Topology and its Applications 160 (2013) Contents lists available at ScienceDirect. Topology and its Applications

Transcription

1 Topology and its Applications 160 (2013) Contents lists available at ScienceDirect Topology and its Applications Universal frames Themba Dube a, Stavros Iliadis a,b,,1, Jan van Mill a,c,d,1, Inderasan Naidoo a a Department of Mathematical Sciences, University of South Africa, P.O. Box 392, 0003 Unisa, South Africa b Department of General Topology and Geometry, Faculty of Mechanics and Mathematics, Moscow State University, Moscow , Russia c Faculty of Sciences, VU University Amsterdam, De Boelelaan 1081A, 1081 HV Amsterdam, The Netherlands d Faculty of Electrical Engineering, Mathematics and Computer Science, TU Delft, Postbus 5031, 2600 GA Delft, The Netherlands article info abstract MSC: 06A06 06D22 54A05 Keywords: attice Frame Universal frame For a class of frames we define the notion of a universal element and prove that in the class of all frames of weight less than or equal to a fixed infinite cardinal number τ there are such elements Elsevier B.V. All rights reserved. 0. Introduction Universal objects play an important role in mathematics. In topology, there is a wealth of such objects that go back a long time. For example, the Cantor set is a universal space for the class of all zero-dimensional separable metrizable spaces, the n-dimensional Nöbeling spaces for the class of all n-dimensional separable metrizable spaces, the Hilbert space for the class of all separable metrizable spaces, etc. There are also many interesting results for isometries. For example, the Urysohn Universal Metric Space and the space C[0, 1] of all continuous functions on [0, 1] with the uniform convergence contain isometrically all separable metric spaces. For details and references on universality, see Iliadis [2]. For basic information about frames or pointfree topology, see Picado and Pultr [4] or Johnstone [3]. The aim of this note is to construct a universal frame in the class of all frames of weight at most a given infinite cardinal number τ. Some but not all frames have a topological space as underlying object, so our result is independent of that of the existence of universal objects in the class of all topological T 0 -spaces * Corresponding author at: Department of General Topology and Geometry, Faculty of Mechanics and Mathematics, Moscow State University, Moscow , Russia. addresses: dubeta@unisa.ac.za (T. Dube), s.d.iliadis@gmail.com (S. Iliadis), j.van.mill@vu.nl (J. van Mill), naidoi@unisa.ac.za (I. Naidoo). 1 The second and the third named authors thank UNISA for generous support /$ see front matter 2013 Elsevier B.V. All rights reserved.

2 T. Dube et al. / Topology and its Applications 160 (2013) of weight τ given in [2]. We employ the general technique of constructing universal objects as given by Iliadis [2]. In so doing, a novel method for constructing new frames involving very set theoretical methods is presented. 1. Universal frames 1.1. Definitions and notation. Recall that a frame is a complete lattice in which x S = {x s: s S for any x and any S. Our notations shall be fairly standard from Picado and Pultr [4]. For instance we denote the top element and the bottom element of by 1 and 0 respectively. As usual, is the rather below relation whilst denotes the completely below relation. For x, y, x y iff x y =1 where x = {t : t x =0 whilst x y iff there is a system {c r : r Q [0, 1] such that c 0 = x, c 1 = y and c r c s whenever r < s.aframe is completely regular if for each x, x = {y : y x and regular frames are those in which each element is the join of elements rather below it. If x x =1 then x is complemented in. Theframe is zero-dimensional provided that each element of is a join of complemented elements (see Banaschewski [1]). A frame homomorphism is a map between frames which preserves finite meets, including the top element, and arbitrary joins, including the bottom element. We denote the class of all frames that are completely regular by CRegFrm, regularby RegFrm, and zero-dimensional by 0-DFrm. A subset B of a complete lattice (or frame) is called base for if each element of is a supremum of a subset of B. Theweight of a complete lattice (or frame) is the minimal cardinal κ for which there exists abaseb for of cardinality κ. An ordinal number is the set of smaller ordinal numbers, and a cardinal number is an initial ordinal number. By ω we denote the least infinite cardinal. By τ we denote a fixed infinite cardinal and by F the set of all non-empty finite subsets of τ. Each mapping ϕ of τ onto a set X is called an indexation of X and will be denoted by where x δ = ϕ(δ), δ τ. X = {x 0,...,x δ,..., 1.2. Definition of a universal complete lattice (or frame). et be a class of (non-empty) complete lattices (or frames). We say that a complete lattice (or frame) T is universal in this class if (a) T and (b) for every there exists a homomorphism of T onto Theorem. In the class of all frames of weight τ there exist universal elements. Proof. Without loss of generality we can suppose that is a set and that its elements are mutually disjoint. For every we denote by B = { a 0,a 1,...,a δ,... an indexed base of of cardinality τ such that a 0 =0anda 1 = 1. We assume without loss of generality that B is closed under taking finite infima. By θ we denote a fixed mapping of the set F into τ such that for every t F we have a θ (t) = { a δ : δ t,

