Journal of Function Spaces

Volume 2017, Article ID 4901762, 7 pages

https://doi.org/10.1155/2017/4901762

## On Topological Properties of Metrics Defined via Generalized “Linking Construction”

Faculty of Mathematics and Computer Science, Adam Mickiewicz University, Ul. Umultowska 87, 61-614 Poznań, Poland

Correspondence should be addressed to Marcin Borkowski; lp.ude.uma@krobm

Received 2 May 2017; Accepted 25 October 2017; Published 21 November 2017

Academic Editor: P. Veeramani

Copyright © 2017 Marcin Borkowski et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We analyze topological properties of metric spaces obtained by using Száz’s construction, which we used to call generalized “linking construction.” In particular, we provide necessary and sufficient conditions for completeness of metric spaces obtained in this way. Moreover, we examine the relation between Száz’s construction and the “linking construction.” A particular attention is drawn to the “floor” metric, the analysis of which provides some interesting observations.

#### 1. Introduction

In the paper [1] the authors described a construction of a large class of hyperconvex metric spaces (for an introduction to the theory of hyperconvex metric spaces we refer the reader to [2] or [3]). This construction is based on using Chebyshev subsets of a normed space endowed with a hyperconvex metric. In particular, it contains two classical hyperconvex metric spaces, endowed with the “river” metric or with the radial metric, as special cases (cf. [4]). Let us also add that the theory of hyperconvex metric spaces is closely connected with the theory of -trees (for more information about -trees the reader can see, e.g., [5–7]).

On the other hand, in paper [8] the authors showed a general “linking construction” yielding hyperconvex spaces. That construction also encompasses the “river” metric and the radial metric. Let us add that a similar concept appears in [9], where it was used to study existence of certain mappings between Banach spaces. A slight generalization of the “linking construction” along with several examples can be found in the paper [10].

An interesting generalization of the two constructions mentioned above was proposed by Száz in the paper [11]. In that construction, instead of a metric projection and a collinearity relation, Száz used an arbitrary function and a suitable relation defined on a set under consideration.

The main goal of this note is to examine some topological properties of metric spaces obtained by using Száz’s construction. The main emphasis will be put on completeness of metric spaces produced by Száz’s construction. In particular, in Section 4 we provide necessary and sufficient conditions for completeness of those metric spaces while in Section 5 we deal with their boundedness.

In Section 6, completing this note, we concentrate on one particular case of Száz’s construction, namely, on the so-called “floor” metric, defined on the set of real numbers . We provide some interesting observations concerning that metric. It turns out, for example, that balls in that metric space need not be connected nor compact. We will give formulae which allow calculating the Kuratowski measure of noncompactness of bounded subsets included in that metric space. Let us emphasize that in general it is not easy to find concrete formulae which would allow calculating the Kuratowski measure of noncompactness. Such formulae for bounded subsets of endowed with the “river” metric or with the radial metric can be found in paper [12].

#### 2. Preliminaries

In this section, we recall the definition of a metric introduced by Száz, next we mention some of its basic properties, and we give some examples.

*Definition 1 (see [11]). *Let be a metric space, be a function, be a metric on , and “” be an equivalence relation on such that implies . Define the function by the formula

It turns out that the above defined function is a metric on .

Proposition 2 (see [11]). *Under the assumptions and notation of Definition 1, is a metric on .*

*Remark 3. *In the sequel we will use the symbol to denote the metric on inherited from .

The following obvious observation will be very useful in the sequel.

Proposition 4. *Under the assumptions and notation of Definition 1 and assuming that , the metric is stronger than .*

*Proof. *It is enough to observe that for any .

In the next examples we mention some particular cases covered by Definition 1.

*Example 1. *(a)Let be a metric space, be a function, “” be the equality relation, and .(b)Let be a metric space, be a function, “” be the relation defined by , and .(c)Let be a metric space, be a function, “” be the equality relation, and be a metric on .

*Example 2. *Let be the space of functions differentiable infinitely many times and be the “maximum” metric: . Let for all . Of course, , so we may deduce that the function defined by the formula is a metric in .

It turns out that there is a strong connection between limit points of and fixed points of the mapping .

Proposition 5. *Under the assumptions and notation of Definition 1 and assuming that “” is the equality relation, each limit point of is a fixed point of .*

*Proof. *Let be a sequence of points of converging to some such that each is distinct from . This means that ; since , it must be the case that .

*Remark 6. *Let us consider a particular case of Example 1(b). Namely, let and let be the “river” metric. Then for and obviously not all limit points of are fixed points of .

