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When lattices of uniformly continuous functions on X determine X
Topology and its Applications 194 (2015) 228–240
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Topology and its Applications www.elsevier.com/locate/topol
When lattices of uniformly continuous functions on X determine X Miroslav Hušek a,b,∗ , Antonio Pulgarín c a b c
Charles University, Sokolovská 83, 18675 Prague, Czech Republic Purkyně University, Klíšká 30, 40096 Ústínad Labem, Czech Republic Univ. of Extremadura, Avda. Universidad s/n, 10071 Cáceres, Spain
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 23 February 2015 Received in revised form 20 August 2015 Accepted 21 August 2015 Available online xxxx MSC: 54C35 54E15 06B99
Several Banach–Stone-like generalizations of Shirota’s result for metrizable uniform spaces are proved. Namely, if complete uniform spaces X, Y have isomorphic lattices U (X), U (Y ) of their real-valued uniformly continuous functions, and both X, Y are either some products of spaces having monotone bases (metrizable or uniformly zero-dimensional), or are locally fine and of non-measurable cardinality, then X and Y are uniformly homeomorphic. © 2015 Published by Elsevier B.V.
Keywords: Lattice of functions Uniform continuity Banach–Stone theorem
1. Introduction S. Banach proved the following theorem in 1932 ([1], p. 170): Pour que deux ensembles métriques, complets et compacts Q et Q1 soient homèomorphes, il faut et il suffit que les espaces E et E1 des fonctions réelles continues dèfinies dans ces ensembles soient isométriques. Banach uses the Mazur–Ulam theorem (every isometry of a normed space onto another normed space is affine, i.e., linear if it maps 0 to 0) and gives a description of the homeomorphism. If U : E → E1 is a surjective isometry mapping 0 to 0, then the homeomorphism h : Q → Q1 is given by the equality U (f )(q1 ) = f (h−1 (q1 ))ε(q1 )
for some ε ∈ E1 , |ε| = 1 .
* Corresponding author at: Charles University, Sokolovská 83, 18675 Prague, Czech Republic. E-mail addresses: Miroslav.Husek@mff.cuni.cz (M. Hušek), [email protected] (A. Pulgarín). http://dx.doi.org/10.1016/j.topol.2015.08.015 0166-8641/© 2015 Published by Elsevier B.V.
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The above theorem does not hold if one takes linear isomorphism instead of isometry. M.H. Stone removed the condition on metrizability 5 years later ([18], p. 469): If R and R∗ are bicompact H-spaces and if M and M∗ are the corresponding function-rings, then the existence of an isometric correspondence f → U (f ) = f ∗ between M and M∗ is equivalent to the existence of a topological equivalence r → ρ(r) = r∗ between R and R∗ , the two correspondences being connected by the relations f (r) = ϕ∗ (r∗ )[f ∗ (r∗ ) − θ∗ (r∗ )], U f = f ∗ , ρ(r) = r∗ , U 0 = θ∗ , U 1 = ϕ∗ , where |ϕ∗ | = 1. If the relations U 0 = 0, U 1 = 1 or, equivalently, the relations θ∗ = 0, ϕ∗ = 1 are satisfied, then the correspondence f → U f = f ∗ is an analytical isomorphism between M and M∗ . From that time, many generalizations appeared on that theme. Usually, more general ranges were used (e.g., Banach spaces) or more general domains (e.g., locally compact or realcompact spaces), or various sets of function spaces with various algebraic structures, or various categories (e.g., metric spaces with Lipschitz maps or uniformly continuous maps), or non-isomorphisms between function spaces were used. For other algebraic structures on C(X) we can mention I. Gelfand and A.N. Kolmogorov in 1939 [6] for rings, A.N. Milgram in 1949 [12] for multiplicative semigroups, I. Kaplansky in 1947 [11] for lattices (all for compact spaces) and, finally the most general result, T. Shirota in 1952 [17] for lattices and realcompact spaces. In the present paper, we shall deal with Banach–Stone type results for uniform spaces as objects, with function spaces composed of uniformly continuous functions and endowed with a lattice structure only. Probably the first result in that direction was the one by T. Shirota from 1952 ([17], Th. 6): Let X be a complete metric space. Then X is determined by the lattice of all uniformly continuous real functions on X. Moreover X is determined by the lattice of all bounded uniformly continuous real functions on it. It was noticed that the Shirota’s proof does not work for unbounded functions. A correct proof of the Shirota’s result has recently been published in [2]: Let X and Y be complete metric spaces and T : U (Y ) → U (X) be a lattice isomorphism. There is a uniform homeomorphism τ : X → Y such that (T f )(x) = t(x, f (τ (x))) (f ∈ U (Y ), x ∈ X) , where t : X × R → R is given by t(x, c) = (T c)(x). Here c is first treated as a real number and then as a constant function on Y . The present authors investigated Banach–Stone-like theorems both for metrizable and non-metrizable uniform spaces in [9] for less general lattice isomorphisms (they were supposed to preserve constant maps). In the present paper we show that most of the Banach–Stone-like results from [9] hold without the additional assumption on lattice isomorphisms. That gives several generalizations of the result from [2]. Our idea is to use the fact (proved in [8]) that existence of a lattice isomorphism between U (X) and U (Y ) implies existence of a lattice isomorphism between U ∗ (X) and U ∗ (Y ), i.e., a homeomorphism between the Samuel compactifications of X and of Y . The first step is to find situations when homeomorphisms between the Samuel compactifications of X and of Y map X onto Y . In such cases we get a proximal homeomorphism between X and Y . The second
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step is to find situations when such a proximal homeomorphism implies a uniform homeomorphism between X and Y (i.e., when X and Y have the same position in their proximity classes). We cannot expect a Banach–Stone theorem to be true for all complete uniform spaces. If we take for X the separable metrizable hedgehog and for Y the Samuel compactification of X, then U (X) = U ∗ (X) = U ∗ (Y ) = U (Y ) but X and Y are not homeomorphic (even |X| < |Y |). Both spaces are complete and proximally fine. 2. Generalities Most of the concepts and terms used in this paper can be found in [5] or [10]. We shall recall some of them used frequently here or not appearing in both cited books. In the sequel, uniform space means separated (Hausdorff) uniform space. For a uniform space X, by U (X) (or U ∗ (X)) we mean the lattice of all (or bounded, resp.) uniformly continuous real-valued functions. If X is a uniform space, then γX denotes its completion and pX its precompact modification. Thus, γpX is the Samuel compactification sX of X. Two uniformities on X have the same precompact modification iff they induce the same proximity, i.e. the relation A B on subsets of X, defined by the property that the sets A, X \ B are separated by a uniformly continuous function (e.g. by a function having the zero value on A and the value 1 on X \B), i.e., the sets A, X \B are far (or not proximal). Two sets are far iff their closures in the Samuel compactification are disjoint. A map between uniform spaces is proximally continuous if it maps proximal sets into proximal sets. Other useful reflections (preserving underlying sets) are eX (the uniform space having for its base all countable uniform coverings of X) and cX (the uniform space having for its base preimages of uniform covers of R by all f ∈ U (X)). We shall also use the topological fine coreflection tf X (the finest uniform space generating the same topology as X does) and the proximally fine coreflection pf X. Proximally fine spaces are the finest elements in the collections of all uniform spaces having the same precompact modification. We recall that uniform spaces having a monotone base (with respect to refinement of covers or with respect to inclusion for vicinities of diagonal) are proximally fine (see [14]) and are uniformly zero-dimensional in case the monotone base has not countable cofinality. It was proved in [7] that any product of uniform spaces with monotone bases is proximally fine, too. We say that a uniform space is R-regular (according Mrówka’s terminology) if it is uniformly homeomorphic to s subspace of power of R, i.e., X = cX. Uniformly realcomplete spaces are the uniform spaces X having complete c-modification (i.e., γcX = cX). Those are the spaces X allowing a uniform injection that preserves topology onto a closed subspace of a power RA of reals. See [9] for their properties. Uniformly zero-dimensional spaces are uniform spaces having partitions for their bases of uniform covers (or equivalences for their bases of uniform vicinities of the diagonal). The precompact modification pX of a uniformly zero-dimensional space X is uniformly zero-dimensional, too (it has a subbase composed of {G, X \ G}, where G is a uniformly clopen set, i.e., a cozero set of a uniformly continuous two-valued function, or equivalently, G G). If X is a uniformly zero-dimensional space, then cX = eX. J. Isbell introduced locally fine uniform spaces X: for any uniform cover U of X and any uniform covers VU of every U ∈ U, the union {VU ; U ∈ U} is a uniform cover of X. Every subfine uniform space (i.e., a subspace of a fine space) is locally fine. J. Pelant proved in [13] that every locally fine space is subfine. Every locally fine space X is RE-space, i.e., every uniformly continuous real-valued function on a subspace of X can be extended to a uniformly continuous function on X. A subset of a uniform space X is called regular cozero set if it is the interior of the closure of a cozero set of a function from U (X) (thus regular cozero sets are regular open sets but not every regular open set is a regular cozero set). The cozero set of f will be denoted as coz f and the corresponding regular cozero set int coz f as cozr f in the sequel. Cauchy filters on X have bases composed of regular cozero sets.