3 2456 T. Dube et al. / Topology and its Applications 160 (2013) with the additional condition that if t = {δ, thenθ (t) =δ. We note that for distinct elements and G in the mappings θ and θ G need not have any relation to each other. Fix s F for a moment. By s we denote the equivalence relation on the family defined as follows: two elements, G are s -equivalent if and only if for all t s such that t we have θ (t) =θ G (t). It is easy to see that if s s, then s s.wesetr={ s : s F. We denote by C( s ) the set of all equivalence classes of the relation s, s F, and put C(R) = { C ( s ) : s F. It is easy to see that the cardinality of the set C( s )islessthanorequaltoτ and, therefore, the cardinality of the set C(R) is also less than or equal to τ. Now we consider the free union U of the elements of, i.e., For every s F, H C( s ), and δ s we put U = {:. A δ (H) = { a δ : H { a 0 =0 : \ H U. Obviously, for every the intersection A δ (H) consists of one element only: the element a δ and the element 0 if / H. We put if H Obviously, the cardinality of B is τ. For every subset B= { A δ (H): δ s, s F, H C ( s). M = { A δj (H j ): j J B we shall define a subset of U, denoted by M. For the definition of this set we first fix an element and denote by M the set of elements a δ of for which there exists j J such that { a δ = Aδj (H j ). Therefore, a δ coincides either with a δ j or with 0. Then, we put M = {sup M :. Observe that the element sup M M in general is not an element of M. Now we consider the set T={ M: M B. The subset M B will be called a generator of M T. The generators of the elements of T are not uniquely determined. It is easy to see that the union of any number of generators of an element of T is also a generator of this element. Therefore, the union of all generators of a certain element of T is the maximal generator of that element. We note that each element A δ (H) B is an element of T, that is B T. Indeed, for the set N = {A δ (H) B, it is easy to see that Ñ = A δ(h). In T we define an order by putting M Ñ for M,Ñ T if and only if for every we have sup M sup N. Obviously, the pair (T, ) isaposet.

4 T. Dube et al. / Topology and its Applications 160 (2013) emma 1. The poset (T, ) is a complete lattice of weight τ. Proof. To prove the lemma it suffices to prove the existence of suprema of arbitrary subsets (Picado and Pultr [4, 4.3.1]), and that the weight of T is at most τ. This last fact will follow once we show that B is a base for T. The existence of suprema. et G = { M λ : λ Λ T. We must prove that the supremum of G exists. Denote by M the union of all generators M λ of all elements of G, i.e., Then, the set M = { M λ : λ Λ. M = {sup M : is an element of T. We shall prove that M =supg. Indeed, consider an element. Then, it is easy to see that M = { M λ : λ Λ and, therefore, sup M =sup( { M λ : λ Λ ) =sup { sup M λ : λ Λ. (1) This relation shows sup M λ sup M,thatis M λ M for every λ Λ. Now, suppose that N is an element of T such that M λ N for every λ Λ, thatis for every and for every λ Λ. Then, sup M λ sup N sup { sup M: λ λ Λ sup N and by relation (1) we have sup M sup N, that is M Ñ proving that M =supg. The set B is a base for T. Consider an arbitrary element M of T and let M = { A δλ (H λ ): λ Λ. We put M λ = {A δλ (H λ ). Then, M λ = A δλ (H λ ) B T and M = { M λ : λ Λ.

5 2458 T. Dube et al. / Topology and its Applications 160 (2013) Consider the set As in the above we can prove that G = { M λ : λ Λ B. M =supg =sup { M λ : λ Λ. This relation shows that B is a base for T and, therefore, the weight of T is at most τ. The proof of the lemma is complete. It follows from emma 1 that T has a bottom and a top. For later use, it will be convenient to have an explicit description of these two elements. The existence of the bottom. et s = {0 F. Consider the set M = { A 0 (H): H C ( s). By the definition of the indexation of B, for every H C( s )wehave A 0 (H) ={0 : H {0 : \ H. Then, for every, M = {0 and, therefore, sup M =sup{0 =0.Thus, M = {0 :. Obviously, by the definition of the order in T for every Ñ Twehave M Ñ. Therefore, the constructed element M T is the least element of the poset (T, ) and will be denoted by 0 T. The existence of the top. et s = {1 F. Consider the set By the definition of the indexation of B, M = { A 1 (H): H C ( s). A 1 (H) ={1 : H {0 : \ H, H C ( s). Since the distinct classes of s are disjoint and the union of all classes is, for every we have M = {1 and, therefore, sup M =sup{1 =1. Then, M = {1 :. Obviously, by the definition of the order in T for every Ñ TwehaveÑ M. Therefore, the constructed element M T is the greatest element of (T, ) and will be denoted by 1 T. emma 2. For every there exists a homomorphism h of T onto. Proof. et be a fixed element of.

