Construction of ball spaces and the notion of continuity
aa r X i v : . [ m a t h . GN ] O c t CONSTRUCTION OF BALL SPACES AND THE NOTIONOF CONTINUITY
REN´E BARTSCH, KATARZYNA KUHLMANN AND FRANZ-VIKTORKUHLMANN
Abstract.
Spherically complete ball spaces provide a framework forthe proof of generic fixed point theorems. For the purpose of theirapplication it is important to have methods for the construction of newspherically complete ball spaces from given ones. Given various ballspaces on the same underlying set, we discuss the construction of newball spaces through set theoretic operations on the balls. A definitionof continuity for functions on ball spaces leads to the notion of quotientspaces. Further, we show the existence of products and coproducts anduse this to derive a topological category associated with ball spaces. Introduction
Ball spaces have been introduced in [2] as a framework for the proof ofgeneric fixed point theorems for functions which in some way are contract-ing. Since then, the development of their theory and their applications hasled to several articles ([1, 3, 4, 5, 6, 7]).A ball space is a pair ( X, B ) consisting of a nonempty set X and anonempty set of distinguished nonempty subsets B of X . If we denote thepower set of X by P ( X ), this means that ∅ 6 = B ⊆ P ( X ) \ {∅} . The elements B of B will be called balls , in analogy to the case of metric (orultrametric) balls. In analogy to the case of ultrametric spaces, we will calla nonempty collection N of balls in B a nest of balls (in B ) if it is totallyordered by inclusion. We will say that ( X, B ) is spherically complete ifthe intersection T N of each nest N of balls in B is nonempty.Ball spaces can be derived from various settings, such as metric spaces,ultrametric spaces, topological spaces (where it is convenient to take theballs to be the nonempty closed sets), partially ordered sets, lattices. Thegeneric fixed point theorems that hold in spherically complete ball spacesthen allow specializations to all of these settings, in which spherically com-pleteness is equivalent to natural completeness or compactness properties Date : October 8, 2018.2010
Mathematics Subject Classification.
Primary 54A05; Secondary 54H25, 12J15,03E20, 18B05. (cf. [4]). In addition, the general framework allows transfers of concepts andapproaches between these various settings (again, cf. [4]).There are many different procedures to construct a new ball space ( X, B )from a given set of ball spaces ( X i , B i ), i ∈ I . We will discuss several ofthem in this paper. Since most of the generic fixed point theorems work forspherically complete ball spaces, it is an important question under whichconditions the spaces ( X i , B i ) being spherically complete implies that so is( X, B ). An easy example for this to work is when we take the union of thesets of balls of two ball spaces on the same set X (see Proposition 3). Incertain cases we may need stronger forms of spherical completeness, whichwe will introduce in Section 2.1. This for instance happens when we aregiven a single ball space ( X, B ) and want to replace B by the set of all finiteunions of balls in B (see Theorem 4). For the study of further operationson the set of balls and the connection with topologies, see [4].Returning to the case of several ball spaces on a given set X , we willdiscuss in Section 3.2 the particularly interesting natural example of or-dered abelian groups and fields. For the case of fields, S. Shelah introducedin [12] the notion of symmetrical completeness which means that forevery Dedekind cut, the cofinality of the left cut set is different from thecoinitiality of the right cut set (as is the case in the reals). He showed thatarbitrarily large symmetrically complete ordered fields exist. With a differ-ent construction idea, the existence result is reproven in [6] and generalizedto the case of ordered abelian groups and ordered sets. For the ordered sets,the existence result in fact follows already from the comprehensive work ofHausdorff. It is also shown in [6] that symmetrical completeness is equiv-alent to the spherical completeness of the induced order ball space, whichis defined as follows. If ( I, < ) is any nonempty totally ordered set, then wetake B o to consist of all closed bounded intervals [ a, b ] with a, b ∈ I , a ≤ b .We call ( I, B o ) the order ball space associated with ( I, < ).On the other hand, non-archimedean ordered abelian groups and fieldscarry a nontrivial natural valuation (cf. Section 2.3), which induces an ul-trametric ball space (cf. Section 2.2). This ball space is always sphericallycomplete when the order ball space is, but the two spaces are distinct. Itshould be noted that the balls in the ultrametric ball space are precisely allcosets of the principal convex subgroups; here a convex subgroup is calledprincipal if it is the smallest among all convex subgroups containing a fixedelement.We can apply Proposition 3 to these two ball spaces, but we cannot applyTheorem 4 as the order ball space and hence also the union of the two ballspaces does not satisfy the necessary requirements. However, in Section 3.2we prove:
ONSTRUCTION OF BALL SPACES AND THE NOTION OF CONTINUITY 3
Theorem 1.
Take a symmetrically complete ordered group or field G and B to be the set of all convex sets in G that are finite unions of closed boundedintervals and ultrametric balls. Then ( G, B ) is spherically complete. We also show that the convex sets appearing in the theorem can always berepresented as two distinct ultrametric balls connected by a closed interval(hence we gave them the nickname “bar-bells”).
Open question:
Does the theorem also hold if the condition “convex” isremoved?
