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Stable set of self map

The attracting set and the inverse limit set are important objects associated to a self-map on a set. We call \emph{stable set} of the self-map the projection of the inverse limit set. It is included in the attracting set, but is not equal in the general case. Here we determine whether or not the equality holds in several particular cases, among which are the case of a dense range continuous function on an Hilbert space, and the case of a substitution over left infinite words.


Introduction
Let X be a set and ϕ : X → X be a self-map. One can associated to ϕ four subsets of X, namely its fixed set Fix(ϕ), its orbit set Orb(ϕ), its stable set Stab(ϕ) and its attracting set Atrac(ϕ), which are defined by These four subsets are the objects of numerous articles. The fixed set of an automorphism of a finitely generated (free) group has been widely studied [3,10,16,27,29]. The orbit set and the attracting set are considered in Dynamic System Theory [7,12]. The stable set has been considered in Mathematical Economy [14,19], in Theoretical Computer Science [13,26], in analysis [22] and in Dynamic System Theory [1]. Moreover, the stable set is implicitly a key tool in one of the proof of Scott Conjecture [15]. In general the inclusions in the sequence ( * ) are not equalities. In this note we address the question of deciding whether or not the latter inclusion is indeed an equality under some various extra structural hypotheses. For instance, we consider the case of dense range continuous self-maps on an Hilbert space, homomorphisms of free groups and substitution over right infinite words. Surprisingly, we have not be able to find any reference where this natural question is even mentioned. One of the motivation for this work is to fill this gap. Another motivation is a better understanding of some families of infinite words that occur as stable-set of a family of substitutions (cf. Section 8 for details). The two following theorems gather our main results: Theorem 1.1. Assume one of the following cases: (1) the set X is an Hilbert space with a countably infinite base, and ϕ is a linear continuous self-map with dense range; (2) the set X is the open unit disc D of the complex plane, and ϕ is an analytic function such that |ϕ(z)| < 1. Then, the equality Stab(ϕ) = Atrac(ϕ) may not hold. Theorem 1.2. Assume one of the following cases: (1) the set X is a compact metric space, and ϕ is a continuous self-map; (2) the set X is a limit group of free groups, and ϕ is a group endomorphism; (3) the set X is the set of finite words (or of left infinite words) over a finite set, and ϕ is a substitution. Then, the equality Stab(ϕ) = Atrac(ϕ) holds.
The paper is organized as it follows, in Section 2 we present easy examples where the equality either holds or does not hold. We also explain the connection between the stable set and the inverse limit of a self-map. In Sections 3 and 4, we focus on the cases considered in Theorem 1.1. In Sections 5, 6 and 7, we turn to the cases considered in Theorem 1.2. We also investigate the question of deciding whether or not the other inclusions of the sequence ( * ) are actually equalities. Finally, in Section 8 we replace the map ϕ by a monoid generated by a given set of selfmaps, and introduce a generalized definition of the notion of a self-map. Then, we provided motivating examples for such a definition.

Example, counter-example and inverse limit
Let us start with a simple counter example to the equality, and obvious cases where the equality holds.
The proof of Lemma 2.2 and 2.3 is obvious, and let to the reader. We end this section by explaining the connection between the inverse limit of a self-map and its stable set. Let us recall the definition of the former.
Definition 2.4. Let P be a poset and (X i ) i∈P be a family of sets. Assume that for each pair (i, j) of P such that i ≤ j we have a map ϕ i,j : X j → X i such that Now, for each index j, we have projection maps π j : lim ← − (X i ) → X j , (x i ) i∈P → x j . If we choose P = N, X i = X and ϕ i,j = ϕ for every i, j, then by definition Stab(ϕ) is equal to π 0 (lim ← − (X i )) ( = π j (lim ← − (X i ))).

