Abstract
Let X be a class of metric spaces and let PX be the set of all f : [0, ∞) → [0, ∞) preserving X, i.e., (Y, f ∘ ρ) ∈ X whenever (Y, ρ) ∈ X. For arbitrary subset A of the set of all metric preserving functions, we show that the equality PX = A has a solution if A is a monoid with respect to the operation of function composition. In particular, for the set SI of all amenable subadditive increasing functions, there is a class X of metric spaces such that PX = SI holds.
2020 Mathematics Subject Classification:
Primary 26A30, Secondary 54E35, 20M20
1 Introduction
The following is a particular case of the concept introduced by Jachymski and Turoboś [].
Definition 1. Let A be a class of metric spaces. Let us denote by PA 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, ∞);
F0, set of functions f ∈ F with f(0) = 0;
Am, set of functions f ∈ F0 with f−1(0) = {0};
SI, set of subadditive increasing f ∈ Am;
M, class of metric spaces;
U, class of ultrametric spaces;
Dis, class of discrete metric spaces;
M2, class of two-points metric spaces;
M1, class of one-point metric spaces.
The main purpose of this article is to provide a solution to the following problems.
Problem 2. Let A ⊆ PM. Find conditions under which the equation
has a solution X ⊆ M.
Problem 3. Let A ⊆ PU. Find conditions under which Equation (1) has a solution X ⊆ U.
In addition, we find all solutions to Equation (1) for A equal to F, F0, or Am and answer the following question.
Question 4. Is there a subclass X of the class M such that
This question was posed as a challenge in [] in a different but equivalent form and it was the original motivation for our research.
The article is organized as follows. The next section contains some necessary definitions and facts from the theories of metric spaces and metric preserving functions.
Section 3 provides some definitions from the semigroup theory and describes solutions to Equation (1), for the cases when A is F, F0, or Am. In addition, we show that PX is always a submonoid of (F, ∘). See Theorems 21, 23, 24, and Proposition 27, respectively.
Section 4 provides solutions to Problems 2 and 3, which are given, respectively, in Theorems 30 and 33. Theorem 32 gives a positive answer to Question 4.
2 Preliminaries on metrics and metric preserving functions
Let X be non-empty 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
(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
holds for all x, y, z ∈ X.
Example 5. Let us denote by the set (0, ∞). Then the mapping
is the ultrametric on introduced by Delhommé et al. [].
Definition 6. Let (X, d) be a metric space. The metric d is discrete if there is k ∈ (0, ∞) such that
for all distinct x, y ∈ X.
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 the class of all discrete metric space by Dis. In addition, for given non-empty set X1, we will denote by DisX1 the subclass of Dis consisting of all metric spaces (X1, d) with discrete d.
Remark 7. All discrete topological spaces can be endowed with a metric which is discrete, but not every metric space with discrete topology is discrete in the sense of Definition 6.
Example 8. Let Mk, for k = 1, 2, be the class of all metric spaces (X, d) satisfying the equality
Then all metric spaces belonging to M1 ∪ M2 are discrete.
Proposition 9. The following statements are equivalent for each metric space (X, d) ∈ M.
(i) (X, d) is discrete.
(ii) Every three-points 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
We write X1 : = {i, j, k} and X2 : = {j, k, l}. Then the spaces (X1, d|X1×X1) and (X2, d|X2×X2) are discrete subspaces of (X, d) by statement (ii). Consequently we have
and
by definition of the class Dis. Now (3) and (4) give us
which contradicts (2).
Remark 10. The standard definition of discrete metric can be formulated as follows: “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, Searcóid []).
Let F be the set of all functions f : [0, ∞) → [0, ∞).
Definition 11. A function f ∈ F is metric preserving (ultrametric preserving) if f ∈ PM (f ∈ PU).
Remark 12. The concept of metric preserving functions can be traced back to Wilson []. Similar problems were considered by Blumenthal []. The theory of metric preserving functions was developed by Borsík, Doboš, Piotrowski, Vallin, and other mathematicians [–]. See also lectures by Doboš [] and the introductory paper by Corazza []. The study of ultrametric preserving functions began by Pongsriiam and Termwuttipong [] and was continued in [, ].
We will say that f ∈ F is amenable if
holds and the set of all amenable functions from F will be denoted by Am. Let us denote the set of all functions f ∈ F satisfying the equality f(0) = 0 by F0. It follows directly from the definition that Am⊊F0 ⊊ F.
Moreover, a function f ∈ F is increasing if the implication
is valid for all x, y ∈ [0, ∞).
The following theorem was proved in [].
