# Generalized Virtually Stable Maps and Their Associated Sequences

Authors: P. Chaoha; S. Iampiboonvatana; J. Intrakul
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We extend the concept of virtual stability of continuous self-maps to arbitrary selfmaps and investigate the structure of sequences associated with uniformly virtually stable selfmaps.
We also obtain a necessary and sufficient condition for a uniformly
virtually stable selfmap to have the largest possible associated sequence. Examples of a uniformly virtually stable selfmap having
the prescribed largest sequence and a uniformly virtually stable
selfmap having no largest sequence are given.


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## Body

## 1. Introduction

Virtual stability of a self-map was first introduced in [1], where it was proved to unify various (continuous) types of nonexpansiveness in fixed-point theory. The main feature of a virtually stable self-map is that its fixed-point set is always a retract of its convergence set, and this fact immediately allows us to connect topological structures (e.g., connectedness and contractibility) of the fixed-point set to those of the convergence set. Although the original definition of virtual stability requires the continuity of a self-map, it seems that the connection between the fixed-point set and the convergence set also remains valid to some degree in discontinuous settings as hinted in [2, Corollary 2.6] and the work on mean-type mappings in [3–5] (see Example 4 below). A careful investigation of this fact will definitely result in a more powerful notion of virtual stability that enables us to study fixed-point sets of interesting discontinuous self-maps (e.g., Suzuki generalized nonexpansive maps [6]). Moreover, since all well-known self-maps in metric fixed-point theory are always uniformly virtually stable with respect to the same sequence (n) of all natural numbers, it is quite natural to ask whether this also holds in general for a uniformly virtually stable self-map. It turns out that the answer to this simple question is highly nontrivial and interestingly involves algebraic properties of a certain object. Therefore, the purpose of this work is to revise the concept of virtual stability as well as to take care of the previos question. We will first redefine the notion of virtual stability to include discontinuous self-maps and investigate some properties of these newly defined virtually stable self-maps and prove the expected connection between the convergence set and the fixed-point set. Then we will show that the sequence associated with a uniformly virtually stable self-map can be made, in some sense, largest possible provided that the set of all associated sequences has a certain structure. We will also see that when the largest sequence exists, it may not be (n) unless the map is continuous. For the sake of completeness, since the existence of the largest possible sequence associated with a given self-map cannot be guaranteed in general, we will provide two constructive examples of uniformly virtually stable self-maps: one having the prescribed largest sequence and the other having no largest sequence.


## 2. Virtual Stability without Continuity

LetX be a (nonempty) Hausdorff space and f:X→X a self-map (that may not be continuous). We will also let 𝒩(x) and 𝒩Y(x) denote the sets of (open) neighborhoods of x in X and in Y⊆X, respectively. Recall that the fixed-point set and the convergence set of f are defined, respectively, to be F(f)={x:f(x)=x} and
(1)C(f)={x:the sequence(fn(x))converges},
where fn denotes the n-th iterate of f. We also define the map f∞:C(f)→X by
(2)f∞(x)=limn→∞fn(x),
for all x∈C(f). Notice that f∞ may not be continuous, and since we do not assume the continuity of f, we only have F(f)⊆f∞(C(f)). However, in this work, we will always assume that
(3)∅≠F(f)=f∞(C(f)).
This condition is quite natural in the sense that it is automatically satisfied whenever f is continuous with F(f)≠∅, and it is also required for the iterative sequence (fn) to be a scheme according to [2, Definition 2.1]. With the above assumption in mind, we are able to previous define the concept of virtual stability just as in [1, Definition 2.1].Definition 1.
