On monoids of metric preserving functions

Let $\mathbf{X}$ be a class of metric spaces and let $\mathbf{P}_{\mathbf{X}}$ be the set of all $f:[0, \infty)\to [0, \infty)$ preserving $\mathbf{X},$ $(Y, f\circ\rho)\in\mathbf{X}$ whenever $(Y, \rho)\in\mathbf{X}.$ For arbitrary subset $\mathbf{A}$ of the set of all metric preserving functions we show that the equality $\mathbf{P}_{\mathbf{X}}=\mathbf{A}$ has a solution iff $\mathbf{A}$ is a monoid with respect to the operation of function composition. In particular, for the set $\mathbf{SI}$ of all amenable subadditive increasing functions there is a class $\mathbf{X}$ of metric spaces such that $\mathbf{P}_{\mathbf{X}}=\mathbf{SI}$ holds, which gives a positive answer to the question of paper [1].


Introduction
The following is a particular case of the concept introduced by Jacek Jachymski and Filip Turoboś in [2].Definition 1.Let A be a class of metric spaces.Let us denote by P A the set of all functions f : [0, ∞) → [0, ∞) such that the implication is valid for every metric space (X, d).
For mappings F : X → Y and Φ : Y → Z we use the symbol F • Φ to denote the mapping We also use the following notation: F, set of functions f : [0, ∞) → [0, ∞); F 0 , set of functions f ∈ F with f (0) = 0; Am, set of amenable f ∈ F; SI, set of subadditive increasing f ∈ Am; M, class of metric spaces; U, class of ultrametric spaces; Dis, class of discrete metric spaces; M 2 , class of two-points metric spaces; M 1 , class of one-point metric spaces.The main purpose of this paper is to give a solution of the following problems.
Problem 2. Let A ⊆ P M .Find conditions under which the equation In addition, we find all solutions to equation (1) for A equal to F, F 0 , or Am and answer the following question.Question 4. Is there a subclass X of the class M such that This question was asked in [1] in a different but equivalent form and it was the original motivation for our research.
The paper is organized as follows.The next section contains some necessary definitions and facts from the theories of metric spaces and metric preserving functions.
In Section 3 we recall some definitions from the semigroup theory and describe solutions to equation (1), for the cases when A is F, F 0 or Am.In addition, we show that P X is always a submonoid of (F, •).See Theorems 20,22,23 and Proposition 26,respectively.Solutions to Problems 2 and 3 are given, respectively, in Theorems 29 and 31 of Section 4. Theorem 30 gives a positive answer to Question 4.

