On semi-open codes and bi-continuing almost everywhere codes
aa r X i v : . [ m a t h . D S ] M a y ON SEMI-OPEN CODES AND BI-CONTINUING ALMOSTEVERYWHERE CODES
D. AHMADI DASTJERDI AND S. JANGJOOYE SHALDEHI
Abstract.
We will show that a system is synchronized if and only if it has acover whose cover map is semi-open. Also, any factor code on an irreduciblesofic shift is semi-open and the image of a synchronized system by a semi-opencode is synchronized. On the other side, right-closing semi-open extension ofan irreducible shift of finite type is of finite type. Moreover, we give conditionson finite-to-one factor codes to be open and show that any semi-open code ona synchronized system is bi-continuing a.e.. We give some sufficient conditionsfor a right-continuing a.e. factor code being right-continuing everywhere. Introduction
A map φ : X → Y is called open map if images of open sets are open and iscalled semi-open (quasi-interior) if images of open sets have non-empty interior.In dynamical systems, they are of interest when they appear as factor map or asit is called factor code in symbolic dynamics. There are many classes of open andspecially semi-open factor maps. For instance, any factor map between minimalcompacta is semi-open [3] and all surjective cellular automata are semi-open [12,Theorem 2.9.3]. As one may expect, there are some strict restrictions for a factormap being open; nonetheless, some important classes do exist. As an example, if X and Y are compact minimal spaces, there are almost one-to-one minimal extensions X ′ and Y ′ by ψ and ψ respectively and an open factor map φ ′ : X ′ → Y ′ suchthat ψ ◦ φ ′ = φ ◦ ψ [2]. Here, our investigation is mainly confined on semi-openfactor codes between shift spaces. For open factor codes between shift spaces see[9, 10]. A stronger notion to a semi-open map is the notion of an irreducible mapwhich will be of our interest as well. A map φ : X → Y is called irreducible if theonly closed set A ⊆ X for which φ ( A ) = φ ( X ) is A = X .A summary of our results is as follows. In section 3, we show that a coded system X is synchronized if and only if there is a cover G = ( G, L ) so that its cover map L ∞ is semi-open and if G = ( G, L ) is an irreducible right-resolving cover for X witha magic word, then L ∞ is semi-open (Theorem 3.7). Also, when G is Fischer cover, L ∞ is irreducible (Theorem 3.8). Furthermore, any factor code on an irreduciblesofic shift is semi-open (Corollary 3.11). Theorem 4.2 implies that a right-closingsemi-open extension of an irreducible shift of finite type is of finite type. Theorem4.3 shows that the image of a synchronized system is again synchronized undera semi-open code and so providing a class of non-semi-open codes, mainly, codesfactoring synchronized systems over non-synchronized ones. Mathematics Subject Classification.
Key words and phrases. semi-open, almost one-to-one, shift of finite type, sofic, synchronized,continuing code.
We also consider bi-continuing and bi-continuing almost everywhere (a.e.) codes.A right-continuing or u-eresolving code is a code which is surjective on each unstableset and it is a natural dual version of a right-closing code and plays a fundamentalrole in the class of infinite-to-one codes [7]. In Theorem 4.13, we give conditions onbi-continuing a.e. factor codes to be open and also by Corollary 4.14 and Theorem4.15, we give some sufficient conditions for a right-continuing a.e. factor codebeing right-continuing everywhere. Finally, we show that any semi-open code on asynchronized system is bi-continuing a.e. (Theorem 4.16).2.