Take two ball spaces ( X, B ) and ( X ′ , B ′ ) and a function f : X → X ′ . Thequestion arises under which conditions f transfers spherical completenessfrom one side to the other. In analogy to the case of topological spaces, wewill call f ball continuous if the preimage of every ball in B ′ is a ball in B . Then the following holds: If f is ball continuous and ( X, B ) is spherically complete, then so is ( X ′ , B ′ ) . This result is part of Theorem 9 in Section 4, in which we study furtherconditions for the transfer of spherical completeness. It provides the neces-sary background for the definition of the notion of “quotient ball space” inthe same section.Having the notion of ball continuity at hand, we define in Section 5 thecategory of ball spaces, where the morphisms are the ball continuous func-tions. This leads to further constructions of new ball spaces from givenones: products and coproducts. We will show that products and coprod-ucts of spherically complete ball spaces are again spherically complete (seeTheorem 13).The final result of our paper (see Section 6) is inspired by the fact thatin many situations the behaviour of ball spaces reminds us of topologicalspaces, and that indeed many inspiring examples come from a topologicalcontext. Hence in order to provide topological methods, we derive from thecategory of ball spaces the category of augmented ball spaces by adding thefull space and the empty set to the sets of balls. We prove:
Theorem 2.
The category of augmented ball spaces is topological.
For definitions and details, see Section 6. Note that this result implies thatthe category can be embedded in a cartesian closed topological category,providing natural function spaces; see e.g. [11].2.
Preliminaries
A hierarchy of ball spaces.
In [4] we introduce and study the fol-lowing hierarchy of spherical completeness properties: S : The intersection of each nest in ( X, B ) is nonempty. S : The intersection of each nest in ( X, B ) contains a ball. BARTSCH, KUHLMANN AND KUHLMANN S : The intersection of each nest in ( X, B ) contains a largest ball. S : The intersection of each nest in ( X, B ) is a ball. S ci : The same as S i , but with “centered system” in place of “nest”.Here, a centered system of balls is a collection of balls such that theintersection of any finite number of balls in the collection is nonempty.2.2. Ball spaces associated with ultrametric spaces and valuations. An ultrametric space is a pair ( X, u ) where X is a set and u : X × X → Γwith Γ a totally ordered set with largest element ∞ , such that: (UM1) u ( x, y ) = ∞ if and only if x = y ; (UM2) u ( x, y ) = u ( y, x ); (UM3) u ( x, z ) ≥ min { u ( x, y ) , u ( y, z ) } .(UM3) is called the ultrametric triangle law . The image uX := u ( X × X ) \ {∞} is called the value set of u . A generalization of the notion of ultrametricspace works with partially ordered value sets uX in place of totally orderedones (see [8, 9, 4, 1]), but we will not need this generalization in the presentpaper.One frequent source of ultrametrics are valuations: if v is a valuation on afield or an abelian group, then one may define u ( a, b ) := v ( a − b ). With thisdefinition, the above axioms are satisfied if v is written as a Krull valuation,that is, it satisfies the following axioms: (V1) vx = ∞ if and only if x = 0; (V2) v ( x − y ) ≥ min { vx, vy } (ultrametric triangle law).Note that for valuations on fields we require in addition that Γ is an orderedabelian group together with ∞ (an element larger than all elements of thegroup) and the following axiom holds: (VH) v ( xy ) = vx + vy (homomorphism law).Take an ultrametric space ( X, u ). For x ∈ X and γ ∈ uX ∪ {∞} , the set B γ ( x ) := { x ∈ X | u ( x, y ) ≥ γ } is called the closed ball of radius γ around x . Further, we define B ( x, y ) := B u ( x,y ) ( x ) = B u ( x,y ) ( y ) , where the latter equality follows from the fact that in an ultrametric spaceevery element of a ball is its center. If X is nonempty, then ( X, B u ) with B u := { B ( x, y ) | x, y ∈ X } is a ball space, which we call the ultrametric ball space .From the ultrametric triangle law it follows that any two balls in B u arealready comparable by inclusion once they have a nonempty intersection.It follows that every centered system of balls in B u is in fact a nest of balls. ONSTRUCTION OF BALL SPACES AND THE NOTION OF CONTINUITY 5
Nonarchimedean orderings and the natural valuation.
Take anordered abelian group (
G, < ). Two elements a, b ∈ G are called archime-dean equivalent if there is some n ∈ N such that n | a | ≥ | b | and n | b | ≥ | a | .The ordered abelian group ( G, < ) is archimedean ordered if all nonzeroelements are archimedean equivalent. If 0 ≤ a < b and na < b for all n ∈ N ,then we say that “ a is infinitesimally smaller than b ”. We denote by va the archimedean equivalence class of a . The set of archimedean equivalenceclasses can be ordered by setting va > vb if and only if | a | < | b | and a and b are not archimedean equivalent, that is, if n | a | < | b | for all n ∈ N . Wewrite ∞ := v a va is a group valuation on G , i.e., itsatisfies (V1) and (V2). By definition,0 ≤ a ≤ b = ⇒ va ≥ vb . The set vG := { vg | = g ∈ G } is called the value set of the valuedabelian group ( G, v ).If (
K, < ) is an ordered field, then we consider the natural valuation onits ordered additive group and define va + vb := v ( ab ). This turns the setof archimedean classes into an ordered abelian group, with neutral element0 := v − va = v ( a − ) . In this way, v becomes a field valuation.As shown in Section 2.2 above, the natural valuation of an ordered abeliangroup or ordered field induces an ultrametric ball space B u .3. Hybrid ball spaces
In this section we investigate the following question:
Given two spherically complete ball spaces ( X, B ) and ( X, B ) on the sameset X , which operations of forming new balls from the balls in B ∪ B willpreserve spherical completeness? A first step is provided by the following proposition; the easy proof is leftto the reader.