6 T. Dube et al. / Topology and its Applications 160 (2013) The definition of h. We define the mapping h setting for every element M T: Obviously, h ( M) is the unique element in M. h ( M) =supm. We note that h ( M) is independent of the generator M of the element M. Since for any elements Ñ and M of T the relation Ñ M means that sup N sup M, the mapping h is order preserving. By the definition of h it follows immediately that h (0 T )=0 and h (1 T )=1. The mapping h is onto. et a. SinceB is a base for, there exists a subset κ τ such that for A = { a δ : δ κ B we have a =supa. For every s F denote by H s δ s κ consider the set A δ (H s ) and put the equivalence class of C( s ) containing. For every M = { A δ ( H s ) : s F,δ s κ. Then, M = {{ a δ = Aδ ( H s ) : s F,δ s κ. The above equality means that A = M. Hence, sup M = a, soh ( M) =a showing that h is indeed onto. The mapping h preserves suprema. et M =sup { M λ : λ Λ. Then, h ( M) ( { =supm =sup M λ : λ Λ ) =sup { sup M λ : λ Λ =sup { h ( M λ ) : λ Λ, that is, h preserve suprema. The mapping h preserves finite infima. eta δj (H j )anda ηi (H i ) be two fixed elements of B such that δ j s j, H j R( s j ), η i s i,andh i R( s i ). We shall define a subset of B, denoted by M (j,i) = A δj (H j )ΔA ηi (H i ) as follows. et s = s j s i. Then, for every, G H C( s )wehaveθ (t) =θ G (t) where t = {δ j,η i.in this case we put θ H (t) =θ (t). By the properties of the elements of the family R it follows that if H is an element of C( s ), then either H H j H i or H (H j H i )=. IfthesetH j H i is empty, then we put M (j,i) = { A 0 (H): H C ( s).

7 2460 T. Dube et al. / Topology and its Applications 160 (2013) If H j H i, then we put M (j,i) = { A δ (H): H C ( s), H H j H i,δ= θ H (t). We note that if s j = s i, H j = H i,andδ j = η i,thenθ (t) =δ j and s = s j. Therefore, in this case M (j,i) = { A δj (H j ) = { A ηi (H i ). It is easy to see that for every A δ (H) M (j,i) we have Now, let Ñ and K be two elements of T. Suppose that and A δ (H) A δj (H j ) and A δ (H) A ηi (H i ). (2) N = { A δj (H j ): j J, H j C ( s j ), δ j s j F K = { A ηi (H i ): i I, H i C ( s i ),η i s i F. Without loss of generality we can suppose that N and K are the maximal generators of Ñ and K, respectively. Consider the set First, we shall prove that M = { M(j,i) = A δj (H j )ΔA ηi (H i ): (j, i) J I. M = Ñ K. (3) Indeed, let. Obviously, for every (j, i) J I we have A δj (H j ) Ñ and A η i (H i ) K. (4) Then, since N and K are maximal generators by the definition of the set M and the relations (2) and (4) for every (j, i) J I we have and, therefore, M (j,i) N and M (j,i) K { M(j,i) :(j, i) J I N and { M(j,i) :(j, i) J I K. This means that sup M =sup( { (M(j,i) ) :(j, i) J I ) sup N (5) and similarly sup M sup K. (6)

8 T. Dube et al. / Topology and its Applications 160 (2013) The relations (5) and (6) imply that M Ñ and M K. Suppose now that for an element P Twehave P Ñ and P K. Since N and K are maximal generators we have P N and P K. et A δ (H) beanelementofp. Then, A δ (H) N and A δ (H) K. Setting A δj (H j )=A ηi (H i )=A δ (H), by the above, we have M (j,i) = {A δ (H). Therefore, A δ (H) M, thatisp M which means that P M. Thus, relation (3) is proved. We shall now prove that First, we shall prove that if a δ j N and a η i K,then sup M =supn sup K. (7) a δ j M. (8) Indeed, let a δ j N and a η i K. Then there exist elements H j C( s j )andh i C( s i )forsome s j,s i F such that H j H i, A δj (H j ) N and A ηi (H i ) K. ett = {δ j,η i, s = s j s i,andh be the element of C( s ) containing. Then, A θh (t)(h) M (j,i) M. Sincea θ H (t) = a θ (t) = a δ j we have a δ j M. Now, let a M. Then, there exist such that a (M (j,i) ), where Therefore, we have δ j s j F, H j C ( s j ), η i s i F and H i C ( s i ) M (j,i) = A δj (H j )ΔA ηi (H i ). a A θh (t)(h) M (j,i), where t = {δ j,η i, s = s j s i,andh is the element of C( s ) containing. This means that The relations (8) and (9) show that and, therefore, a = a δ j. (9) M = { a δ j : a δ j N,a η i K sup M =sup { a δ j : a δ j N,a η i K.

