On Topological Properties of Metrics Defined via Generalized ( Linking Construction )

We analyze topological properties ofmetric 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.


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, R 2 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 R-trees (for more information about Rtrees the reader can see, e.g., [5][6][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 socalled "floor" metric, defined on the set of real numbers R. 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 R 2 endowed with the "river" metric or with the radial metric can be found in paper [12].

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.
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 .
In the next examples we mention some particular cases covered by Definition 1.
(c) Let (, ) be a metric space,  :  →  be a function, "∼" be the equality relation, and  be a metric on .
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 .

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   :  ×  → [0, +∞) 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.

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  = | () ).
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 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  > 0 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 (  , (  )) → 0. Fix  > 0 and let  be large enough so that (  , (  )) < (1/5) for  ≥  and (  ,   ) < ( The first term on the right-hand side above tends to zero and the latter two terms are equal to zero, and hence (  , ) → 0. ( as (  , ) → 0, it would be the case that also (  , (  )) → 0, which is a contradiction.Assume now that (  , (  )) → 0. 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 (  , (  ))   0 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 .
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  fl R, 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 (, 0) = 1 + | + 1|, if  ̸ = 0. Let us consider the sequence (1/) ∈N .Clearly, it is -Cauchy, but not -convergent.
Let us notice that, putting   = 1/, we obtain a Cauchy sequence such that ((  )) is also a Cauchy sequence and (  , (  )) = 0, but while (  , 0) → 0, neither the equality 0 = (0) 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  fl (0, 1), let  be the Euclidean metric and  = | () ,  be the identity, and let "∼" be the equality relation.Then  is the Euclidean metric on (0, 1).Again, considering the sequence (1/) shows that the assumptions of Theorem 10 are not satisfied.Example 6.Let us consider the "river" metric.Here,  = R 2 ,  is the Euclidean metric, (, ) = (, 0), the relation "∼" is defined by the formula ( 1 ,  1 ) ∼ ( 2 ,  2 ) ⇔  1 =  2 , 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  = R 2 , let  be the Euclidean metric, (, ) = (0, 0) for all ,  ∈ R (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 (  , 0)   0 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 (  , 0) → 0, then (2) holds, because (0, 0) = (0, 0).

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 .
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 𝑔 ̸
= | () , then the boundedness of  does not have to imply the boundedness of .Indeed, consider  = R, let  : R → R 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 3), but (, ) is not.

The (Floor) Metric
For now, we will concentrate on one particular case of Definition 1, given in the following example.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 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.
We will now describe balls in the metric space (R, ).We will start with the case when the center is an integer.

Proposition 22 (form of open sets). A subset of R is 𝜑-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 R 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  > 0 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  ∉ Z, and the proof is complete.
Corollary 23.A subset of R 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.Now we are going to establish the formula which allows calculating the Kuratowski measure of noncompactness of bounded subsets of R 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  ⊂ [,  + 1) for some  ∈ Z.The Kuratowski measure of noncompactness of , considered as a subset of (R, ), is given by the formula  () = 2 max ({ −  |  is the limit point of } ∪ {0}) . ( (Let us note that for the sake of this proposition and its corollary, when we talk of a limit point of some subset of R, 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 () = 0. From now on we will assume that  is infinite.
Let us define  fl sup{ −  |  is the limit point of } ∈ [0, 1].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 now prove that actually () = 2; that is, there is no finite covering of  with sets of diameter strictly less than 2.Choose  ∈ (0, 2); of course, there are infinitely many points of  in the interval  fl [ + (1/2),  + 1).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 (R, ).The Kuratowski measure of noncompactness of  is given by the formula (15) 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.

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  fl R, let  be the Euclidean metric and  = | () , let  be a constant mapping (say, () fl 0 for any  ∈ R), and let  ∼  if and only if, ⌊⌋ = ⌊⌋.Let us consider the sequence (  ) defined by the formula   = 1 − 1/.It is obvious that it is -Cauchy.If it were -convergent to some  ∈ R, Proposition 4 would imply its -convergence to ; hence  = 1.But (1 − 1/, 1Once again, let us notice that both the sequences (  ) and ((  )) are -Cauchy, (  , (  )) = |1 − 1/|   0, 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.