Background and Notations
Let A be a non-empty finite set. The full A -shift denoted by A Z , is the collectionof all bi-infinite sequences of symbols from A . A block (or word) over A is afinite sequence of symbols from A . The shift map on A Z is the map σ where σ ( { x i } ) = { y i } is defined by y i = x i +1 . The pair ( A Z , σ ) is called the full shift andany closed invariant set of that is called a shift space over A .Denote by B n ( X ) the set of all admissible n -words and let B ( X ) = S ∞ n =0 B n ( X )be the language of X . For u ∈ B ( X ), let the cylinder [ u ] be the set { x ∈ X : x [ l,l + | u |− = u } . For l ≥ | u | = 2 l + 1, [ u ] is called a central l + 1 cylinder.Let A and D be alphabets and X a subshift over A . For m, n ∈ Z with − m ≤ n ,define the ( m + n + 1) -block map Φ : B m + n +1 ( X ) → D by(2.1) y i = Φ( x i − m x i − m +1 ...x i + n ) = Φ( x [ i − m,i + n ] )where y i is a symbol in D . The map φ = Φ [ − m,n ] ∞ : X → D Z defined by y = φ ( x )with y i given by 2.1 is called the code induced by Φ. If m = n = 0, then φ is called1 -block code and φ = Φ ∞ . An onto code φ : X → Y is called a factor code .A point x in a shift space X is doubly transitive if every block in X appears in x infinitely often to the left and to the right. Let φ : X → Y be a factor code. Ifthere is a positive integer d such that every doubly transitive point of Y has exactly d pre-images under φ , then we call d the degree of φ .A code φ : X → Y is called right-closing (resp. right-continuing ) if whenever x ∈ X , y ∈ Y and φ ( x ) is left asymptotic to y , then there exists at most (resp. atleast) one x ∈ X such that x is left asymptotic to x and φ ( x ) = y . A left-closing (resp. left-continuing ) code is defined similarly. If φ is both left and right-closing(resp. continuing), it is called bi-closing (resp. bi-continuing ). An integer n ∈ Z + is called a (right-continuing) retract of a right-continuing code φ : X → Y if,whenever x ∈ X and y ∈ Y with φ ( x ) ( −∞ , = y ( −∞ , , we can find x ∈ X suchthat φ ( x ) = y and x ( −∞ , − n ] = x ( −∞ , − n ] [9].Let G be a directed graph and V (resp. E ) the set of its vertices (resp. edges)which is supposed to be countable. An edge shift , denoted by X G , is a shift spacewhich consist of all bi-infinite sequences of edges from E . A graph G is called locallyfinite , if it has finite out-degree and finite in-degree at any vertex. Recall that X G is locally compact if and only if G is locally finite.Let v ∼ w be an equivalence relation on V whenever there is a path from v to w and vice versa. For an equivalence class, consider all vertices together with alledges whose endpoints are in that equivalence class. Thus a subgragh called the(irreducible) component of G associated to that class arises; and if it is non-empty,then an irreducible subshift of X G , called (irreducible) component of X G is defined. N SEMI-OPEN CODES AND BI-CONTINUING ALMOST EVERYWHERE CODES 3
There is no subshift of X G which is irreducible and contains a component of X G properly.A labeled graph G is a pair ( G, L ) where G is a graph and L : E → A its labeling.Associated to G , a space X G = closure {L ∞ ( ξ ) : ξ ∈ X G } = L ∞ ( X G )is defined and G is called a presentation (or cover ) of X G . When G is a finite graphand hence compact, X G = L ∞ ( X G ) is called a sofic shift . Call F ( I ) = { u : u is thelabel of some paths starting at I } the follower set of I .An irreducible sofic shift is called almost-finite-type (AFT) if it has a bi-closingpresentation [11]. A (possibly reducible) sofic shift that has a bi-closing presentationis called an almost Markov shift. A shift space X has specification with variablegap length (SVGL) if there exists N ∈ N such that for all u, v ∈ B ( X ), there exists w ∈ B ( X ) with uwv ∈ B ( X ) and | w | ≤ N .A word v ∈ B ( X ) is synchronizing if whenever uv, vw ∈ B ( X ), we have uvw ∈B ( X ). An irreducible shift space X is a synchronized system if it has a synchronizingword. Let G = ( G, L ) be a labeled graph. A word w ∈ B ( X G ) is a magic word ifall paths in G presenting w terminate at the same vertex.A labeled graph G = ( G, L ) is right-resolving if for each vertex I of G the edgesstarting at I carry different labels. A minimal right-resolving presentation of asofic shift X is a right-resolving presentation having the fewest vertices among allright-resolving presentations of X . It is unique up to isomorphism [11, Theorem3.3.18] and is called the Fischer cover of X .Now we review the concept of the Fischer cover for a not necessarily sofic systemdeveloped in [8]. Let x ∈ B ( X ) and call x + = ( x i ) i ∈ Z + (resp. x − = ( x i ) i< ) the right (resp. left) infinite X -ray . For x − , its follower set is defined as ω + ( x − ) = { x + ∈ X + : x − x + is a point in X } . Consider the collection of all follower sets ω + ( x − ) as the set of vertices of a graph X + . There is an edge from I to I labeled a if and only if there is a X -ray x − such that x − a is a X -ray and I = ω + ( x − ), I = ω + ( x − a ). This labeled graph is called the Krieger graph for X . If X is asynchronized system with synchronizing word α , the irreducible component of theKrieger graph containing the vertex ω + ( α ) is called the (right) Fischer cover of X .The entropy of a shift space X is defined by h ( X ) = lim n →∞ (1 /n ) log |B n ( X ) | .A component X of a shift space X is called maximal if h ( X ) = h ( X ).3. open, Semi-open and irreducible Codes We start with some necessary lemmas.
Lemma 3.1. [5, Lemma 1.4]
Assume φ : X → Y and ψ : Y → Z are surjectionssuch that ψ ◦ φ is semi-open. Then, ψ is semi-open as well. Moreover, if ψ isirreducible, then also φ is semi-open. A continuous map φ : X → Y is almost 1-to-1 if the set of the points x ∈ X such that φ − ( φ ( x )) = { x } is dense in X . Lemma 3.2.
