aa r X i v : . [ m a t h . GN ] A ug A GLANCE INTO THE ANATOMY OF MONOTONIC MAPS
RAUSHAN BUZYAKOVA
Abstract.
Given an autohomeomorphism on an ordered topological spaceor its subspace, we show that it is sometimes possible to introduce a newtopology-compatible order on that space so that the same map is monotonicwith respect to the new ordering. We note that the existence of such a re-ordering for a given map is equivalent to the map being conjugate (topologi-cally equivalent) to a monotonic map on some homeomorphic ordered space.We observe that the latter cannot always be chosen to be order-isomorphicto the original space. Also, we identify other routes that may lead to similaraffirmative statements for other classes of spaces and maps. Introduction
It is one of classical problems of various areas of topology if a given continuousmap on a topological space with perhaps a richer structure has nice propertiesrelated to this rich structure. For clarity of exposition, let us agree on terminol-ogy. An autohomeomorphism on a topological space X is any homeomorphismof X onto itself. An open interval with end points a and b of a linearly or-dered set L will be denoted by ( a, b ) L . If it is clear what ordered set is underconsideration, we simply write ( a, b ). The same concerns other types of inter-vals. Linearly ordered topological spaces are abbreviated as LOTS and theirsubspaces as GO-spaces. We will mostly be concerned with GO-spaces. It isdue to ˇCech ([4]) that a Hausdorff space X is a GO-space if and only if a familyof convex sets with respect to some ordering on X is a basis for the topology of X . Given a GO-space X , an order ≺ on X is said to be GO-compatible if somecollection of ≺ -convex subsets of h X, ≺i is a basis for the topology of X . Notethat if X is a LOTS, a GO-compatible order on X need not witness the factthat X is a LOTS. We will be concerned with the following general problem. Problem 1.1.
Let X be a GO-space and let f be an autohomeomorphism on X . What conditions on X and/or f guarantee that X has a GO-compatibleordering with respect to which f is monotonic? Mathematics Subject Classification.
Key words and phrases. monotonic map, ordered topological spaces, topologically equiv-alent maps.
Since monotonicity is an order-dependent concept, we will specify with respectto which ordering a map is monotonic. If no clarification is given, the assumedorder is the original one and should be clear from the context. Since our dis-cussion will be around Problem 1.1, we will isolate the target property into adefinition.
Definition 1.2.
An autohomeomorphism f on a GO-space is potentially mono-tonic if there exists a GO-compatible order on X with respect to which f ismonotonic. Definition 1.2 is equivalent to the following definition:
Definition 1.3. (Equivalent to 1.2) An autohomeomorphism f on a GO-spaceis potentially monotonic if there exists a GO-space Y , a homeomorphism h : X → Y , and a monotonic autohomeomorphism m on Y such that f = h − ◦ m ◦ h . To see why these two definitions are equivalent, let f be an autohomeomorphismon a GO-space X . Assume f is potentially monotonic by Definition 1.2. Fixa GO-compatible order ≺ on X with respect to which X is monotonic. Put Y = h X, ≺i , h = id X (the identity map), and m = f . Clearly, f = id − X ◦ m ◦ id X .Hence, f is potentially monotonic with respect to Definition 1.3. We now assumethat f is potentially monotonic with respect to Definition 1.3. Fix Y, f, m asin the definition. The order on Y induces an order ≺ on X as follows: a ≺ b if and only if h ( a ) < h ( b ). Since h is a homeomorphism, ≺ is compatible withthe GO-topology of X . Next, let us show that f is ≺ -monotonic. We have a ≺ b is equivalent to h ( a ) < h ( b ). By the choice of m , the latter is equivalentto m ◦ h ( a ) < m ◦ h ( b ). By the definition of ≺ , the latter