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We shall need the following useful result from [2]: If L : U (X) → U (Y ) is a lattice isomorphism with L(0) = 0 and for some f, g ∈ U (X) we have f = g on cozr h, then L(f ) = L(g) on cozr (L(h)). 3. Banach–Stone-like theorems for uniform spaces The results we are interested in are those of a following form: if uniform spaces X and Y satisfy certain conditions and their lattices U (X) and U (Y ) are isomorphic, then X and Y are uniformly homeomorphic. Clearly, one condition is completeness of both spaces. We do not know a characterization of points or of uniform covers of X by means of the lattice U (X) and we must go via Samuel compactifications. It was proved in [8] that existence of a lattice isomorphism between U (X) and U (Y ) guarantees a homeomorphism between sX and sY . Then we must find conditions under which the homeomorphism maps X onto Y in order to get a proximal homeomorphism between X and Y . Other conditions are needed to have a uniform homeomorphism. In general, one cannot use any homeomorphism sX → sY , it should be related to a lattice isomorphism U (X) → U (Y ). Take, for instance, topologically fine spaces X = N ∪ {ξ}, Y = N ∪ {η} as topological subspaces of βN, where ξ, η are different free ultrafilters on N of the same type (i.e., there is a permutation p of N with p(ξ) = η). Then sX = sY = βN and the identity mapping βN → βN does not map X into Y . Both X, Y are countable, uniformly zero-dimensional, topologically fine (thus both locally and proximally fine) and complete spaces. Nevertheless, except for one case, our next special situations allow to use any homeomorphism sX → sY to get a proximal homeomorphism X → Y . Therefore, we conjecture there may be more results using special homeomorphisms generated by isomorphisms between lattices of uniformly continuous functions. 3.1. Metrizable spaces We shall now prove a Banach–Stone-like result for products of complete metrizable spaces. The case of complete metrizable uniform spaces is special. As mentioned at the beginning, the first correct proof of Shirota theorem was given by F. Cabello Sánchez and J. Cabello Sánchez in [2]. A different proof appears in [8] as a corollary to the main result. It uses Cauchy filters and a modification of Čech’s method from [3] showing that a homeomorphism βX → βY maps X onto Y provided X, Y are metrizable topological spaces. That procedure does not use lattice properties of function spaces. We can give now a quite different approach using Baire property and any homeomorphism between sX, sY without referring to the lattices U (X), U (Y ). Theorem 3.1. Let X, Y be complete metric spaces and h : sX → sY be a homeomorphism onto. Then h X is a uniform homeomorphism of X onto Y . Proof. Since Y is metrizable and complete, it is Gδ in sY , i.e., Y = Gn , where Gn are open sets in sY . The preimages Hn = h−1 (Gn ) are open dense sets in sX. Since X is metrizable and complete, it is Baire and, thus, H = X ∩ Hn is a dense set in X. The restriction h H : H → Y between the metric spaces H and Y is proximally continuous, thus uniformly continuous (since metric spaces are proximally fine). Since Y is complete, there is a uniformly continuous extension of h H to X → Y that must coincide with h : X → sY because of their continuity. We have just proved that h(X) ⊂ Y and h X : X → Y is uniformly continuous. Using h−1 one can prove the other inclusion h(X) ⊃ Y . 2 We shall now use the previous approach to products of metric spaces. The metric or metrizable spaces used in the next part of this subsection are considered as uniform spaces.
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Let X = ΠI Xi be a product of complete metrizable spaces, (M, d) be a complete metric space and g : sX → sM be a continuous mapping. As in the preceding proof, M is Gδ in sM but now we need special n in sM having M for their intersection, namely unions of extensions of basic uniform covers open sets G of M . A collection V star-refines a collection U if {stV V ; V ∈ V} refines U, where stV A = {V ∈ V, A ∩V = ∅}. So, for every V ∈ V one can find U ∈ U such that every V ∈ V meeting V is contained in U . In the sequel, is the maximal open subset of sM with G ∩ M = G. If U is an open cover of the for an open set G in M , G ; U ∈ U}. We start with the following observation. space M , we denote U = {U star-refines U. Lemma 3.2. If U, V are open covers of M (open in M ) such that V star-refines U, then V contains the star st V . Indeed, if V meets Proof. Let V ∈ V and let U ∈ U contain the star stV V . Then U V the V then V meets V and is, thus, contained in U ; consequently, V is contained in U since we take for U maximal open set in sM containing U . 2 We denote by Gn the uniform cover of M formed by the open balls Bx,4−n , x ∈ M , of radius 4−n n = Gn . It is convenient to notice that if 0 < t < r then B x,t ⊂ B x,r since Bx,r is a proximal and G neighborhood of Bx,t in M and, thus, in sM . By the previous lemma, Gn+1 star-refines Gn and Gn = M since M is complete. n ) are open dense subsets of sX and their intersections with X forms a n = g −1 (G The preimages H decreasing sequence {Hn } of dense open sets in X and its intersection H is dense in X (since X is Baire as a product of metric Baire spaces). Clearly, the restriction of g to H is a proximally continuous mapping into M . Unlike the situation in the proof of Theorem 3.1 we are not able to say that g is uniformly continuous on H because H is proximally fine since we do not know if H is proximally fine. We shall prove uniform continuity of g H by using factorizations on subproducts. In [7] a procedure shows how a proximally continuous map on a subspace of a product depends on less number of coordinates. Since that publication is not normally available, we shall repeat details modified to our situation. Lemma 3.3. For every n the set In = {i ∈ I; ∃ xi , y i ∈ Hn+1 such that prI\{i} (xi ) = prI\{i} (y i ) and p,4−n for no p ∈ M } is finite. g(xi ), g(y i ) ∈ B Proof. Assume that the set In is infinite. By the result from [4,19], there exists a countable J ⊂ In such p,4−(n+1) for no p ∈ M and all a ∈ A, b ∈ B. Now, the that, for A = {xi }J , B = {y i }J we have g(a), g(b) ∈ B sets g(A), g(B) are far in M and, thus, the sets A, B are far in X. So, there is a finite set K ⊂ I such that the projections of A, B to ΠK Xi are far. But those projections are not disjoint, since we can find i0 ∈ J \ K and then prK (xi0 ) = prK (y i0 ). This contradiction proves our assertion. 2 We shall denote J = n In . Our special situation allows to show that if we change a point of H in finitely many coordinates outside J then the new point is still in H and the value of g does not change. Lemma 3.4. If x ∈ H, y ∈ X and prI\F (x) = prI\F (y) for some finite set F ⊂ I \ J, then y ∈ H and g(x) = g(y). Proof. It suffices to prove the assertion for a singleton F = {i} and then to use induction. Suppose that n . We can find neighborhoods Ux , Uy in X of y∈ / H, which means that there exists n such that g(y) ∈ / G p,4−(n+3) for some p ∈ M . Since H is dense in X x, y, resp., such that prI\{i} (Ux ) = prI\{i} (Uy ), g(Ux ) ⊂ B we can find points u ∈ Ux , v ∈ Uy ∩ H with prI\{i} (u) = prI\{i} (v). Since i ∈ / In+2 and u ∈ Hn+3 we have
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q,4−(n+2) for some q ∈ M . Thus the set B q,4−(n+2) meets the set B p,4−(n+3) and, therefore, the g(u), g(v) ∈ B point g(v) belongs to Bp,4−(n+1) . The point y is in the closure of such found points v, which implies that p,4−(n+1) . But the last set is contained in B p,4−(n) and, thus g(y) ∈ G n and that is g(y) is in the closure of B a contradiction. Since i ∈ / In we have d(g(x), g(y)) < 2.4−n for any n, i.e., g(x) = g(y).
2
Lemma 3.5. If x ∈ H, y ∈ X and prJ (x) = prJ (y) then g(x) = g(y). Proof. Take x, y satisfying the conditions of our assertion. For any finite set F ⊂ I \ J take yF with prF (yF ) = prF (y), prI\F (yF ) = prI\F (x). By the previous lemmas, yF ∈ H, g(x) = g(yF ). Clearly, {yF ; F ∈ [I \ J]<ω } converges to y, which implies g(y) = g(x). 2 Proposition 3.6. The restriction of the mapping g to X maps X into M . Proof. By the previous lemma, the restriction of g to H = pr−1 J (prJ (H)) depends on J and maps H into M . Consequently, there is a mapping h : prJ (H) → M such that h ◦ prJ = g H . Proximal continuity of g implies proximal continuity of h because prJ : H → prJ (H) is a proximal retraction. Since, prJ (H) is metrizable, thus proximally fine, the mapping h is uniformly continuous and has a uniformly continuous extension h : ΠJ Xi → M since M is complete and prJ (H) is dense in ΠJ Xi . The composition h ◦ prJ is uniformly continuous and must coincide with the restriction of g to X. Consequently, g(X) ⊂ M . 2
Theorem 3.7. If two products of complete metrizable spaces have homeomorphic Samuel compactifications, they are uniformly homeomorphic. Proof. Let X = ΠI Xi , Y = ΠJ Yj be products of complete metrizable spaces and h : sX → sY be a homeomorphism. We denote by pr j the extension of prj : Y → Yj to sY → sYj , By the previous Proposition 3.6 we have ( prJ ◦h)(X) ⊂ Yj , which implies h(X) ⊂ Y . Using h−1 we get the other inclusion. 2 Corollary 3.8. If X, Y are products of complete metrizable spaces and U (X), U (Y ) are lattice isomorphic, then X and Y are uniformly homeomorphic. 3.2. Uniformly 0-dimensional spaces For non-metrizable spaces or their products we do not know another approach than to proceed “pointwise”, i.e., to consider images of points x ∈ X in sY . Since in our final applications we assume the spaces to be complete, we must find properties distinguishing points of γX and of sX \ γX. A classical procedure is to use filters. This procedure has the advantage of combining local topological properties and global uniform properties. If H, U are collections of subsets of X, we denote HU = {V ⊂ U; V ∈ H}. We shall now apply that situation to filters H in X and partitions U of X – then HU is a filter on U and HU = {{U ∈ U; U ∩ H = ∅}; H ∈ H} . Also in this subsection we can work with any homeomorphism sX → sY . Lemma 3.9. Let X be a uniformly zero-dimensional space, ξ ∈ sX and H be the trace on X of neighborhoods of ξ. Then 1. HU is an ultrafilter for any uniform partition U of X; 2. ξ ∈ sX \ γX iff there exists a uniform partition U of X such that the ultrafilter HU is free.