Let φ : X → Y and ψ : Y → Z be two factor codes such that ψ ◦ φ isirreducible. Then, φ and ψ are irreducible.Proof. If φ or ψ is not irreducible, then it is not almost 1-to-1 [14, Theorem 10.2]and so ψ ◦ φ cannot be almost 1-to-1 either, a contradiction. (cid:3) D. AHMADI DASTJERDI AND S. JANGJOOYE SHALDEHI Figure 1.
The cover of X . Figure 2.
The Fischer cover of the golden shift.
Lemma 3.3. [10, Lemma 1.2]
A code φ : X → Y between shift spaces is open ifand only if for each l ∈ N , there is k ∈ N such that whenever x ∈ X , y ∈ Y and φ ( x ) [ − k, k ] = y [ − k, k ] , we can pick x ∈ X with x [ − l, l ] = x [ − l, l ] and φ ( x ) = y . So if φ is open, then for any l ∈ N we can find k ∈ N such that the image ofa central 2 l + 1 cylinder in X consists of central (2 k + 1) cylinders in Y . We saythat φ has a uniform lifting length if for each l ∈ N , there exists k satisfying theabove property such that sup l | k − l | < ∞ . Let us formally state a definition for asemi-open code. Definition 3.4.
A code φ : X → Y between shift spaces is semi-open if for each l ∈ N , there is k ∈ N such that the image of a central 2 l + 1 cylinder in X containsa central (2 k + 1) cylinder in Y .Note that the definition of uniform lifting length extends naturally to semi-opencodes. Lemma 3.5.
Let X and Y be shift spaces with Y irreducible. If φ : X → Y is asemi-open code, then it is onto and so a factor code.Proof. Since φ is semi-open, φ ( X ) contains a non-empty open set U ⊆ Y . There-fore, since ( Y, σ ) is topologically transitive, ∪ ∞ n = −∞ σ n ( U ) = Y [13]. On the otherhand, φ ( X ) is a σ -invariant subset of Y . Thus φ is onto. (cid:3) Our next results give situations where the factor code on covers are semi-open.In all of them, irreducibility is an important hypothesis. The following exampleshows that non-semi-openness may occur quite easily in reducible covers.
Example 3.6.
Let X be a shift space presented by Figure 1 and Y the golden shiftgiven by Figure 2. Define Φ : B ( X ) → { , } as,Φ( w ) = (cid:26) w = 1 , , and let φ : X → Y be the code induced by Φ. Since [2] = 2 ∞ , φ ([2]) = 0 ∞ and thisshows that φ is not semi-open. Note that φ is a finite-to-one code. N SEMI-OPEN CODES AND BI-CONTINUING ALMOST EVERYWHERE CODES 5
Theorem 3.7.
A coded system X is synchronized if and only if there is a cover G = ( G, L ) so that L ∞ is semi-open. In fact, any irreducible right-resolving coverof a synchronized system with a magic word is semi-open.Proof. Let X be synchronized with a magic word α and let G = ( G, L ) be anirreducible right-resolving cover for X . Suppose [ π ] = [ e − l · · · e l ] ∈ B ( X G ) is acentral 2 l +1 cylinder in X G with L ( e i ) = a i , − l ≤ i ≤ l ; thus L ∞ ([ π ]) ⊆ [ a − l · · · a l ].Choose λ ∈ B ( X G ) such that L ( λ ) = α . Since X G is irreducible, there are twopaths γ and ξ in G such that π ′ = λξπγλ ∈ B ( X G ). Now set w = L ( π ′ ) = α L ( ξ ) a − l · · · a l L ( γ ) α and note that by the fact that α is magic and G is right-resolving, if w = L ( π ′′ ), then π ′′ = π ′ . So [ w ] ⊆ L ∞ ([ π ′ ]) and we are done.For the converse let G = ( G, L ) be a cover for X with L ∞ semi-open. Since X G is a Polish space, L ∞ ( X G ) is analytic and so there is an open set O ⊆ X and a firstcategory set P ⊆ X so that L ∞ ( X G ) = O △ P . Therefore, L ∞ ( X G ) = ∪ n ∈ Z σ n ( O △ P ) ⊇ ∪ n ∈ Z σ n ( O ) − ∪ n ∈ Z σ n ( P ) . Since L ∞ is semi-open, O is non-empty. This means that the first union on right isa non-empty open dense set. As a result, L ∞ ( X G ) is residual and X synchronized[8, Theorem 1.1]. (cid:3) When G is the Fischer cover for a synchronized system, we will have a strongerconclusion. Theorem 3.8. If X is a synchronized system with the Fischer cover G = ( G, L ) ,then L ∞ is irreducible.Proof. Let F be a proper closed subset of X G . Then, F ′ = X G \ F is open and thereis a doubly transitive point x ∈ F ′ . But the degree of L ∞ is one [11, Proposition9.1.6] and so L ∞ ( x )
6∈ L ∞ ( F ) which means that L ∞ ( F ) = X . (cid:3) For sofics, when considering finite covers, a rather weaker hypothesis holds thesame conclusion as Theorem 3.7.