Preliminaries on metrics and metric preserving functions
Let X be nonempty set.A function d : X × X → [0, ∞) is said to be a metric on the set X if for all x, y, z ∈ X we have: is the ultrametric on R + 0 introduced by C. Delhommé, C. Laflamme, M. Pouzet, and N. Sauer in [3].
In what follows we will say that a metric space (X, d) is discrete if d is a discrete metric on X.We will denote by Dis the class of all discrete metric space.In addition, for given nonempty set X 1 , we will denote by Dis X 1 the subclass of Dis consisting of all metric spaces (X 1 , d) with discrete d. (i) (X, d) is discrete.
(ii) Every three-point subspace of (X, d) is discrete.
Proof.The implication (i) ⇒ (ii) is evidently valid.Suppose that (ii) holds but (X, d) ∈ Dis.Then there are some different points i, j, k, l ∈ X such that (2) Write X 1 := {i, j, k} and X 2 := {j, k, l}.Then the spaces (X 1 , d| X 1 ×X 1 ) and (X 2 , d| X 2 ×X 2 ) are discrete subspaces of (X, d) by statement (ii).Consequently we have Remark 9.The standard definition of discrete metric can be formulated as: "The metric on X is discrete if the distance from each point of X to every other point of X is one."(See, for example, [4, p. 14].) Let F be the set of all functions f : [0, ∞) → [0, ∞).
We will say that f ∈ F is amenable iff holds and will denote by Am the set of all amenable functions from F.
Let us denote by F 0 the set of all functions f ∈ F satisfying the equality f (0) = 0.It follows directly from the definition that Am F 0 F.Moreover, a function f ∈ F is increasing iff the implication is valid for all x, y ∈ [0, ∞).
The following theorem was proved in [22].
Theorem 12.A function f ∈ F is ultrametric preserving if and only if f is increasing and amenable.
Recall that a function f ∈ F is said to be subadditive if holds for all x, y ∈ [0, ∞).Let us denote by SI the set of all subadditive increasing functions f ∈ Am.Corollary 36 of [1] implies the following result.
Proposition 14.The equality Remark 15.The metric preserving functions can be considered as a special case of metric products (= metric preserving functions of several variables).See, for example, [31,32,33,34,35,36].An important special class of ultrametric preserving functions of two variables was first considered in 2009 [37].
3. Preliminaries on semigroups.Solutions to Let us recall some basic concepts of semigroup theory, see, for example, "Fundamentals of Semigroup Theory" by John M. Howie [38].
The following simple lemmas are well known.
Lemma 18.Let T be a submonoid of a monoid (S, * ) and let V ⊆ T. Then V is a submonoid of (S, * ) if and only if V is a submonoid of T.
Lemma 19.Let T 1 and T 2 be submonoids of a monoid (S, * ).Then the intersection T 1 ∩ T 2 also is a submonoid of (S, * ).
The next theorem describes all solutions to the equation P X = F. Theorem 20.The following statements are equivalent for every X ⊆ M.
(i) X is the empty subclass of M.
(ii) The equality (5) Proof.(i) ⇒ (ii).Let X be the empty subclass of M. Definition 1 implies the inclusion F ⊇ P X .Let us consider an arbitrary f ∈ F. To prove equality (5) it suffices to show that f ∈ P X .Let us do it.Since X is empty, the membership relation (X, d) ∈ X is false for every metric space (X, d).Consequently, the implication is valid for every (X, d) ∈ M. It implies f ∈ P X by Definition 1. Equality (5) follows.
(ii) ⇒ (i).Let (ii) hold.We must show that X is empty.Suppose contrary that there is a metric space (X, d) ∈ X. Since, by definition, we have X = ∅, there is a point In particular, we have (6) f (0) = c > 0.
Equality (5) implies that f • d is a metric on X.Thus, we have which contradicts (6).Statement (i) follows.
Remark 21.Theorem 20 becomes invalid if we allow the empty metric space to be considered.The equality holds if the nonempty class X contains only the empty metric space.
Let us describe now all possible solutions to P X = F 0 .
Theorem 22.The equality Proof.Let X ⊆ M 1 be nonempty.Equality (7) holds iff (8) P X ⊇ F 0 and (9) Let us prove the validity of (8).Let f ∈ F 0 be arbitrary.Since every (X, d) ∈ X is an one-point metric space, we have f • d = d for all (X, d) ∈ X by positivity property of metric spaces, Inclusion (8) follows.
Since X is nonempty, there is (X 0 , d 0 ) ∈ X.Let x 0 be a (unique) point of X 0 .Since f 0 belongs to P X , the function f 0 • d 0 is a metric on X 0 .Now, using (10), we obtain which contradicts the positivity property of metric spaces.Inclusion (9) follows.Let (7) hold.We must show that X is a nonempty subclass of M 1 .If X is empty, then (11) P X = F holds by Theorem 20.Equality (11) contradicts equality (7).Hence, X is nonempty.To complete the proof we must show that Let us consider the constant function for every t ∈ [0, ∞).Then f 0 belongs to F 0 .Hence, for every (X, d) ∈ X, the mapping d 0 := f 0 •d is a metric on X.Now (13) implies d 0 (x, y) = 0 for all x, y ∈ X and (X, d) ∈ X.Hence, card(X) = 1 holds, because the metric space (X, d 0 ) is one-point by positivity property.Inclusion (12) follows.The proof is completed.
The next theorem gives us all solutions to the equation P X = Am.Let us prove (18).Inclusion ( 18) holds iff we have To see it we only note that Am ⊆ F 0 .Let us consider the case when card(X 1 ) 2. Since (X 1 , d 1 ) is discrete by ( 14), Definition 6 implies that there is Now, (20) follows from ( 16) and (21).
Let us prove (19).To do it we must show that every f ∈ P X is amenable.
Suppose contrary that f belongs to P X but the equality (22) f (t 1 ) = 0 holds with some t 1 ∈ (0, ∞).By statement (i) we can find (Y, ρ) ∈ X such that (15) and ρ(x, y) = t 1 hold for all distinct x, y ∈ Y. Now f ∈ P X and (Y, ρ) ∈ X imply that f • ρ is a metric on Y. Consequently, for all distinct x, y ∈ Y, we have which contradicts (22).The validity of (19) follows.
(ii) ⇒ (i).Let X satisfy equality (4).Since Am = F holds, the class X is nonempty by Theorem 20.Moreover, using Theorem 22 we see that X contains a metric space (X, d) with card(X) 2, because Am = F 0 .
If the inequality card(Y ) 2 holds for every (Y, ρ) ∈ X, then all metric spaces belonging to X are discrete (see Example 7).Using the definitions of Dis and Am, it is easy to prove that for each (X 1 , d 1 ) ∈ Dis and every (X 1 , d) ∈ Dis X 1 there exists f ∈ Am such that d = f • d 1 .Hence to complete the proof it suffices to show that every (X, d) ∈ X is discrete when ( 23) card(X) 3.
Let us consider arbitrary (X, d) ∈ X satisfying (23).Suppose that (X, d) ∈ Dis.Then by Proposition 8 there are distinct a, b, c, ∈ X such that Let c 1 and c 2 be points of (0, ∞) such that (25) Now we can define f ∈ Am as ( 26) Equality (17) implies that f • d is a metric on X.Consequently, we have by triangle inequality.Now using (24) and (26) we can rewrite (27) as which contradicts (15).It implies (X, d) ∈ Dis.The proof is completed.
Corollary 24.The equalities Remark 25.The equality is known, see, for example, Remark 1.2 in paper [13].This paper contains also "constructive" characterizations of the smallest bilateral ideal and the largest subgroup of the monoid P M .
Proposition 26.Let X be a subclass of M. Then P X is a submonoid of (F, •).
Proof.It follows directly from Definition 1 that holds and that the identity mapping id : [0, ∞) → [0, ∞) belongs to P X .Hence, by Lemma 18, it is suffices to prove for all f, g ∈ P X .
Let us consider arbitrary f ∈ P X and g ∈ P X .Then, using Definition 1, we see that (X, g • d) belongs to X for every (X, d) ∈ X.Consequently, (29) (X, f holds.Since the composition of functions is always associative, the equality holds for every (X, d) ∈ X.Now (28) follows from ( 29) and (30).
The above proposition implies the following corollary.
Corollary 27.If the equation has a solution, then A is a submonoid of F.
The following example shows that the converse of Corollary 27 is, generally speaking, false.