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Proof. Take any uniform partition U of X and let U = U1 ∪ U2 . Then { U1 , U2 } is a uniform cover of pX. Because H is Cauchy on pX, it must contain one of the two sets of the uniform cover, which means that one of the sets Ui , i = 1, 2, belongs to HU . Consequently, HU is an ultrafilter on U. Clearly, if ξ ∈ γX, then the filter HU is fixed for every uniform partition U of X. Assume now ξ ∈ sX \γX. Then H cannot be a Cauchy filter on X and, thus, there exists a uniform partition U of X such that no member of U belongs to H. Thus, the collection HU is a free ultrafilter on U. 2 Using the fact that free ultrafilters cannot have a monotone base, we can prove the next result. Lemma 3.10. If x ∈ X has a monotone neighborhood base in X and Y is uniformly zero-dimensional then for any homeomorphism sX → sY we have h(x) ∈ γY . Proof. The point x has a monotone base of its neighborhoods in sX, too; denote it by N . The image h(N ) is a monotone base of the neighborhood system of h(x) in sY , and its trace H to Y is a Cauchy filter in pY . According Lemma 3.9, HU is an ultrafilter on U for every uniform partition U of Y . Having monotone bases, those ultrafilters must be fixed, which implies h(x) ∈ γY . 2 Lemma 3.10 implies easily the next assertion. Proposition 3.11. Let Y be a complete uniformly zero-dimensional uniform space and X be a uniform space having monotone neighborhood bases at all points. Then any homeomorphism sX → sY maps X into Y . Now we get a Banach–Stone-like theorem for sums of uniform spaces having monotone uniform bases. By sum we mean coproduct in the categorical sense. Recall that uniform spaces with monotone bases are proximally fine and, thus, sums of such spaces are proximally fine, too (proximally fine spaces form a coreflective subclass of uniform spaces). Theorem 3.12. Let X, Y be sums of complete uniformly zero-dimensional spaces having monotone uniform bases. If the lattices U (X) and U (Y ) are isomorphic, then X and Y are uniformly homeomorphic. We want to prove the previous assertion for products instead of coproducts. We shall start with finite products and, then, reduce the infinite case to a finite one. We must generalize Lemma 3.10 to points having a neighborhood base indexed by a finite product of monotone orders. If we say that a neighborhood base U is indexed by an ordered set (A, ≤), we have in mind that U = {Ua }A and a ≤ b implies Ua ⊃ Ub . Since every monotone order has a well-ordered cofinal subset corresponding to a regular cardinal, our assumption is equivalent to indexing a base by a finite product of regular cardinals. Moreover, we may assume that those cardinals are different since a finite product of monotonically ordered sets having cofinal sets of the same regular cardinality has again a cofinal set of that cardinality. Lemma 3.13. If x ∈ X has a neighborhood base indexed by a finite product of monotonically ordered sets and Y is uniformly zero-dimensional, then for any homeomorphism h : sX → sY we have h(x) ∈ γY . Proof. Let x have a base indexed by κ1 × κ2 × . . . × κn , n > 1, where κ1 < κ2 < . . . < κn . The point x has a base indexed by κ1 × κ2 × . . . × κn also in sX, thus the same is true for h(x) in sY and for the trace of a neighborhood base of h(x) in Y . Denote that last base by B = {B{αi } }, where αi ∈ κi and, thus, B{αi } ⊃ B{βi } provided αi ≤ βi for each i ≤ n. If h(x) ∈ sY \ γY , the collection BU is a base of a free ultrafilter H on U for a uniform partition U of Y . The sets H{αi } = {U ∈ U; U ∩ B{αi } = ∅} form a base of
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H indexed by κ1 × κ2 × . . . × κn . We want to show that it is not possible for a free ultrafilter to have such a base. Since no free ultrafilter has a base of regular cardinality κ such that each of its member has a cardinality larger or equal to κ (otherwise we can construct two disjoint sets meeting each member of the base), we may assume that always |H{αi } | < κn . Then for any a = {αi } ∈ κ1 × . . . × κn−1 the collection {H{a,α} }α∈κn is a decreasing collection of length κn of sets of cardinalities less than κn . Consequently, there exists αa ∈ κn such that H{a,αa } = H{a,α} for any α ≥ αa . So, we may take a new base of H, namely {H{a,αa } } for a ∈ κ1 × . . . × κn−1 . This new base is indexed by κ1 × . . . × κn−1 . Repeating the process we finish with a monotone base of H indexed by κ1 , which is not possible since the ultrafilter H is free. 2 Corollary 3.14. Let Y be a uniformly zero-dimensional uniform space and X be a product of finitely many uniform spaces having monotone neighborhood bases. Then any homeomorphism sX → sY maps X into γY . Using a factorization theorem for functions on products, we are able to transfer the previous results to infinite products of uniformly zero-dimensional spaces. We shall use the following factorization theorem from [7] (Proposition 3). By uniform pseudocharacter of a uniform space Q we mean the least cardinal η such that the diagonal ΔQ of Q is intersection of η many uniform neighborhoods of ΔQ . The factorized map need not be proximally continuous, in general. Proposition 3.15. Let P be a subspace of a product Πκ Pα of uniform spaces, κ ≥ ω and f : P → Q be a proximally continuous mapping into a uniform space Q. If uniform pseudocharacter of Q is less than cof(κ), then f depends on less than κ coordinates. Now we can prove the substantial assertion of this part. Proposition 3.16. Let X be a product of uniformly zero-dimensional spaces having monotone bases and Y be uniformly zero-dimensional. Then any homeomorphism h : sX → sY maps X into γY . Proof. Let X = ΠI Xi , where Xi are uniformly zero-dimensional spaces having monotone bases and |I| ≥ ω. Let x ∈ X and suppose ξ = h(x) ∈ sY \ γcY . According Lemma 3.9 there is a free ultrafilter Pξ on a uniform partition U of Y generated by the neighborhoods of ξ, i.e., x ) = ∅}; U x is a neighborhood of x in sX} . Pξ = {{U ∈ U; U ∩ h(U There exists a uniformly continuous map g on Y onto a uniformly discrete space D such that U = {g −1 (d); d ∈ D}. We take its continuous extension g : sY → βD and the restriction f of g ◦ h : sX → βD to G = X ∩ h−1 ( g −1 (D). The mapping f : G → D is proximally continuous and G is an open dense subset of X. By Proposition 3.15, f depends on a set J ⊂ I with |J| < |I|. That means there is a mapping f0 : prJ (G) → D such that f = f0 ◦ (prJ G). We shall show that G depends on J, i.e., if x ∈ G, y ∈ X and prJ (x) = prJ (y) then y ∈ G. Find a canonical neighborhood U × V ⊂ G of x, where U is a neighborhood of prJ (x) in prj (G). There exists a canonical neighborhood of y of a form U ×W and a net {ya } in G ∩(U ×W ) converging to y. We can find a net {xa } in U × V converging to x and such that prJ (ya ) = prJ (xa ) for each a. Thus f (xa ) = f (ya ) = da ∈ D for each a. Since f is continuous and xa → x, the points da must coincide starting from some index. It follows that ( g ◦ h)(y) ∈ D and, thus, y ∈ G. Since G depends on J, the projection G → prJ (G) is a proximal quotient and, consequently, the mapping f0 is proximally continuous. We can repeat the process and finally get that f and G depend on a finite set J. Thus, the sets prJ (f −1 (d)), d ∈ D, form an open partition of prJ (G).