Theorem 3.9.
Suppose X is sofic and G = ( G, L ) an irreducible finite cover.Then, L ∞ is semi-open.Proof. Suppose L ∞ is not semi-open. So there is a cylinder [ u ] in X G such that L ∞ ([ u ]) does not have any central cylinder. Without loss of generality, assume that | u | = 1. Then, there must exist another edge u ′ ∈ E ( G ) such that L ( u ′ ) = L ( u );otherwise, [ L ( u )] ⊆ L ∞ ([ u ]) and this contradicts our assumption. But to ruleout [ L ( u )] ⊆ L ∞ ([ u ]) completely, one must have either F ( L ( u )) ( F ( L ( u ′ ))or P ( L ( u )) ( P ( L ( u ′ )). Assume the former and suppose f ∈ F ( u ). Then,there must be an edge f ′ ∈ F ( u ′ ) such that L ( f ′ ) = L ( f ) and yet another path f ′′ ∈ F ( u ′ ) such that L ( f ′′ ) F ( I ) where I = t ( u ).Now choose any path u = u w u ′ , t ( u ) = t ( u ′ ) = I and observe that sincewe cannot have [ L ( u )] ⊆ L ∞ ([ u ]), there is another path u ′ labeled as u whoselast edge is u ′′ with L ( u ′′ ) = L ( u ′ ) = L ( u ) such that I = t ( u ′′ ) has at least onepath more than I . Construct I n inductively. Then, the outer edges of I n increasesby n which is absurd for a finite graph such as G . (cid:3) The following example shows that Theorem 3.9 does not hold for an infiniteirreducible cover.
D. AHMADI DASTJERDI AND S. JANGJOOYE SHALDEHI · · · I − I − I − I I I I · · · Figure 3.
A cover of the full shift on { , , } which is not semi-open. Example 3.10.
Let G = ( G, L ) be the cover shown in Figure 3 and e ∈ E ( G ) with i ( e ) = I and t ( e ) = I . Note that X G is a full-shift on { , , } . One way to seethis is that for r ∈ N and m i ∈ N and any a j ∈ { , , } , we have a path labeled a m a m · · · a m r r . This implies that for any periodic point p ∞ ∈ { , , } Z , we havea path π such that L ( π ) = p . The conclusion follows since the periodic points aredense in our space.Now suppose L ∞ ([ e ]) contains a central 2 n + 1 cylinder [ b − n · · · b n ] with b = 0.By the fact that our system is a full-shift, the point x = · · · n +2 b − n · · · . · · · b n · · · ∈ [ b − n · · · b n ]; however x
6∈ L ∞ ([ e ]). This is because, we have i ( b − n ) = I i , − n + 1 ≤ i ≤ n + 1. For instance, if i = n + 1, then b − = b − = · · · b − n = 0 and if0 ≤ i < n + 1, then at least one b j equals 2. Now let A j = { i ∈ N : 2 i ∈ P ( I j ) } for j ∈ Z where P ( I j ) is the collection of labels of paths terminating at I j . Then, A − = A = ∅ and max A j = j for j ∈ N and max A j = − j − j ∈ {− n : n ≥ } .So if a path π is labeled 2 n +2 b − n · · · .
0, then t ( π ) = I . This means that φ is notsemi-open. Corollary 3.11.
Suppose X is an irreducible sofic and let φ : X → Y be a factorcode. Then, φ is semi-open.Proof. Let G = ( G, L ) be the Fischer cover of X . Then, φ ◦L ∞ is an irreducible finitecover for Y and by Theorem 3.9, it is semi-open. Now the proof is a consequenceof Lemma 3.1. (cid:3) Theorem 3.12.
Let T : X → X and S : Y → Y be transitive homeomorphismsand φ : X → Y a factor map. Suppose there is an open set U such that for anyopen set V ⊆ U , int ( φ ( V )) = ∅ . Then, φ is semi-open.Proof. Let U be an arbitrary open set. Since T is transitive, there is an integer n such that U ∩ T − n ( U ) is a non-empty open set. By hypothesis, there is open set V ⊆ φ ( U ∩ T − n ( U )) and for that S n ( V ) ⊆ S n φ ( U ∩ T − n ( U )) = φT n ( U ∩ T − n ( U )) = φ ( T n ( U ) ∩ U ) ⊆ φ ( U ) . and since S is homeomorphism, S n ( V ) is open. (cid:3) An immediate consequence of this theorem is that if X and Y are subshifts with T and S shift maps, then image of any cylinder under φ has empty interior if andonly if there is just one cylinder in X with empty interior under φ . Example 3.10is a good example to see this fact.4. Open vs semi-open codes
The next two theorems say that [10, Proposition 3.3] and [10, Proposition 3.4]that were stated for open codes are actually valid for semi-open codes as well. The
N SEMI-OPEN CODES AND BI-CONTINUING ALMOST EVERYWHERE CODES 7 main ingredient in the proof is to have an open set lying in the image of any openset which is supplied by any semi-open map.