#### 3. Relation to the “Linking Construction”

In this short section we are going to compare Száz’s construction with the so-called linking construction considered in paper [8].

*Definition 7. *Let be a metric space and be a family of metric spaces. Let us assume that for each and that for any (i.e., each pair of is either disjoint or intersects at ). Let us define and let be defined by the formula (Let us note that the five cases in the above formula are not pairwise disjoint, but if more than one applies, the result is in fact the same in either case.)

*Remark 8. *The above metric can be also realized using Száz’s construction in the following way.

Let be defined by the formula For any , we declare that if or for some . Finally, we define .

In order to check that and coincide, one has to examine several cases. As an example, let us consider one of them, namely, and . Then, either (which can only happen if , and in that case ), or . Then we obtain . On the other hand, .

#### 4. Completeness

In this section we are going to prove a sufficient and necessary condition for the metric to be complete (with the natural assumption that ).

Throughout this section we will frequently make use of the fact that for any points , the obvious inequalities hold. In particular, if some sequence is -Cauchy (resp., -convergent), that is, it is a Cauchy (respectively, convergent) sequence with respect to the metric , then it is -Cauchy (resp., -convergent to the same limit).

Also, we will sometimes say shortly that some points of are “equivalent” to mean that they are equivalent with respect to the relation “”.

Finally, we will often assume that all but finitely many of the terms (i.e.,* almost all* terms) of some sequence are pairwise equivalent with respect to the relation “”; in such cases, in order to shorten the considerations, we will silently drop enough of the beginning terms so that the rest of them are equivalent.

Let us now present a way to express the fact that a sequence is -Cauchy in terms of the metric and the relation “.”

Proposition 9. *With the assumptions and notation of Definition 1, with the additional assumption that , a sequence of points in is -Cauchy if and only if is -Cauchy, and *(i)*almost all ’s are equivalent with respect to the relation “,” or*(ii)*.*

*Proof. *Let the sequence be -Cauchy. It is then -Cauchy. If almost all of its terms are equivalent, the proof is complete. Assume the contrary. Fix and choose large enough so that if . For any there exists some such that . We have then Assume now that the sequence is -Cauchy. If almost all of its terms are equivalent, the metrics and coincide for those terms and hence is -Cauchy. Assume now that . Fix and let be large enough so that for and for . Then, if , we obtain

*Theorem 10. With the assumptions and notation of Definition 1, if we assume additionally that , the metric is complete if and only if the following two conditions are true for any -Cauchy sequence such that is also -Cauchy.(1)If and almost all ’s are equivalent, then there exists some equivalent to the ’s such that .(2)If , then for some such that the alternative, or almost all ’s are equivalent to , is true.*

*Proof. ****Sufficiency*. Let be -Cauchy and hence -Cauchy. Let us consider two cases.*(I) Almost All **’s Are Equivalent*. This means that the sequence is constant and hence -Cauchy. If , then for some equivalent to the ’s. This means that and coincide for ’s and and hence . If on the other hand , then for some . Moreover, since is constant, denoting we obtain and hence . If is equivalent to ’s, and coincide again on ’s and and hence . Otherwise we have and we obtain The first term on the right-hand side above tends to zero and the latter two terms are equal to zero, and hence .*(II) It Is Not the Case That Almost All **’s Are Equivalent*. Proposition 9 implies that . Since is -Cauchy and , the sequence is also -Cauchy. This means that for some . We have the estimate and the proof is complete.*Necessity*. Assume that the metric is complete. Let be a -Cauchy sequence such that is also -Cauchy.

If and almost all ’s are equivalent, this means that and coincide for ’s and hence the sequence is -Cauchy—hence -convergent and hence -convergent to some . If were not equivalent to ’s, we would have as , it would be the case that also , which is a contradiction.

Assume now that . From Proposition 9 we know that the sequence is -Cauchy, hence -convergent and hence -convergent to some . For the sake of contradiction, assume that and that we can find a pair of nonequivalent terms arbitrarily far in the sequence . We may thus pass to a subsequence of ’s not equivalent to . Since and , we obtain a contradiction with the -convergence of the sequence .

*As a corollary, we are going to show that some very natural properties of and imply completeness of .*

*Corollary 11. With the assumptions and notation of Definition 1, if we assume additionally that , the mapping is continuous (with respect to the metric ), the space is complete, and the equivalence classes of the relation “” are closed with respect to the metric ; then the space is complete.*

*Proof. *Assume that the sequences and are -Cauchy. If and almost all ’s are equivalent, then is -convergent to some due to -completeness of and is equivalent to ’s due to the equivalence classes of the relation “” being closed with respect to the metric .