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The sets U, D, {f −1 (d); d ∈ D} and {prJ (f −1 (d)); d ∈ D} are isomorphic with the isomorphisms formed by g, f −1 and prJ . Using those isomorphisms, the ultrafilter Pξ is isomorphic to a free ultrafilter P on {prJ (f −1 (d)); d ∈ D} described as P = {{prJ (f −1 (d)); prJ (f −1 (d)) ∩ prJ (Ux ) = ∅}; Ux is a neighborhood of x in X} . Since the sets prJ (Ux ) form a base of neighborhoods of prJ (x) in ΠJ Xi , the free ultrafilter P has a base indexed by a finite product of cardinals, which is impossible by Lemma 3.13. This contradiction proves that h(x) ∈ γY . 2 As a corollary we get a requested Banach–Stone-like theorem. Theorem 3.17. Let both X, Y be products of uniformly zero-dimensional complete spaces having monotone bases. Each of the following conditions implies that X, Y are uniformly homeomorphic. 1. The Samuel compactifications of X, Y are homeomorphic. 2. The lattices U ∗ (X), U ∗ (Y ) are isomorphic. 3. The lattices U (X), U (Y ) are isomorphic. Proof. By the result in [8], the condition 3 implies the condition 2. Both conditions 1 and 2 are equivalent according to Kaplansky theorem. It remains to show the condition 1 implies that X, Y are uniformly homeomorphic. Let h : sX → sY be a homeomorphism. By the preceding theorem we have h(X) ⊂ Y , h−1 (Y ) ⊂ X and both restrictions h1 , h−1 of h or h−1 to X or to Y , resp., are uniformly continuous on 1 the restrictions of the uniformities on sX, sY , i.e. on pX, pY , therefore proximally continuous on X, Y . Since both spaces X, Y are proximally fine (see [7]), both maps h1 : X → Y , h−1 1 : Y → X are uniformly continuous with respect to uniformities on X, Y . 2 3.3. Uniformly realcomplete spaces The procedure used for uniformly zero-dimensional spaces cannot be transferred directly to general uniform spaces. If we take HU as defined in the previous subsection for a uniform cover U and a Cauchy filter H in pX, we get a filter but not having convenient maximality property. If we take those subsets of a uniform cover U the unions of which belong to a filter H, we need not get a filter, but the collection HU has some maximality property. It seems to us that maximality property can be used to get, e.g., Lemma 3.10 without requiring zero-dimensionality (i.e., no point of sX \ γX has a monotone base of neighborhoods). If the cover U is finite-dimensional, it is a finite union of disjoint collections and the union of one of them must belong to H. We may then proceed similarly as for partitions. Nevertheless, there is one problem. When we start with a complete uniform space X, we can use finite-dimensional uniform covers only and the uniformity generated by them need not be complete. We do not know, except for one nontrivial case, nice conditions for complete X implying completeness of the modification of finite-dimensional covers. The exception is cX (see Theorem 3.26). For that reason we shall describe the next procedure for uniformly realcomplete spaces only. Clearly, ξ ∈ sX \ γcX iff there exists a positive function f ∈ U (X) such that f(ξ) = ∞. Take such a point ξ together with such a function f . By Uξ we denote the neighborhood base of ξ in sX composed of all regular cozero sets in sX containing ξ, and by Uξ the trace of Uξ on X. For i = 1, . . . , 4 define An,i = (n + (i − 1)/4 − 1/10, n + i/4 + 1/10), Un,i = intf −1 (An,i ) (thus each Un,i is a regular cozero set). Then {An,i ; i = 1, . . . , 4, n ∈ ω} is a uniform cover of [0, ∞) and so {Un,i ; i = 1, . . . , 4, n ∈ ω} is a uniform cover of X. The family { n Un,i }i is a finite uniform cover of X, so there
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must be some i such that n Un,i ∈ ξ. There exists index i such that n Un,i ∈ / ξ (take i such that An.i ∩ Am,i = ∅ for any n, m). Define Pξ = {P ⊂ ω; {Un,i ; n ∈ P } ∈ Uξ }. The collection Pξ depends on f , the above cover of R and on the choice of i . In the sequel, those choices will be either irrelevant or specified. Lemma 3.18. If ξ ∈ sX \ γcX then Pξ is a free ultrafilter on ω. Proof. We use the notation from the preceding paragraph. It is trivial to see that if P ∈ Pξ , P ⊂ Q ⊂ ω then Q ∈ Pξ . Now let P, Q ∈ Pξ . Then n∈P Un,i ∩ n∈Q Un,i ∈ ξ and the intersection coincides with n∈P ∩Q Un,i , which implies P ∩ Q ∈ Pξ . Take any partition {P1 , P2 } of ω and Wk = n∈Pk {Un,j ; j = i }, V = n Un,i . The cover {W1 , W2 , V } is a uniform cover of pX and so, some of those sets belongs to Uξ . Since V ∈ / Uξ , there exists k ∈ {1, 2} such that Wk ∈ Uξ . Consequently, Wk ∩ n Un,i = {Un,i ; n ∈ Pk } ∈ Uξ , which entails Pk ∈ Pξ . 2 Lemma 3.19. If h : sX → sY is a homeomorphism and x ∈ X has a monotone neighborhood base in X then h(x) ∈ γcY . / γcY . The filter Uh(x) has a monotone base, hence also the ultrafilter Ph(x) has Proof. Assume h(x) ∈ a monotone base. Since only fixed ultrafilters have monotone bases, we came to a contradiction with Lemma 3.18. 2 Corollary 3.20. If the uniform space X has monotone neighborhood bases at all its points and h : sX → sY is a homeomorphism, then h(X) ⊂ γcY . In the next Banach–Stone-like theorem one can again use any homeomorphism between sX, sY without referring to the lattices U (X), U (Y ). Theorem 3.21. Let X, Y be uniformly realcomplete and proximally fine spaces with monotone neighborhood bases at all their points. If X and Y have homeomorphic Samuel compactifications, then X and Y are uniformly homeomorphic. Proof. By the previous corollary there exists a proximal homeomorphism between γcX = cX and γcY = cY , thus between X and Y . Since both X, Y are proximally fine, the proximal homeomorphism is a uniform homeomorphism. 2 The special case of Theorem 3.21 of uniform spaces having monotone uniform bases is weaker than the corresponding Theorems 3.12 and 3.1. The procedure above works if we know that a trace of the neighborhood filter of x ∈ X on a disjoint collection {Gn } of open sets with x ∈ Gn \ Gn cannot be a free ultrafilter. There is another approach using extension of maps. In this case we have to use the special homeomorphism λ : sX → sY generated sX by the isomorphism L : U (X) → U (Y ). By [8], both L and λ are related by the property λ(cozr f ) = sY
cozr (Lf )
for every f ∈ U (X), f ≥ 0.
Theorem 3.22. If Y is a locally fine space then λ(X) ⊂ γcY . Proof. Assume that for some x ∈ X we have ξ = λ(x) ∈ sY \ γcY . Since ξ is not Cauchy in cY , there is the situation described before Lemma 3.18. We have a uniformly discrete sequence {cozr gn } with ξ ∈ cozr gn \ cozr gn . Take fn = L−1 (gn ) and denote by ϕn the L-image of the constant function on X
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having value n. The function on cozr gn coinciding with ϕn on cozr gn is uniformly continuous and can be extended to a uniformly continuous function h on Y (both assertions follow from the fact that Y is locally fine). Now L−1 (h) coincides with n on cozr fn , which contradicts the facts that it is bounded on a neighborhood of x and x ∈ cozr (fn ). 2 Corollary 3.23. If X, Y are locally fine and uniformly realcomplete and have isomorphic lattices U (X), U (Y ), they are proximally homeomorphic. The following special case is formulated as a theorem since it belongs to Banach–Stone-like results. Theorem 3.24. Let X, Y be locally fine, proximally fine and uniformly realcomplete. If the lattices U (X) and U (Y ) are isomorphic then X, Y are uniformly homeomorphic. Realcompact topological space endowed with its fine uniformity is uniformly realcomplete and, trivially, locally fine and proximally fine. So, Shirota theorem for topological spaces follows from the previous assertion. Corollary 3.25. If X, Y are realcompact topological spaces having isomorphic lattices C(X), C(Y ), then X and Y are homeomorphic. Complete uniform spaces need not be uniformly realcomplete. To formulate the previous results for complete spaces instead of uniformly realcomplete spaces, we need a uniform version of another Shirota theorem (a completely regular space is realcompact provided it has a complete uniformity and no closed discrete subspace of Ulam measurable cardinality). There are several results in that direction. A space X is called star-finite if it has a base of star-finite uniform covers. A space X has inversion property (or is inversion-closed) if 1/f ∈ U (X) provided f ∈ U (X), f (x) = 0 for every x ∈ X. Ulam measurable cardinal is the first uncountable two-valued measurable cardinal. The next results are proved in [10, Thm. VII.18] (item 1), [15] (item 2) and [16, 2.8] (item 3). In fact, the last quoted paper contains a result that X is complete iff cX is complete and every minimal cX-Cauchy filter has countable intersection property. Proposition 3.26. Let X be a complete uniform space having no uniformly discrete subspace of Ulam measurable cardinality. Then cX is complete in either of the following cases: 1. X is locally fine; 2. X is star-finite; 3. X has inversion property. Theorem 3.27. Let X, Y be locally fine, proximally fine and complete having no uniformly discrete subspace of Ulam measurable cardinality. If the lattices U (X) and U (Y ) are isomorphic then X, Y are uniformly homeomorphic. We conjecture that a similar result holds if one assumes that X is inversion-closed instead of being locally fine. For star-finite spaces it may be a different situation. The assumption on non-measurability of covering numbers of X, Y is substantial. If we take for X a uniformly discrete space of measurable cardinality and Y = γcX, then both X, Y are complete, zerodimensional and locally fine, sX = sY = βX. The lattices U (X) and U (Y ) are isomorphic when we assign to f ∈ U (X) its uniformly continuous extension onto γcX. Thus λ : sX → sY is identity and λ(X) Y . Moreover, X and Y are not homeomorphic. Notice that X is not uniformly realcomplete.