Proposition 3. If ( X, B ) and ( X, B ) are S ball spaces, then so is theball space ( X, B ∪ B ) . The same holds with S or S in place of S . Note that the assertion may become false if we replace S by S . Indeed,the intersection of a nest in B may properly contain a largest ball whichdoes not remain the largest ball contained in the intersection in B ∪ B .Having obtained B = B ∪ B , the next question is how to create newballs from the balls in B without losing spherical completeness. The resultsof taking unions and intersections are discussed in [4]. In the next section,we present a particular case. BARTSCH, KUHLMANN AND KUHLMANN
Closure under finite unions of balls.
Take a ball space ( X, B ). By f-un( B ) we denote the set of all unions offinitely many balls in B . In [4], the following theorem is proven: Theorem 4. If ( X, B ) is an S c ball space, then so is ( X, f-un( B )) . For the convenience of the reader, we repeat the proof here. We need alemma.
Lemma 5. If S is a maximal centered system of balls in f-un( B ) (that is, nosubset of f-un( B ) properly containing S is a centered system), then there is asubset S of S which is a centered system in B and has the same intersectionas S . Proof: It suffices to prove the following: if B , . . . , B n ∈ B such that B ∪ . . . ∪ B n ∈ S , then there is some i ∈ { , . . . , n } such that B i ∈ S .Suppose that B , . . . , B n ∈ B \ S . By the maximality of S this impliesthat for each i ∈ { , . . . , n } , S ∪ { B i } is not centered. This in turn meansthat there is a finite subset S i of S such that T S i ∩ B i = ∅ . But then S ∪ . . . ∪ S n is a finite subset of S such that \ ( S ∪ . . . ∪ S n ) ∩ ( B ∪ . . . ∪ B n ) = ∅ . This yields that B ∪ . . . ∪ B n / ∈ S , which proves our assertion. (cid:3) Proof of Theorem 4:
Take a centered system S ′ of balls in f-un( B ).Centered systems of balls in a ball space are inductively ordered by inclu-sion. Hence there is a maximal centered system S of balls in f-un( B ) whichcontains S ′ . By Lemma 5 there is a centered system S of balls in B suchthat T S = T S ⊆ T S ′ . Since ( X, B ) is an S c ball space, we have that T S = ∅ , which yields that T S ′ = ∅ . This proves that ( X, f-un( B )) is anS c ball space. (cid:3) This theorem becomes false if “ S c ” is replaced by “ S ”. Example 6.
For every i ∈ N we let p i be the i -th prime. For every i ∈ N we define a ball B i ⊂ R by B i := (cid:18) , p i (cid:19) \ (cid:26) p ji | j ∈ N with p ji > p i +1 (cid:27) , and set B := { B i | i ∈ N } . Then for i = j , B i and B j are not comparableby inclusion. Therefore, B admits no nests with more than one ball and isthus spherically complete. But B i ∪ B i +1 = (cid:18) , p i (cid:19) since all the real numbers in (0 , p i ) that are missing in B i are elements of B i +1 . Further, ( B i ∪ B i +1 ) i ∈ N is a nest in ( R , f-un( B )). As it has emptyintersection, ( R , f-un( B )) is not spherically complete. ONSTRUCTION OF BALL SPACES AND THE NOTION OF CONTINUITY 7
This leads us to the following question:
Under which other conditions than S c is spherical completeness preservedunder taking finite unions? We discuss one special case in the next section, starting with a ball spacethat is not S c .3.2. Bar-bells.
As already mentioned in the Introduction, a natural exam-ple of algebraic structures on which two distinct ball spaces appear naturallyare ordered abelian groups and ordered fields. On the one hand such a struc-ture (
G, < ) admits a natural valuation which is nontrivial if the ordering isnonarchimedean (cf. Section 2.3). This gives rise to an ultrametric space,from which in turn we can derive the ball space ( G, B u ) where B u consists ofall closed ultrametric balls (cf. Section 2.2). On the other hand, one can con-sider the order ball space ( G, B o ) where B o consists of all nonempty closedbounded intervals. In [6] it is shown that if ( G, < ) is symmetrically com-plete, then ( G, B o ) is spherically complete, and that if ( G, B o ) is sphericallycomplete, then so is ( G, B u ). Hence if ( G, < ) is symmetrically complete,then ( G, B ) is spherically complete for B = B u ∪ B o by Proposition 3.While ( G, B u ) is always an S c ball space once it is spherically complete(cf. [4]), ( G, B o ) and ( G, B ) are S c if and only if G = R (with the canonicalordering). Hence if G = R , we cannot apply Theorem 4 here.A bar-bell is a subset of G obtained from a nonempty closed boundedinterval [ a, b ] by joining it with ultrametric balls centered in a and in b ; itcan thus be written as B α ( a ) ∪ [ a, b ] ∪ B β ( b ) with α, β ∈ uG and a ≤ b . Lemma 7.
For every symmetrically complete ordered abelian group or field,the ball space of bar-bells is spherically complete.