Theorem 13. A function f ∈ F is ultrametric preserving if and only if f is increasing and amenable.
Remark 14. Theorem 13 was generalized in [] to the special case of the so-called ultrametric distances. These distances were introduced by Priess-Crampe and Ribenboim [] and were studies by different researchers [–].
Recall that a function f ∈ F is said to be subadditive if
holds for all x, y ∈ [0, ∞). Let us denote the set of all subadditive increasing functions f ∈ Am by SI.
In the next proposition, we restate the equivalence between statements (i) and (ii) of Corollary 36 [].
Proposition 15. The equality
holds.
Remark 16. The metric preserving functions can be considered as a special case of metric products (= metric preserving functions of several variables). See, for example, [–]. An important special class of ultrametric preserving functions of two variables was first considered in 2009 [].
3 Preliminaries on semigroups. Solutions to FX = A for A = F, F0, and Am
Let us recall some basic concepts of semigroup theory, see, for example, “Fundamentals of Semigroup Theory” by Howie [].
A semigroup is a pair (S, *) consisting of a non-empty set S and an associative operation * : S × S → S, which is called the multiplication on S. A semigroup S = (S, *) is a monoid if there is e ∈ S such that
for every s ∈ S.
Definition 17. 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.
Example 18. The semigroups (F, ∘), (Am, ∘), (PM, ∘), and (PU, ∘) are monoids, and the identical mapping id:[0, ∞) → [0, ∞), id(x) = x for every x ∈ [0, ∞) is the identity of these monoids.
The following simple lemmas are well-known.
Lemma 19. 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 20. Let T1 and T2 be submonoids of a monoid (S, *). Then the intersection T1 ∩ T2 also is a submonoid of (S, *).
The next theorem describes all solutions to the equation PX = F.
Theorem 21. The following statements are equivalent for everyX ⊆ M.
(i) Xis the empty subclass ofM.
(ii) The equality
holds.
Proof. (i) ⇒ (ii). Let X be the empty subclass of M. Definition 1 implies the inclusion F⊇PX. Let us consider an arbitrary f ∈ F. To prove equality (5), it is suffice to show that f ∈ PX. 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 ∈ PX 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 x0 ∈ X. Consequently, d(x0, x0) = 0 holds. Let c ∈ (0, ∞) and let f : [0, ∞) → [0, ∞) be a constant function,
for every t ∈ [0, ∞). In particular, we have
Equality (5) implies that f ∘ d is a metric on X. Thus, we have
which contradicts (6). Statement (i) follows.
Remark 22. Theorem 21 becomes invalid if we allow the empty metric space to be considered. The equality
holds if the non-empty class X contains only the empty metric space.
Let us describe now all possible solutions to PX = F0.
Theorem 23. The equality
holds if and only ifXis a non-empty subclass ofM1.
Proof. Let X ⊆ M1 be non-empty. Equality (7) holds if
and
Here, we prove the validity of (8). Let f ∈ F0 be arbitrary. Since every (X, d) ∈ X is a one-point metric space, we have f ∘ d = d for all (X, d) ∈ X by the positivity property of metric spaces, Inclusion (8) follows.
Here, we prove (9). The inclusion PX ⊆ F follows from Definition 1. Thus, if (9) does not hold, then there is f0 ∈ F such that f0 ∈ PX,
Since X is non-empty, there is (X0, d0) ∈ X. Let x0 be a (unique) point of X0. Since f0 belongs to PX, the function f0 ∘ d0 is a metric on X0. 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 non-empty subclass of M1. If X is empty, then
holds by Theorem 21. Equality (11) contradicts equality (7). Hence, X is non-empty. To complete the proof, we must show that
Let us consider the constant function f0 : [0, ∞) → [0, ∞) such that
for every t ∈ [0, ∞). Then f0 belongs to F0. Hence, for every (X, d) ∈ X, the mapping d0 : = f0 ∘ d is a metric on X. Now (13) implies d0(x, y) = 0 for all x, y ∈ X and (X, d) ∈ X. Hence, card(X) = 1 holds, because the metric space (X, d0) is one-point by the positivity property. Inclusion (12) follows. The proof is completed.
The next theorem gives us all solutions to the equation PX = Am.
Theorem 24. The following statements are equivalent for everyX ⊆ M.
(i) The inclusion
holds, and there is (Y, ρ) ∈ Xwith
and we have
for every (X1, d1) ∈ X.
(ii) The equality
holds.