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4. Problems In the preceding section we found some general cases when a lattice isomorphism L : U (X) → U (Y ) entails a proximal homeomorphism X → Y . To get a uniform homeomorphism X → Y , the spaces X and Y must have, in some sense, the same position in their proximity classes. The case when both spaces are the coarsest members of their proximity classes is trivial, since then both spaces are precompact, thus compact because of completeness. In all our cases we took the other easiest situation, namely when both spaces are the finest members of their proximity classes, i.e., they are proximally fine. What about other situations? A corresponding transfer to uniform spaces of Shirota theorem for topological spaces (if X, Y are realcompact spaces and C(X), C(Y ) are lattice isomorphic, then X, Y are homeomorphic) would be the result for R-generated uniform spaces. We were able to prove special cases only and the next question is open. Problem 1. Is it true that if X, Y are complete R-generated spaces with lattice isomorphic U (X), U (Y ) then X and Y are uniformly (or proximally) homeomorphic? We have a Banach–Stone-like theorems for products of complete metrizable spaces and for products of complete spaces having monotone bases that are uniformly zero-dimensional if metrizable. We conjecture that the first mentioned case can be included in the second one after we remove the condition on zero-dimensionality for metrizable spaces. Problem 2. Is it true that if X, Y are products of complete spaces having monotone bases, with lattice isomorphic U (X), U (Y ) then X and Y are uniformly (or proximally) homeomorphic? As we already mentioned after Theorem 3.27 we do not know what is the situation for inversion-closed spaces. Problem 3. Is it true that if X, Y are complete inversion-closed spaces with lattice isomorphic U (X), U (Y ) then X and Y are uniformly (or proximally) homeomorphic? Acknowledgements The first author acknowledges the support of the grant GACR P201/12/0290. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]
S. Banach, Théorie des opérations linéaires, Monogr. Mat., vol. 1, 1932, Warszawa. F. Cabello Sánchez, J. Cabello Sánchez, Lattices of uniformly continuous functions, Topol. Appl. 160 (2013) 50–55. E. Čech, On bicompact spaces, Ann. Math. 38 (1937) 823–844. V.A. Efremovič, Geometry of proximity (Russian), Mat. Sb. 31 (1952) 189–200. R. Engelking, General Topology, Heldermann Verlag, Berlin, 1989. I. Gelfand, A.N. Kolmogoroff, On rings of continuous functions on a topological space, Dokl. Akad. Nauk SSSR 22 (1939) 11–15. M. Hušek, Factorization of Mappings (Products of Proximally Fine Spaces), Seminar Uniform Spaces 1973–1974, Math. Inst. Academy, Prague, 1974, pp. 173–190. M. Hušek, Lattices of uniformly continuous functions determine sublattices of bounded functions, Topol. Appl. 182 (2015) 71–76. M. Hušek, A. Pulgarín, Banach–Stone-like theorems for lattices of uniformly continuous functions, Quaest. Math. 35 (2012) 417–430. Isbell, Uniform Spaces, Math. Surv., vol. 12, Amer. Math. Soc., Rhode Island, 1964. I. Kaplansky, Lattices of continuous functions, Bull. Am. Math. Soc. 53 (1947) 617–623. A.N. Milgram, Multiplicative semigroups of continuous functions, Duke Math. J. 16 (1949) 377–383. J. Pelant, Locally fine uniformities and normal covers, Czechoslov. Math. J. 37 (1987) 181–187. I. Ramm, A.S. Švarc, Geometry of proximity, uniform geometry and topology (Russian), Mat. Sb. 33 (1953) 157–180.
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[15] G.D. Reynolds, M.D. Rice, Completeness and covering properties of uniform spaces, Q. J. Appl. Math. Oxf. 29 (1978) 367–374. [16] M.D. Rice, Subcategories of uniform spaces, Trans. Am. Math. Soc. 201 (1975) 305–314. [17] T. Shirota, A generalization of a theorem of I. Kaplansky, Osaka Math. J. 4 (1952) 121–132. [18] M.H. Stone, Applications of the theory of Boolean rings to general topology, Trans. Am. Math. Soc. 41 (1937) 375–481. [19] V. Vilhelm, C. Vitner, Continuity in metric spaces (Czech), Čas. Pěst. Mat. 77 (1952) 147–173.
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Topology and its Applications www.elsevier.com/locate/topol
When lattices of uniformly continuous functions on X determine X Miroslav Hušek a,b,∗ , Antonio Pulgarín c a b c
Charles University, Sokolovská 83, 18675 Prague, Czech Republic Purkyně University, Klíšká 30, 40096 Ústínad Labem, Czech Republic Univ. of Extremadura, Avda. Universidad s/n, 10071 Cáceres, Spain
a r t i c l e
i n f o
a b s t r a c t
Article history: Received 23 February 2015 Received in revised form 20 August 2015 Accepted 21 August 2015 Available online xxxx MSC: 54C35 54E15 06B99
Several Banach–Stone-like generalizations of Shirota’s result for metrizable uniform spaces are proved. Namely, if complete uniform spaces X, Y have isomorphic lattices U (X), U (Y ) of their real-valued uniformly continuous functions, and both X, Y are either some products of spaces having monotone bases (metrizable or uniformly zero-dimensional), or are locally fine and of non-measurable cardinality, then X and Y are uniformly homeomorphic. © 2015 Published by Elsevier B.V.
Keywords: Lattice of functions Uniform continuity Banach–Stone theorem
1. Introduction S. Banach proved the following theorem in 1932 ([1], p. 170): Pour que deux ensembles métriques, complets et compacts Q et Q1 soient homèomorphes, il faut et il suffit que les espaces E et E1 des fonctions réelles continues dèfinies dans ces ensembles soient isométriques. Banach uses the Mazur–Ulam theorem (every isometry of a normed space onto another normed space is affine, i.e., linear if it maps 0 to 0) and gives a description of the homeomorphism. If U : E → E1 is a surjective isometry mapping 0 to 0, then the homeomorphism h : Q → Q1 is given by the equality U (f )(q1 ) = f (h−1 (q1 ))ε(q1 )
for some ε ∈ E1 , |ε| = 1 .
* Corresponding author at: Charles University, Sokolovská 83, 18675 Prague, Czech Republic. E-mail addresses: Miroslav.Husek@mff.cuni.cz (M. Hušek), [email protected] (A. Pulgarín). http://dx.doi.org/10.1016/j.topol.2015.08.015 0166-8641/© 2015 Published by Elsevier B.V.