Theorem 4.1.
Let X be a shift of finite type, Y an irreducible sofic shift and φ : X → Y a finite-to-one semi-open code. Then, X is non-wandering and allcomponents are maximal. Example 3.6 shows that semi-openness is required in the hypothesis of the abovetheorem.
Theorem 4.2.
Let φ : X → Y be a right-closing semi-open code from a shift space X to an irreducible shift of finite type Y . Then, X is a non-wandering shift of finitetype. By Theorem 3.7, unlike open codes [10, Corollary 4.3], a right-closing semi-openextension of an irreducible strictly almost Markov (resp. non-AFT sofic) shift isnot necessarily strictly almost Markov (resp. non-AFT sofic).We have some similarities between open and semi-open codes. For instance, acoded system X is SFT if and only if there is a cover G = ( G, L ) so that L ∞ isopen which can be deduced from [11, Proposition 3.1.6] and [10, Proposition 2.3].A similar result for semi-open codes is Theorem 3.7. Also, SFT is invariant underopen codes as synchronized systems under semi-open codes: Theorem 4.3.
Suppose X is a synchronized system and φ : X → Y a semi-opencode. Then, Y is synchronized.Proof. The Fischer cover of X will be a cover for Y whose cover map is semi-open;this fact and the conclusion is supplied by Theorem 3.7. (cid:3) Note that any coded system Y can be a factor of some synchronized systems [6,Proposition 4.1]. In particular, if Y is not synchronized, then by the above theorem,the associated code is not semi-open. Example 4.4.
Unlike codes between irreducible sofic shifts which are all semi-opens (Corollary 3.11), there can be a non-semi-open code between two synchro-nized systems. For an example, let Y ′ be a non-synchronized system on A ′ = { , , . . . , k − } generated by a prefix code C ′ Y ′ = { ua ′ , w , . . . } , a ′ ∈ A ′ asin [6, Proposition 4.1]. This means if v ∈ C ′ Y ′ , then no other element of C ′ Y ′ starts with v and we know that any coded system has a prefix code [6]. Let φ ′ : X ′ → Y ′ be the one-block code between the synchronized system X ′ gen-erated by C ′ X ′ = { uz, w , . . . } , z
6∈ A ′ ∪ { k } = { , , . . . , k } and φ ′ induced by theblock map Φ ′ defined as(4.1) Φ ′ ( x ) i = (cid:26) x i x i ∈ A ′ ,a ′ x i = z. By Theorem 4.3, φ ′ is not semi-open.Now let X and Y be the synchronized systems generated by C X = C ′ X ′ ∪ { k } and C Y = C ′ Y ′ ∪ { k } respectively. Here, k is a synchronizing word for both systems.Also, define φ : X → Y as φ ′ when restricted to X ′ and in other places by Φ( k ) = k .Then, since φ ′ is not semi-open, φ is not semi-open either.Jung in [10] shows that when φ is a finite-to-one code from a shift of finite type X to an irreducible sofic shift Y , then φ is open if and only if it is constant-to-one.For semi-open codes we have D. AHMADI DASTJERDI AND S. JANGJOOYE SHALDEHI
Theorem 4.5.
Let φ be a finite-to-one code from a shift of finite type X to anirreducible sofic shift Y . If φ is semi-open, then it will be constant-to-one a.e..Proof. By Theorem 4.1, X is non-wandering and all components are maximal.Since each component is SFT and φ is finite-to-one, the restriction of φ to eachcomponent is constant-to-one a.e. [11, Theorem 9.1.11]. So φ is constant-to-onea.e. as well. (cid:3) Remark . Example 3.6 shows that the converse of Theorem 4.5 does not holdnecessarily. However, when X is an irreducible shift of finite type and φ : X → Y afinite-to-one factor code, then, φ is semi-open and constant-to-one a.e. (Corollary3.11 and [11, Theorem 9.1.11]).Let φ : X → Y be an open code between shift spaces. If φ is bi-closing and Y irreducible, then φ is constant-to-one [10, Corollary 2.8]. The following exampleshows that this result is not necessarily true for semi-open codes. Example 4.7.