Proof: Take a nest N of bar-bells. W.l.o.g. we may assume that N =( B i ) i<κ where κ is the cofinality of the nest and that i < j < κ ⇒ B j ( B i .We write B i = B α i ( a i ) ∪ [ a i , b i ] ∪ B β i ( b i )with α i , β i ∈ uG and a i ≤ b i .If there is a nest of ultrametric balls that has an intersection which is asubset of the intersection of N , then we are done. Hence assume that thereis no such nest. We will construct a sequence ( i ν ) ν<κ that is cofinal in κ ,such that(1) B i ν +1 ( [ a i ν , b i ν ] ⊆ B i ν . Then \ i<κ B i = \ ν<κ B i ν = \ ν<κ [ a i ν , b i ν ]and we are done again. BARTSCH, KUHLMANN AND KUHLMANN
We take i = 0. Assume that for some ν < κ we have already chosen i µ forall µ ≤ ν such that (1) holds with µ in place of ν . By our assumption theremust be some i ν +1 < κ , i ν +1 > i ν , such that B α iν +1 ( a i ν +1 ) B α iν ( a i ν )and B β iν +1 ( b i ν +1 ) B β iν ( b i ν ). Then B α iν +1 ( a i ν +1 ) ∩ B α iν ( a i ν ) = ∅ and B β iν +1 ( b i ν +1 ) ∩ B β iν ( b i ν ) = ∅ . Since B i ν +1 ⊂ B i ν , we must have that a i ν < B i ν +1 < b i ν . Therefore, (1) holds.Now assume that λ < κ is a limit ordinal and that we have already chosen i ν and constructed B ′ i ν for all ν < λ such that (1) holds. Then we chooseany i ∈ I that is larger than all i ν (which exists since λ is smaller than thecofinality κ ) and set i λ := i and proceed as above with ν = λ . (cid:3) Any finite union S of closed bounded intervals and ultrametric balls thatis convex is a bar-bell. This is seen as follows. Suppose that the union ofthe intervals [ a i , b i ], 1 ≤ i ≤ m , and the balls B α j ( c j ), 1 ≤ j ≤ n , is convex.Since ultrametric balls with equal centers are comparable by inclusion, bylisting only the largest ones we may assume that all c j are distinct. Further,we can add the singleton ball B ∞ ( a i ) or B ∞ ( b i ) to the balls B α j ( c j ) in case a i , or b i respectively, is not contained in any of the balls. Let min ≤ j ≤ n c j = c j and max ≤ j ≤ n c j = c j . Then S = B α j ( c j ) ∪ [ c j , c j ] ∪ B α j ( c j ).From this fact together with Lemma 7 we obtain Theorem 1.Let us collect a few special properties of the ball space B of convex finiteunions of closed bounded intervals and ultrametric balls, which could behelpful in answering the question stated after Theorem 1.1) All of its balls can be expressed by a union of at most 3 balls from thetwo generating ball spaces B u and B o .2) Every ball in B u ∩ B o is a singleton.3) Both B u and B o are closed under finite nonempty intersections.4) If N is a nest in B l, then there is a nest N ′ in B u or in B o such that T N ′ ⊆ T N .An attempt to generalize the notion of convexity to arbitrary ball spacescould be to call a finite collection of balls pseudo convex if it is of theform { B , . . . , B n } with B i ∩ B i +1 = ∅ for 1 ≤ i < n . Open question:
Does the closure of a ball space under unions of finitepseudo convex collections of balls always preserve spherical completeness,even if the ball space fails to be S c ? If not, what other conditions couldensure this? Ball continuity and quotient ball spaces
Take two ball spaces ( X, B ) and ( X ′ , B ′ ) and a function f : X → X ′ . Wewill call f ball continuous if the preimage of every ball in B ′ is a ball in ONSTRUCTION OF BALL SPACES AND THE NOTION OF CONTINUITY 9 B , that is,(2) { f − ( B ′ ) | B ′ ∈ B ′ } ⊆ B . We will call f ball closed if the image of every ball in B is a ball in B ′ . Lemma 8.
Take three ball spaces ( X, B X ) , ( Y, B Y ) and ( Z, B Z ) , and func-tions f : X → Y and g : Y → Z . Then g ◦ f is ball continuous if f and g are. Likewise, g ◦ f is ball closed if f and g are. Proof: For the first assertion, note that for every B ∈ B Z , ( g ◦ f ) − ( B ) = f − ( g − ( B )). The second assertion is obvious. (cid:3) The next theorem gives conditions for functions to preserve sphericalcompleteness in one or the other ditrection.
Theorem 9.