Let now . Again, is -convergent to some . By continuity of , we have . Finally, and hence .

*The following three examples show that while the above theorem gives only a sufficient and not necessary conditions for completeness of , releasing either one of these three conditions may result in an incomplete space.*

*Example 3. *If is discontinuous, the metric may be incomplete. To see this, let us define , let be the Euclidean metric and , let the relation “” be the equality relation, and let We have , if are simultaneously equal or not equal to zero, and , if . Let us consider the sequence . Clearly, it is -Cauchy, but not -convergent.

Let us notice that, putting , we obtain a Cauchy sequence such that is also a Cauchy sequence and , but while , neither the equality nor the equivalence of almost all ’s holds. In other words, the second assumption of Theorem 10 is not satisfied.

*Example 4. *If is not complete, the metric may be incomplete. To see this, let , let be the Euclidean metric and , be the identity, and let “” be the equality relation. Then is the Euclidean metric on . Again, considering the sequence shows that the assumptions of Theorem 10 are not satisfied.

*Example 5. *If the equivalence classes of the relation “” are not closed with respect to the metric , the metric may be incomplete. To see this, let us define , let be the Euclidean metric and , let be a constant mapping (say, for any ), and let if and only if, . Let us consider the sequence defined by the formula . It is obvious that it is -Cauchy. If it were -convergent to some , Proposition 4 would imply its -convergence to ; hence . But .

Once again, let us notice that both the sequences and are -Cauchy, , and all ’s are equivalent. However, the sequence is not -convergent to a point in the same equivalence class.

*Now, we are going to illustrate Theorem 10 and Corollary 11 with the following two examples.*

*Example 6. *Let us consider the “river” metric. Here, , is the Euclidean metric, , the relation “” is defined by the formula , and . It is easy to see that all the conditions of Corollary 11 are satisfied and hence the “river” metric is complete.

*Example 7. *Let us now consider the radial metric. Put , let be the Euclidean metric, for all (hence is trivial), and define the relation “” by the formula Let us notice that while is complete and is -continuous, the equivalence classes of the relation “” are not closed with respect to the metric . Despite that, the radial metric is complete. Indeed, let us take any -Cauchy sequence . Obviously, is a constant sequence and hence -Cauchy. If and almost all ’s lie on the same ray, this means that is -convergent to some nonzero point on the same ray and (1) is satisfied. If on the other hand , then (2) holds, because .

*5. Boundedness*

*5. Boundedness*

*In this very short section we are going to discuss the problem of boundedness of the metric defined in Definition 1.*

*Proposition 12. Under the assumptions and notation of Definition 1, if the metrics and are bounded, then so is . On the other hand, if is bounded, then so is . Moreover, if and is bounded, then so is .*

*Proof. *Let us first assume that and are bounded by some constant . Then, for any , we obtain .

Assume now that is bounded. Let and be two distinct points in . Then , so is bounded. Furthermore, if , then for any we either have (and in this case, ) or (and in this case ).

*Remark 13. *Let us notice that, in the last part of the above proposition, we assume that * restricted to * is equal to and hence bounded, and from that we infer the boundedness of * on the whole *.

*Remark 14. *If , then the boundedness of does not have to imply the boundedness of . Indeed, consider , let be defined by , let “” be the equality relation, and let be the Euclidean metric and the discrete metric. Then is bounded (in fact, has the diameter of ), but is not.

*6. The “Floor” Metric*

*6. The “Floor” Metric*

*For now, we will concentrate on one particular case of Definition 1, given in the following example.*

*Example 8. *Let , , “” be the equality relation, and . Applying Definition 1 we obtain a certain metric on , which in what follows will be denoted by .

*Remark 15. *By setting , that is, letting be the smallest integer not less than , we obtain a metric with similar properties—in fact, it is isometric to the one given here. A family of metrics with similar properties can be generated by functions of the form or .

*We are now going to establish some properties of the metric defined in Example 8.*

*Proposition 16. The metric defined in Example 8 is neither homogeneous, nor translation invariant, nor metrically convex.*

*Proof. *To see that is not homogeneous, let us observe that , but . To see that it is not translation invariant, let us observe that , but . To see that it is not metrically convex, let us observe first that . Now if , we have . Indeed, if , then , and if (and ), then .