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The above theorem does not hold if one takes linear isomorphism instead of isometry. M.H. Stone removed the condition on metrizability 5 years later ([18], p. 469): If R and R∗ are bicompact H-spaces and if M and M∗ are the corresponding function-rings, then the existence of an isometric correspondence f → U (f ) = f ∗ between M and M∗ is equivalent to the existence of a topological equivalence r → ρ(r) = r∗ between R and R∗ , the two correspondences being connected by the relations f (r) = ϕ∗ (r∗ )[f ∗ (r∗ ) − θ∗ (r∗ )], U f = f ∗ , ρ(r) = r∗ , U 0 = θ∗ , U 1 = ϕ∗ , where |ϕ∗ | = 1. If the relations U 0 = 0, U 1 = 1 or, equivalently, the relations θ∗ = 0, ϕ∗ = 1 are satisfied, then the correspondence f → U f = f ∗ is an analytical isomorphism between M and M∗ . From that time, many generalizations appeared on that theme. Usually, more general ranges were used (e.g., Banach spaces) or more general domains (e.g., locally compact or realcompact spaces), or various sets of function spaces with various algebraic structures, or various categories (e.g., metric spaces with Lipschitz maps or uniformly continuous maps), or non-isomorphisms between function spaces were used. For other algebraic structures on C(X) we can mention I. Gelfand and A.N. Kolmogorov in 1939 [6] for rings, A.N. Milgram in 1949 [12] for multiplicative semigroups, I. Kaplansky in 1947 [11] for lattices (all for compact spaces) and, finally the most general result, T. Shirota in 1952 [17] for lattices and realcompact spaces. In the present paper, we shall deal with Banach–Stone type results for uniform spaces as objects, with function spaces composed of uniformly continuous functions and endowed with a lattice structure only. Probably the first result in that direction was the one by T. Shirota from 1952 ([17], Th. 6): Let X be a complete metric space. Then X is determined by the lattice of all uniformly continuous real functions on X. Moreover X is determined by the lattice of all bounded uniformly continuous real functions on it. It was noticed that the Shirota’s proof does not work for unbounded functions. A correct proof of the Shirota’s result has recently been published in [2]: Let X and Y be complete metric spaces and T : U (Y ) → U (X) be a lattice isomorphism. There is a uniform homeomorphism τ : X → Y such that (T f )(x) = t(x, f (τ (x))) (f ∈ U (Y ), x ∈ X) , where t : X × R → R is given by t(x, c) = (T c)(x). Here c is first treated as a real number and then as a constant function on Y . The present authors investigated Banach–Stone-like theorems both for metrizable and non-metrizable uniform spaces in [9] for less general lattice isomorphisms (they were supposed to preserve constant maps). In the present paper we show that most of the Banach–Stone-like results from [9] hold without the additional assumption on lattice isomorphisms. That gives several generalizations of the result from [2]. Our idea is to use the fact (proved in [8]) that existence of a lattice isomorphism between U (X) and U (Y ) implies existence of a lattice isomorphism between U ∗ (X) and U ∗ (Y ), i.e., a homeomorphism between the Samuel compactifications of X and of Y . The first step is to find situations when homeomorphisms between the Samuel compactifications of X and of Y map X onto Y . In such cases we get a proximal homeomorphism between X and Y . The second
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step is to find situations when such a proximal homeomorphism implies a uniform homeomorphism between X and Y (i.e., when X and Y have the same position in their proximity classes). We cannot expect a Banach–Stone theorem to be true for all complete uniform spaces. If we take for X the separable metrizable hedgehog and for Y the Samuel compactification of X, then U (X) = U ∗ (X) = U ∗ (Y ) = U (Y ) but X and Y are not homeomorphic (even |X| < |Y |). Both spaces are complete and proximally fine. 2. Generalities Most of the concepts and terms used in this paper can be found in [5] or [10]. We shall recall some of them used frequently here or not appearing in both cited books. In the sequel, uniform space means separated (Hausdorff) uniform space. For a uniform space X, by U (X) (or U ∗ (X)) we mean the lattice of all (or bounded, resp.) uniformly continuous real-valued functions. If X is a uniform space, then γX denotes its completion and pX its precompact modification. Thus, γpX is the Samuel compactification sX of X. Two uniformities on X have the same precompact modification iff they induce the same proximity, i.e. the relation A B on subsets of X, defined by the property that the sets A, X \ B are separated by a uniformly continuous function (e.g. by a function having the zero value on A and the value 1 on X \B), i.e., the sets A, X \B are far (or not proximal). Two sets are far iff their closures in the Samuel compactification are disjoint. A map between uniform spaces is proximally continuous if it maps proximal sets into proximal sets. Other useful reflections (preserving underlying sets) are eX (the uniform space having for its base all countable uniform coverings of X) and cX (the uniform space having for its base preimages of uniform covers of R by all f ∈ U (X)). We shall also use the topological fine coreflection tf X (the finest uniform space generating the same topology as X does) and the proximally fine coreflection pf X. Proximally fine spaces are the finest elements in the collections of all uniform spaces having the same precompact modification. We recall that uniform spaces having a monotone base (with respect to refinement of covers or with respect to inclusion for vicinities of diagonal) are proximally fine (see [14]) and are uniformly zero-dimensional in case the monotone base has not countable cofinality. It was proved in [7] that any product of uniform spaces with monotone bases is proximally fine, too. We say that a uniform space is R-regular (according Mrówka’s terminology) if it is uniformly homeomorphic to s subspace of power of R, i.e., X = cX. Uniformly realcomplete spaces are the uniform spaces X having complete c-modification (i.e., γcX = cX). Those are the spaces X allowing a uniform injection that preserves topology onto a closed subspace of a power RA of reals. See [9] for their properties. Uniformly zero-dimensional spaces are uniform spaces having partitions for their bases of uniform covers (or equivalences for their bases of uniform vicinities of the diagonal). The precompact modification pX of a uniformly zero-dimensional space X is uniformly zero-dimensional, too (it has a subbase composed of {G, X \ G}, where G is a uniformly clopen set, i.e., a cozero set of a uniformly continuous two-valued function, or equivalently, G G). If X is a uniformly zero-dimensional space, then cX = eX. J. Isbell introduced locally fine uniform spaces X: for any uniform cover U of X and any uniform covers VU of every U ∈ U, the union {VU ; U ∈ U} is a uniform cover of X. Every subfine uniform space (i.e., a subspace of a fine space) is locally fine. J. Pelant proved in [13] that every locally fine space is subfine. Every locally fine space X is RE-space, i.e., every uniformly continuous real-valued function on a subspace of X can be extended to a uniformly continuous function on X. A subset of a uniform space X is called regular cozero set if it is the interior of the closure of a cozero set of a function from U (X) (thus regular cozero sets are regular open sets but not every regular open set is a regular cozero set). The cozero set of f will be denoted as coz f and the corresponding regular cozero set int coz f as cozr f in the sequel. Cauchy filters on X have bases composed of regular cozero sets.
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We shall need the following useful result from [2]: If L : U (X) → U (Y ) is a lattice isomorphism with L(0) = 0 and for some f, g ∈ U (X) we have f = g on cozr h, then L(f ) = L(g) on cozr (L(h)). 3. Banach–Stone-like theorems for uniform spaces The results we are interested in are those of a following form: if uniform spaces X and Y satisfy certain conditions and their lattices U (X) and U (Y ) are isomorphic, then X and Y are uniformly homeomorphic. Clearly, one condition is completeness of both spaces. We do not know a characterization of points or of uniform covers of X by means of the lattice U (X) and we must go via Samuel compactifications. It was proved in [8] that existence of a lattice isomorphism between U (X) and U (Y ) guarantees a homeomorphism between sX and sY . Then we must find conditions under which the homeomorphism maps X onto Y in order to get a proximal homeomorphism between X and Y . Other conditions are needed to have a uniform homeomorphism. In general, one cannot use any homeomorphism sX → sY , it should be related to a lattice isomorphism U (X) → U (Y ). Take, for instance, topologically fine spaces X = N ∪ {ξ}, Y = N ∪ {η} as topological subspaces of βN, where ξ, η are different free ultrafilters on N of the same type (i.e., there is a permutation p of N with p(ξ) = η). Then sX = sY = βN and the identity mapping βN → βN does not map X into Y . Both X, Y are countable, uniformly zero-dimensional, topologically fine (thus both locally and proximally fine) and complete spaces. Nevertheless, except for one case, our next special situations allow to use any homeomorphism sX → sY to get a proximal homeomorphism X → Y . Therefore, we conjecture there may be more results using special homeomorphisms generated by isomorphisms between lattices of uniformly continuous functions. 3.1. Metrizable spaces We shall now prove a Banach–Stone-like result for products of complete metrizable spaces. The case of complete metrizable uniform spaces is special. As mentioned at the beginning, the first correct proof of Shirota theorem was given by F. Cabello Sánchez and J. Cabello Sánchez in [2]. A different proof appears in [8] as a corollary to the main result. It uses Cauchy filters and a modification of Čech’s method from [3] showing that a homeomorphism βX → βY maps X onto Y provided X, Y are metrizable topological spaces. That procedure does not use lattice properties of function spaces. We can give now a quite different approach using Baire property and any homeomorphism between sX, sY without referring to the lattices U (X), U (Y ). Theorem 3.1. Let X, Y be complete metric spaces and h : sX → sY be a homeomorphism onto. Then h X is a uniform homeomorphism of X onto Y . Proof. Since Y is metrizable and complete, it is Gδ in sY , i.e., Y = Gn , where Gn are open sets in sY . The preimages Hn = h−1 (Gn ) are open dense sets in sX. Since X is metrizable and complete, it is Baire and, thus, H = X ∩ Hn is a dense set in X. The restriction h H : H → Y between the metric spaces H and Y is proximally continuous, thus uniformly continuous (since metric spaces are proximally fine). Since Y is complete, there is a uniformly continuous extension of h H to X → Y that must coincide with h : X → sY because of their continuity. We have just proved that h(X) ⊂ Y and h X : X → Y is uniformly continuous. Using h−1 one can prove the other inclusion h(X) ⊃ Y . 2 We shall now use the previous approach to products of metric spaces. The metric or metrizable spaces used in the next part of this subsection are considered as uniform spaces.