9 2462 T. Dube et al. / Topology and its Applications 160 (2013) Since is a frame we have sup M =sup { a δ j : a δ j N sup { a ηi : η i K =supn sup K proving relation (7). Relations(3) and (7) prove that the mapping h preserves finite infima. Hence we are done. emma 3. The complete lattice (T, ) isaframe. Proof. We must prove that Ñ sup { Kλ : λ Λ =sup { Ñ K λ : λ Λ (10) for all Ñ, K λ T. Without loss of generality we can suppose that N and K λ, λ Λ, are the maximal generators of Ñ and K λ, respectively. Suppose that and N = { A δj (H j ): j J, δ j s j F, H j C ( s j ) K λ = { A ηi(λ) (H i(λ) ): i(λ) I(λ),η i(λ) s i(λ) F, H i(λ) C ( s i(λ)). Then, the set K = { K λ : λ Λ = { A ηi(λ) (H i(λ) ): i(λ) I(λ),η i(λ) s i(λ) F, H i(λ) C ( s ) i(λ),λ Λ is a generator of the element sup{ K λ : λ Λ T and, therefore, K =sup { Kλ : λ Λ. By the relation (4) for every λ Λ the set M λ = { Aδj (H j )ΔA ηi(λ) (H i(λ) ): j J, δ j s j F, H j C ( s ) j,i(λ) I(λ), η i(λ) s i(λ) F, H i(λ) C ( s ) i(λ) is a generator of the element Ñ K λ T. Hence, the set M = { M λ : λ Λ = { A δj (H j )ΔA ηi(λ) (H i(λ) ): j J, δ j s j F, H j C ( s ) j,i(λ) I(λ), η i(λ) s i(λ) F, H i(λ) C ( s ) i(λ),λ Λ is a generator of the element sup{ñ K λ : λ Λ T. Therefore, M =sup { Ñ K λ : λ Λ. Thus, relation (10) takes the form

10 T. Dube et al. / Topology and its Applications 160 (2013) Suppose that K = K, where Ñ K = M. K = { A δj (H j ): j J,δ j s j F, H j C ( s j ) is the maximal generator of the element K T (therefore, containing K). The set P = { A δj (H j )ΔA δj (H j ): j J, δ j s j F, H j C ( s j ),j J, δ j s j F, H j C ( s j ) is a generator of the element Ñ K T and, therefore, P = Ñ K = Ñ K. Since K K we have M P. Therefore, M Ñ K = Ñ K. Thus, to prove the lemma it suffices to prove that or equivalently Ñ K M sup P sup M for every. For this purpose it suffices to prove that if and a P,thena sup M.et a P. Then, there exist A δj (H j ) N and A δj (H j ) K such that a A θh (t)(h), where H is the element of C( s ) containing, s = s j s j,andt = {δ j,δ j. Therefore, a = a θ (t) = a δ j a δ j. If a δ j = a η i(λ) for some λ Λ and i(λ) I(λ), then a M.Thus, sup { a δ j a δ j : j J, j J =sup { a δ j : j J sup { a δ j : j J =sup { a δ j : j J sup { a η i(λ) : λ Λ, i(λ) I(λ) proving the lemma. emmas 1 3 prove the theorem Some problems. Consider the class of all frames having a given property P. It is interesting for what properties P this class contains universal elements. For example:

11 2464 T. Dube et al. / Topology and its Applications 160 (2013) Are there universal elements in the class CRegFrm? 2. Are there universal elements in the class RegFrm? 3. Are there universal elements in the class 0-DFrm? References [1] B. Banaschewski, Universal zero-dimensional compactifications, in: Proc. Int. Conf. on Categorical Topology, Prague 1988, World Sci. Pub., Singapore, 1989, pp [2] S.D. Iliadis, Universal Spaces and Mappings, North-Holland Mathematics Studies, vol. 198, Elsevier Science B.V., Amsterdam, [3] P.T. Johnstone, Stone Spaces, Cambridge Studies in Advanced Math., vol. 3, Camb. Univ. Press, [4] J. Picado, A. Pultr, Frames and ocales, Frontiers in Mathematics, Birkhäuser/Springer Basel AG, Basel, 2012.