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

Authors: Marcin Borkowski; Dariusz Bugajewski; Adam Burchardt
Journal: Journal of Function Spaces
 (2017)

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## 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.


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## Body

## 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, R2 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 R-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 Section4 we provide necessary and sufficient conditions for completeness of those metric spaces while in Section 5 we deal with their boundedness.In Section6, 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 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 R2 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(X,d) be a metric space, T:X→X be a function, g be a metric on T(X), and “~” be an equivalence relation on X such that x~y implies T(x)=T(y). Define the function φ:X×X→[0,+∞) by the formula (1)φx,y≔dx,y,ifx~y,dx,Tx+gTx,Ty+dTy,y,otherwise.It turns out that the above defined functionφ is a metric on X.Proposition 2 (see [11]).
Under the assumptions and notation of Definition1, φ is a metric on X.Remark 3.
In the sequel we will use the symbold|T(X) to denote the metric on T(X) inherited from d.The following obvious observation will be very useful in the sequel.Proposition 4.
Under the assumptions and notation of Definition1 and assuming that g=d|T(X), the metric φ is stronger than d.Proof.
It is enough to observe thatd(x,y)≤φ(x,y) for any x,y∈X.In the next examples we mention some particular cases covered by Definition1.Example 1.
(a)
Let(X,d) be a metric space, T:X→X be a function, “~” be the equality relation, and g=d|T(X).
(b)
Let(X,d) be a metric space, T:X→X be a function, “~” be the relation defined by x~y⇔T(x)=T(y), and g=d|T(X).
(c)
Let(X,d) be a metric space, T:X→X be a function, “~” be the equality relation, and g be a metric on X.Example 2.
LetC[a,b]∞ be the space of functions f:[a,b]→R differentiable infinitely many times and d be the “maximum” metric: d(f,g)=supx∈[a,b]⁡f(x)-g(x). Let T(f)=f′ for all f∈C[a,b]∞. Of course, T:C[a,b]∞→C[a,b]∞, so we may deduce that the function φ:C[a,b]∞×C[a,b]∞→R+ defined by the formula (2)φ=0,iff=g,df,f′+df′,g′+dg′,f,otherwise,is a metric in C[a,b]∞.It turns out that there is a strong connection between limit points of(X,φ) and fixed points of the mapping T.Proposition 5.
Under the assumptions and notation of Definition1 and assuming that “~” is the equality relation, each limit point of (X,φ) is a fixed point of T.Proof.
Let(xn) be a sequence of points of X converging to some x∈X such that each xn is distinct from x. This means that d(x,T(x))≤d(x,T(x))+g(T(x),T(xn))+d(T(xn),xn)=φ(x,xn); since limn→∞⁡φ(x,xn)=0, it must be the case that d(x,T(x))=0.Remark 6.
Let us consider a particular case of Example1(b). Namely, let X=R2 and let φ be the “river” metric. Then T((x,y))=(x,0) for (x,y)∈R2 and obviously not all limit points of (X,φ) are fixed points of T.


## 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(Xc,dc) be a metric space and (Xi,di)i∈I be a family of metric spaces. Let us assume that Xc∩Xi=xi for each i∈I and that Xi∩Xj⊂xi for any i,j∈I (i.e., each pair of Xi′s is either disjoint or intersects at xi). Let us define X=Xc∪⋃i∈IXi and let φa:X×X→[0,+∞) be defined by the formula (3)φax,y=dcx,y,ifx,y∈Xc,dix,y,ifx,y∈Xi,dcx,xi+dixi,y,ifx∈Xc,y∈Xi,dix,xi+dcxi,y,ifx∈Xi,y∈Xc,dix,xi+dcxi,xj+djxj,y,ifx∈Xi,y∈Xj,i≠j.(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.LetT:X→X be defined by the formula (4)Tx=xi,ifx∈Xifor  somei∈I,x,otherwise.For any x,y∈X, we declare that x~y if x=y or x,y∈Xi∖xi for some i∈I. Finally, we define g=d|T(X)=φa.In order to check thatφ and φa coincide, one has to examine several cases. As an example, let us consider one of them, namely, x∈Xc and y∈Xi. Then, either x~y (which can only happen if x=y=xi, and in that case φ(x,y)=φa(x,y)=0), or x≁y. Then we obtain φa(x,y)=dc(x,xi)+di(xi,y)=φa(x,xi)+φa(xi,y). On the other hand, φ(x,y)=d(x,T(x))+g(T(x),T(y))+d(T(y),y)=d(x,x)+g(x,xi)+d(xi,y)=φa(x,xi)+φa(xi,y).