Vector spaces and linear maps
Here we consider the case of a linear self-map on a vector space. One can see without difficulty that the equality Stab(ϕ) = Atrac(ϕ) does not hold in general for infinite dimensional vector spaces: Consider Example 2.1 and its notations; let V be the vector space with base X, and extend ϕ from X to V by linearity. Nevertheless, there is a case where the equality always holds. Note that we may have Stab(ϕ) = Atrac(ϕ) even if the sequence (Ker(ϕ n )) does not stabilize. The above result applies in particular when V is a finite dimensional vector space.
Among infinite dimensional vector spaces, the case of Hilbert spaces is of importance. The notion of a linear continuous self-map with dense range on an infinite dimensional Hilbert space is crucial in Operator Theory. If H is an Hilbert space (with a countably infinite base) and T : H → H is a linear continuous self-map, then we know by [11,28] that Stab(T ) is non-empty, since H is a complete metric space. A simple example of such a map is the following: Let H be the Hilbert space ℓ 2 (C) and fix an Hilbert base : (e k ) k≥0 . We define T : H → H by T ( λ k e k ) = λ k+1 k+1 e k . This map is clearly not injective, continuous and compact with dense range. One can see that T is not surjective since the element 1 k+1 e k does not have a pre-image. Finally, the kernel of ϕ is generated by e 1 , hence, Atrac(T ) = Stab(T ) by Proposition 3.3 below. Proposition 3.3. Let V be an infinite dimensional vector space, and ϕ : V → V be a linear map such that dim(Ker(ϕ)) = 1. Then, Proof. Since dim(Ker(ϕ)) = 1, either Ker(ϕ) is included in Atrac(ϕ), or ϕ is injective on some Im(ϕ n ). In the latter case, we apply Lemma 2.3. Assume the former case holds. Then we claim that ϕ(Atrac(ϕ)) = Atrac(ϕ). Clearly, ϕ(Atrac(ϕ)) is included in Atrac(ϕ). Conversely, assume ϕ(x) belongs to Atrac(ϕ). For every positive integer n, let x n be such that ϕ n (x n ) = ϕ(x). Then ϕ n−1 (x n ) − x belong to Ker(ϕ) and, therefore, belongs to Im(ϕ n−1 ). Thus x belongs to Im(ϕ n−1 ). The claim follows, and we can apply Lemma 2.2 to conclude.
One may wonder whether or not the equality Atrac(T ) = Stab(T ) holds for every linear continuous self-map with dense range on a countably infinite-dimensional Hilbert space. Actually, it is not the case, even if it is not so easy to obtain a counterexample (see Proposition 3.5). We build it up using Example 3.2.
Proof. n = 1 n = 2 n = 3 n = 4 · · · n k = 1 1 2 Let H be the Hilbert space ℓ 2 (C) and fix an Hilbert base (e k ) k≥0 . We define the linear mapT : We are going to prove that The reader may note that for each k the restriction ofT to the Hilbert subspace of H generated by the set {e α(k,n) | n ≥ 1} ∪ {e 0 } is closed to the definition of the map T : H → H of Example 3.2. In the sequel, we denote by f k the element a n e n of H defined by a α(k,n) = 1 (n+1)···(n+k) , a α(j,n) = 0 for j = k and a 0 = 0.

Proof. The mapT is linear by definition. It is continuous since
It has dense range because all the e i belong to the image ofT . Finally, the last part of the proposition is a direct consequence of the two following lemmas.
Lemma 3.6. Let g = a n e n belong to H. Then, Proof. This is a direct computation using the definition ofT .
Assume it is the case and that T ( a n e n ) = f k,k . As in Lemma 3.6, a computation proves that a α(k,n) = 1 + a α(k,1) (n+1)···(n+k) for every n ≥ k + 1. This is impossible, since a n e n lies in H. For r in {1, · · · , m}, this impliesT r (h) = a r e 0 + m−r j=1 +∞ k=j λ k,j+r f k,j where a r = +∞ k=1 a λ k,r k 2 . Indeed, for m = 1, the Part (a) of the claim is equivalent to Lemma 3.6(ii); Part (b) is a consequence of the above computation: e 0 and all the f k,1 belong to Im(T ) except f 1,1 . We deduce the claim for m ≥ 2 by an easy induction: Part (a) follows from equalitiesT (f k,j ) = f k,j−1 for j ≤ k and Lemma 3.6(i), Part (b) follows from similar arguments to the case m = 1. Now, let g be as in the lemma, and set f 0,1 = e 0 . The element g belongs to Im( +∞ k=j λ k,j f k,j with λ m,m = λ m = 0, then h does not belong to Im(T ).