A fixed pointx of f is said to be virtuallyf-stable if for each U∈𝒩(x), there exist V∈𝒩(x) together with an increasing sequence (an) of positive integers such that fan(V)⊆U for all n∈ℕ. We simply call fvirtually stable if every fixed point of f is virtually stable. Moreover, we will call a fixed point x of funiformly virtuallyf-stable with respect to an increasing sequence (an) of positive integers if for each U∈𝒩(x), there exists V∈𝒩(x) such that fan(V)⊆U for all n∈ℕ. When every fixed point of f is uniformly virtually f-stable with respect to the same sequence, we will simply call funiformly virtually stable.Clearly, a uniformly virtually stable self-map is always virtually stable, and a continuously (uniformly) virtually stable self-map as defined in [1] is also (uniformly) virtually stable according to our new definition. Moreover, if f is uniformly virtually stable with respect to (an), it is immediately uniformly virtually stable with respect to any subsequence of (an).Let us recall that, when(X,d) is a metric space, a self-map f:X→X is called (1)
virtually nonexpansive [7] if f is continuous and the family {fn} of all iterates of f is equicontinuous on F(f) (equivalently, on C(f)),(2)
uniformly quasi-Lipschitzian if there is L>0 such that
(4)d(fn(x),p)≤Ld(x,p),for all(x,p,n)∈X×F(f)×ℕ.Notice that a virtually nonexpansive self-map is always continuous and uniformly virtually stable with respect to the sequence(n) of all natural numbers, while a uniformly quasi-lipschitzian self-map may not be continuous in general.Proposition 2.
Supposef is a self-map on a metric space (X,d). (1)
Iffis continuous and uniformly virtually stable with respect to(n), then it is virtually nonexpansive.(2)
Iffis uniformly quasi-lipschitzian, then it is uniformly virtually stable with respect to(n).Proof.
Supposef is a self-map on a metric space (X,d). (1)
Letp∈F(f) and ϵ>0. Then there is H∈𝒩(p) such that fn(H)⊆B(p;ϵ) for all n∈ℕ. Thus, for δ>0 whose B(p;δ)⊆H, we have fn(B(p;δ))⊆fn(H)⊆B(p;ϵ) for all n∈ℕ. This implies that {fn} is equicontinuous on F(f) and hence f is virtually nonexpansive.(2)
Suppose there isL>0 such that d(fn(x),p)≤Ld(x,p) for all (x,p,n)∈X×F(f)×ℕ. It follows that fn(B(p;r/L))⊆B(p;r) for all p∈F(f), n∈ℕ and r>0. Thus, f is uniformly virtually stable with respect to (n).It is not difficult to see that the class of uniformly quasi-lipschitzian self-maps includes various important (possibly discontinuous) self-maps in metric fixed-point theory such as nonexpansive maps, Kannan maps [8], Suzuki generalized nonexpansive maps [6], quasi-nonexpansive maps, and even asymptotically quasi-nonexpansive maps. Hence, the previous proposition immediately implies that those maps are all uniformly virtually stable with respect to (n). The following theorem directly extends [1, Theorem 2.6] to include discontinuous maps.Theorem 3.
Suppose thatX is a regular space. If f is virtually stable and fk is continuous for some k∈ℕ, then f∞ is continuous and hence F(f) is a retract of C(f).Proof.
Letx∈C(f) and U∈𝒩F(f)(f∞(x)). Since X is regular, so is F(f) and then there is W∈𝒩(f∞(x)) such that W∩F(f)⊆W¯∩F(f)⊆U. By virtual stability of f, fan(V)⊆W for all n∈ℕ, for some sequence (an) and V∈𝒩(f∞(x)). Also there is, by the fact that V∈𝒩(f∞(x)), n0∈ℕ such that fn(x)∈V for all n≥n0. Then A=f-kn0(V)∩C(f)∈𝒩C(f)(x) and for each a∈A,
(5)f∞(a)=limn→∞fn(a)=limn→∞fan(fkn0(a))∈W¯∩F(f)⊆U.
Thus, f∞ is continuous and hence F(f) is a retract of C(f).Example 4.
Letp∈ℕ and p≥2, and let I⊆ℝ be an interval. Consider the subspace Ip of ℝp equipped with the maximum norm ∥(x1,…,xp)∥=max{|x1|,…,|xp|}. Since the maximum norm on ℝp also induces the usual topology, Ip is a regular space. Recall that a self-map f:Ip→Ip is called mean-type if f=(M1,…,Mp), where each Mi:Ip→ℝ is a mean on I; that is, for each (x1,…,xp)∈Ip,
(6)min{x1,…,xp}≤Mi(x1,…,xp)≤max{x1,…,xp}.