## 4. Completeness

In this section we are going to prove a sufficient and necessary condition for the metricφ to be complete (with the natural assumption that g=d|T(X)).Throughout this section we will frequently make use of the fact that for any pointsx,y∈X, the obvious inequalities d(x,y)≤φ(x,y)≤d(x,T(x))+d(T(x),T(y))+d(T(y),y) hold. In particular, if some sequence is φ-Cauchy (resp., φ-convergent), that is, it is a Cauchy (respectively, convergent) sequence with respect to the metric φ, then it is d-Cauchy (resp., d-convergent to the same limit).Also, we will sometimes say shortly that some points ofX 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 (xn) 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 isφ-Cauchy in terms of the metric d and the relation “~.”Proposition 9.
With the assumptions and notation of Definition1, with the additional assumption that g=d|T(X), a sequence (xn) of points in X is φ-Cauchy if and only if (xn) is d-Cauchy, and (i)
almost allxn’s are equivalent with respect to the relation “~,” or(ii)
d(xn,T(xn))→0.Proof.
Let the sequence(xn) be φ-Cauchy. It is then d-Cauchy. If almost all of its terms are equivalent, the proof is complete. Assume the contrary. Fix ϵ>0 and choose N large enough so that φ(xm,xn)<ϵ if m,n≥N. For any n≥N there exists some m≥N such that xm≁xn. We have then (5)dxn,Txn≤dxn,Txn+dTxn,Txm+dTxm,xm=φxn,xm<ϵ.
Assume now that the sequence(xn) is d-Cauchy. If almost all of its terms are equivalent, the metrics d and φ coincide for those terms and hence (xn) is φ-Cauchy. Assume now that d(xn,T(xn))→0. Fix ϵ>0 and let N be large enough so that d(xn,T(xn))<1/5ϵ for n≥N and d(xm,xn)<1/5ϵ for m,n≥N. Then, if m,n≥N, we obtain (6)φxm,xn≤dxm,Txm+dTxm,Txn+dTxn,xn≤dxm,Txm+dTxm,xm+dxm,xn+dxn,Txn+dTxn,xn≤ϵ.Theorem 10.
With the assumptions and notation of Definition1, if we assume additionally that g=d|T(X), the metric φ is complete if and only if the following two conditions are true for any d-Cauchy sequence (xn) such that (T(xn)) is also d-Cauchy.(1)
Ifd(xn,T(xn))↛0 and almost all xn’s are equivalent, then there exists some x∈X equivalent to the xn’s such that d(xn,x)→0.(2)
Ifd(xn,T(xn))→0, then d(xn,x)→0 for some x∈X such that the alternative, x=T(x) or almost all xn’s are equivalent to x, is true.Proof.
⁡
Sufficiency. Let (xn) be φ-Cauchy and hence d-Cauchy. Let us consider two cases.
(I) Almost All
x
n
’s Are Equivalent. This means that the sequence (T(xn)) is constant and hence d-Cauchy. If d(xn,T(xn))↛0, then d(xn,x)→0 for some x equivalent to the xn’s. This means that d and φ coincide for xn’s and x and hence φ(xn,x)→0. If on the other hand d(xn,T(xn))→0, then d(xn,x)→0 for some x∈X. Moreover, since (T(xn)) is constant, denoting y=T(xn) we obtain d(xn,y)→0 and hence y=x. If x is equivalent to xn’s, d and φ coincide again on xn’s and x and hence φ(xn,x)→0. Otherwise we have x=T(x) and we obtain (7)φxn,x≤dxn,Txn+dTxn,Tx+dTx,x. The first term on the right-hand side above tends to zero and the latter two terms are equal to zero, and hence φ(xn,x)→0.
(II) It Is Not the Case That Almost All
x
n
’s Are Equivalent. Proposition 9 implies that d(xn,T(xn))→0. Since (xn) is d-Cauchy and d(xn,T(xn))→0, the sequence (T(xn)) is also d-Cauchy. This means that d(xn,x)→0 for some x=T(x). We have the estimate (8)φxn,x≤dxn,Txn+dTxn,Tx+dTx,x≤dxn,Txn+dTxn,xn+dxn,x+dx,Tx+dTx,x⟶0 and the proof is complete.
Necessity. Assume that the metric φ is complete. Let (xn) be a d-Cauchy sequence such that (T(xn)) is also d-Cauchy.
Ifd(xn,T(xn))↛0 and almost all xn’s are equivalent, this means that d and φ coincide for xn’s and hence the sequence (xn) is φ-Cauchy—hence φ-convergent and hence d-convergent to some x∈X. If x were not equivalent to xn’s, we would have (9)φxn,x=dxn,Txn+dTxn,Tx+dTx,x; as φ(xn,x)→0, it would be the case that also d(xn,T(xn))→0, which is a contradiction.