is equivalent to h − ◦ m ◦ h ( a ) ≺ h − ◦ m ◦ h ( b ). Since f = h − ◦ m ◦ h , we conclude that f ( a ) ≺ f ( b ).One may wonder if the property in Definition 1.2 is equivalent to the property ofbeing topologically equivalent to a monotonic map with respect to the existingorder. Recall that homeomorphisms f, g : X → X are topologically equivalent (or conjugate)if there exists a homeomorphism t : X → X such that t ◦ f = g ◦ t .A supported explanation will be given later in Remark 2.9 that a map canbe potentially monotonic but not topologically equivalent to a monotonic map(with respect to the existing order). It is clear, however, from Definition 1.3 thata map topologically equivalent to a monotonic map is potentially monotonic. Inour arguments, given a monotonic function f on a GO-space L and an x ∈ L ,we will make a frequent use of the set { f n ( x ) : n ∈ Z } . In literature, similarlydefined sets are often referred to as the orbit of x under f . We will also refer to Glance into the Anatomy of Monotonic Maps 3 this set as the f -orbit of x . Similarly, the f -orbit of a set A ⊂ X is the collection { f n ( A ) : n ∈ Z } . By looking at the behavior of monotonic maps on the reals,we quickly observe that the orbit of each point under such maps exhibits verystrong properties. Namely, the following holds. Proposition 1.4.
Let f be a fixed-point free monotonic autohomeomorphismon a GO-space L and x ∈ L . Then there exists an open neighborhood I of x suchthat f n ( I ) ∩ f m ( I ) = ∅ for any distinct integers n and m and { f n ( I ) : n ∈ Z } is a discrete family.Proof. Without loss of generality, we may assume that f is strictly increasing.Fix x ∈ L . We have two cases.Case ( x is isolated): Let us show that I = { x } is as desired. By strictmonotonicity, f n ( x ) = f m ( x ) for distinct integers n and m . To showthat S = { f n ( { x } ) : n ∈ Z } is a discrete family of sets, fix y ∈ L . If y = f n ( x ) for some n , then { y } is an open neighborhood of y that meetsexactly one element of the collection, namely, f n ( x ). Assume y = f n ( x )for any n and y is a limit point for S . By monotonicity, lim n →∞ f n ( x ) = y or lim n →∞ f − n ( x ) = y . By continuity, f ( y ) = y , contradicting the fact that f is fixed-point free.Case ( x is not isolated): Since f is an increasing homeomorphism, the in-tervals ( x, f ( x )) and ( f − ( x ) , x ) are not empty. Pick and fix a ∈ ( f − ( x ) , x ). Since f is strictly increasing, f ( a ) ∈ ( x, f ( x )). Let usshow that I = ( a, f ( a )) is as desired. First due to monotonicity, f n ( I ) = ( f n ( a ) , f n +1 ( a )). Therefore, f n ( I ) misses f m ( I ) for distinct n, m ∈ Z . The fact that { f n ( I ) : n ∈ Z } is a discrete collection isproved as in the previous case. (cid:3) We will next isolate the necessary condition identified in Proposition 1.4 into aproperty.
Definition 1.5.
Let f : X → X be a map and A ⊂ X . The f -orbit of A isstrongly discrete if there exists an open neighborhood U of A such that { f n ( U ) : n ∈ Z } is a discrete collection and f n ( U ) ∩ f m ( U ) = ∅ for distinct n, m . The f -orbit of x ∈ X is strongly discrete if the f -orbit of { x } is strongly discrete. In this note we will present partial results addressing Problem 1.1. At the endof our study we will identify a few questions that may have a good chance foran affirmative resolution.
R. Buzyakova
In notation and terminology we will follow [3]. In particular, if ≺ is an orderon L and A, B ⊂ L , by A ≺ B we denote the fact that a ≺ b for any a ∈ A and b ∈ B . 2. Study
One may wonder if our introduction of the concepts of strongly discrete orbitsis really necessary. Can we use the requirement of being ”period-point free”instead? The next example shows that a periodic-point free automohomeomor-phism even on a nice space need not have strongly discrete orbits.