Unit open disc and analytic function
As explained at the end of Section 2, the stable set is deeply connected with the inverse limit. The latter has been investigated in the case of an analytic self-map over the unit open disc [22,23]. Here we focus on this particular case.
Proposition 4.1. [21] Consider an analytic function ϕ on the open unit disc D of the complex plane such that |ϕ(z)| < 1. Then, the set Atrac(ϕ) and Stab(ϕ) may be distinct.
Proof. The following example has been provided to us by P. Poggi-Corradini. It is closed to Example 2.1. Let H = {z ∈ C | Re(z) > 0} \ {2 n | n ∈ N}, and let f : H → H be the multiplication by 1 2 . LetS be the universal covering of H, and choosex 0 ∈S that projects down to x 0 = 3 4 . The self-map f lifts to an analytic self-mapφ ofS. The pre-image ofx 0 underφ is in one-to-one correspondence with all the equivalent classes of paths in H that start at x 0 and end at x 1 = 3/2. Label the pre-images by (x 1,j ). Now for every j consider the set of pre-images under (1/2) j and call them (a k,j ). These are points that project fromS down to 3 · 2 j−1 . Draw a slit of the form {(3 * 2 j−1 , y) | y ≥ 0} on each sheet ofS containing one of the a ′ k,j s. Draw a similar vertical half-slit from every pre image of the a k,j 's. Call the obtained surface S. The mapφ restrict to a self-map of S, and by the Uniformization Theorem [24, Chap. IV], the surface S is conformally equivalent to D. Therefore, the restriction ofφ gives rise to an analytic self-map ϕ of D. Now, by construction, the point of D corresponding tox 0 belongs to Atrac(ϕ), but not to Stab(ϕ).

Compact metric spaces and continuous functions
Inverse limit of continuous function on a compact metric space has also been considered [2]. Here we focus on this particular case. Proof. The set Atrac(ϕ) is a non-empty compact subset as an intersection of nested compact subsets of X. Assume x belongs to Atrac(ϕ), and for each n consider a finite sequence x n,0 , · · · , x n,n such that x n,0 = x and ϕ(x n,m ) = x n,m−1 for m ∈ {1, · · · , n}. Since X is compact, each infinite sequence (x n,m ) n possesses a subsequence that converge to an element y m of X. Clearly we have y 0 = x and ϕ(y m+1 ) = y m for every m. Thus x belongs to Stab(ϕ).

Free groups and homomorphims
We recall that every subgroup of a free group is a free group [18, Cor. 2.9] and that finite rank free groups are hopfian, that is every surjective self-homomorphism is indeed an automorphism [18, Th. 2.13]. Proposition 6.1. Let X = F m be the free group of rank m, and consider a group endomorphism ϕ : F m → F m . Then,

Stab(ϕ) = Atrac(ϕ).
It is easy to see that the set Orb(ϕ) is not equal to Stab(ϕ) in general (just consider some inner automorphism of F n ).
Proof. The main argument is like in [15,Theorem. 1]: the sequence rk(ϕ n (F m )) eventually stabilizes, and the restrictions ϕ : ϕ n (F m ) → ϕ n+1 (F m ) is surjective. Since free groups are hopfian, it follows that the restriction of ϕ to ϕ n (F m ) is injective for sufficiently large n. Then, we apply Lemma 2.3.
The above result can be extended without difficulty to limit groups of free groups. Let us briefly recall how they are defined [6]. Fix a positive integer n. A marked group is a pair (G, S) where G is a group and S : F n → G is an homomorphism that is onto. The space of marked groups G n is the set of marked groups where two marked groups (G, S) and (G ′ , S ′ ) are identified when there exists a group isomorphism ϕ : G → G ′ such that S ′ = ϕ • S. The set G n can be identified with a subset of the set {0, 1} Fn of subsets of F n (an equivalence class of pairs (G, S) is identified with the kernel of S). The product topology on {0, 1} Fn makes G n a compact space. A limit group of free groups is a group G such there exists a pair (G, S) in the closure of the free groups in G n (that is in the closure of the set of pairs (F m , S)). Proposition 6.2. Let X be a limit group of free groups and consider a group endomorphism ϕ : X → X. Then,