Example 28. Let us define
and id is the identical mapping of [0, ∞).The equalities Then using Theorem 20, we see that X 1 is nonempty because A 1 = F holds.Let (X 1 , d 1 ) be an arbitrary metric space from A 1 .Since X 1 is Since A is a monoid, the membership relations f ∈ A and g ∈ A imply g • f ∈ A. Hence, we have by (35).Now (Y, f • ρ) ∈ X follows from ( 38) and (39).Let us prove (37).Let g 1 belong to P X and let (X, d) be the same as in (35).Then (X, g 1 • d) belongs to X and, using (35), we can find The last equality implies for all x, y ∈ X.Consequently, g 1 (t) = f 1 (t) holds for every t ∈ [0, ∞) by (34).Thus, we have g 1 = f 1 .That implies g 1 ∈ A. Inclusion (37) follows.The proof is completed.
Let us turn now to Question 4. Proposition 14 and Lemma 19 give us the following result.The next theorem is an ultrametric analog of Theorem 29 and it gives us a solution to Problem 3. Theorem 31.Let A be a nonempty subset of the set P U of all ultrametric preserving functions.Then the following statements are equivalent.
(i) The equality P X = A has a solution X ⊆ U.
(iii) A is a submonoid of (P U , •).
A proof of Theorem 31 can be obtained by a simple modification of the proof of Theorem 29.We only note that the ultrametric space defined in Example 5 satisfies the equality (34) with X = R + 0 and d = d + .

Two conjectures
Conjecture 32.The equality P X = A has a solution X ⊆ M for every submonoid A of the monoid Am.
Example 28 shows that we cannot replace Am with F in Conjecture 32, but we hope that the following is valid.
Conjecture 33.For every submonoid A of the monoid F there exists X ⊆ M such that P X and A are isomorphic submonoids.

Problem 3 .
Let A ⊆ P U .Find conditions under which equation (1) has a solution X ⊆ U.

Example 5 .
(i) d(x, y) 0 with equality if and only if x = y, the positivity property; (ii) d(x, y) = d(y, x), the symmetry property; (iii) d(x, y) d(x, z) + d(z, y), the triangle inequality.A metric space (X, d) is ultrametric if the strong triangle inequality d(x, y) max{d(x, z), d(z, y)} holds for all x, y, z ∈ X.Let us denote by R + 0 the set (0, ∞).Then the mapping d

Example 7 .Proposition 8 .
Let M k , for k = 1, 2, be the class of all metric spaces (X, d) satisfying the equality card(X) = k.Then all metric spaces belonging to M 1 ∪ M 2 are discrete.The following statements are equivalent for each metric space (X, d) ∈ M.
A semigroup S = (S, * ) is a monoid if there is e ∈ S such that e * s = s * e = s for every s ∈ S. Definition 16.Let (S, * ) be a semigroup and ∅ = T ⊆ S. Then T is a subsemigroup of S if a, b ∈ T ⇒ a * b ∈ T. If (S, * ) is a monoid with the identity e, then T is a submonoid of S if T is a subsemigroup of S and e ∈ T.

Theorem 30 .
There is X ⊆ M such that (42) P X = SI.Proof.By Proposition 26, the monoids (P M , •) and (P U , •) are submonoids of (F, •).The equality (43) SI = P M ∩ P U holds by Proposition 14.Using (43) and Lemma 19 with T 1 = P M , T 2 = P U and S = F we see that SI also is a submonoid of F. Consequently, Theorem 29 with A = SI implies that there is X ⊆ M such that (42) holds.