Proof. (i) ⇒ (ii). Let (i) hold. Equality (17) holds if
and
Here, we prove (18). Inclusion (18) holds if we have
for all f ∈ Am and (X1, d1) ∈ X. Relation (20) follows from Theorem 23 if (X1, d1) ∈ M1. To see it we only note that Am ⊆ F0. Let us consider the case when
Since (X1, d1) is discrete by (14), Definition 6 implies that there is k1 ∈ (0, ∞) satisfying
for all distinct x, y ∈ X1. Let f ∈ Am be arbitrary. Then f(k1) is strictly positive and
holds for all distinct x, y ∈ X1. Thus, f ∘ d1 is a discrete metric on X1, i.e., we have
Now, Equation (20) follows from Equations (16, 21).
Here, we prove (19). To prove, we must show that every f ∈ PX is amenable.
Suppose contrary that f belongs to PX but the equality
holds with some t1 ∈ (0, ∞). By statement (i) we can find (Y, ρ) ∈ X such that (15) and
hold for all distinct x, y ∈ Y. Now f ∈ PX 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 (17). Since Am ≠ F holds, the class X is non-empty by Theorem 21. Moreover, using Theorem 23, we see that X contains a metric space (X, d) with card(X) ⩾ 2, because Am ≠ F0.
If the inequality
holds for every (Y, ρ) ∈ X, then all metric spaces belonging to X are discrete (see Example 8). Using the definitions of Dis and Am, it is easy to prove that for each (X1, d1) ∈ Dis and every (X1, d) ∈ DisX1 there exists f ∈ Am such that d = f ∘ d1. Hence to complete the proof, it is suffice to show that every (X, d) ∈ X is discrete when
Let us consider arbitrary (X, d) ∈ X satisfying (23). Suppose that (X, d) ∉ Dis. Then by Proposition 9 there are distinct a, b, c, ∈ X such that
Let c1 and c2 be points of (0, ∞) such that
Now we can define f ∈ Am as
Equality (17) implies that f ∘ d is a metric on X. Consequently, we have
by triangle inequality. Now using Equations (24, 26) we can rewrite Equation (27) as
which contradicts Equation (25). It implies (X, d) ∈ Dis. The proof is completed.
Corollary 25. The equalities
hold.
Remark 26. The equality
is known, see, for example, Remark 1.2 in the article []. This article also contains “constructive” characterizations of the smallest bilateral ideal and the largest subgroup of the monoid PM.
Proposition 27. LetXbe a subclass ofM. ThenPXis a submonoid of (F, ∘).
Proof. It follows directly from Definition 1 that
holds and that the identity mapping id:[0, ∞) → [0, ∞) belongs to PX. Hence, by Lemma 19, it is suffice to prove
for all f, g ∈ PX.
Let us consider arbitrary f ∈ PX and g ∈ PX. Then, using Definition 1, we see that (X, g ∘ d) belongs to X for every (X, d) ∈ X. Consequently,
holds. Since the composition of functions is always associative, the equality
holds for every (X, d) ∈ X. Now Equation (28) follows from Equations (29, 30).
The above proposition implies the following corollary.
Corollary 28. If the equation
has a solution, thenAis a submonoid ofF.
The following example shows that the converse of Corollary 28 is, generally speaking, false.
Example 29. Let us define A1 ⊆ F as
where f1 ∈ F is defined such that
and id is the identical mapping of [0, ∞). The equalities f1 ∘ f1 = id, f1 ∘ id = f1 = id ∘ f1 show that A1 is a submonoid of (F, ∘). Suppose that there is X1 ⊆ M satisfying the equality
Then using Theorem 21, we see that X1 is non-empty because A1 ≠ F holds. Let (X1, d1) be an arbitrary metric space from A1. Since X1 is non-empty, we can find x1 ∈ X1. Then (32) implies that f1 ∘ d1 is metric on X1. Now using (31), we obtain
which contradicts the positivity property of metrics.
4 Submonoids of monoids PM and PU
The following theorem provides a solution to Problem 2.
Theorem 30. LetAbe a non-empty subset of the setPMof all metric preserving functions. Then the following statements are equivalent.
(i) The equation
has a solutionX ⊆ M.
(ii) Ais a submonoid of (F, ∘).
(iii) Ais a submonoid of (PM, ∘).
Proof. (i) ⇒ (ii). Suppose that there is X ⊆ M such that (33) holds.
Then A is a submonoid of (F, ∘) by Proposition 27.
(ii) ⇒ (iii). Let A be a submonoid of (F, ∘). By Proposition 27, the monoid (PM, ∘) also is a submonoid of (F, ∘). Then using the inclusion A ⊆ PM, we obtain that A is a submonoid of (PM, ∘) by Lemma 19.
