On Some Properties of Hyperconvex Spaces

Abstract

We are going to answer some open questions in the theory of hyperconvex metric spaces. We prove that in complete -trees hyperconvex hulls are uniquely determined. Next we show that hyperconvexity of subsets of normed spaces implies their convexity if and only if the space under consideration is strictly convex. Moreover, we prove a Krein-Milman type theorem for -trees. Finally, we discuss a general construction of certain complete metric spaces. We analyse its particular cases to investigate hyperconvexity via measures of noncompactness.

1. Introduction

It is hard to believe that although hyperconvex metric spaces have been investigated for more that fifty years, some basic questions in their theory still remain open (let us recall that hyperconvex metric spaces were introduced in [1] (see also [2]), but from formal point of view it has to be emphasized that the notion of hyperconvexity was investigated earlier by Aronszajn in his Ph.D. thesis [3] which was never published). The main purpose of this paper is to answer some of these questions.

Let us begin with the notion of hyperconvex hull which was introduced by Isbell in [4] (see Definition 2.7). This notion is more difficult to investigate than the classical notion of convex hull, since the former one is not uniquely determined (see Proposition 2.8). In Section 3 we are going to prove that in hyperconvex metric spaces with the unique metric segments property, hyperconvex hulls are uniquely determined. Let us recall that such hyperconvex spaces were characterized by Kirk (see [5]) as complete -trees (see Theorem 2.15). This led to a surprising application of the theory of hyperconvex spaces to graph theory (see [6]).

Another interesting question is about the relation between the notion of convexity and hyperconvexity (cf. Remark 4.1). In particular, it is inspired by the following Sine's remark [7, page 863], stated without a proof: "The term hyperconvex does have some unfortunate aspects. First, a hyperconvex subset of even (with the norm) need not be convex. Also convex sets can fail to be hyperconvex (but for this one must go to at least )." It turns out that all hyperconvex subsets of a given normed space are convex if and only if the space in question is strictly convex; this fact is proved in Section 4.

In Section 5 we turn our attention to the classical Krein-Milman theorem (see [8]). We prove that a bounded complete -tree is a convex hull of its extremal points (note that a similar result, but with the assumption of compactness, is proved in [9]). Hence, in particular, such a property holds for bounded hyperconvex metric spaces with unique metric segments.

Let us denote by and the Kuratowski and Hausdorff measures of noncompactness, respectively, (see [10, 11] for the definition and basic properties). It was noticed by Espínola (see [12]) that if a metric space is hyperconvex, then for all its bounded subsets . The question is about the inverse implication. More precisely, assume that for every bounded subset of a given metric space . Does this equality imply that is hyperconvex? (Obviously, we mean nontrivial cases, i.e., we exclude spaces in which every bounded set is relatively compact.) In Sections 6 and 7 we introduce a few metric spaces which are not hyperconvex, but for all their bounded subsets. Hence the answer to the above question is negative. Let us emphasize that the metrics considered in Sections 6 and 7 are extensions and generalizations of commonly known radial metric and river metric, which were proved in [13] to be hyperconvex.

Let us notice that in general it is not easy to provide explicit formulae which would allow to evaluate the measures of noncompactness in particular spaces. We are going to state such formulae for the metric spaces considered in Sections 6 and 7.

Let us emphasize that another motivation to consider those metrics comes from the real world. Let us consider an example of the transmission of phone signals, when one person (say, ) calls another (say, ), assuming there are two base transceiver stations (say, and ). We may have two cases. If and are in the range of one of the BTS's, say , then the signal is first transmitted from to and then from to —even if and are "close" to each other. If and are located in the ranges of and respectively, then the signal is transmitted from to , then from to and finally from to . Hence we have the metric considered in Definition 7.4.

In Section 8 we provide a general scheme to construct metrics similar to these of Sections 6 and 7. This scheme is a generalization of a construction from [14].

For completeness, in Section 2 we collect some basic definitions and facts used in the sequel.