Take two ball spaces ( X, B ) and ( X ′ , B ′ ) , and a function f : X → X ′ .a) If f is ball continuous and ( X, B ) is spherically complete, then so is ( X ′ , B ′ ) .b) Assume that f is surjective and (3) B ⊆ { f − ( B ′ ) | B ′ ∈ B ′ } . If ( X ′ , B ′ ) is spherically complete, then so is ( X, B ) .c) Assume that f is surjective or ( X ′ , B ′ ) is an S ball space, and that (4) B = { f − ( B ′ ) | B ′ ∈ B ′ } . Then ( X, B ) is spherically complete if and only if ( X ′ , B ′ ) is.d) If (4) holds and f is surjective, then (5) B ′ = { f ( B ) | B ∈ B} and f induces an isomorphism of the partially ordered sets B and B ′ .e) f : X → X ′ is ball closed and finite-to-one, and if ( X ′ , B ′ ) is sphericallycomplete, then so is ( X, B ) . Proof: a): Take a nest N ′ = ( B ′ i ) i ∈ I of balls in ( X ′ , B ′ ). Set N =( f − ( B ′ i )) i ∈ I . By assumption, we have that f − ( B ′ i ) ∈ B for all i ∈ I , Forany i, j ∈ I we have that B ′ i ⊆ B ′ j or B ′ j ⊆ B ′ i , hence also f − ( B ′ i ) ⊆ f − ( B ′ j )or f − ( B ′ j ) ⊆ f − ( B ′ i ) . This proves that N is a nest in B . As ( X, B ) isspherically complete, T N is nonempty. Since f ( T N ) ⊆ B ′ i for all i ∈ I , itfollows that f ( T N ) ⊆ T N ′ , which shows that the latter is nonempty.b): Take a nest N = ( B i ) i ∈ I of balls in ( X, B ). Set N ′ = ( f ( B i )) i ∈ I . Byassumption, we have that every B i is the preimage of a ball B ′ i in B ′ , henceby surjectivity of f , f ( B i ) = B ′ i ∈ B ′ for all i ∈ I . For any i, j ∈ I we havethat B i ⊆ B j or B j ⊆ B i , hence also f ( B i ) ⊆ f ( B j ) or f ( B j ) ⊆ f ( B i ) .This proves that N ′ is a nest in B ′ . As ( X ′ , B ′ ) is spherically complete, T N ′ is nonempty. Take x ′ ∈ T N ′ . Then x ′ ∈ B ′ i for all i ∈ I . Pick somepreimage x ∈ X of x ′ under f ; this is possible since f is assumed surjective. Then x ∈ f − ( B ′ i ) = B i for all i ∈ I . Hence x ∈ T N , showing that thisintersection is nonempty.c): If f is surjective, everything follows from assertions a) and b). If f isnot surjective, but ( X ′ , B ′ ) is an S ball space, then we have to modify theprevious proof in order to show that some x ′ ∈ T N ′ has a preimage in X .Since ( X ′ , B ′ ) is S , T N ′ contains a ball B ′ . By (4), f − ( B ′ ) is a ball in B and hence nonempty. So there is an x ′ ∈ T N ′ which has a preimage in X .d): A surjective function f : X → X ′ induces an inclusion preservingbijection between all subsets of X ′ and their preimages. Condition (4)implies (5) and that the restriction of f to B ′ is a bijection between B ′ and B .e): Take a nest N = ( B i ) i ∈ I of balls in ( X, B ). Set N ′ = ( f ( B i )) i ∈ I .By assumption, each f ( B i ) is a ball in B ′ , and since f preserves inclusionbetween balls, N ′ is a nest. As ( X ′ , B ′ ) is spherically complete, T N ′ isnonempty. Take x ′ ∈ T N ′ . Then x ′ ∈ f ( B i ) for all i ∈ I . Amongthe finitely many preimages of x ′ under f there must be at least one thatis contained in all B i . This element then lies in T N , showing that theintersection is nonempty. (cid:3) Take any function f : X → X ′ . We define functions ϕ f : P ( X ) → P ( X ′ ) and ψ f : P ( X ′ ) → P ( X )by setting ϕ f ( S ) := f ( S ) for each set S ⊆ X and ψ f ( S ′ ) := f − ( S ′ ) foreach set S ′ ⊆ X ′ . If S ⊆ P ( X ) and S ′ ⊆ P ( X ′ ), then in accordance withour general use for functions, ϕ f ( S ) = { ϕ f ( S ) | S ∈ S} and ψ f ( S ′ ) = { ψ f ( S ′ ) | S ′ ∈ S ′ } . If ( X, B ) is a ball space, then B is just a nonemptysubset of P ( X ) \ {∅} . We observe that then ϕ f ( B ) is a nonempty subsetof P ( X ′ ) \ {∅} . This shows that ϕ f sends ball spaces on X to ball spaceson X ′ . Similarly, if f is surjective, then ψ f sends ball spaces on X ′ to ballspaces on X .Every nest of balls in B is also a ball space on X , but it has the specialproperty of being totally ordered by inclusion. So we are interested in thequestion when this property is preserved by ϕ f and ψ f . The following is acorollary to the previous proof. Corollary 10.
Take two ball spaces ( X, B ) and ( X ′ , B ′ ) and a function f : X → X ′ .a) If f is ball continuous and N ′ is a nest of balls in B ′ , then ψ f ( N ′ ) is anest of balls in B .b) If f is ball closed and N is a nest of balls in B , then ϕ f ( N ) is a nest ofballs in B ′ . Take two ball spaces ( X, B ) and ( X ′ , B ′ ). If there is a surjective function f : X → X ′ such that (4) holds, then we call ( X ′ , B ′ ) a quotient ballspace of ( X, B ) (induced by the function f ). Note that B is the coarsest of ONSTRUCTION OF BALL SPACES AND THE NOTION OF CONTINUITY 11 all ball spaces S on X such that f is a ball continuous function from ( X, S )to ( X ′ , B ′ ), and that B ′ is the finest of all ball spaces S ′ on X ′ such that f is a ball continuous function from ( X, B ) to ( X ′ , S ′ ).5. Products and coproducts
The category of ball spaces consists of all ball spaces as objects andthe ball continuous functions between them as morphisms. In this sectionwe show that products and coproducts exist in this category, and we willdescribe them explicitly.