Let Y be the even shift with its Fischer cover G = ( G, L ) in Figure2. Note that L ∞ is bi-resolving and so Y is an AFT. By Theorem 3.7, L ∞ issemi-open. But |L − ∞ ( y ) | = 1 for all y = 0 ∞ and equals 2 for y = 0 ∞ . So L ∞ is notconstant-to-one.Let X , Y and Z be shift spaces and φ : X → Z , φ : Y → Z the codes. Then,the fiber product of ( φ , φ ) is the triple (Σ , ψ , ψ ) whereΣ = { ( x, y ) ∈ X × Y : φ ( x ) = φ ( y ) } and ψ : Σ → X is defined by ψ ( x, y ) = x ; similarly for ψ . Let φ be onto.Then, if ψ is open, then so is φ [10, Lemma 2.4]. We do not know this result forsemi-openness. However, we have the following. Theorem 4.8.
Let (Σ; φ , φ ) be the fiber product of φ : X → Z and φ : Y → Z .If ψ and φ are semi-open and ψ is onto, then ψ and φ are semi-open.Proof. If ψ and φ are semi-open, then Lemma 3.1 implies that φ is semi-open.To see that ψ is semi-open let C be the central 2 l + 1 cylinder in Σ and let m be the coding length of φ . Choose p ≥ l so that ψ ( C ) contains a central 2 p + 1cylinder and let x ∈ [ x − p · · · x p ] ⊆ ψ ( C ). Also, let q ≥ l so that φ ([ x − p · · · x p ])contains a central 2 q + 1 cylinder and let z ∈ [ z − q · · · z q ] ⊆ φ ([ x − p · · · x p ]). Then, z = φ ( x ) where x ∈ [ x − p · · · x p ]. So there is y ∈ Y such that ( x, y ) ∈ C .Now we claim that [ y − q − m · · · y q + m ] ⊆ ψ ( C ). For y ′ [ − q − m, q + m ] = y [ − q − m, q + m ] ,we have φ ( y ′ ) [ − q, q ] = φ ( y ) [ − q, q ] which means that φ ( y ′ ) ∈ [ z − q · · · z q ]. Sothere is x ′ ∈ [ x − p · · · x p ] such that φ ( x ′ ) = φ ( y ′ ) and this in turn means that( x ′ , y ′ ) ∈ Σ. But y ′ [ − l, l ] = y [ − l, l ] and so in fact ( x ′ , y ′ ) ∈ C . Thus ψ ( C ) contains[ y − q − m · · · y q + m ] and the claim is established. (cid:3) Lifting the semi-open code between synchronized systems to a codebetween their Fischer covers.
We usually read most of the properties of asynchronized system from its Fischer cover, so lifting the codes between systems tocodes between their respective Fischer covers could be helpful.
Lemma 4.9.
Let X i , i = 1 , be a synchronized system with the Fischer cover G i = ( G i , L i ) . Then, a code f : X → X lifts to the Fischer covers. That is, thereis a code F : X G → X G such that L ∞ ◦ F = f ◦ L ∞ . N SEMI-OPEN CODES AND BI-CONTINUING ALMOST EVERYWHERE CODES 9
Proof. If t ∈ X is doubly transitive, then f ( t ) is doubly transitive in X andso by the fact that the degree of L i ∞ is one [11, Proposition 9.1.6], F = L − ∞ ◦ f ◦ L ∞ makes sense on doubly transitive points. Pick x ∈ X G and choose asequence { x n } n ∈ N of doubly transitive points converging to x . We will show thatlim n →∞ F ( x n ) exists. First note that F ( x n ) is a doubly transitive point for all n ∈ N . Let u = L ( π ) be a synchronizing word for X and w any word such that uwu ∈ B ( X ). If π terminates at a vertex I , then since L ∞ is right-resolving, thereis a unique path in G with initial vertex I representing w . Also, any word occursin a doubly transitive point infinitely often to the left and to the right; therefore,arbitrary central paths of F ( x n ) and F ( x m ) are the same for sufficiently large m and n and since we are on paths, lim n →∞ F ( x n ) exists. Set F ( x ) := lim n →∞ F ( x n ).Now we claim that F is continuous. Let { z n } n be a sequence in X G such that z n → z . It is sufficient to show that F ( z n ) → F ( z ).For all n , suppose { w ( n ) m } m is a sequence of doubly transitive points in X G suchthat w ( n ) m → z n . Then, we have lim ( m, n ) →∞ w ( n ) m = z. So by definition, F ( z ) = lim ( m, n ) →∞ F ( w ( n ) m ). On the other hand, for all n , F ( z n ) =lim m →∞ F ( w ( n ) m ) and this proves the claim and we are done. (cid:3) Theorem 4.10.