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Let X = ΠI Xi be a product of complete metrizable spaces, (M, d) be a complete metric space and g : sX → sM be a continuous mapping. As in the preceding proof, M is Gδ in sM but now we need special n in sM having M for their intersection, namely unions of extensions of basic uniform covers open sets G of M . A collection V star-refines a collection U if {stV V ; V ∈ V} refines U, where stV A = {V ∈ V, A ∩V = ∅}. So, for every V ∈ V one can find U ∈ U such that every V ∈ V meeting V is contained in U . In the sequel, is the maximal open subset of sM with G ∩ M = G. If U is an open cover of the for an open set G in M , G ; U ∈ U}. We start with the following observation. space M , we denote U = {U star-refines U. Lemma 3.2. If U, V are open covers of M (open in M ) such that V star-refines U, then V contains the star st V . Indeed, if V meets Proof. Let V ∈ V and let U ∈ U contain the star stV V . Then U V the V then V meets V and is, thus, contained in U ; consequently, V is contained in U since we take for U maximal open set in sM containing U . 2 We denote by Gn the uniform cover of M formed by the open balls Bx,4−n , x ∈ M , of radius 4−n n = Gn . It is convenient to notice that if 0 < t < r then B x,t ⊂ B x,r since Bx,r is a proximal and G neighborhood of Bx,t in M and, thus, in sM . By the previous lemma, Gn+1 star-refines Gn and Gn = M since M is complete. n ) are open dense subsets of sX and their intersections with X forms a n = g −1 (G The preimages H decreasing sequence {Hn } of dense open sets in X and its intersection H is dense in X (since X is Baire as a product of metric Baire spaces). Clearly, the restriction of g to H is a proximally continuous mapping into M . Unlike the situation in the proof of Theorem 3.1 we are not able to say that g is uniformly continuous on H because H is proximally fine since we do not know if H is proximally fine. We shall prove uniform continuity of g H by using factorizations on subproducts. In [7] a procedure shows how a proximally continuous map on a subspace of a product depends on less number of coordinates. Since that publication is not normally available, we shall repeat details modified to our situation. Lemma 3.3. For every n the set In = {i ∈ I; ∃ xi , y i ∈ Hn+1 such that prI\{i} (xi ) = prI\{i} (y i ) and p,4−n for no p ∈ M } is finite. g(xi ), g(y i ) ∈ B Proof. Assume that the set In is infinite. By the result from [4,19], there exists a countable J ⊂ In such p,4−(n+1) for no p ∈ M and all a ∈ A, b ∈ B. Now, the that, for A = {xi }J , B = {y i }J we have g(a), g(b) ∈ B sets g(A), g(B) are far in M and, thus, the sets A, B are far in X. So, there is a finite set K ⊂ I such that the projections of A, B to ΠK Xi are far. But those projections are not disjoint, since we can find i0 ∈ J \ K and then prK (xi0 ) = prK (y i0 ). This contradiction proves our assertion. 2 We shall denote J = n In . Our special situation allows to show that if we change a point of H in finitely many coordinates outside J then the new point is still in H and the value of g does not change. Lemma 3.4. If x ∈ H, y ∈ X and prI\F (x) = prI\F (y) for some finite set F ⊂ I \ J, then y ∈ H and g(x) = g(y). Proof. It suffices to prove the assertion for a singleton F = {i} and then to use induction. Suppose that n . We can find neighborhoods Ux , Uy in X of y∈ / H, which means that there exists n such that g(y) ∈ / G p,4−(n+3) for some p ∈ M . Since H is dense in X x, y, resp., such that prI\{i} (Ux ) = prI\{i} (Uy ), g(Ux ) ⊂ B we can find points u ∈ Ux , v ∈ Uy ∩ H with prI\{i} (u) = prI\{i} (v). Since i ∈ / In+2 and u ∈ Hn+3 we have
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q,4−(n+2) for some q ∈ M . Thus the set B q,4−(n+2) meets the set B p,4−(n+3) and, therefore, the g(u), g(v) ∈ B point g(v) belongs to Bp,4−(n+1) . The point y is in the closure of such found points v, which implies that p,4−(n+1) . But the last set is contained in B p,4−(n) and, thus g(y) ∈ G n and that is g(y) is in the closure of B a contradiction. Since i ∈ / In we have d(g(x), g(y)) < 2.4−n for any n, i.e., g(x) = g(y).
2
Lemma 3.5. If x ∈ H, y ∈ X and prJ (x) = prJ (y) then g(x) = g(y). Proof. Take x, y satisfying the conditions of our assertion. For any finite set F ⊂ I \ J take yF with prF (yF ) = prF (y), prI\F (yF ) = prI\F (x). By the previous lemmas, yF ∈ H, g(x) = g(yF ). Clearly, {yF ; F ∈ [I \ J]<ω } converges to y, which implies g(y) = g(x). 2 Proposition 3.6. The restriction of the mapping g to X maps X into M . Proof. By the previous lemma, the restriction of g to H = pr−1 J (prJ (H)) depends on J and maps H into M . Consequently, there is a mapping h : prJ (H) → M such that h ◦ prJ = g H . Proximal continuity of g implies proximal continuity of h because prJ : H → prJ (H) is a proximal retraction. Since, prJ (H) is metrizable, thus proximally fine, the mapping h is uniformly continuous and has a uniformly continuous extension h : ΠJ Xi → M since M is complete and prJ (H) is dense in ΠJ Xi . The composition h ◦ prJ is uniformly continuous and must coincide with the restriction of g to X. Consequently, g(X) ⊂ M . 2
Theorem 3.7. If two products of complete metrizable spaces have homeomorphic Samuel compactifications, they are uniformly homeomorphic. Proof. Let X = ΠI Xi , Y = ΠJ Yj be products of complete metrizable spaces and h : sX → sY be a homeomorphism. We denote by pr j the extension of prj : Y → Yj to sY → sYj , By the previous Proposition 3.6 we have ( prJ ◦h)(X) ⊂ Yj , which implies h(X) ⊂ Y . Using h−1 we get the other inclusion. 2 Corollary 3.8. If X, Y are products of complete metrizable spaces and U (X), U (Y ) are lattice isomorphic, then X and Y are uniformly homeomorphic. 3.2. Uniformly 0-dimensional spaces For non-metrizable spaces or their products we do not know another approach than to proceed “pointwise”, i.e., to consider images of points x ∈ X in sY . Since in our final applications we assume the spaces to be complete, we must find properties distinguishing points of γX and of sX \ γX. A classical procedure is to use filters. This procedure has the advantage of combining local topological properties and global uniform properties. If H, U are collections of subsets of X, we denote HU = {V ⊂ U; V ∈ H}. We shall now apply that situation to filters H in X and partitions U of X – then HU is a filter on U and HU = {{U ∈ U; U ∩ H = ∅}; H ∈ H} . Also in this subsection we can work with any homeomorphism sX → sY . Lemma 3.9. Let X be a uniformly zero-dimensional space, ξ ∈ sX and H be the trace on X of neighborhoods of ξ. Then 1. HU is an ultrafilter for any uniform partition U of X; 2. ξ ∈ sX \ γX iff there exists a uniform partition U of X such that the ultrafilter HU is free.
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Proof. Take any uniform partition U of X and let U = U1 ∪ U2 . Then { U1 , U2 } is a uniform cover of pX. Because H is Cauchy on pX, it must contain one of the two sets of the uniform cover, which means that one of the sets Ui , i = 1, 2, belongs to HU . Consequently, HU is an ultrafilter on U. Clearly, if ξ ∈ γX, then the filter HU is fixed for every uniform partition U of X. Assume now ξ ∈ sX \γX. Then H cannot be a Cauchy filter on X and, thus, there exists a uniform partition U of X such that no member of U belongs to H. Thus, the collection HU is a free ultrafilter on U. 2 Using the fact that free ultrafilters cannot have a monotone base, we can prove the next result. Lemma 3.10. If x ∈ X has a monotone neighborhood base in X and Y is uniformly zero-dimensional then for any homeomorphism sX → sY we have h(x) ∈ γY . Proof. The point x has a monotone base of its neighborhoods in sX, too; denote it by N . The image h(N ) is a monotone base of the neighborhood system of h(x) in sY , and its trace H to Y is a Cauchy filter in pY . According Lemma 3.9, HU is an ultrafilter on U for every uniform partition U of Y . Having monotone bases, those ultrafilters must be fixed, which implies h(x) ∈ γY . 2 Lemma 3.10 implies easily the next assertion. Proposition 3.11. Let Y be a complete uniformly zero-dimensional uniform space and X be a uniform space having monotone neighborhood bases at all points. Then any homeomorphism sX → sY maps X into Y . Now we get a Banach–Stone-like theorem for sums of uniform spaces having monotone uniform bases. By sum we mean coproduct in the categorical sense. Recall that uniform spaces with monotone bases are proximally fine and, thus, sums of such spaces are proximally fine, too (proximally fine spaces form a coreflective subclass of uniform spaces). Theorem 3.12. Let X, Y be sums of complete uniformly zero-dimensional spaces having monotone uniform bases. If the lattices U (X) and U (Y ) are isomorphic, then X and Y are uniformly homeomorphic. We want to prove the previous assertion for products instead of coproducts. We shall start with finite products and, then, reduce the infinite case to a finite one. We must generalize Lemma 3.10 to points having a neighborhood base indexed by a finite product of monotone orders. If we say that a neighborhood base U is indexed by an ordered set (A, ≤), we have in mind that U = {Ua }A and a ≤ b implies Ua ⊃ Ub . Since every monotone order has a well-ordered cofinal subset corresponding to a regular cardinal, our assumption is equivalent to indexing a base by a finite product of regular cardinals. Moreover, we may assume that those cardinals are different since a finite product of monotonically ordered sets having cofinal sets of the same regular cardinality has again a cofinal set of that cardinality. Lemma 3.13. If x ∈ X has a neighborhood base indexed by a finite product of monotonically ordered sets and Y is uniformly zero-dimensional, then for any homeomorphism h : sX → sY we have h(x) ∈ γY . Proof. Let x have a base indexed by κ1 × κ2 × . . . × κn , n > 1, where κ1 < κ2 < . . . < κn . The point x has a base indexed by κ1 × κ2 × . . . × κn also in sX, thus the same is true for h(x) in sY and for the trace of a neighborhood base of h(x) in Y . Denote that last base by B = {B{αi } }, where αi ∈ κi and, thus, B{αi } ⊃ B{βi } provided αi ≤ βi for each i ≤ n. If h(x) ∈ sY \ γY , the collection BU is a base of a free ultrafilter H on U for a uniform partition U of Y . The sets H{αi } = {U ∈ U; U ∩ B{αi } = ∅} form a base of