Theorem 11.
The category of ball spaces admits products and coproducts.
For the proof of the theorem, we will explicitly construct these objects.Take ball spaces ( Y i , B i ), i ∈ I .
1) Products.
We set X = Q i ∈ I Y i and denote by p i : X → Y i theprojection from X to Y i . We set(6) B := (Y i ∈ I B i ⊆ X | for some k ∈ I, B k ∈ B k and ∀ i = k : B i = Y i ) . Since the sets B i are nonempty, it follows that B 6 = ∅ , and as no ball in any B i is empty, it follows that no ball in B is empty.The definition of B yields that all projections are ball continuous. Wehave to show that for any ball space ( Z, B Z ) and ball continuous functions f i : Z → Y i there is a unique ball continuous function g : Z → X such that p i ◦ g = f i for all i ∈ I . The latter forces g ( z ) = ( f i ( z )) i ∈ I for all z ∈ Z ; thisensures uniqueness. It remains to show that g is ball continuous. A ball B ∈ B is of the form Q j ∈ I S j as in (6). Then g − ( B ) = f − k ( B k ), which is aball in B Z since f k is ball continuous. This shows that g is ball continuous.
2) Coproducts.
Take ball spaces ( Y i , B i ), let X be the disjoint union˙ S i ∈ I Y i and denote by ι i : Y i → X the canonical embedding of Y i in X . Weset(7) B := ([ i ∈ I ι i ( B i ) | ∀ i ∈ I : B i ∈ B i ) . For the same reasons as before,
B 6 = ∅ and no ball in B is empty.For all j ∈ I we have that ι − j ( S i ∈ I ι i ( B i )) = B j , so each ι j is ballcontinuous. We have to show that for any ball space ( Z, B Z ) and ballcontinuous functions f i : Y i → Z there is a unique ball continuous function g : X → Z such that g ◦ ι i = f i for all i ∈ I . The latter forces g ( x ) = f i ( y )when y ∈ Y i with x = ι i ( y ); this ensures uniqueness. It remains to show that g is ball continuous. Take a ball B ∈ B Z . Then g − ( B ) = S i ∈ I ι i ( f − i ( B )).This is a ball in B because f − i ( B ) ∈ B i for all i ∈ I as all f i are ballcontinuous. Notation:
We will denote the product defined above by Q i ∈ I ( Y i , B i ), andthe coproduct by ` i ∈ I ( Y i , B i ). Further, we may rewrite S i ∈ I ι i ( B i ) as˙ S i ∈ I B i . Theorem 12.
Take ball spaces ( Y i , B i ) , i ∈ I . Then the following assertionshold:a) Q i ∈ I ( Y i , B i ) is spherically complete if and only if all ball spaces ( Y i , B i ) , i ∈ I , are spherically complete.b) If at least one of the ball spaces ( Y i , B i ) , i ∈ I , is spherically complete,then ` i ∈ I ( Y i , B i ) is spherically complete.c) The following are equivalent:i) all ball spaces ( Y i , B i ) , i ∈ I , are S ,ii) Q i ∈ I ( Y i , B i ) is S ,ii) ` i ∈ I ( Y i , B i ) is S .The same holds with S and with S in place of S . Proof: Let us first observe the following. If I and J are some indexsets and B i,j , i ∈ I , j ∈ J are arbitrary sets, then(8) \ j ∈ J Y i ∈ I B i,j = Y i ∈ I \ j ∈ J B i,j and \ j ∈ J ˙ [ i ∈ I B i,j = ˙ [ i ∈ I \ j ∈ J B i,j a): First assume that all ball spaces ( Y i , B i ), i ∈ I , are spherically complete.Take a nest ( Q i ∈ I B i,j ) j ∈ J in Q i ∈ I ( Y i , B i ). Then there is some k ∈ I suchthat B i,j = Y i for all i ∈ I with i = k and all j ∈ J , and ( B k,j ) j ∈ J is a nestof balls in B k . Since ( Y k , B k ) is spherically complete, T j ∈ J B k,j = ∅ . As T j ∈ J B i,j = Y i for i = k , we find that T j ∈ J Q i ∈ I B i,j = Q i ∈ I T j ∈ J B i,j = ∅ .This proves that Q i ∈ I ( Y i , B i ) is spherically complete.Now assume that Q i ∈ I ( Y i , B i ) is spherically complete. We have alreadyseen that all projections are ball continuous. Hence it follows from part a)of Theorem 9 that for every i ∈ I , ( Y i , B i ) is spherically complete.b): Assume that at least one of the ball spaces ( Y i , B i ), i ∈ I , is sphericallycomplete. Take a nest ( ˙ S i ∈ I B i,j ) j ∈ J in ` i ∈ I ( Y i , B i ). Since the unions aredisjoint, ˙ S i ∈ I B i,j ⊆ ˙ S i ∈ I B i,j implies that B i,j ⊆ B i,j for all i ∈ I . Thisshows that for each i ∈ I , ( B i,j ) j ∈ J is a nest in B i . If ( Y k , B k ) is sphericallycomplete, then ∅ 6 = \ j ∈ J B k,j ⊆ ˙ [ i ∈ I \ j ∈ J B i,j = \ j ∈ J ˙ [ i ∈ I B i,j . c): We keep the notations of the proofs of a) and b). If all ball spaces( Y i , B i ), i ∈ I , are S , then for each i ∈ I , T j ∈ J B i,j contains a ball B ′ i and hence T j ∈ J Q i ∈ I B i,j = Q i ∈ I T j ∈ J B i,j contains the ball Q i ∈ I B ′ i . If allspaces are even S , then the B ′ i can be taken as maximal balls in B