Example 2.1.
There exist a periodic-point free autohomeomorphism f of thespace of rationals Q and a point q ∈ Q such that the f -orbit of q is not stronglydiscrete.Proof. Example [1, Example 2.5 ] provides a construction of a fixed point auto-homeomorphism f on the rationals that satisfies the hypothesis of Lemma [1,Lemma 2.4]. For convenience, the cited hypotheses is copied next:Hypothesis of Lemma [1, Lemma 2.4]: ”Suppose f : Q → Q is not an identitymap and p ∈ Q satisfy the following property:(*) ∀ n > ∃ m > such that f m +1 (( p − /n, p + 1 /n ) Q ) meets f − m (( p − /n, p +1 /n ) Q ) .” Clearly an f that satisfies the above hypothesis fails having a strong f -orbit at p . (cid:3) For our next affirmative result we need a technical statement that incorporatesour general strategy for showing that a map is potentially monotonic.
Lemma 2.2.
Let L be a GO-space and f : L → L an autohomeomorphism.Suppose that O is a collection of clopen subsets of L with the following properties: (1) The f -orbit of each O ∈ O is strongly discrete. (2) f n ( O ) ∩ f m ( O ′ ) = ∅ for distinct O, O ′ ∈ O and n, m ∈ Z . (3) { f n ( O ) : n ∈ Z , O ∈ O} is a cover of L .Then there exists a GO-compatible order ≺ on L with respect to which f isstrictly increasing.Proof. By < we denote some ordering with respect to which L is a generalizedordered space. Enumerate elements of O as { O α : α < |O|} . We will define ≺ in three stages. Glance into the Anatomy of Monotonic Maps 5
Stage 1:
For each O ∈ O and n ∈ ω \ { } , define ≺ on f n ( O ) and f − n ( O )recursively as follows:Step 0: Put ≺ | O = < | O .Assumption: Assume that ≺ is defined on f k ( O ) and on f − k ( O ) forall k = 0 , , ..., n − n : If x, y ∈ f n ( O ), put x ≺ y if and only if f − ( x ) ≺ f − ( y ).This is well defined since f − ( x ) , f − ( y ) are in f n − ( O ) and ≺ isdefined on f n − ( O ) by assumption. Similarly, if x, y ∈ f − n ( O ), put x ≺ y if and only if f ( x ) ≺ f ( y ). Stage 2:
For any O ∈ O and any n, m ∈ Z such that n < m , put f n ( O ) ≺ f m ( O ). Stage 3:
For any α < β < |O| and any n, m ∈ Z , put f n ( O α ) ≺ f m ( O β ).The next two claims show that ≺ is as desired. Claim 1. ≺ is compatible with the GO-topology of L .Proof of Claim . To prove the claim, for each O ∈ O , let T O be the collection ofall < -convex open subsets of L that are subsets of O . Since < coincides with ≺ one every O ∈ O , we conclude that every element in T O is ≺ -convex. By theconstructions at Stage 1, f n ( O ) is ≺ -convex. Since f is an autohomeomorphism,the collection { f n ( I ) : I ∈ T O , O ∈ O} is a basis for the topology of L andconsists of open ≺ -convex sets. The claim is proved. Claim 2. f is increasing with respect to ≺ .Proof of Claim . Pick distinct x and y . If x, y ∈ S n f n ( O ) for some O ∈ O , thenapply Stages 1 and 2. Otherwise, apply Stage 3. (cid:3) Remark to Lemma 2.2.
Note that if h L, < i is a LOTS and each O in theargument of the lemma has both extremities or each O has neither extremity,then h L, ≺i is a LOTS too. The converse of Lemma 2.2 for fixed-point free autohomeomorphisms on zero-dimensional GO-spaces holds too (Lemma 2.4). To prove the converse, we needthe following quite technical statement. Recall that given a continuous self-map f : X → X , a closed set A ⊂ X is an f -color if A ∩ f ( A ) = ∅ . For a review ofmajor results on colors of continuous maps, we refer the reader to [6]. Proposition 2.3.