(iii) ⇒ (i). Let A be a submonoid of (PM, ∘). We must prove that (33) has a solution X ⊆ M.
Let (X, d) be a metric space such that
Thus, a metric space (Y, ρ) belongs to X if and only if Y = X and there is f ∈ A such that ρ = f ∘ d.
We claim that Equation (33) holds if X is defined by Equality (35). To prove it, we note that Equation (33) holds if
and
Here, we prove Inclusion (36). This inclusion holds if for every f ∈ A and each (Y, ρ) ∈ X we have (Y, f ∘ ρ) ∈ X. Let us consider arbitrary (Y, ρ) ∈ X and f ∈ A. Then, using Equation (35), we can find g ∈ A such that
Since A is a monoid, the membership relations f ∈ A and g ∈ A imply g ∘ f ∈ A. Hence, we have
by Equality (35). Now (Y, f ∘ ρ) ∈ X follows from Equations (38, 39).
Here, we prove Inclusion (37). Let g1 belongs to PX and let (X, d) be the same as in (35). Then (X, g1 ∘ d) belongs to X and, using (35), we can find f1 ∈ A such that
Equality (40) implies
for all x, y ∈ X. Consequently, g1(t) = f1(t) holds for every t ∈ [0, ∞) by Equation (34, 41). Thus, we have g1 = f1. That implies g1 ∈ A. Inclusion (37) follows. The proof is completed.
Remark 31. A reviewer of the article noted that condition (34) can be neatly expressed in terms of center distances which stems from article [].
Let us turn now to Question 4. Proposition 15 and Lemma 20 provide the following result.
Theorem 32. There isX ⊆ Msuch that
Proof. By Proposition 27, the monoids (PM, ∘) and (PU, ∘) are submonoids of (F, ∘). The equality
holds by Proposition 15. Using Equality (43) and Lemma 20 with T1 = PM, T2 = PU, and S = F, we see that SI also is a submonoid of F. Consequently, Theorem 30 with A = SI implies that there is X ⊆ M such that (42) holds.
The next theorem is an ultrametric analog of Theorem 30 and it gives us a solution to Problem 3.
Theorem 33. LetAbe a non-empty subset of the setPUof all ultrametric preserving functions. Then the following statements are equivalent.
(i) The equationPX = Ahas a solutionX ⊆ U.
(ii) Ais a submonoid of (F, ∘).
(iii) Ais a submonoid of (PU, ∘).
A proof of Theorem 33 can be obtained by a simple modification of the proof of Theorem 30. We only note that the ultrametric space defined in Example 5 satisfies equality (34) with and d = d+.
5 Two conjectures
Conjecture 34. The equation
has a solutionX ⊆ Mfor every submonoidAof the monoidAm.
Example 29 shows that we cannot replace Am with F in Conjecture 34, but we hope that the following is valid.
Conjecture 35. For every submonoidAof the monoidF, there existsX ⊆ Msuch thatPXandAare isomorphic submonoids.
Statements
Data availability statement
The original contributions presented in the study are included in the article/supplementary material, further inquiries can be directed to the corresponding author.
Author contributions
VB: Writing – original draft, Writing – review & editing. OD: Methodology, Project administration, Supervision, Validation, Writing – original draft, Writing – review & editing.
Funding
The author(s) declare financial support was received for the research, authorship, and/or publication of this article. VB was partially supported by a grant from the Simons Foundation (Award 1160640, Presidential Discretionary-Ukraine Support Grants, VB). OD was supported by grant 359772 of the Academy of Finland.
Acknowledgments
We express our deep gratitude toward the reviewers for the careful study of our paper. Their elaborate, constructive criticism allowed us to improve this paper and suggested further directions of the research.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher’s note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
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Summary
Keywords
metric preserving function, monoid, subadditive function, ultrametric space, ultrametric preserving function
Citation
Bilet V and Dovgoshey O (2024) On monoids of metric preserving functions. Front. Appl. Math. Stat. 10:1420671. doi: 10.3389/fams.2024.1420671
Received
20 April 2024
Accepted
20 May 2024
Published
14 June 2024
Volume
10 - 2024
Edited by
Kateryna Buryachenko, Humboldt University of Berlin, Germany
Reviewed by
Ravindra K. Bisht, National Defence Academy, India
Filip Turoboś, Lodz University of Technology, Poland
Updates
Copyright
© 2024 Bilet and Dovgoshey.
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*Correspondence: Viktoriia Bilet viktoriiabilet@gmail.com
Disclaimer
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.