2. Preliminaries

In what follows we will denote the Euclidean metric on by and a "maximum" norm on any suitable space by .

Let us begin with some classical definitions and facts.

Definition 2.1.

Let be a metric space. We call a set a metric segment(joining the points) if there exists an isometric embedding such that and .

Definition 2.2 (see [1, page 410, Definition  1]).

We call a metric space (X, d) hyperconvex, if any family of closed balls with centers at 's and radii of 's, respectively, such that for any has a nonempty intersection.

Hyperconvex spaces possess—among others—the following properties.

Proposition 2.3 (see [1, page 417, Theorem  1']).

A hyperconvex space is complete.

Proposition 2.4 (see [1, page 423, Theorem  9]).

A nonexpansive retract (i.e., a retract by a nonexpansive retraction) of a hyperconvex space is hyperconvex.

Proposition (see [1, page 422, Corollary  4]).

Each hyperconvex metric space is an absolute nonexpansive retract, that is, it is a nonexpansive retract of any metric space it is isometrically embedded in. In particular, hyperconvex spaces are absolute retracts.

The following theorem gives a characterization of hyperconvex real Banach spaces.

Theorem 2.6 (Nachbin-Kelley, see [15, 16]).

A real Banach space is hyperconvex if and only if it is isometrically isomorphic to some space of all real continuous functions on a Hausdorff, compact and extremally disconnected topological space with the norm.

Now let us state the definition of a hyperconvex hull. We will not need the general version of this notion, investigated by Isbell in [4]; instead, the notion of a hyperconvex hull of a subset of a hyperconvex space will suffice for our considerations.

Definition 2.7 (see, e.g., [17, page 408]).

Let be a nonempty subset of a hyperconvex space . We call a hyperconvex hull of (in) if , the set is hyperconvex (as a metric subspace) and there exists no hyperconvex such that .

A hyperconvex hull always exists, but needs not to be unique. It is, however, unique up to an isometry. To be more precise, the following holds.

Proposition 2.8 (cf. [17, page 408, Proposition  5.6]).

Each nonempty subset of a hyperconvex metric space possesses a hyperconvex hull. If and are hyperconvex spaces, , are isometric and is an isometry, then for any hyperconvex hulls , of and respectively, the isometry extends to an isometry .

In what follows, we will also need the definitions of total and strict convexity.

Definition 2.9 (see, e.g. [1, page 407] and [18, page 6, Definition  2.1]).

A metric space is called totally convex if for any two points and for all such that there exists a point satisfying the equalities and . If this point is unique for all possible combinations of , we call the spaceconvex and denote this point by .

Remark 2.10 (see [1, page 410]).

A hyperconvex space is totally convex.

Remark 2.11 (see, e.g., [18, page 7]).

For normed spaces, the above definition of strict convexity (Definition 2.9) coincides with the usual one.

Proposition 2.12 (see, e.g., [18, page 7]).

In a strictly convex metric space, intersection of any family of totally convex subsets is itself totally convex.

The above proposition lets us define the notion of a convex hull in any strictly convex metric space in a natural way.

Definition 2.13.

Let be a nonempty subset of a strictly convex metric space . The convex hull of A (in is the set

(2.1)

When the underlying space is obvious from the context, we will usually write instead of .

Now, let us recall the definition of an -tree.

Definition 2.14 (see, e.g., [5, page 68, Definition  1.2]).

A metric space is called an -tree, if the following conditions are satisfied:

(1)any two points are joined by a unique metric segment (denoted by );

(2)if and then ;

(3)for any there exists such that .

(Let us note that (3) follows from (1); it is, however, useful to have it among the basic properties of -trees.) We will also use the notation for an open metric segment joiningand and for a left-open one.

Theorem 2.15 (see [5, Theorem  3.2]).

For a metric space the following conditions are equivalent:

(1) is a complete -tree;

(2) is hyperconvex and any two points in are joined by a unique metric segment.