Let L be a zero-dimensional GO-space, f : L → L a fixed-point free monotonic homeomorphism, and x ∈ L . Then there exists a maximalconvex clopen set I ⊂ L containing x such that the following hold: (1) S n ∈ Z f n ( I ) is clopen and convex. (2) f n ( I ) ∩ f m ( I ) = ∅ for any distinct integers n and m . (3) f n ( I ) is a maximal clopen convex f -color for any n ∈ Z . R. Buzyakova
Proof.
We may assume that f is strictly increasing. If L is locally compact, byzero-dimensionality there exists a ∈ L such that x < a and { y ∈ L : y < a } is clopen and non-empty. Put I = [ a, f ( a )). Let us show that I is as desired.By monotonicity, { f n ( I ) : n ∈ Z } = { ..., [ f − ( a ) , a ) , [ a, f ( a )) , [ f ( a ) , f ( a )) , ... } .Enlarging any interval in this sequence would make that interval meet its image.Therefore, (3) is met. Visual inspection of the sequence is a convincing evidencethat the union S n f n ( I ) is convex. The union is also open as the union of opensets. Since f is fixed-point free, f n ( I )’s form a discrete collection, and hence, theunion is closed. By our choice, f n +1 ( I ) = [ f n +1 ( a ) , f n +2 ( a )), which guaranteesthat (2) is met.We now assume that L is not locally compact. Let dL be the smallest orderedcompactification of h L, < i . Since f is a monotonic autohomeomorphism, f hasa unique continuous extension ˜ f : dL → dL . Let F ⊂ dL be the set of all fixedpoints of ˜ f . Since f is fixed-point free, F is a subset of dL \ L . Since L is notlocally compact, there exists a ∈ ( dL \ L ) \ F . Put I = [ a, ˜ f ( a )] dL ∩ L . Clearly, I is as desired. (cid:3) Lemma 2.4.
Let L be a zero-dimensional GO-space and let f : L → L be afixed-point free monotonic autohomeomorphism. Then there exists a collection O of convex clopen subsets of L with the following properties: (1) The f -orbit of each O ∈ O is strongly discrete. (2) f n ( O ) ∩ f m ( O ′ ) = ∅ for distinct O, O ′ ∈ O and n, m ∈ Z . (3) { f n ( O ) : n ∈ Z , O ∈ O} is a cover of L .Proof. Without loss of generality, we may assume that f is strictly increasing.We will construct O = { O α } α recursively. Assume that O β is constructed foreach β < α and the following properties hold:P1: S n ∈ Z f n ( O β ) is clopen and convex.P2: f n ( O β ) ∩ f m ( O β ) = ∅ for any distinct integers n and m .P3: f n ( O β ) is a maximal clopen convex f -color for any n ∈ Z .Note that P1 and P2 imply the following:P4: The f -orbit of O β is strongly discrete. Construction of O α : Let L α = L \ S { f n ( O β ) : β < α, n ∈ Z } . If L α is empty,then the recursive construction is complete and O = { O β : β < α } . Otherwise,we have f − ( f ( L α )) = L α . Let us show that L α is clopen in L . Firstly, it isclosed as the complement of the union of open sets. To show that it is open,fix x ∈ L α . Let I be as in Proposition 2.3 for given x, f, L . If x were a limitpoint for L \ L α , then it would have contained some f n ( O β ) for β < α and n ∈ Z , which contradicts property P3. Hence, I is an open neighborhood of x Glance into the Anatomy of Monotonic Maps 7 contained in L α . Since properties (1)-(3) of I in the conclusion of Proposition2.3 coincide with the properties P1-P3, we can put O α = I .The family O = { O α } α is as desired by construction. (cid:3) Lemmas 2.2 and 2.4 form the following criterion.