Finite words, infinite words and substitutions
Let S be a finite set. We denote by S * and S ω the free monoid of finite words and the set of right infinite words, respectively, on the alphabet S. We denote by ε the trivial word. A substitution is a map from S to S * . It is clear that a substitution ϕ : S → S * extends to an endomorphism of monoid ϕ * : S * → S * defined by ϕ * (s 1 · · · s n ) = ϕ(s 1 ) · · · ϕ(s n ). We say that a letter s in S is mortal (for the substitution) if there exists a positive integer n such that ϕ n * (s) is the empty word. The set of mortal letters is denoted by M ϕ . When s is not mortal, we say that it is immortal. The map ϕ is said to be non-erasing if the set M ϕ is empty. When ϕ is non-erasing, it can also be extended to a map ϕ ω : S ω → S ω defined by ϕ ω (s 1 s 2 · · · ) = ϕ(s 1 )ϕ(s 2 ) · · · . A special case occurs when S = {s 1 , s 2 } and ϕ extends to an automorphism of F 2 . The elements of Stab(ϕ ω ) are special kinds of the so-called sturmian words [13]. In the erasing case, we can still extend ϕ to a map a map from S ω to S ω ∪ S * that restricts to a self-map ϕ ω : S ω ∞ → S ω ∞ , where S ω ∞ is the set of infinite words in S ω that contains an infinite number of immortal letters. When no confusion is possible, we will write ϕ for ϕ * and ϕ ω . Proposition 7.1. Let S be a finite set and ϕ : S → S * be a substitution. Then, (i) Orb(ϕ * ) = Stab(ϕ * ) = Atrac(ϕ * ); (ii) Orb(ϕ ω ) = Stab(ϕ ω ) = Atrac(ϕ ω ); (iii) When ϕ is non-erasing, then Orb(ϕ * ) is the free monoid {s ∈ S | ϕ(s) ∈ S} * . Remark 7.2. (i) In Proposition 7.1, Point (i) is a consequence of Point (ii). Indeed, consider an extra letter t and the set T = S ∪{t}. Define the mapφ : T → T * byφ(s) = ϕ(s) for s in S andφ(t) = t. Let ψ : S * → T ω ∞ be the map defined by ψ(w) = wttt · · · . It is immediate that W belongs to Orb(ϕ * ), Stab(ϕ * ) and Atrac(ϕ * ) if and only if ψ(W ) belongs to Orb(φ ω ), Stab(φ ω ) and Atrac(φ ω ), respectively.
(ii) It is well-known that S ω is compact for its standard metric defined [20,Cor. 3.13]. A non-erasing substitution is continuous. By Proposition 5.1, the sets Stab(ϕ ω ) and Atrac(ϕ ω ) are equal in this case.
To prove Proposition 7.1 we need to introduce some notation. Consider a substitution ϕ : S → S * . The mortality exponent exp(ϕ * ) of ϕ * is the least integer m such that ϕ m * (s) is the empty word for every letter in M ϕ . One has exp(ϕ * ) ≤ Card(M ϕ ) ≤ Card(S). When W belongs to S * , we denote by ℓ(W ) and by ℓ ∞ (W ) its length and its number of immortal letters, respectively. We will need the following classical lemma [26].