In what follows, we will also use the classical notions of Chebyshev subset of a metric space, a metric projection onto such a set (which we will denote by ), Kuratowski and Hausdorff measures of noncompactness (which we will denote by and , resp.), and the radial and river metrics (which we will denote by and , resp.). The reader may find the relevant definitions, for instance, in the papers [11, 19, 20].

3. -Trees

Let us begin this section with the following three simple propositions, which will enable us to characterize -trees as exactly these hyperconvex spaces in which hyperconvex hulls are unique.

Proposition 3.1.

A hyperconvex hull of a two-point subset of a hyperconvex metric space is a metric segment joining and .

Proof.

It is enough to consider as a subset of and apply the uniqueness (up to isometry) of hyperconvex hulls (Proposition 2.8).

Proposition 3.2.

-trees are strictly convex.

Proof.

Let be an -tree. Assume that , , , , and for . Then But we have for and therefore , which is a contradiction.

Proposition 3.3.

For a subset of an -tree, the following conditions are equivalent:

(1) is hyperconvex;

(2) is closed and totally convex.

Proof.

For , it is enough to use Proposition 2.3 and Remark 2.10. On the other hand, if a subset of an -tree is closed and totally convex, it is a complete sub--tree of . Indeed, it is enough to show that for each , the metric segment . But in view of the strict convexity of , we have . Now, in view of Theorem 2.15, is hyperconvex.

A natural question to ask is: in which hyperconvex metric spaces the hyperconvex hulls are unique? The following theorem answers this question.

Theorem 3.4.

Let be a hyperconvex metric space. The following conditions are equivalent:

(1)for each there exists exactly one hyperconvex hull of in ;

(2) is an -tree.

Proof.

Necessity follows easily from Proposition 3.1 and Theorem 2.15. Sufficiency. Let be a subset of an -tree . Notice that . Using Propositions 3.2, 2.12 and 3.3, we arrive at the conclusion that is the hyperconvex hull of in .

4. Normed Spaces

In the first part of this section we will give an answer to the following question: In which spaces closed and convex subsets are hyperconvex?

Remark 4.1.

Note that the question whether all closed and convex subsets of some normed space are hyperconvex makes sense only in spaces which are themselves hyperconvex, so we will now restrict our attention to such spaces.

Theorem 4.2 (see [21, page 474, Theorem  1]).

If is a two-dimensional real normed space, then each nonempty, closed, and convex subset of is a nonexpansive retract of .

Corollary 4.3.

Each nonempty, closed and convex subset of endowed with any hyperconvex norm is hyperconvex.

Remark 4.4.

Notice that "any hyperconvex norm on " means essentially (i.e., up to an isometric isomorphism) the maximum norm; this follows from Theorem 2.6 and can also be proved using a geometric argument (see [19, Theorem  4.1]).

Theorem 4.5.

Let be a hyperconvex normed space. If is not isometrically isomorphic to or , then there exists a two-dimensional linear subspace of which is not hyperconvex.

Proof.

Since is not isometrically isomorphic to , its dimension must be at least 2. Further, since the only (up to an isometric isomorphism) two-dimensional hyperconvex space is , we may assume . By Theorem 2.6 we may assume that is the space for some Hausdorff, compact and extremally disconnected topological space . Since , the space has at least three points, so includes a copy of . This means that it is enough to prove the theorem in case of with the "maximum" norm.

For simplicity, we will construct an affine non-hyperconvex subspace of ; by an appropriate translation one can obtain a linear one. Let . Consider the following three balls in : , , . Since the corresponding balls in intersect only at the space is not hyperconvex.

Corollary 4.3 and Theorem 4.5 yield the following characterization.

Corollary 4.6.

Let be a real normed space. The following conditions are equivalent:

(1)each nonempty, closed, and convex subset of is hyperconvex;

(2) is isometrically isomorphic to or .

We will now turn our attention to the problem of describing the spaces in which hyperconvexity implies convexity. We will start with an observation suggested to us by Grzybowski [22].

Proposition 4.7.

If a real normed space is strictly convex, then all its hyperconvex subsets are one-dimensional.