Notice that f may not be continuous in general. However, if f satisfies
(*)|Mi(x1,…,xp)-zi|≤max{|x1-z1|,…,|xp-zp|}
for all (x1,…,xp)∈Ip, (z1,…,zp)∈F(f) and i=1,…,p, it is always quasi-nonexpansive (with respect to the maximum norm), and hence uniformly virtually stable (with respect to the sequence (n)). For if (x1,…,xp)∈Ip and (z1,…,zp)∈F(f), one has
(7)∥f(x1,…,xp)-f(z1,…,zp)∥=max{|Mi(x1,…,xp)-zi|:i=1,…,p}≤max{|x1-z1|,…,|xp-zp|}[by(*)]=∥(x1,…,xp)-(z1,…,zp)∥.
In particular, the mean-type mappingf:(0,∞)2→(0,∞)2 defined by f(x,y)=((x+y)/2, 2xy/(x+y)) satisfies (*), and hence it is uniformly virtually stable. Since f is also continuous, F(f) is a retract of C(f) by the previous theorem. In fact, it follows from [4] that F(f)={(x,x):x>0}, C(f)=(0,∞)2, and f∞(x,y)=(xy,xy) for (x,y)∈(0,∞)2. Therefore, F(f) is clearly a retract of C(f).Example 5.
Consider thatℝ2 equipped with the maximum norm, and a discontinuous self-map f:ℝ2→ℝ2 given by
(8)f(x,y)={(0,1)if(x,y)=(0,3);(x,|x|)otherwise.
It is straightforward to verify that f is quasi-nonexpansive (hence, uniformly virtually stable) and f2 is continuous. Therefore, by the previous theorem, f∞ is continuous and hence F(f) is a retract of C(f). Notice also that F(f)={(x,|x|):x∈ℝ} is not convex.The following examples show that if we drop the continuity assumption of some iteratefk in Theorem 3, the map f∞ may or may not be continuous.Example 6.
Letf:[0,1]→[0,1] be defined by
(9)f(x)={xifx<12;1otherwise.
Observe that f is uniformly virtually stable and fn=f is not continuous for all n∈ℕ. Moreover, F(f)=[0,1/2)∪{1} and C(f)=[0,1]. Hence f∞ is not continuous.Example 7.
Consider thatℝ2 equipped with the supremum norm. Define f:ℝ2→ℝ2 by
(10)f(x,y)={(0,3n-1)if(x,y)=(0,3n)for somen∈ℕ;(x,|x|)otherwise.
It is straightforward to verify that f is quasi-nonexpansive (hence, uniformly virtually stable) and fn is not continuous for all n∈ℕ. However, f∞ is continuous because f∞(x,y)=(x,|x|) for all (x,y)∈ℝ2.The next theorem gives a condition allowing us to enlarge the sequence associated with a given uniformly virtually stable self-map. As a result, we immediately obtain a refinement and a generalization of [1, Theorem 2.17].Definition 8.
An increasing sequence(an) of natural numbers is said to satisfy the sup-finite condition if supn(an+1-an):=sup{an+1-an:n∈ℕ}<∞.Theorem 9.
If, for somek∈ℕ, fk is continuous on F(f) and f is uniformly virtually stable with respect to (ank) where (an) satisfies the sup-finite condition, then f is uniformly virtually stable with respect to (nk).Proof.
Lets=supn(an+1-an), p∈F(f), and U∈𝒩(p). Because f is uniformly virtually stable with respect to (ank), there is W∈𝒩(p) such that fank(W)⊆U for all n∈ℕ. By continuity at p of fik, i∈ℕ∪{0}, there exists V∈𝒩(p) for which fik(V)⊆W∩U for all i=0,…,max{s,a1}. Now let n∈ℕ. If n≤a1, fnk(V)⊆W∩U⊆U, whereas if n>a1, there is j∈ℕ such that n=aj+i for some i=0,…,s and so
(11)fnk(V)=fajk(fik(V))⊆fajk(W∩U)⊆fajk(W)⊆U.Corollary 10.