(a) is fixed by ϕ * . (ii) If ℓ ∞ (V 2 ) = 0 then for every n one has ℓ(ϕ n * (a)) ≥ n + 1, and there exists an infinite word − → ϕ ω * (a) which is fixed by ϕ ω and such that for every n ≥ exp(ϕ * ) the finite word ϕ n * (a) is one of its (left) prefixes. Proof. Fix an integer m ≥ q = exp(ϕ * ). Then, one has ϕ m and is fixed by ϕ ω ; furthermore ϕ m * (a) is one of its left prefix. We are now ready to prove Proposition 7.1.
We turn now to the proof of (ii). It is in the same spirit than the above one, but more technical. Let W belong to Atrac(ϕ ω ) and assume it does not belong to Orb(ϕ ω ). Denote by k the number of immortal letters of ϕ, that is k = Card(S)− Card(M ϕ ). Write W = U V where U is the maximal periodic prefix ( for ϕ * ) of W with period equal or lower than k. The word U exists since W is not periodic and k is fixed. The word V lies also in Atrac(ϕ ω ) and not in Orb(ϕ ω ). Thus, we can assume without restriction that U is the empty word, and W = V . We choose a sequence (V n ) in S ω ∞ such that ϕ n ω (V n ) = W for each n. We set W n,i = ϕ n−i ω (V n ) for 0 ≤ i ≤ n. In particular, W n,n = V n and W n,0 = W . Let s n,i denote the first (left) immortal letter of W n,i . For each n, there exists s n in S such that Card{0 ≤ i ≤ n | s n,i = s n } ≥ λ n = E( n−1 k ) + 1. Since S is finite, there exists an immortal letter s and an increasing map ψ : N → N such that s ψ(n) = s for every n. By hypothesis, each s n,i is the first immortal letter of the word W n,i . It follows there exists a minimal positive integer r such that ϕ r (s) = V 1 sV 2 with ℓ ∞ (V 1 ) = 0. We must have 1 ≤ r ≤ k. Furthermore, for every n there exists i n in {0, · · · , k − 1} such that s ψ(n),in+jr = s for all j in {0, · · · , λ ψ(n) − 1}. We deduce there exists ι in {0, · · · , k − 1} and ψ 1 : N → N with s ψ1(n),ι+jr = s for every n in N and every j in {0, · · · , λ ψ1(n) − 1}. Assume ℓ ∞ (V 2 ) = 0, and consider n such that λ ψ1(n) ≥ exp(ϕ * ) ≥ exp(ϕ r * ). By Lemma 7.3, ϕ i * (s) is fixed by ϕ r * for i ≥ exp(ϕ r * )r. Therefore, ϕ λ ψ 1 (n) r+ι * (s) is fixed by ϕ r * . But the word ϕ λ ψ 1 (n) r+ι * (s) is a non-trivial prefix of ϕ ω (W ψ1(n),λ ψ 1 (n) r+ι ), that is of W : a contradiction since r ≤ k. Assume ℓ ∞ (V 2 ) = 0. For every n such that λ ψ1(n) ≥ exp(ϕ * ) ≥ exp(ϕ r * ), the word ϕ ι * (ϕ λ ψ 1 (n) r * (s)) is a prefix of W and its length is greater than λ ψ1(n) . Thus, by Lemma 7.3, the word W is equal to ϕ ι ω − −− → (ϕ r * ) ω (a) and is therefore fixed by ϕ r * , again a contradiction. This substitution is non-erasing, and is called the Thue-Morse substitution. It turns out that ϕ ω has a unique fixed point W = abba · · · , the wellknown Thue-Morse infinite word. By the above remark, Fix(ϕ ω ) = Orb(ϕ ω ) = Stab(ϕ ω ) = Atrac(ϕ ω ) = {W }.