By the hypothesis of the above lemma, (1) f is irreducible if and only if F is. (2) If F is semi-open, then f is semi-open. When G and G are locally-finite,the converse is also true.Proof. (1) is a consequence of Lemma 3.2 and Theorem 3.8.For the proof of (2), let F be semi-open and U an open subset of X and set g := L ∞ ◦ F (= f ◦ L ∞ ) . Then, there is an open set V ⊆ g ( L − ∞ ( U )). But g ( L − ∞ ( U )) ⊆ f ( U ) and so V ⊆ f ( U ) and consequently f is semi-open.For the converse suppose f is semi-open and let U be an open subset of X G .Since by Theorem 3.7, g is semi-open, we may assume that there is V , an opensubset of X , such that g ( U ) = V ; otherwise, replace U with U ∩ g − ( V ), V ⊆ g ( U ).Also, by the fact that X G i is locally compact, we may assume that U is a compactsubset of X G and so X G \ F ( U ) an open subset of X G . Furthermore, we choose U so small so that X G \ F ( U ) = ∅ . By these assumptions, we will show that W = F ( U ) is open in X G . Suppose the contrary. Thus there is w ∈ W which isa boundary point of W and then indeed w ∈ ∂F ( U ). Choose a sequence of doublytransitive points { x n } in X G \ F ( U ) approaching w . Since X G \ F ( U ) is open,such a sequence exists and lim n →∞ L ∞ ( x n ) = L ∞ ( w ) ∈ V. But L ∞ is one-to-one on the set of doubly transitive points which means that L ∞ ( x n ) ∈ X \ V and this in turn implies that our open set V contains a boundarypoint which is absurd. (cid:3) Now let X be a non-SFT sofic shift with the Fischer cover G = ( G, L ) and let f = L ∞ and F = Id : X G → X G . Then, F is open while f cannot be open [10,Proposition 2.3]. So Theorem 4.10 does not hold for open maps.4.2. On the bi-continuing codes and bi-closing codes.
A code φ : X → Y is called right-continuing almost everywhere (a.e.) if whenever x is left transitivein X and φ ( x ) is left asymptotic to a point y ∈ Y , then there exists x ∈ X suchthat x is left asymptotic to x and φ ( x ) = y . Similarly we have the notions calledleft-continuing a.e. and bi-continuing a.e.If φ : X → Y is open with a uniform lifting length, then it is bi-continuing(everywhere) with a bi-retract [9, Lemma 2.1]. The following shows that a semi-open code with a uniform lifting length is not necessarily bi-continuing a.e. with abi-retract. Example 4.11.
This example was constructed in [9] to show that a continuingcode may not have a retract; we use it for our prementioned purpose.Let X be a shift space on the alphabet { , , , } defined by forbidding { n n ≥ } , and let Y be the full 3-shift { , , } Z . Define φ = Φ ∞ : X → Y byletting Φ(1) = 1 and Φ( a ) = a for all a = 1. The map φ is bicontinuing and ithas (left continuing) retract [9]. Now for n ∈ N , consider a left transitive point x = · · · n . ∞ and pick y = · · · n . ∞ ∈ Y so that φ ( x ) is left asymptotic to y .One can readily see that φ does not have right continuing a.e. retract.We show that φ is semi-open with a uniform lifting length. Let [ a − n · · · a n ]be a central 2 n + 1 cylinder in X and Φ( a i ) = b i , − n ≤ i ≤ n . Thus if a i =1, then φ ([ a − n · · · a n ]) = [ b − n · · · b n ]; otherwise, [1 b − n · · · b n ⊆ [ b − n · · · b n ⊆ φ ([ a − n · · · a n ]). So φ is semi-open with a uniform lifting length as required.If X is an irreducible sofic shift, then by [11, Corollary 4.4.9],(4.2) h ( Y ) < h ( X ) , Y is a proper subsystem of X. This condition is sufficient to have double transitivity a totally invariant propertyfor finite-to-one factor codes. That is,
Lemma 4.12.
Suppose X is compact and φ : X → Y a finite-to-one factor code.If either X satisfies (4.2) or φ is irreducible, then x ∈ X is doubly transitive if andonly if φ ( x ) is.Proof. When X satisfies (4.2), it is a consequence of [1, Theorem 3.4]. So suppose φ is irreducible. If φ ( x ) is doubly transitive, but x is not, then the proof of [11,Lemma 9.1.13] implies that there will be a proper subshift Z of X with φ ( Z ) = Y ;and this contradicts the irreducibility of φ . (cid:3) Theorem 4.13.