Assume now thatd(xn,T(xn))→0. From Proposition 9 we know that the sequence (xn) is φ-Cauchy, hence φ-convergent and hence d-convergent to some x∈X. For the sake of contradiction, assume that x≠T(x) and that we can find a pair of nonequivalent terms arbitrarily far in the sequence (xn). We may thus pass to a subsequence of xn’s not equivalent to x. Since (10)φxn,x=dxn,Txn+dTxn,Tx+dTx,x and x≠T(x), we obtain a contradiction with the φ-convergence of the sequence (xn).As a corollary, we are going to show that some very natural properties ofd and T imply completeness of φ.Corollary 11.
With the assumptions and notation of Definition1, if we assume additionally that g=d|T(X), the mapping T is continuous (with respect to the metric d), the space (X,d) is complete, and the equivalence classes of the relation “~” are closed with respect to the metric d; then the space (X,φ) is complete.Proof.
Assume that the sequences(xn) and (T(xn)) are d-Cauchy. If d(xn,T(xn))↛0 and almost all xn’s are equivalent, then (xn) is d-convergent to some x∈X due to d-completeness of X and x is equivalent to xn’s due to the equivalence classes of the relation “~” being closed with respect to the metric d.
Let nowd(xn,T(xn))→0. Again, (xn) is d-convergent to some x∈X. By continuity of T, we have T(xn)→T(x). Finally, d(x,T(x))≤d(x,xn)+d(xn,T(xn))+d(T(xn),T(x))→0 and hence x=T(x).The following three examples show that while the above theorem gives only a sufficient and not necessary conditions for completeness ofφ, releasing either one of these three conditions may result in an incomplete space.Example 3.
IfT is discontinuous, the metric φ may be incomplete. To see this, let us define X≔R, let d be the Euclidean metric and g=d|T(X), let the relation “~” be the equality relation, and let (11)Tx≔x,ifx≠0,-1,ifx=0. We have φ(x,y)=x-y, if x,y are simultaneously equal or not equal to zero, and φ(x,0)=1+x+1, if x≠0. Let us consider the sequence (1/n)n∈N. Clearly, it is φ-Cauchy, but not φ-convergent.
Let us notice that, puttingxn=1/n, we obtain a Cauchy sequence such that (T(xn)) is also a Cauchy sequence and d(xn,T(xn))=0, but while d(xn,0)→0, neither the equality 0=T(0) nor the equivalence of almost all xn’s holds. In other words, the second assumption of Theorem 10 is not satisfied.Example 4.
If(X,d) is not complete, the metric φ may be incomplete. To see this, let X≔0,1, let d be the Euclidean metric and g=d|T(X), T be the identity, and let “~” be the equality relation. Then φ is the Euclidean metric on 0,1. Again, considering the sequence (1/n) 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 d, the metric φ may be incomplete. To see this, let us define X≔R, let d be the Euclidean metric and g=d|T(X), let T be a constant mapping (say, T(x)≔0 for any x∈R), and let x~y if and only if, x=y. Let us consider the sequence (xn) defined by the formula xn=1-1/n. It is obvious that it is φ-Cauchy. If it were φ-convergent to some x∈R, Proposition 4 would imply its d-convergence to x; hence x=1. But φ(1-1/n,1)=2n-1/n→n→∞2≠0.
Once again, let us notice that both the sequences(xn) and (T(xn)) are d-Cauchy, d(xn,T(xn))=|1-1/n|↛0, and all xn’s are equivalent. However, the sequence (xn) is not d-convergent to a point in the same equivalence class.Now, we are going to illustrate Theorem10 and Corollary 11 with the following two examples.Example 6.
Let us consider the “river” metric. Here,X=R2, d is the Euclidean metric, T(x,y)=(x,0), the relation “~” is defined by the formula (x1,y1)~(x2,y2)⇔x1=x2, and g=d|T(X). 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. PutX=R2, let d be the Euclidean metric, T(x,y)=(0,0) for all x,y∈R (hence g is trivial), and define the relation “~” by the formula (12)x1,y1~x2,y2⟺x1=y1=x2=y2=0,orx1,y1,x2,y2≠0,0and∃λ>0:x2,y2=λx1,y1. Let us notice that while d is complete and T is d-continuous, the equivalence classes of the relation “~” are not closed with respect to the metric d. Despite that, the radial metric is complete. Indeed, let us take any d-Cauchy sequence (xn). Obviously, (T(xn)) is a constant sequence and hence d-Cauchy. If d(xn,0)↛0 and almost all xn’s lie on the same ray, this means that (xn) is d-convergent to some nonzero point on the same ray and (1) is satisfied. If on the other hand d(xn,0)→0, then (2) holds, because (0,0)=T(0,0).


