denote by q : N → C (2 N , X ) themap S q S . For any p ∈ N , denote by δ p : C (2 N , X ) → X the evaluation mapat p . Then, the map p δ p ◦ q is continuous from 2 N into C ( N , X ). So theset E := { δ p ◦ q ; p ∈ N } is a compact subset of C ( N , X ). Moreover, for any z = δ p ◦ q ∈ E and all S ∈ N we have U N ( z ) ( S ) = S ( z ( S )) = S ( q S ( p )) = q S ( U ( p )) = δ U ( p ) ◦ q ( S );and this show that E is invariant under the map U N . Finally, the evaluation mapat any S ∈ N maps E onto X because q S is onto. (cid:3) OME UNIVERSALITY RESULTS FOR DYNAMICAL SYSTEMS 13 Common generalized extensions
In this section, we show that something similar to Theorem 4.7 does hold truewith 2 N in place of N N , but with a weaker notion of “factor”.For any compact metric space X , let K ( X ) be the space of all nonempty compactsubsets of X endowed with the Haudorff metric. If T : X → X is a map, wedenote again by T : K ( X ) → K ( X ) the map induced by T on K ( X ); that is, T ( M ) = { T ( x ); x ∈ M } for every M ∈ K ( X ). Obviously, ( X, T ) is equivalent to asubsystem of ( K ( X ) , T ) thanks to the embedding X ∋ x
7→ { x } ∈ K ( X ). However,in general ( X, T ) is not a factor of ( K ( X ) , T ). For example, this cannot be the caseif T is onto (so that X is a fixed point of T in K ( X )) and T has no fixed point in X .We shall say that ( X, T ) is a a generalized factor of a dynamical system (
E, S )if there exists an upper semi-continuous mapping π : E → K ( X ) such that thefollowing conditions hold: • { x } ∈ π ( E ) for every x ∈ X ; • π ( S ( u )) = T ( π ( u )) for all u ∈ E .(Recall that π : E → K ( X ) is upper semicontinuous if the relation x ∈ π ( u ) isclosed in E × X ; equivalently, if for any open set O ⊆ X , the set { u ∈ E : π ( u ) ⊆ O } is open in E .)It is obvious from the definition that “true” factors are generalized factors. Also,we have the expected transitivity property: Remark 5.1.
If (
X, T ) is a generalized factor of (
Y, S ) and (
Y, S ) is a generalizedfactor of (
Z, R ), then (
X, T ) is a generalized factor of (
Z, R ). Proof.
Let π : Y → K ( X ) witness that T is a generalized factor of S and let π ′ : Z →K ( Y ) witness the fact that S is a generalized factor of R . Define π ∗ : K ( Y ) → K ( X )by π ∗ ( M ) := S { π ( y ) : y ∈ M } . Then, π ∗ is a well-defined upper semicontinuousmapping. Now, the mapping γ : Z → K ( X ) defined by γ := π ∗ ◦ π ′ is is easilychecked to be again upper semicontinuous, and shows that T is a generalized factorof R . (cid:3) Theorem 5.2.
Let X be a compact metric space. Given any σ -compact family F ⊆ C ( X, X ) , one can find a map U F : 2 N → N such that every T ∈ F is ageneralized factor of U F . The proof will use Corollary 4.9, Lemma 4.11 and the following variant of Lemma4.10.
Lemma 5.3.
Let N be a compact subset of C (2 N , N ) . Then, there exists map U N : 2 N → N such that every S ∈ N is generalized factor of U N .Proof. We assume of course that
N 6 = ∅ . Let us denote by Z be the set of allcompact subsets of N × N whose projection on the first coordinate is all of N .This is a closed subset of K ( N × N ). Moreover, Z is 0-dimensional because N × N is, and it is easily seen to have no isolated points. Hence, Z is homeomorphic to2 N .We define U N : Z → Z in the following manner. Let A ∈ Z and for each S ∈ N ,let A S be the vertical cross section of A at S , i.e., A S = { x ∈ N : ( S, x ) ∈ A } . We define the set U N ( A ) ⊆ N × N by its vertical cross sections: U N ( A ) S := S ( A S )for every S ∈ N . In other words, U N ( A ) = [ S ∈N { S } × S ( A S ) = { ( S, S ( x )); ( S, x ) ∈ A } . Let us check that U N ( A ) ∈ Z . That the projection of U N ( A ) on the first coor-dinate is all of N is obvious because A S = ∅ for every S ∈ N . Let (cid:8) ( S n , u n ) (cid:9) be asequence in U N ( A ) converging to some point ( S, u ) ∈ N × N . Then u n = S n ( x n )for some x n ∈ N such that ( S n , x n ) ∈ A . By compactness, we may assume that x n converges to some x ∈ N . Then ( S, x ) ∈ A because A is closed in N × N ;and u = S ( x ) because u n → u and S n → S uniformly. So u ∈ S ( A S ), i.e. ( S, u ) ∈ U N ( A ). Thus, we see that U N ( A ) is a closed and hence compact subset of N × N .The same kind of reasoning shows that U N : Z → Z is continuous. One has tocheck that if A n → A in K ( N × N ) then: (i) for every ( S, u ) ∈ U N ( A ), one canfind a sequence (cid:8) ( S n , u n ) (cid:9) with ( S n , u n ) ∈ U N ( A n ) such that ( S n , u n ) → ( S, u );and (ii) for every sequence (cid:8) ( S n k , u n k ) (cid:9) with ( S n k , u n k ) ∈ U N ( A n k ) converging tosome ( S, u ) ∈ N × N , we have ( S, u ) ∈ U N ( A ). Both points follow easily from thedefinition of U N and the fact that A n → A .Now, fix S ∈ N and consider π S : Z → K (2 N ) defined by π S ( A ) := A S . This isan upper semicontinuous mapping; and π S U N = Sπ S by the very definition of U N .Moreover, { x } = π S ( N × { x } ) for every x ∈ N . This shows that S is a generalizedfactor of U N . (cid:3) Proof of Theorem 5.2.
One can now proceed exactly as in the proof of Theorem4.7 (using Lemma 5.3 instead of Lemma 4.10), keeping in mind that factors aregeneralized factors and that generalized factors of generalized factors are againgeneralized factors. (cid:3)
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E-mail address , Darji: [email protected]
E-mail address , Matheron: [email protected] (Darji)
Department of Mathematics, University of Louisville, Louisville, KY 40292,USA. (Matheron)(Matheron)