Theorem 2.5.
Let f be a fixed-point free autohomeomorphism on a zero-dimensional GO-space X . Then f is potentially monotonic if and only if thereexists a collection O of convex clopen subsets of L with the following properties: (1) The f -orbit of each O ∈ O is strongly discrete. (2) f n ( O ) ∩ f m ( O ′ ) = ∅ for distinct O, O ′ ∈ O and n, m ∈ Z . (3) { f n ( O ) : n ∈ Z , O ∈ O} is a cover of L . We next put one part (Lemma 2.2) of the above criterion to a good use.
Theorem 2.6.
Let X be a zero-dimensional subspace of the reals and let f : X → X be an autohomeomorphism with strongly discrete orbits at all points.Then there exists a GO-compatible order ≺ on X such that f is ≺ -monotonic.Proof. To prove the statement, we will construct a collection O as in the hy-pothesis of Lemma 2.2. Fix a countable cover F = { F n : n ∈ ω } of X so thateach F i is clopen and has strongly discrete f -orbit. Step : Put O = F . Assumption:
Assume that O k is defined for k < n , clopen, and hasstrongly discrete f -orbit. In addition, assume that S m ∈ Z f m ( O i ) misses S m ∈ Z f m ( O j ), whenever i = j and i, j < n . Step n : Let i n be the smallest index such that F i n is not covered by { f m ( O i ) : i < n, m ∈ Z } . Put O n = F i n \ ∪{ f m ( O i ) : i < n, m ∈ Z } .Construction is complete. The collection O = { O n : n ∈ ω } has properties (1)and (2) in the hypothesys of Lemma 2.2 by construction. To show (3), thatis, the equality X = ∪{ f m ( O i ) : i ∈ ω, m ∈ Z } , fix any x ∈ X . Since F isa cover of X , there exists n such that x ∈ F n . If x is not in f m ( O i ) for some i < n and m ∈ ω , then F n is the first element in F that meets the constructionrequirements at step n . Therefore, x ∈ O n . (cid:3) Corollary 2.7.
Every periodic-point free bijection on Z is potentially mono-tonic. R. Buzyakova
In contrast with Corollary 2.7, we next observe that not every periodic-pointfree bijection on Z is topologically equivalent to a monotonic map. Example 2.8.
There exists a periodic-point free bijection on Z that is not topo-logically equivalent to a monotonic map.Proof. First observe that every monotonic bijection on Z is a shift. Therefore,any bijection on Z that is topologically equivalent to a monotonic map is alsotopologically equivalent to a shift. It is observed in [2, Example 1.2] that if abijection f on Z has infinitely many points with mutually disjoint orbits, thensuch a map is not topologically equivalent to a shift. Thus, any such fixed-point free map is an example of a potentially monotonic map on Z that is nottopologically equivalent to a monotonic map. (cid:3) Remark 2.9.
Corollary 2.7 and Example 2.8 imply that the property of beingpotentially monotonic does not imply the property of being topologically equiva-lent to a monotonic map (with respect to the existing order).
We can strengthen Corollary 2.7 as follows.
Theorem 2.10.