Proof.

Let be at least two-dimensional. Therefore there exist three noncollinear points . Put and let , . It is clear that and similarly for other distances. But is strictly convex, so we have and , so . It must be therefore , which finishes the proof.

Corollary 4.8.

If a real normed space is strictly convex, then all its hyperconvex subsets are convex.

Proof.

From Proposition 4.7 we know that hyperconvex subsets of are one dimensional; but from Proposition 2.5 we infer that hyperconvex sets are connected, which for one-dimensional sets is equivalent to their convexity.

To prove the inverse implication, we will need a simple lemma.

Lemma 4.9 (see [23, page 44, Lemma  15.1]).

Let be a metric space and be such that . If there exist metric segments: joining the points and and joining the points and , then is a metric segment joining the points and.

Now we are ready to prove the following theorem.

Theorem 4.10.

If all hyperconvex subsets of a real normed space are convex, then is strictly convex.

Proof.

Assume that is not strictly convex; we will construct a nonconvex, hyperconvex subset of . There exist points and positive numbers such that , and the equalities and hold. From Lemma 4.9, both sets and , where means an affine segment with endpoints , , are metric segments joining and (and hence hyperconvex sets). They cannot be, however, both convex, so at least one of them is the desired counterexample.

Again, combining Corollary 4.8 and Theorem 4.10, we obtain the following characterization of strictly convex normed spaces.

Theorem 4.11.

A normed space is strictly convex if and only if each its hyperconvex subset is convex.

5. Krein-Milman Type Theorem

In this short section, we will show that a Krein-Milman type theorem holds for -trees. It turns out that instead of compactness we only need a weaker boundedness condition.

For completeness, let us state the definition of an extremal point in the setting of -trees.

Definition 5.1.

Let be a subset of an -tree . We call a point an extremal point of if no open metric segment included in contains .

Theorem 5.2.

A complete and bounded -tree is a convex hull of the set of its extremal points.

Proof.

It is enough to show that each point of lies on a metric segment joining some two extremal points of . Let . We may assume that is not extremal; let . The family of all metric segments having as one of its endpoints satisfies the assumptions of the Kuratowski-Zorn lemma. Let and be maximal metric segments containing the respective given metric segments. We will first show that and are extremal points.

If, say, were not extremal, we would have for some , . Let and If , we would have and , so but , so —contradiction. This means that or ; assume . Now , so , which contradicts the maximality of .

Now let us show that . We will prove that . Assume and . Let . Choose such that . We have and hence ; analogously, . This means that and ; but —contradiction.

Since closed and convex subsets of an -tree are hyperconvex (Proposition 3.3), Corollary 4.6 might give the impression that -trees are somehow similar to 1- or 2-dimensional vector spaces and that completeness and boundedness of an -tree imply its compactness. As the following example shows, this analogy is misleading.

Example 5.3.

Let be with the radial metric. It is easy to see that is an -tree and so is , which is both complete and bounded, but not compact.

6. Hyperconvexity and Measures of Noncompactness

Let us begin this section with the following definition.

Definition 6.1.

Let be some point in the Euclidean plane. Let us define a function as follows:

(6.1)

for all . If , we will write instead of .

It is easy to prove the following lemma.

Lemma 6.2.

is a complete metric space.

We will call the function (resp., ) introduced in Definition 6.1, the modified radial metric (resp., centered at).

Remark 6.3.

The topology of with the metric is strictly stronger than the topology of the same space induced by the radial metric.

Lemma 6.4.

The space with the metric is not hyperconvex.

Proof.

Let us consider two closed balls and Then

(6.2)

but

(6.3)

This shows that the metric fails to be hyperconvex.

Now we are going to examine the measures of noncompactness in the space . For this purpose we are going to use a similar approach as in the case of the measures of noncompactness in with the radial metric (cf. [20, Theorem  4]). First let us introduce the following definition.

Definition.