The stable set of a monoid of self-maps
The original motivation for this note is to have a better understanding of the connection between a monoid of self-maps and what we call below its stable set. More precisely we are interested by the stable set of infinite words associated with a given finite family of substitutions. The seminal case we have in mind is the one of episturmian morphisms and its stable set, which is the set of infinite epistumian words [13]. The object of this last section is to introduce the notion of a stable set of a monoid of self-maps and to provide motivating examples.
In the sequel, we fix a non-empty set X, and denote by F (X) the set of self-maps of X. If F is included in F (X), we denote by F * and by F ⋆ the free monoid on the alphabet F and the sub-monoid of F (X) generated by F , respectively. The notion of an inverse limit was recalled in Section 2. We remark that for any non-empty subset F of F (X), the free monoid F * is a poset for the prefix order: ϕ 1 · · · ϕ n ≤ ϕ ′ 1 · · · ϕ ′ m if n ≤ m and ϕ i = ϕ ′ i for 1 ≤ i ≤ n. Definition 8.1. Consider a non-empty subset F of F (X). Denote by f → f the canonical surjective morphism from F * to F ⋆ . If ϕ ≤ ϕ ′ in F * , we denote by ψ ϕ,ϕ ′ the element of F ⋆ such that ϕ ′ = ϕ•ψ ϕ,ϕ ′ . Let (X ϕ , ψ ϕ,ϕ ′ ) be the projective system defined by X ϕ = X for ϕ ∈ F * . The stable set Stab(F ) of F is defined by It is immediate that the above definition generalized the definition of the stable set of a self-map. When F = {ϕ}, the sets Stab({ϕ}) and Stab(ϕ) are equal. As far as we know, this notion of stable set has not been considered before. However, interesting sets of infinite words occur as stable sets: Example 8.2. Let Σ be a finite set and denote by Σ ω the set of left infinite words over Σ. For a in Σ, denote by L a : Σ ω → Σ ω and R a : Σ ω → Σ ω the substitutions defined by L a (b) = ab, R a (b) = ba for b = a in Σ and L a (a) = R a (a) = a. Set F = {L a , R a , a ∈ Σ}. The elements of the monoid generated by F are called episturmian morphisms [13]. The elements of Stab(F ) are the so-called infinite episturmian words [17,Theorem 3.10], that have been introduced independently of (and before) the episturmian morphisms [9]. Example 8.3. Let Σ = {1, 2, · · · , n} and denote by Σ ω the set of left infinite words over Σ. We say that a word W is 2-square free if any two consecutive letters are distinct. For each 2-square free word W = a 1 a 2 a 3 · · · in Σ ω , we define the map ψ W : Σ ω → Σ ω by ψ W (x 1 x 2 x 3 · · · ) = a 1 · · · a 1 x1 a 2 · · · a 2 x2 a 3 · · · a 3 x3 · · · Given a word V in Σ ω , there is at most one pair (W, ∆(V )) in (Σ ω ) 2 that verifies ψ W (∆(V )) = V . The map ∆ is the well-known run-length encoding used in compressing data algorithms [5]. The so-called Kolakoski word 2211212211 · · · is the unique fixed point of ψ 21212··· in {1, 2} ω . If for F we consider the set of maps ψ W , where the words W are 2-square free in Σ ω , then the elements of Stab(F ) are by definition the so-called smooth words over Σ considered in [4,5].
The basic property of stable-sets of self-maps stated in Lemma 2.2 extends without difficulties to the wider context of stable sets of self-map monoids: Definition 8.4. Let F be included in F (X) and Y be a subset of X. We say that Y is stabilized by F if ϕ∈F ϕ(Y ) = Y.
Lemma 8.5. The set Stab(F ) is stabilized by F , and every set stabilized by F is included in Stab(F ).
Proof. The proof is immediate.
Let us finish this note with a question and a comment.
Question. Consider a non-empty subset F of F (X), and denote by f → f the canonical surjective morphism from F * to F ⋆ . Defined the attracting set of F to be As in the case of a single self-map, it is clear that Stab(F ) is included in Atrac(F ). When is this inclusion an equality ?
We remark that in the case of Example 8.3, it is trivial that the equality holds by run-length encoding map. In general, answering to this question seems more complicated. The reader can note that when F is finite, then for every element x 0 in Atrac(F ) the exists a right infinite word f 1 f 2 · · · in F ω and a sequence (x n ) in X such that for every n one has x 0 = f 0 (f 1 (· · · f n (x n ))). However, this does not mean that we can choose x n−1 = f n (x n ).