Suppose thatX is a metric space. If f is continuous and uniformly virtually stable with respect to (an), where (an) satisfies the sup-finite condition, then f is virtually nonexpansive. In particular, C(f) is a Gδ-set when X is complete.Proof.
By settingk=1 in the previous theorem, f is uniformly virtually stable with respect to (n) and hence virtually nonexpansive by Proposition 2 (1). For, if X is complete, by [9, Theorem 1.2], C(f) is then a Gδ-set.Corollary 11.
Suppose thatX is a complete metric space. If, for some k∈ℕ, f is uniformly virtually stable with respect to (nk) and fk is continuous, then C(fk) is a Gδ-set.Proof.
Letg=fk. Then g is uniformly virtually stable with respect to (n). Thus, C(fk)=C(g) is a Gδ-set by the previous corollary.


## 3. Sequences Associated with Uniformly Virtually Stable Self-Maps

In view of Proposition2 (2) and Theorem 9, the sequence (an) associated with a given uniformly virtually stable self-map f seems to have an implicit structure. In fact, some questions may naturally arise.(i)
Can we replace(an) with the larger (in some sense) sequence (an′) so that f is still uniformly virtually stable with respect to (an′)?(ii)
If(an) can be enlarged to (an′), can we make (an′) largest possible? and how does it look like?(iii)
Does the largest possible sequence(an′) satisfy the sup-finite condition?(iv)
Is the largest possible sequence(an′) always (n)?As we will see later on in this section, it turns out that the answers to these questions are interestingly nontrivial and require some careful considerations in both continuous and discontinuous settings.Throughout this section,X is a Hausdorff space and f is a self-map of X. Every sequence is assumed to be an increasing sequence of natural numbers, so it is uniquely determined by its image. We then always identify a sequence (an) with the infinite set {an:n∈ℕ}, and with this identification in mind, the following notations become natural: (i)
(an)⊆(bn) if (an) is a subsequence of (bn);(ii)
(an)∪{c} represents the sequence whose image is {an:n∈ℕ}∪{c};(iii)
(an+c) represents the sequence whose image is {an+c:n∈ℕ};(iv)
(an)∪(bn) represents the sequence whose image is {an:n∈ℕ}∪{bn:n∈ℕ}, and so (an)∪⋯∪(bn) represents the sequence whose image is {an:n∈ℕ}∪⋯∪{bn:n∈ℕ};(v)
(an)⊕(bn) represents the sequence whose image is {ai+bj:i,j∈ℕ}, and ⊕i=1k(an)=(an)⊕⋯⊕(an)︸kcopies of(an).For a nonempty subsetA of ℕ, the natural number d satisfying (1)
d∣a for all a∈A and(2)
ifc∣a for all a∈A, then d∣cwill be calledthe greatest common divisor of A and denoted by gcdA. Also, we let 〈A〉 denote the set of all finite sums of elements in A; that is,
(12)〈A〉={x:x=x1+⋯+xkfor somex1,…,xk∈A}.
Notice that A⊆〈A〉 and if 1∈A, we have 〈A〉=ℕ. As usual, gcd(an) will represent gcd{an:n∈ℕ}, and 〈an〉 will represent at the same time both the set 〈{an:n∈ℕ}〉 and the sequence whose image is 〈{an:n∈ℕ}〉. We may simply write 〈a1,…,an〉 for 〈{a1,…,an}〉.Moreover, for a uniformly virtually stable self-mapf, we let
(13)𝒮f={(an):fis  uniformly  virtually  stablewith  respect  to(an)}.
Clearly, 𝒮f is partially ordered by ⊆, and a maximal element in (𝒮f,⊆) will be called a maximal sequence associated withf. The following are some basic properties of 𝒮f.Proposition 12.
Supposef is uniformly virtually stable with respect to (an). one has the following: (1)
If(bn)⊆(an), then(bn)∈𝒮f.(2)
If(bn)∈𝒮f, then(an)∪(bn)∈𝒮f.(3)
⊕i=1k(an)∈𝒮ffor anyk∈ℕ.(4)
(an+c),(an)∪{c}∈𝒮ffor anyc∈〈an〉.Proof.