Let be a bounded subset of . We say that satisfies

(1) condition, if for every , there exist infinitely many pairwise distinct points such that ;

(2) condition, if for every , there exist infinitely many pairwise distinct points such that .

Let us put .

Using above conditions we can prove the following theorem

Theorem 6.6.

For any bounded subset of with the metric we have and .

Proof.

If there exists no nonnegative number satisfying either or , then clearly consists of a finite number of points. Hence in this case.

Now consider a bounded set such that there exists a satisfying or condition. To prove that , let us first show that . For this, consider a covering of such that

(6.4)

for some . Consider the sets , where . Then for every there exists a and such that . Since for every . Hence .

Next we prove that . Obviously, if

(6.5)

then is contained in the closed ball of center and radius . So in this case .

Let

(6.6)

then according to Definition 6.5, for every there exist at most finitely many points with the property . Hence . Moreover,

(6.7)

Since is arbitrary, we get in this case. Finally, we get . This implies and

Example 6.7.

Using the previous formulae, we can calculate that in we have ; in particular, the closed unit ball is noncompact.

Remark.

It is known (see [12, page 135] for the details) that if a space is hyperconvex, then for any of its bounded subset , the following equality holds

(6.8)

The above theorem shows that even in the nontrivial cases (i.e., in cases, when bounded sets are not necessarily relatively compact), the above equality does not have to imply that the space in question is hyperconvex.

Definition 6.1 can be slightly modified. Namely, let us introduce the following definition.

Definition 6.9.

Let us define a function as follows:

(6.9)

for all .

Remark 6.10.

It can be easily checked that is a complete metric space. Its topology is also stronger than the topology of with the radial metric. On the other hand this topology is obviously equivalent to the topology induced by the metric .

Lemma 6.11.

The space with the metric is not hyperconvex.

Proof.

Let us consider two closed balls and Then

(6.10)

It shows that the metric fails to be hyperconvex.

For the measures of noncompactness in the space of bounded subsets in the space we have similar formulas to those given in Theorem 6.6.

Definition 6.12.

Let D be a bounded subset of with the metric . We say that satisfies

(1) condition, if for every , there exist infinitely many pairwise distinct points such that ;

(2) condition, if for every , there exist infinitely many pairwise distinct points such that .

Let us put .

Theorem 6.13.

For any bounded subset of with the metric one has and

The proof of Theorem 6.13 is similar to the proof of Theorem 6.6 and therefore we omit it.

The metric we are going to consider to the end of this section is, roughly speaking, like between the radial metric and the river metric. We will call it a modified river metric.

Definition 6.14.

Let . Define a function as follows:

(6.11)

for all If we will write instead of .

The following fact can be easily checked.

Lemma 6.15.

is a complete metric space.

Remark 6.16.

The topology of is strictly stronger than the topology of induced by the river metric.

It is interesting to consider a closed ball where and Such a ball consists of two disjoint closed sets (a square and a segment) which, in particular, means that it is not connected.

Lemma 6.17.

The space with the metric is not hyperconvex.

Proof.

Let us consider two closed balls and Then but . This shows that is not hyperconvex.

To evaluate the measures of noncompactness of any bounded subset of one can use a similar approach as in the case of (cf. Definition 6.12 and Theorem 6.13).

In connection with Remark 6.8 let us notice that as well as are also examples of metric spaces such that for any bounded subset or , but those spaces are not hyperconvex.

7. Generalized Modified Radial and River Metrics

The metric spaces as well as are special cases of a general construction provided in [19]. More precisely, let be a normed space and its Chebyshev subset.

Definition 7.1.

Let be a Chebyshev set in a normed space and let be any metric defined on . Let us define by the formula

(7.1)

The above defined function is a metric (see [19, Lemma  3.1]). Now, the following question can be risen. Is it possible to consider two disjoint Chebyshev sets, instead of one Chebyshev set , in such a way to get a variant of the metric defined above? The following two examples show that in the case of classical hyperconvex metrics: the radial metric as well as the river metric, this problem seems not to be easy.

Example 7.2.