Suppose X and φ satisfy the hypothesis of Lemma 4.12 and Y aSFT. If φ is bi-continuing a.e. with a bi-retract, then it will be open with a uniformlifting length.Proof. Suppose that φ is bi-continuing a.e. with a bi-retract n ∈ N . We mayassume that our SFT system Y is an edge shift and φ is a 1-block code. Let l ≥ u ∈ B l +1 ( X ). Choose x ∈ [ u ] to be a left transitive point and pick a doublytransitive point y ∈ Y with y [ − l − n, l + n ] = φ ( x ) [ − l − n, l + n ] . The point ˆ y given byˆ y i = (cid:26) φ ( x ) i i ≤ ,y i i > . N SEMI-OPEN CODES AND BI-CONTINUING ALMOST EVERYWHERE CODES 11 is a doubly transitive point in Y . Since n is a right continuing a.e. retract and φ ( x ) ( −∞ , l + n ] = ˆ y ( −∞ , l + n ] , there is x ∈ X such that x ( −∞ , l ] = x ( −∞ , l ] and φ ( x ) =ˆ y . By Lemma 4.12, x is a doubly transitive point and we have φ ( x ) [ − l − n, ∞ ) =ˆ y [ − l − n, ∞ ) = y [ − l − n, ∞ ) . By the fact that n is also a left continuing a.e. retract,there is x ∈ X such that x [ − l, ∞ ) = x [ − l, ∞ ) and φ ( x ) = y . Note that x [ − l, l ] = x [ − l, l ] = x [ − l, l ] which means that x is in [ u ].Now choose an arbitrary y ′ ∈ [ φ ( x ) [ − l − n, l + n ] ] and pick a sequence of doublytransitive points { y m } m in [ φ ( x ) [ − l − n, l + n ] ] such that y m → y ′ and let x m be apoint in [ u ] with φ ( x m ) = y m . Thus for all m , y m ∈ φ ([ u ]). But φ ([ u ]) is closedand so y ′ ∈ φ ([ u ]) which implies that [ φ ( x ) [ − l − n, l + n ] ] ⊆ φ ([ u ]). This shows that φ is semi-open.It remains to prove that φ is open. Let again u = u − l · · · u l and let x ′ ∈ [ u ] andchoose a doubly transitive point x such that x [ − l − n, l + n ] = x ′ [ − l − n, l + n ] . Since φ is a 1-block code, φ ( x ) [ − l − n, l + n ] = φ ( x ′ ) [ − l − n, l + n ] . Also since Y is an edge shift,[ φ ( x ′ ) [ − l − n, l + n ] ] = [ φ ( x ) [ − l − n, l + n ] ]. By above, [ φ ( x ) [ − l − n, l + n ] ] ⊆ φ ([ u ]). But x ′ was arbitrary and so φ is open with uniform lifting length. (cid:3) By Theorem 4.13 and [9, Lemma 2.1], we have the following:
Corollary 4.14.
Let X and φ satisfy the hypothesis of Lemma 4.12 and Y a SFT.If φ is bi-continuing a.e. with a bi-retract, then it will be bi-continuing (everywhere)with a bi-retract. There is another situation where a bi-continuing a.e. code implies bi-continuing.In fact, Ballier interested in sofics in [4], states and proves the next theorem forwhen X is an irreducible sofic and k = 0. His proof exploits only the irreducibilityof X which is provided here. Hence we have Theorem 4.15.
Suppose X is an irreducible shift space and Y an irreducible SFT.A right-continuing a.e. factor map φ : X → Y with retract k is right-continuing(everywhere) with retract k . The next theorem gives a sufficient condition for a factor code being bi-continuinga.e..
Theorem 4.16.
Let X be synchronized and φ : X → Y a semi-open code. Then, φ is bi-continuing a.e..Proof. Let x = · · · x − x x · · · ∈ X be a left transitive point and y ∈ Y such that φ ( x ) ( −∞ , = y ( −∞ , . Since x is left transitive, any word in X happens infinitelymany often on the left of x . So without loss of generality, assume that x is thesynchronizing word.By semi-openness, there is a cylinder l [ u ] contained in φ ( [ x ]) where l [ u ] denotesthe set { y ∈ Y : y [ l, l + | u |− = u } . Since x is left transitive, there exist infinitelymany k > σ − k ( φ ( x )) ∈ l [ u ], or equivalently, φ ( x ) ∈ σ k ( l [ u ]) = l − k [ u ].Choose a sufficiently large k so that l − k + | u | − < φ ( x ) ( −∞ , = y ( −∞ , , y ∈ l − k [ u ] or in fact y ∈ l − k [ v ] where v = y l − k . . . y − y = u l − k . . . u l − k + | u |− y l − k + | u | . . . y − y . We have l − k [ v ] ⊆ l − k [ u ] ⊆ φ ( − k [ x ]). Set W = φ − ( l − k [ v ]) ∩ [ x ] and note that W is an open set containing x and y ∈ φ ( W ). Therefore, there must be z ∈ W such that φ ( z ) = y and define x i = (cid:26) x i i ≤ ,z i i ≥ . Since x is a synchronizing word, x ∈ X and so we have a x which is left-asymptoticto x and φ ( x ) = y . This means φ is right-continuing a.e.; and similarly, it is left-continuing a.e.. (cid:3) References
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