Letx∈F(f) and U∈𝒩(x).(1)
It is straightforward from the definition of virtual stability.(2)
There existV,W∈𝒩(x) such that fan(V)⊆U and fbn(W)⊆U for all n∈ℕ. Hence fcn(V∩W)⊆U for all n∈ℕ where (cn)=(an)∪(bn).(3)
By induction, it suffices to show that(an)⊕(an)∈𝒮f. There are, by virtual stability of f, V,W∈𝒩(x) such that fan(V)⊆U and fan(W)⊆V for all n. This implies that fan+am(W)⊆fan(V)⊆U for all n,m∈ℕ.(4)
By lettingc∈〈an〉, we have c=ai1+⋯+aik∈⊕i=1k(an) for some k∈ℕ, and hence, (an+c)⊆⊕i=1k+1(an)∈𝒮f and (an)∪{c}⊆⋃j=1k‍⊕i=1j(an)∈𝒮f by (1) and (2).Remark 13.
From (2) in the previous proposition, if(an) and (bn) are maximal elements in (𝒮f,⊆), then (an)=(an)∪(bn)=(bn). This implies that a maximal sequence associated with f, if exists, is always unique.Before considering the existence, we will first investigate the structure of the maximal sequence associated with a given self-map.Lemma 14.
Supposef is uniformly virtually stable. For a sequence (an), one has (1)
gcd(an)=gcd〈an〉=gcd{ai:i≤m} for some m∈ℕ,(2)
there areb∈〈an〉 and m∈ℕ such that kgcd(an)+b∈〈a1,…,am〉 for all k≥0,(3)
if(an)∈𝒮f and (an) satisfies the sup-finite condition, then 〈an〉∈𝒮f.Proof.
(1) The fact thatgcd(an)=gcd〈an〉 follows directly from the definition. Now, let d=min{gcd{ai:i≤k}:k∈ℕ}. Then, d=gcd{ai:i≤m} for some m∈ℕ. If d∤an for some n>m, then gcd{a1,…,an}<d, a contradiction. Also, for any c>0 satisfying c∣an for all n∈ℕ, we have c∣ai for all 1≤i≤m, and hence, c≤d. This implies gcd(an)=gcd{ai:i≤m}.
(2) Ifgcd(an)=a1, the proof is clear. So, we assume that gcd(an)≠a1. By (1), gcd(an)=gcd{ai:i≤m} for some m∈ℕ. Then, for some partition {P,N} of {1,…,m}, gcd(an)=∑i∈P‍niai-∑i∈N‍niai where ni≥0. Set A=∑i∈P‍niai and B=∑i∈N‍niai. Notice that B≠0 because gcd(an)≠a1, and lB∈〈a1,…,am〉 for all l∈ℕ. Then for each l∈ℕ and 0≤r≤B,
(14)rgcd(an)+B2+lB=rA+(B-r+l)B∈〈a1,…,am〉.
Since for k≥0, we have k=qB+r for some q≥0 and 0≤r<B,
(15)kgcd(an)+(B2+B)=rgcd(an)+B2+(1+qgcd(an))B∈〈a1,…,am〉.
Therefore, the result follows by letting b=B2+B.
(3) Assume that(an)∈𝒮f and Δ:=supn(an+1-an)<∞. By (2), there is b∈〈an〉 for which kgcd(an)+b∈〈an〉 for all k≥0. By Proposition 12, we have
(16)(bn):=(an)∪(an+gcd(an)+b)∪(an+2gcd(an)+b)∪⋯∪(an+Δgcd(an)+b)∈𝒮f,
and hence (cn):=(bn)∪{c∈〈an〉:c≤a1+b}∈𝒮f. The proof will immediately follow by Proposition 12 (1) once we can show that 〈an〉⊆(cn). To see this, let c∈〈an〉. The case of c≤a1+b is clear. Otherwise, am+b≤c<am+1+b for some m≥1. Since gcd(an)∣c-am-b, we have c=am+b+rgcd(an) for some 0≤r<am+1-am≤Δ. This implies that c∈(an+rgcd(an)+b)⊆(bn)⊆(cn).Theorem 15.