Let be a fixed segment in and the perpendicular bisector of dividing the whole plane into two open half-planes and . Let us define a function as follows:

(7.2)

for all , where , are the radial metrics on the plane centered at and , respectively. Then this is not a metric. Indeed it does not satisfy the triangle inequality in the following case.

Let us consider three points such that ; and are collinear; and are collinear; and Then

Example 7.3.

Let and , where , be two points in . Let be the perpendicular bisector of ; it divides the whole plane into two open half-planes and . Let us define a function as follows:

(7.3)

for all , where denotes the river metric. Then this is not a metric. Indeed, it does not satisfy the triangle inequality in the following case. Let and let us take three points Then, by the definition and but , which shows

However, it appears that all the metrics introduced in Section 6 (Definitions 6.1, 6.9 and 6.14) are appropriate to define new metrics using the idea described at the beginning of this section.

Let us begin with the following definition

Definition 7.4.

Let be a segment in and be the perpendicular bisector of dividing the whole plane into two open half-planes and . Let us define a function as follows:

(7.4)

for all , where and denote modified radial metrics centered at and respectively.

Let us note that if and both are in , then , so is well-defined.

Lemma 7.5.

is a complete metric space.

Proof.

It is easy to check that is a metric. Now to verify that it is complete, let us consider a Cauchy sequence in the space . Then there exists such that for all , the points belong to the same closed half-plane or . Hence, by Lemma 6.2, is convergent, which completes the proof.

Remark 7.6.

It is clear that the topologies of induced by the metric and the modified radial metric are not comparable.

Lemma 7.7.

The space with the metric is not hyperconvex.

Proof.

For convenience consider and consider two closed balls and . Then but .

Remark 7.8.

It is easy to evaluate the Kuratowski and Hausdorff measures of noncompactness of bounded sets in with the metric .

Indeed, let us consider a bounded set D in with this metric. Then we can write as the union of two sets and , where

(7.5)

Then, by the maximum property of the measures of noncompactness, we get

(7.6)

To evaluate and it is enough to apply formulas similar to the one given in Theorem 6.6.

Remark 7.9.

It is clear that in Definition 7.4 one can replace by respectively, (cf. Definition 6.9) getting again a complete metric space which is not hyperconvex.

Now, using the metric from Definition 6.14, let us introduce the following metric.

Definition 7.10.

Let be a fixed segment in parallel to the -axis and perpendicular bisector of dividing the whole plane into two open half-planes and Let us define a function as follows:

(7.7)

for all where and denote the metrics from Definition 6.14.

One can prove the following lemma.

Lemma 7.11.

is a complete metric space.

The proof of this Lemma is similar to the proof of Lemma 7.5 and therefore we omit it.

Remark 7.12.

The metric is a variant of the metric defined in Definition 6.14. The topologies induced by these metrics are not comparable. The space is not hyperconvex, either. Finally, to find the Kuratowski and the Hausdorff measures of noncompactness of bounded sets in with the metric , it is enough to use the same approach as in Remark 7.8.

In Definitions 7.4 and 7.10, we considered two Chebyshev sets. Now one can think of the following question. Is it possible to increase the number of suitably chosen Chebyshev sets? The answer is "yes." Let us introduce the following definition.

Definition 7.13.

Let us consider the square in with vertices: , where Denote , , , , . Let be the "maximum" metric on . By , , and so forth, we will mean a metric defined as in Definition 7.4, but using , , and so forth, instead of , and so forth. Denote the four open quadrants by , , and . Let us define a function as follows:

(7.8)

and eight more similar expressions involving , , and for all , where and denote the closed quadrants and and denote the modified radial metrics defined in Definition 6.1.

Lemma 7.14.

is a complete metric space.

Proof.

To prove that is a metric on is straightforward, although quite long, so we omit this proof. To prove that is complete, let us consider a Cauchy sequence in the space Then for every there exists such that for every It means there exists such that for every belongs to the same closed quadrant, because if and were in different quadrants (without loss of generality suppose and ), then

(7.9)

So, if we choose then which contradicts that is a Cauchy sequence. Hence almost all the terms of any Cauchy sequence must be in the same closed quadrant. Thus by Lemma 6.2, is convergent, which shows that the space is complete.