*Proposition 17. The metric defined in Example 8 is complete.*

*Let us notice that we can easily see that the assumptions of Theorem 10 are satisfied, although—since the mapping is discontinuous at integer points—we cannot apply Corollary 11. We shall, however, show how to prove the completeness directly.*

*Proof. *Let be a -Cauchy sequence. If it has a constant subsequence, there is nothing to prove. If this is not the case, we may assume (passing to a subsequence if necessary) that all ’s are pairwise distinct. We may also assume that for some . We will prove that . Assume for the sake of contradiction that . That means that , which implies —contradiction.

*We will now describe balls in the metric space . We will start with the case when the center is an integer.*

*Remark 18. *If , we define and similarly with other kinds of intervals.

*Proposition 19. Let and . Then the ball centered at with radius with respect to the metric defined in Example 8 has the form *

*Proof. *Let us consider several cases.(i). Then and hence , so .(ii). Then and hence , so .(iii). Then , so .(iv). Since , also and hence . We obtain now .(v); then obviously .(vi). Then and .(vii). Then .

*Let us now consider a ball with an arbitrary center.*

*Proposition 20. Let and . Then the ball centered at with radius with respect to the metric defined in Example 8 has the form .*

*Proof. *Let . If , there is nothing to prove. Let . Then . Since the left-hand side of this expression equals , we are done.

Let now . Again we may assume , and similar computation shows that .

*Remark 21. *In a similar way we may prove that if and , then and that if and , then .

*Now we describe the form of open subsets of the metric space .*

*Proposition 22 (form of open sets). A subset of is -open if and only if it is a disjoint union of maximal intervals (i.e., such that the union of any two or more of these intervals is not an interval itself) such that any of these intervals whose right endpoint is an integer is right-open.*

*Proof. *It is known that any subset of can be represented by a disjoint union of maximal intervals (even in a unique way). It is thus enough to prove that a necessary and sufficient condition for an interval to be -open is that if its right endpoint is an integer, the interval in question is right-open.

Let us notice that an open ball with a noninteger center and sufficiently small radius is a singleton and that an open ball with an integer center and sufficiently small radius is the interval . Let now be an interval with left endpoint and right endpoint . If , then clearly lies in the interior of . If , then must be an interior point of . If , then is in the interior of iff , and the proof is complete.

*Corollary 23. A subset of is closed with respect to the metric if and only if it is a disjoint union of maximal intervals (i.e., such that the union of any two or more of these intervals is not an interval itself) such that any of these intervals whose left endpoint is integer is left-closed.*

*Remark 24. *Balls in need not be connected nor compact. For instance, is the union of two open sets and . On a similar note, , and these two intervals are closed with respect to the metric .

Consider now the ball . Since it can be covered by an infinite family of disjoint open sets, namely, , it is not compact.

*Now we are going to establish the formula which allows calculating the Kuratowski measure of noncompactness of bounded subsets of endowed with the “floor” metric (for the definition and the basic properties of the Kuratowski measure of noncompactness we refer the reader to [13] or to [14]).*

*Proposition 25. Let for some . The Kuratowski measure of noncompactness of , considered as a subset of , is given by the formula (Let us note that for the sake of this proposition and its corollary, when we talk of a limit point of some subset of , we mean the limit point with respect to the Euclidean topology and not the topology generated by the metric . The same holds for limits of sequences.)*

*Proof. *First observe that has a limit point if, and only if, it is infinite; if is finite, obviously . From now on we will assume that is infinite.

Let us define . Let us observe first that this supremum is actually attained, since a limit of a sequence of limit points of any given set is also its limit point.

Let us first prove that . If , this claim is obvious, since . Let us assume that and choose any such that . The family is a finite covering of by sets of diameter no greater than .

Let us now prove that actually ; that is, there is no finite covering of with sets of diameter strictly less than . Choose ; of course, there are infinitely many points of in the interval . Choose two distinct points ; we have . Any finite covering of would have to contain a set containing more than one point of —but this means that the diameter of this set would be greater than .

*Corollary 26. Let be a nonempty, bounded subset of . The Kuratowski measure of noncompactness of is given by the formula *

*Proof. *It is enough to observe that the outer maximum in the above formula is well-defined, since only finitely many of the terms involved may be positive and the rest are equal to zero. Then, it follows from the maximum property of the measure of noncompactness.

*Conflicts of Interest*

*Conflicts of Interest*

*The authors declare that there are no conflicts of interest regarding the publication of this paper.*

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