Supposef is uniformly virtually stable and (an) is the maximal sequence associated with f. One has(1)
(an)=〈an〉,(2)
(an)=〈A〉 for some finite subset A of (an),(3)
iff is continuous, then (an)=(n).Proof.
(1) It suffices to show that〈an〉⊆(an). Let c∈〈an〉. By Proposition 12, (an)∪{c}∈𝒮f, and hence, c∈(an) by maximality of (an).
(2) From (1), we have(an)=〈an〉. Then by Lemma 14 (2), there exist l,m∈ℕ such that kgcd(an)+al∈〈a1,…,am〉 for all k≥0. By setting A={ai:i≤max{l,m}}, we clearly have 〈A〉⊆〈an〉=(an). On the other hand, for each i∈ℕ, notice that (i)
ifi≤max{l,m}, then ai∈A, and(ii)
ifi>max{l,m}, then ai=al+kgcd(an)∈〈a1,…,am〉⊆〈A〉 for some k∈ℕ.
Therefore,(an)⊆〈A〉.
(3) Suppose thatf is continuous. It follows that (an)∪{1}∈𝒮f, and by maximality of (an), 1∈(an). Therefore, by (1), we have (an)=〈an〉=(n).The next theorem gives a necessary and sufficient condition for the existence of the maximal sequence.Theorem 16.
Supposef is uniformly virtually stable. Then the maximal sequence associated with f exists if and only if there is (an)∈𝒮f satisfying the sup-finite condition.Proof.
(
⇒
) If (an) is maximal, then (na1)⊆(an) and hence
(17)supn(an+1-an)≤supn((n+1)a1-na1)=a1<∞.(⇐) Assume that Δ:=supn(an+1-an)<∞ for some (an)∈𝒮f. By Lemma 14 (3), 〈an〉∈𝒮f. If a1=1, we are done. Suppose that a1>1. For 0≤i<a1=min〈an〉, let
(18)Ai={m:m≡imoda1and〈an〉∪{m}∈𝒮f},
and mi=minAi if Ai≠∅, otherwise mi=a1. Then {c:〈an〉∪{c}∈𝒮f}=A0∪⋯∪Aa1-1. By Proposition 12 (2), 〈an〉∪{mi:0≤i<a1}∈𝒮f and again by Lemma 14 (3), since supn(bn+1-bn)≤supn(an+1-an)<∞ where (bn):=〈〈an〉∪{mi:0≤i<a1}〉, (bn)∈𝒮f. Notice that (bn) is maximal because whenever (bn)∪{c}∈𝒮f for some c, there is 0≤i<a1 such that c∈Ai⊆A0∪⋯∪Aa1-1⊆(bn).Towards the end of this section, we will construct some interesting examples including a uniformly virtually stable self-map whose maximal sequence is prescribed as well as a uniformly virtually stable self-map having no maximal sequence.Example 17.
Fork∈ℕ∪{0}, let θk=π∑i=0k‍(1/2i). Then π≤θk<2π for all k. Consider the sequence (an)=〈an〉. Set δ0=a1 and δn=an+1-an for any n∈ℕ. Define f:ℂ→ℂ by
(19)f(z)=f(reiϕ)={(r+1)eiϕifϕ=θk,0<r<δk-1forsomek;(r-⌊r⌋)ei(π+ϕ/2)ifϕ=θk,r≥δk-1forsomek;0otherwise.
Then, f can be illustrated in Figure 1 and it is uniformly virtually stable with respect to the maximal sequence (an).
To see this, note thatF(f)={0}, π+θk/2=π(1+∑i=1k+1‍(1/2i))=π∑i=0k+1(1/2i)=θk+1 for all k≥0, and for each n∈ℕ and 0<r<1,
(20)fan(reiθ0)=fδn-1+⋯+δ0(reiθ0)=fδn-1+⋯+δ1(reiθ1)=⋯=reiθn.