## 5. Boundedness

In this very short section we are going to discuss the problem of boundedness of the metric defined in Definition1.Proposition 12.
Under the assumptions and notation of Definition1, if the metrics d and g are bounded, then so is φ. On the other hand, if φ is bounded, then so is g. Moreover, if g=d|T(X) and φ is bounded, then so is d.Proof.
Let us first assume thatd and g are bounded by some constant M. Then, for any x,y∈X, we obtain φ(x,y)≤d(x,T(x))+g(T(x),T(y))+d(T(y),y)≤3M.
Assume now thatφ is bounded. Let x′=T(x) and y′=T(y) be two distinct points in T(X). Then g(x′,y′)≤d(x,x′)+g(x′,y′)+d(y′,y)=φ(x,y), so g is bounded. Furthermore, if g=d|T(X), then for any x,y∈X we either have x≁y (and in this case, d(x,y)≤d(x,T(x))+d(T(x),T(y))+d(T(y),y)=φ(x,y)) or x~y (and in this case d(x,y)=φ(x,y)).Remark 13.
Let us notice that, in the last part of the above proposition, we assume thatdrestricted toT(X) is equal to g and hence bounded, and from that we infer the boundedness of don the wholeX.Remark 14.
Ifg≠d|T(X), then the boundedness of φ does not have to imply the boundedness of d. Indeed, consider X=R, let T:R→R be defined by T(x)=x, let “~” be the equality relation, and let d be the Euclidean metric and g the discrete metric. Then φ is bounded (in fact, (X,φ) has the diameter of 3), but (X,d) is not.