Let f be a periodic-point free bijection on Z . Then there existan ordering ≺ and a binary operation ⊕ on Z such that Z ′ = h Z , ⊕ , ≺i is adiscrete ordered topological group and f is a shift in Z ′ .Proof. Let M be a minimal subset of Z with respect to the property that the f -orbit of M covers Z .If | M | = n , enumerate the elements of M by Z n . Clearly, Z n × l Z is anordered discrete topological group with the component-wise addition. Definea bijection h : Z → Z n × l Z by letting g ( f k ( n i )) = ( i, k ). Since any elementof Z is in the f -orbit of exactly one element of M , the correspondence is well-defined and is a bijection. Since g is a homeomorphism, we will next abusenotation and will identify f k ( x i ) with ( i, k ). Let us apply f to ( i, k ). We have f ( f k ( x i )) = f k +1 ( x i ), and the latter is identified with ( i, k + 1). Therefore, f isa shift by (0 ,
1) in Z ′ .If M is infinite, enumerate its elements by integers as M = { n i : i ∈ Z } .Define h : Z → Z × l Z by letting g ( f k ( n i )) = ( i, k ). Argument similar to the Z n case shows that the ordering on Z induced by h is as desired. (cid:3) Note that the above statement does not hold for continuous periodic-point freebijections on the rationals. Indeed, as shown in [1, Example 2.5] there existsa continuous periodic-point free bijection on Q with a point with non-strongly Glance into the Anatomy of Monotonic Maps 9 discrete fiber. The mentioned example [1, Example 2.5] is constructed to satisfythe hypothesis of [1, Lemma 2.4], which is a stronger case of not having discretefibers. Nonetheless, the following takes place.
Theorem 2.11.
A fixed-point free autohomeomorphism f : Q → Q is poten-tially monotonic if and only if f is topologically equivalent to a shift.Proof. ( ⇒ ) Since f is potentially monotonic, there exists a collection O as inthe conclusion of Lemma 2.4. The argument of Theorem 2.3 in [1] shows thatthat f with such a collection is topologically equivalent to a non-trivial shift.( ⇐ ) It is proved in [2, Theorem 2.8] that a periodic-point free homeomorphism h on Q is topologically equivalent to a shift if and only if one can introduce a groupoperation ⊕ on Q compatible with the topology of Q so that the topologicalgroup h Q , ⊕i is continuously isomorphic to Q and h is a shift with respect tonew operation. Clearly such an h Q , ⊕i is an ordered topological group, andhence, any shift is monotonic. Therefore, f is potentially monotonic. (cid:3) Recall that given a continuous selfmap f : X → X , the chromatic number of f is the least number of f -colors needed to cover X . Theorem 2.12.
Let f be a fixed-point free autohomeomorphism on a zero-dimensional GO-space L . If f is potentially monotonic, then the chromaticnumber of f is .Proof. ( ⇒ ) Since the chromatic number of f is a purely topological propertynot attached to an order, we may assume that f is strictly monotonic. Let O be as in the conclusion of 2.4 for the given f and L . Put A = ∪{ f n ( O ) : n is an even integer, O ∈ O} and B = ∪{ f n ( O ) : n is an odd integer, n ∈ O} .Clearly, { A, B } is cover of L by colors. (cid:3) Theorem 2.12 and Remark 2.9 prompt the following question.
Question 2.13.
Let f be a periodic point free homeomorphism on a zero-dimensional GO-space L . Let the chromatic number of f be 2. Is f potentiallymonotonic? Theorem 2.6 prompts the following question.
Question 2.14.
Let X be a GO-space and let f : X → X be an autohomeo-morphism with strongly discrete orbits at all points.Is f potentially monotonic?What if X is hereditarily paracompact? References [1] R. Buzyakova,
On Monotonic Fixed-Point Free Bijections on Subgroups of R , AppliedGeneral Topology, Vol 17, No 2 (2016), 83–91.[2] R. Buzyakova and J. West, Three Questions on Special Homeomorphisms on Subgroupsof R and R ∞ , Questions and Answers in General Topology, Volume 36, Number 1 (2018)pp. 1-8[3] R. Engelking, General Topology , PWN, Warszawa, 1977.[4] H. Bennet and D. Lutzer,
Linearly Ordered and Generalized Ordered Spaces , Encyclope-dia of General Topology, Elsevier, 2004.[5] D. Lutzer,
Ordered Topological Spaces, Surveys in General Topology , G. M. Reed., Aca-demic Press, New York (1980), 247-296.[6] J. van Mill,
The infinite-dimensional topology of function spaces , Elsevier, Amsterdam,2001.
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