To show the virtual stability of f with respect to (an), it suffices to prove that fan(B(0;r))⊆B(0;r) for all 0<r<1. Let 0<r<1. For each z=seiϕ∈B(0;r),
(21)|fan(z)|≤|fan(seiθk)|=|fanfak(seiθ0)|=|fam(seiθ0)|=|seiθm|=s<r
for some k, and am=an+ak. Thus, we get the claim. Moreover (an) is maximal because whenever b∉(an), either b<a1 or am<b<am+1 for some m∈ℕ (and so b=am+k for some 1≤k<δm) holds, which the first case implies that fb(reiθ0)=(r+b)eiθ0∉B(0;1) for all 0<r<1, while the latter implies that
(22)|fb(reiθ0)|=|fkfam(reiθ0)|=|fk(reiθm)|=|(r+k)eiθm|=r+k>1
for all 0<r<1; that is, fb(reiθ0)∉B(0;1). Thus (an)∪{b}∉𝒮f for all b∉(an), which induces the maximality of (an) as desired.Figure 1
The mapf in Example 17.Lemma 18.
LetA1={2n:n∈ℕ} and for m>1,
(23)Am={2n1+⋯+2nm:ni∈ℕ}-⋃i<m‍Ai.
Then {An:n∈ℕ} forms a partition of 2ℕ:={2n:n∈ℕ}, and An is infinite for each n∈ℕ.Proof.
Following from the fact that2n+2n=2n+1 for all n∈ℕ, if ∑i=1m‍2ni∈Am, then ni≠nj for all i≠j. Also for each n∈ℕ, since 2n=∑i=1n‍2∈⋃i=1n‍Ai, we have 2n∈Ai for some i≤n. Hence, {An:n∈ℕ} forms a partition of 2ℕ. For the next implication, notice that A1 is clearly infinite. Now, assume that n>1. We first claim that 2n-2∈An-1. Since 2n-2∈2ℕ, we have 2n-2∈Ak for some k∈ℕ. It follows that k<n and for if 2n-2=∑i=1k‍2ni, then ni<n and ∑i=1k‍2ni=2n-2=∑i=1n-1‍2i, and hence k=n-1. Next, we claim that 2n+m+2n-2∈An for all m∈ℕ. Let m∈ℕ. By the previous claim, we have 2n+m+(2n-2)∈⋃i=1n‍Ai. Hence, 2n+m+2n-2∈Ak for some 1≤k≤n; that is, 2n+m+2n-2=∑i=1k‍2ni. Since ∑i=1n+m-1‍2i=2n+m-2<2n+m+2n-2<2n+m+1, we must have 2ni=2n+m for some 1≤i≤k. Thus, 2n+m+2n-2∈An as claimed. Finally, since {2n+m+2n-2:m∈ℕ}⊆An, then An is infinite as desired.Example 19.
For eachm∈ℕ, let Am be defined as in the previous lemma, p1=3 and for each n>1, let pn be the smallest prime number such that pn>max{n2+n,pn-1}. For examples, p2=7 and p3=13. Clearly, pn≠pm whenever n≠m. Now, let X=[0,2]×[0,2] be a subspace of ℝ2 equipped with the maximum norm, 0:=(0,0)∈X and let B(a;r) denote the open ball {x∈X:∥x-a∥<r}.
The goal of this example is to find a uniformly virtually stable self-map having no maximal sequence. This can be done by constructing a uniformly virtually stable self-mapf:X→X with respect to (2n) but not (2n). For if (an) is the maximal sequence associated with such an f, we must have (2n)⊆(an) and hence (2n)=〈2n〉⊆〈an〉=(an), but this contradicts the property that f is not uniformly virtually stable with respect to (2n).
To obtain the desired self-map, it suffices to require the mapf:X→X to satisfy the following three conditions: (C1)
F(f)={0},(C2)
f2n(B(0;1/(i+1)))⊆B(0;1/i) for each i,n∈ℕ,(C3)
For eachk∈ℕ, there exists a∈2ℕ such that fa(1/(k+2),0)∉B(0;1).
Notice that (C1) and (C2) imply the virtual stability off with respect to (2n), while (C1) and (C3) imply that, for each k∈ℕ, there exists a∈2ℕ such that fa(B(0;1/(k+1)))⊈B(0;1), and hence f is not uniformly virtually stable with respect to (2n).