i ∪ ONSTRUCTION OF BALL SPACES AND THE NOTION OF CONTINUITY 13 { Y i } contained in T j ∈ J B i,j and it follows that Q i ∈ I B ′ i is a maximal ballcontained in Q i ∈ I T j ∈ J B i,j . If all spaces are S , then for each i ∈ I , T j ∈ J B i,j is a ball in B i ∪ { Y i } and hence T j ∈ J Q i ∈ I B i,j = Q i ∈ I T j ∈ J B i,j isa ball. This proves that i) implies ii) in all three cases.In view of the equality T j ∈ J ˙ S i ∈ I B i,j = ˙ S i ∈ I T j ∈ J B i,j the above argu-ments can be readily adapted to prove that i) implies iii) in all three cases.Now assume that Q i ∈ I ( Y i , B i ) is S . Take k ∈ I and a nest of balls N =( B j ) j ∈ J in ( Y k , B k ). We define the nest Q i ∈ I ( Y i , B i ) as in the proof of part a).By assumption, the intersection T j ∈ J Q i ∈ I B i,j = Q i ∈ I T j ∈ J B i,j contains aball Q i ∈ I B ′ i . It follows that B ′ k is a ball in B k that is contained in T j ∈ J B k,j .If Q i ∈ I ( Y i , B i ) is even S , then Q i ∈ I B ′ i can be assumed to be a maximalball contained in T j ∈ J Q i ∈ I B i,j , which implies that B ′ k is a maximal ballcontained in T j ∈ J B k,j . If Q i ∈ I ( Y i , B i ) is S , then T j ∈ J Q i ∈ I B i,j is ball,which implies that T j ∈ J B k,j is a ball in B k . We have proved that ii) impliesi) in all three cases.Replacing “ Q i ∈ I ” by “ ˙ S i ∈ I ” in the proof we just gave renders a proof forthe fact that iii) implies i) in all three cases. (cid:3) At first glance it may be surprising that for ` i ∈ I ( Y i , B i ) to be sphericallycomplete it suffices that just one of the ball spaces ( Y i , B i ) is spherically com-plete, while for ` i ∈ I ( Y i , B i ) to be S , S or S , all ball spaces ( Y i , B i ) musthave the same property. This is because for T j ∈ J ˙ S i ∈ I B i,j to be nonemptyit suffices that T j ∈ J B i,j is nonempty for at least one i ∈ I , whereas for itto contain a ball, all T j ∈ J B i,j must contain a ball.From the previous theorem we immediately obtain the following result. Theorem 13.
The categories of spherically complete ball spaces, of S ballspaces, of S ball spaces and of S ball spaces, all of them with ball continuousfunctions as their morphisms, admit products and coproducts. Let us conclude this section by giving an example which shows that theconverse in part b) of Theorem 12 is in general not true. (However, it canbe shown to be true when all B i are countable.) Example 14.
Take X = ω , B to be the set of all final segments in ω , X = ω , and B to be the set of all final segments in ω . Then neither( X , B ) nor ( X , B ) is spherically complete as in both cases, the intersec-tion over all final segments is empty. Note, however, that the intersectionover every countable nest of balls in ( X , B ) is nonempty, and thus thesame is true in the coproduct of the two ball spaces. On the other hand,if N = ( ˙ S i ∈{ , } B i,j ) j ∈ J is an uncountable nest in ` i ∈{ , } ( Y i , B i ), then wemay w.l.o.g. assume that J = ω and that j < j implies B ,j ⊆ B ,j and B ,j ⊆ B ,j . Since B is countable, the balls B ,j must eventually becomestationary, i.e., equal to one and the same ball B ∈ B , which is thencontained in the intersection of the nest N . The topological category of augmented ball spaces
If ( X, B ) is a ball space and A = B ∪ {∅ , X } , or if X is an arbitrary(not necessarily nonempty) set and A = {∅ , X } , then we call ( X, A ) an augmented ball space . Take two ball spaces ( X, B ) and ( X ′ , B ′ ) and aball continuous function f : X → X ′ . Since f − ( ∅ ) = ∅ and f − ( X ′ ) = X , f also satisfies(9) { f − ( A ′ ) | A ′ ∈ A ′ } ⊆ A , where A ′ = B ′ ∪ {∅ , X ′ } . Therefore we will also call f a ball continuousfunction from ( X, A ) to ( X ′ , A ′ ). Note that f is always ball continuouswhen A ′ = {∅ , X } .We define the category of augmented ball spaces to consist of all aug-mented ball spaces as objects, with the ball continuous functions betweenthem as morphisms.A category C is called topological if(1) For all X ∈ | Set | and all families ( f i , ( X i , ξ i )) i ∈ I , indexed by aclass I , of C -objects ( X i , ξ i ) and functions f i : X → X i there existsa unique initial C -object ( X, ξ ) on the set X , i.e., an object ( X, ξ )s.t. for all objects (