For the convenience of the reader, let us present a figure of a closed ball in where and .

Obviously, the following lemma holds.

Lemma 7.15.

The space with the metric is not hyperconvex.

Proof.

For convenience let us consider and and two closed balls and Then but

Remark 7.16.

It is easy to evaluate the Kuratowski and Hausdorff measures of noncompactness of bounded sets in Indeed, one can use a similar approach as in Remark 7.8.

8. Linking Construction

In this section we will give a slight generalization of the so-called linking construction described by Aksoy and Maurizi in [14] and show how this generalization includes the metrics of Section 7. Notice that a similar concept appears in [24], where it is used to study existence of certain mappings between Banach spaces.

Definition 8.1 (cf. [14, page 221, Theorem  2.1]).

Let be a metric space and a collection of pairwise disjoint metric spaces, each disjoint with . Let be an arbitrary function and let be a function satisfying for each . Define for . Let . Define the function by the formula

(8.1)

Theorem 8.2 (cf. [14, page 221, Theorem  2.1]).

The function defined above is a metric on . If all the metric spaces , for are hyperconvex, then so is .

Remark 8.3.

The paper [14] contains the above theorem only for hyperconvex spaces. It is obvious that is a metric also in the general case.

Remark 8.4.

The authors of the paper [14] applied their version of Theorem 8.2 to obtain the hyperconvexity of the metric of Definition 7.1 (see [14, Theorem  2.2]). Let us notice that an identical result was given in an earlier work [19].

Proposition 8.5.

The metric from Definition 8.1 is complete if all the spaces and are complete.

Proof.

Let be a Cauchy sequence in . We will show that has a convergent subsequence. If has infinitely many terms in , we are done. If has infinitely many terms in some , it must be convergent in to some ; if , the proof is complete, and if , it is easily seen that in as . Therefore we may assume that includes only a finite number (possibly zero) of points from and each . Define by

(8.2)

Observe that ; for if that were not the case, there would exist a subsequence and an such that each would lie in different and ; this would mean that for all —contradiction with being Cauchy.

Now notice that for , so the sequence is also Cauchy and hence convergent to some We have as and the proof is complete.

Remark 8.6.

To evaluate the Kuratowski and Hausdorff measures of noncompactness of bounded sets in with the metric , when the set is finite, we use following procedure.

Let us consider a bounded set in with the metric . Then we can write as the following union:

(8.3)

Then, by the maximum property of the measures of noncompactness, we get

(8.4)

Example 8.7.

Notice that the metric from Definition 7.4 can be obtained as a special case of Definition 8.1. Indeed, put and . For each , define

(8.5)

for and for .

In a similar way, other metrics from Sections 6 and 7 are special cases of Definition 8.1. As an example, let us provide a way to construct the metric from Definition 6.14.

Example 8.8.

Let and for . Define the metric by the formula

(8.6)

For each , let be the metric defined by . Further, let be an identity mapping and . It is easily seen that applying Definition 8.1 we obtain the metric space .

At the beginning of Section 7 we posed a question whether it is possible to construct a metric analogous to that from Definition 7.1, but with more than one Chebyshev subset. In all our examples, however, these subsets were singletons. Let us now show an example of two similar metrics constructed using two disjoint Chebyshev subsets consisting of more than one point.

Example 8.9.

Define the following two Chebyshev subsets of the Euclidean plane: and . Put . Let and . Let and be metric projections and define by the formula

(8.7)

For each , let . Let the identity map and be defined by for . The metrics on and 's are inherited from . Applying Definition 8.1 we obtain a certain metric on . Let us notice that it is not complete; taking and for , and as before we obtain another metric, this time complete. Let us finish by observing that since , and hence , is disconnected, in both cases cannot be hyperconvex.