## 6. The “Floor” Metric

For now, we will concentrate on one particular case of Definition1, given in the following example.Example 8.
LetX≔R,  d(x,y)=g(x,y)=x-y, “~” be the equality relation, and T(x)=x. Applying Definition 1 we obtain a certain metric on R, which in what follows will be denoted by φ.Remark 15.
By settingT(x)=⌈x⌉, that is, letting T(x) be the smallest integer not less than x, 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 T(x)=⌊ax+b⌋ or T(x)=⌈ax+b⌉.We are now going to establish some properties of the metric defined in Example8.Proposition 16.
The metric defined in Example8 is neither homogeneous, nor translation invariant, nor metrically convex.Proof.
To see thatφ is not homogeneous, let us observe that φ(1/2,1)=3/2, but φ(1,2)=1. To see that it is not translation invariant, let us observe that φ(0,1/2)=1/2, but φ(1/2,1)=3/2. To see that it is not metrically convex, let us observe first that φ(0,1)=1. Now if z∈R∖0,1, we have φ(0,z)+φ(z,1)>1. Indeed, if z>1, then φ(0,z)>1, and if z<1 (and z≠0), then φ(z,1)>1.Proposition 17.
The metric defined in Example8 is complete.Let us notice that we can easily see that the assumptions of Theorem10 are satisfied, although—since the mapping T is discontinuous at integer points—we cannot apply Corollary 11. We shall, however, show how to prove the completeness directly.Proof.
Let(xn) be a φ-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 xn’s are pairwise distinct. We may also assume that xn∈[k,k+1) for some k∈Z. We will prove that limn→∞⁡xn=k. Assume for the sake of contradiction that ∃ε>0∀N∈N∃n≥N:xn∈[k+ε,k+1). That means that ∃ε>0∀N∈N∃m,n≥N:xm,xn∈[k+ε,k+1), which implies φ(xm,xn)≥2ε—contradiction.We will now describe balls in the metric space(R,φ). We will start with the case when the center is an integer.Remark 18.
Ifa≥b, we define a,b≔∅ and similarly with other kinds of intervals.Proposition 19.
Letx∈Z and r>0. Then the ball centered at x with radius r with respect to the metric defined in Example 8 has the form (13)Bx,r=x-r,x-r+r-r∪x-r+1,x+r.Proof.
Let us consider several cases.(i)
y<x-r. Then y≤y<x-r and hence r<r+1≤x-y, so φ(y,x)=y-y+x-y>x-y>r.(ii)
x-r≤y<x-r+r-r. Then y=x-r and hence y-y<r-r, so φ(y,x)=y-y+x-y<r.(iii)
x-r+r-r≤y<x-r+1. Then y=x-r, so φ(x,y)=y-y+x-y=y-y+r≥r-r+r=r.(iv)
x-r+1≤y<x. Since x-rr+1≤y, also x-r+1≤y and hence x-y+1≤r. We obtain now φ(y,x)=y-y+x-y<x-y+1≤r≤r.(v)
y=x; then obviously φ(y,x)<r.(vi)
x<y<x+r. Then x≤y and φ(y,x)=y-x<r.(vii)
x+r≤y. Then φ(y,x)=y-x≥r.Let us now consider a ball with an arbitrary center.Proposition 20.
Letx∈R and r>0. Then the ball centered at x with radius r with respect to the metric defined in Example 8 has the form B(x,r)=B(x,r-(x-x))∪x.Proof.
Lety∈B(x,r). If y=x, there is nothing to prove. Let y≠x. Then y-y+y-x<r-(x-x). Since the left-hand side of this expression equals φ(y,x), we are done.
Let nowy∈B(x,r-(x-x)). Again we may assume y≠x, and similar computation shows that y∈B(x,r).Remark 21.
In a similar way we may prove that ifx∈R and r≥0, then B¯(x,r)=B¯(x,r-(x-x))∪x and that if x∈Z and r≥0, then B¯(x,r)=[x-r,x-r+r-r]∪[x-r+1,x+r].Now we describe the form of open subsets of the metric space(R,φ).Proposition 22 (form of open sets).
A subset ofR 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 ofR 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 φ-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 centerx and sufficiently small radius is a singleton x and that an open ball with an integer center x and sufficiently small radius ϵ>0 is the interval [x,x+ϵ). Let now I be an interval with left endpoint a and right endpoint b. If x∈(a,b), then clearly x lies in the interior of I. If a∈I, then a must be an interior point of I. If b∈I, then b is in the interior of I iff b∉Z, and the proof is complete.Corollary 23.
A subset ofR is closed with respect to the metric φ 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(R,φ) need not be connected nor compact. For instance, B1,3/2=0,1/2∪[1,5/2) is the union of two open sets 0,1/2 and [1,5/2). On a similar note, B¯1,3/2=[0,1/2]∪[1,5/2], and these two intervals are closed with respect to the metric φ.
Consider now the ballB¯0,1/2=[0,1/2]. Since it can be covered by an infinite family of disjoint open sets, namely, [0,1/4]∪{{x}∣x∈1/4,1/2}, it is not compact.Now we are going to establish the formula which allows calculating the Kuratowski measure of noncompactness of bounded subsets ofR 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.
LetA⊂[k,k+1) for some k∈Z. The Kuratowski measure of noncompactness of A, considered as a subset of (R,φ), is given by the formula(14)αA=2max⁡x-k∣xis  the  limit  point  ofA∪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 φ. The same holds for limits of sequences.)Proof.
First observe thatA has a limit point if, and only if, it is infinite; if A is finite, obviously α(A)=0. From now on we will assume that A is infinite.
Let us definea≔sup⁡{x-k∣xis  the  limit  point  ofA}∈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 first prove thatα(A)≤2a. If a=1, this claim is obvious, since α(A)≤diam⁡[k,k+1)=2. Let us assume that a∈[0,1) and choose any ϵ>0 such that a+1/2ϵ<1. The family {[k,k+a+1/2ϵ)}∪{x∣x∈A∩[k+a+1/2ϵ,k+1)} is a finite covering of A by sets of diameter no greater than 2a+ϵ.
Let us now prove that actuallyα(A)=2a; that is, there is no finite covering of A with sets of diameter strictly less than 2a. Choose d∈0,2a; of course, there are infinitely many points of A in the interval I≔[k+1/2d,k+1). Choose two distinct points x,y∈I; we have φ(x,y)=(x-k)+(y-k)>d. Any finite covering of A would have to contain a set containing more than one point of I—but this means that the diameter of this set would be greater than d.Corollary 26.
LetA be a nonempty, bounded subset of (R,φ). The Kuratowski measure of noncompactness of A is given by the formula (15)αA=2maxk∈Z⁡max⁡x-k∣xis  the  limit  point  ofA∩k,k+1∪0.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.



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*Source: 4901762-2017-11-21.xml*