We are now ready to give the explicit description of such anf. Consider the following subsets of X: T1={(1/(k+2),0):k∈ℕ},T2={(1,1/pkl):k,l∈ℕ  and  l+1∈Am for some m≤k},T3={(1/(k+2-m),1/pkl):k,l∈ℕ and l∈Am for some m≤k},T4={(1,1/pkl):k,l∈ℕand either l∈Am or l+1∈Am for some m>k}.
Observe that(O1)
Ti∩Tj=∅ whenever i≠j. All cases are trivial except for the case of T2∩T4, where it follows from the fact that {Am:m∈ℕ} forms a partition of 2ℕ;(O2)
∥(x,y)∥=x for all (x,y)∈⋃i=14‍Ti, because for each k,m,l∈ℕ with m≤k, we have pkl≥pk>k2+k≥k+1 and hence 1/pkl<1/(k+1)≤1/(k+2-m).
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.Clearly,f is well-defined by (O1). The following properties are also satisfied.(P1)
f(T1)⊆T2, f(T2)⊆T3 and f(T3∪T4)⊆T2∪T4. So, ⋃i=14‍Ti is f-invariant. Moreover, since pi’s are all distinct primes, we have pik=pjl if and only if (i,k)=(j,l), and hence (π2∘f)|⋃i=14‍Ti is injective, where π2 denotes the second-coordinate projection π2(x,y)=y.(P2)
For eachk,l∈ℕ, one can verify that
(25)fl(1k+2,0)={(1k+2-m,1pkl)ifl∈Amfor somem≤k;(1,1pkl)otherwise.
In particular, we have the following.
(i)
f2l(1/(k+2),0)=(1/(k+2-1),1/pk2l)=(1/(k+1),1/pk2l) since 2l∈(2n)=A1 and 1≤k.(ii)
fc(1/(k+2),0)=(1,1/pkc), where c=minAk+1, since c∈Ak+1 and k+1>k.(P3)
For eachk,l,m,n∈ℕ with m<k and l∈Am, we have m+1≤k, fl(1/(k+2), 0)=(1/(k+2-m),1/pkl) by (P2), and 2n+l∈Aj⊆⋃i=1m+1‍Ai for some j≤m+1. Again by (P2), it follows that
(26)(1k+2-j,1pk2n+l)=f2n+l(1k+2,0)=f2n(1k+2-m,1pkl).Clearly, (C1) is implied by (P1). To prove (C2), leti,n∈ℕ, and it suffices to consider the following two cases.(1)
If(x,y)=(1/(k+2),0)∈T1∩B(0;1/(i+1)), then 1/(k+2)<1/(i+1), and hence
(27)∥f2n(x,y)∥=∥f2n(1k+2,0)∥=(P2)∥(1k+1,1pk2n)∥=(O2)1k+1<1i.(2)
If(x,y)=(1/(k+2-m),1/pkl)∈T3∩B(0;1/(i+1)), then l∈Am for some m≤k, and 1/(k+2-m)=(O2)∥(1/(k+2-m),1/pkl)∥<1/(i+1). If m=k, we have 1/2=1/(k+2-m)<1/(i+1), a contradiction. Thus m<k, and by (P3), there is some j≤m+1 such that
(28)∥f2n(1k+2-m,1pkl)∥=(P3)∥(1k+2-j,1pk2n+l)∥=(O2)1k+2-j≤1k+2-(m+1)<1i.From above cases, it follows that f2n(B(0;1/(i+1)))⊆B(0;1/i).Finally, for (C3), letk∈ℕ and c=minAk+1∈2ℕ. Then by (P2), we have fc(1/(k+2),0)=(1,1/pkk+1)∉B(0;1).Remark 20.
From the previous example, we can easily check thatf is uniformly virtually stable with respect to (an)=A1∪⋯∪Am for any m∈ℕ, just by showing that fan(B(0;1/(k+m)))⊆B(0;1/k) for any k,n∈ℕ. Hence, Proposition 12 (2) does not hold in general for an arbitrary union of sequences in 𝒮f.



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*Source: 237858-2013-06-23.xml*