Y, η ) ∈ | C | and maps g : Y → X the followingholds: g ∈ [( Y, η ) , ( X, ξ )] C ⇔ ∀ i ∈ I : f i ◦ g ∈ [( Y, η ) , ( X i , ξ i )] C ( Y, η ) g / / ( X, ξ ) f i / / ( X i , ξ i )That is: arbitrary initial structures exist.(2) Fibre-smallness: For all X ∈ | Set | , the class of C -objects on X isa set.(3) On sets with at most one element exists exactly one C -structure.Proof of Theorem 2:(1) The category admits initial objects. Take augmented ball spaces( Y i , A i ), a set X , and functions f i : X → Y i , i ∈ I . We set(10) A := { f − i ( A i ) | i ∈ I, A i ∈ A i } . Observe that ∅ = f − i ( ∅ ) ∈ A and X = f − i ( Y i ) ∈ A since ( Y i , A i ) is anaugmented ball space; hence also ( X, A ) is an augmented ball space.The definition of A yields that all f i as functions from ( X, A ) to ( Y i , A i )are ball continuous. We have to show that for any ball space ( Z, A Z ) andfunction g : Z → X we have that g : ( Z, A Z ) → ( X, A ) is ball continuous ifand only if for all i ∈ I , f i ◦ g : ( Z, A Z ) → ( Y i , A i ) is ball continuous.If g is ball continuous, then by Lemma 8 and our remark at the startif this section, so are all f i ◦ g . For the converse, assume that all f i ◦ g are ball continuous. Take A ∈ A . Then by definition, there is some i ∈ I and some A i ∈ A i such that A = f − i ( A i ). Since f i ◦ g is ball continuous, ONSTRUCTION OF BALL SPACES AND THE NOTION OF CONTINUITY 15 g − ( A ) = g − ( f − i ( A i )) = ( f i ◦ g ) − ( A i ) ∈ A Z . This shows that g is ballcontinuous.(2) Every ball structure on X is a subset of P ( X ) and so it is an elementof P ( P ( X )).(3) Every set with at most one element carries a unique augmented ballspace structure. Indeed, if the set is empty, then this is ( ∅ , {∅} ). If X is asingleton, then ( X, { X } ) is a ball space, so by definition, ( X, {∅ , X } ) is anaugmented ball space; this is the only augmented ball space structure on X . (cid:3) Note that the definition (10) used in the proof also yields an initial objectin the category of all ball spaces where the morphisms are assumed to be thesurjective ball continuous functions. But if one of the f i ’s is not surjective,it can happen that there is a ball B i ∈ A i such that B i ∩ f i ( X ) = ∅ , so that f − i ( B i ) = ∅ .The definitions of product and coproduct can be taken over from thecategory of ball spaces. If we use the construction described in (6) to derivean augmented ball space ( X, A ) from augmented ball spaces ( Y i , A i ), then A will now contain ∅ (we take B i = ∅ ∈ A i for some i ∈ I ) and X (we take B i = Y i ∈ A i for all i ∈ I ).Further, if we apply the construction described in (7) to derive an aug-mented ball space ( X, A ) from augmented ball spaces ( Y i , A i ), then A willagain contain ∅ (we take B i = ∅ ∈ A i for all i ∈ I ) and X (we take A i = Y i ∈ A i for all i ∈ I ).Observe that now also A ′ := { ι i ( A i ) | i ∈ I, A i ∈ A i } renders all embeddings ι i ball continuous since ∅ ∈ A i . However, in allcases where A ′ differs from the set A obtained from (7), examples can beconstructed that show that this is not a coproduct.Since the category of augmented ball spaces is topological, it must alsoadmit final objects (cf. [10]). We can present them explicitly. Take aug-mented ball spaces ( Y i , A i ), a set X , and functions f i : Y i → X , i ∈ I . Weset(11) A := { A ⊆ X | ∀ i ∈ I : f − i ( A ) ∈ A i } . Then all f i as functions from ( Y i , A i ) to ( X, A ) are ball continuous. Further, ∅ ∈ A since f − i ( ∅ ) = ∅ ∈ A i for all i ∈ I , and X ∈ A since f − i ( X ) = Y i ∈A i for all i ∈ I . It remains to show that for any ball space ( Z, A Z ) andfunction g : X → Z we have that g : ( X, A ) → ( Z, A Z ) is ball continuous ifand only if for all i ∈ I , g ◦ f i : ( Y i , A i ) → ( Z, A Z ) is ball continuous.If g is ball continuous, then by Lemma 8 and our remark at the start ifthis section, so are all g ◦ f i . For the converse, assume that all g ◦ f i areball continuous and take A ∈ A Z . Then f − i ( g − ( A )) = ( g ◦ f i ) − ( A ) ∈ A i for all i ∈ I by continuity. Hence by (11), g − ( A ) ∈ A , showing that g isball continuous.We observe that this definition does not work in the category of ballspaces. For example, consider ball spaces ( X, { B } ) and ( X, { B } ) with B = B and f and f the identity function of X . Then there is no subsetof X with preimage B under f and B under f , so (11) renders the emptyset when ∅ and the whole set are not balls. References [1] Kuhlmann, F.-V. – Kubis, K.:
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