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H indexed by κ1 × κ2 × . . . × κn . We want to show that it is not possible for a free ultrafilter to have such a base. Since no free ultrafilter has a base of regular cardinality κ such that each of its member has a cardinality larger or equal to κ (otherwise we can construct two disjoint sets meeting each member of the base), we may assume that always |H{αi } | < κn . Then for any a = {αi } ∈ κ1 × . . . × κn−1 the collection {H{a,α} }α∈κn is a decreasing collection of length κn of sets of cardinalities less than κn . Consequently, there exists αa ∈ κn such that H{a,αa } = H{a,α} for any α ≥ αa . So, we may take a new base of H, namely {H{a,αa } } for a ∈ κ1 × . . . × κn−1 . This new base is indexed by κ1 × . . . × κn−1 . Repeating the process we finish with a monotone base of H indexed by κ1 , which is not possible since the ultrafilter H is free. 2 Corollary 3.14. Let Y be a uniformly zero-dimensional uniform space and X be a product of finitely many uniform spaces having monotone neighborhood bases. Then any homeomorphism sX → sY maps X into γY . Using a factorization theorem for functions on products, we are able to transfer the previous results to infinite products of uniformly zero-dimensional spaces. We shall use the following factorization theorem from [7] (Proposition 3). By uniform pseudocharacter of a uniform space Q we mean the least cardinal η such that the diagonal ΔQ of Q is intersection of η many uniform neighborhoods of ΔQ . The factorized map need not be proximally continuous, in general. Proposition 3.15. Let P be a subspace of a product Πκ Pα of uniform spaces, κ ≥ ω and f : P → Q be a proximally continuous mapping into a uniform space Q. If uniform pseudocharacter of Q is less than cof(κ), then f depends on less than κ coordinates. Now we can prove the substantial assertion of this part. Proposition 3.16. Let X be a product of uniformly zero-dimensional spaces having monotone bases and Y be uniformly zero-dimensional. Then any homeomorphism h : sX → sY maps X into γY . Proof. Let X = ΠI Xi , where Xi are uniformly zero-dimensional spaces having monotone bases and |I| ≥ ω. Let x ∈ X and suppose ξ = h(x) ∈ sY \ γcY . According Lemma 3.9 there is a free ultrafilter Pξ on a uniform partition U of Y generated by the neighborhoods of ξ, i.e., x ) = ∅}; U x is a neighborhood of x in sX} . Pξ = {{U ∈ U; U ∩ h(U There exists a uniformly continuous map g on Y onto a uniformly discrete space D such that U = {g −1 (d); d ∈ D}. We take its continuous extension g : sY → βD and the restriction f of g ◦ h : sX → βD to G = X ∩ h−1 ( g −1 (D). The mapping f : G → D is proximally continuous and G is an open dense subset of X. By Proposition 3.15, f depends on a set J ⊂ I with |J| < |I|. That means there is a mapping f0 : prJ (G) → D such that f = f0 ◦ (prJ G). We shall show that G depends on J, i.e., if x ∈ G, y ∈ X and prJ (x) = prJ (y) then y ∈ G. Find a canonical neighborhood U × V ⊂ G of x, where U is a neighborhood of prJ (x) in prj (G). There exists a canonical neighborhood of y of a form U ×W and a net {ya } in G ∩(U ×W ) converging to y. We can find a net {xa } in U × V converging to x and such that prJ (ya ) = prJ (xa ) for each a. Thus f (xa ) = f (ya ) = da ∈ D for each a. Since f is continuous and xa → x, the points da must coincide starting from some index. It follows that ( g ◦ h)(y) ∈ D and, thus, y ∈ G. Since G depends on J, the projection G → prJ (G) is a proximal quotient and, consequently, the mapping f0 is proximally continuous. We can repeat the process and finally get that f and G depend on a finite set J. Thus, the sets prJ (f −1 (d)), d ∈ D, form an open partition of prJ (G).
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The sets U, D, {f −1 (d); d ∈ D} and {prJ (f −1 (d)); d ∈ D} are isomorphic with the isomorphisms formed by g, f −1 and prJ . Using those isomorphisms, the ultrafilter Pξ is isomorphic to a free ultrafilter P on {prJ (f −1 (d)); d ∈ D} described as P = {{prJ (f −1 (d)); prJ (f −1 (d)) ∩ prJ (Ux ) = ∅}; Ux is a neighborhood of x in X} . Since the sets prJ (Ux ) form a base of neighborhoods of prJ (x) in ΠJ Xi , the free ultrafilter P has a base indexed by a finite product of cardinals, which is impossible by Lemma 3.13. This contradiction proves that h(x) ∈ γY . 2 As a corollary we get a requested Banach–Stone-like theorem. Theorem 3.17. Let both X, Y be products of uniformly zero-dimensional complete spaces having monotone bases. Each of the following conditions implies that X, Y are uniformly homeomorphic. 1. The Samuel compactifications of X, Y are homeomorphic. 2. The lattices U ∗ (X), U ∗ (Y ) are isomorphic. 3. The lattices U (X), U (Y ) are isomorphic. Proof. By the result in [8], the condition 3 implies the condition 2. Both conditions 1 and 2 are equivalent according to Kaplansky theorem. It remains to show the condition 1 implies that X, Y are uniformly homeomorphic. Let h : sX → sY be a homeomorphism. By the preceding theorem we have h(X) ⊂ Y , h−1 (Y ) ⊂ X and both restrictions h1 , h−1 of h or h−1 to X or to Y , resp., are uniformly continuous on 1 the restrictions of the uniformities on sX, sY , i.e. on pX, pY , therefore proximally continuous on X, Y . Since both spaces X, Y are proximally fine (see [7]), both maps h1 : X → Y , h−1 1 : Y → X are uniformly continuous with respect to uniformities on X, Y . 2 3.3. Uniformly realcomplete spaces The procedure used for uniformly zero-dimensional spaces cannot be transferred directly to general uniform spaces. If we take HU as defined in the previous subsection for a uniform cover U and a Cauchy filter H in pX, we get a filter but not having convenient maximality property. If we take those subsets of a uniform cover U the unions of which belong to a filter H, we need not get a filter, but the collection HU has some maximality property. It seems to us that maximality property can be used to get, e.g., Lemma 3.10 without requiring zero-dimensionality (i.e., no point of sX \ γX has a monotone base of neighborhoods). If the cover U is finite-dimensional, it is a finite union of disjoint collections and the union of one of them must belong to H. We may then proceed similarly as for partitions. Nevertheless, there is one problem. When we start with a complete uniform space X, we can use finite-dimensional uniform covers only and the uniformity generated by them need not be complete. We do not know, except for one nontrivial case, nice conditions for complete X implying completeness of the modification of finite-dimensional covers. The exception is cX (see Theorem 3.26). For that reason we shall describe the next procedure for uniformly realcomplete spaces only. Clearly, ξ ∈ sX \ γcX iff there exists a positive function f ∈ U (X) such that f(ξ) = ∞. Take such a point ξ together with such a function f . By Uξ we denote the neighborhood base of ξ in sX composed of all regular cozero sets in sX containing ξ, and by Uξ the trace of Uξ on X. For i = 1, . . . , 4 define An,i = (n + (i − 1)/4 − 1/10, n + i/4 + 1/10), Un,i = intf −1 (An,i ) (thus each Un,i is a regular cozero set). Then {An,i ; i = 1, . . . , 4, n ∈ ω} is a uniform cover of [0, ∞) and so {Un,i ; i = 1, . . . , 4, n ∈ ω} is a uniform cover of X. The family { n Un,i }i is a finite uniform cover of X, so there
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must be some i such that n Un,i ∈ ξ. There exists index i such that n Un,i ∈ / ξ (take i such that An.i ∩ Am,i = ∅ for any n, m). Define Pξ = {P ⊂ ω; {Un,i ; n ∈ P } ∈ Uξ }. The collection Pξ depends on f , the above cover of R and on the choice of i . In the sequel, those choices will be either irrelevant or specified. Lemma 3.18. If ξ ∈ sX \ γcX then Pξ is a free ultrafilter on ω. Proof. We use the notation from the preceding paragraph. It is trivial to see that if P ∈ Pξ , P ⊂ Q ⊂ ω then Q ∈ Pξ . Now let P, Q ∈ Pξ . Then n∈P Un,i ∩ n∈Q Un,i ∈ ξ and the intersection coincides with n∈P ∩Q Un,i , which implies P ∩ Q ∈ Pξ . Take any partition {P1 , P2 } of ω and Wk = n∈Pk {Un,j ; j = i }, V = n Un,i . The cover {W1 , W2 , V } is a uniform cover of pX and so, some of those sets belongs to Uξ . Since V ∈ / Uξ , there exists k ∈ {1, 2} such that Wk ∈ Uξ . Consequently, Wk ∩ n Un,i = {Un,i ; n ∈ Pk } ∈ Uξ , which entails Pk ∈ Pξ . 2 Lemma 3.19. If h : sX → sY is a homeomorphism and x ∈ X has a monotone neighborhood base in X then h(x) ∈ γcY . / γcY . The filter Uh(x) has a monotone base, hence also the ultrafilter Ph(x) has Proof. Assume h(x) ∈ a monotone base. Since only fixed ultrafilters have monotone bases, we came to a contradiction with Lemma 3.18. 2 Corollary 3.20. If the uniform space X has monotone neighborhood bases at all its points and h : sX → sY is a homeomorphism, then h(X) ⊂ γcY . In the next Banach–Stone-like theorem one can again use any homeomorphism between sX, sY without referring to the lattices U (X), U (Y ). Theorem 3.21. Let X, Y be uniformly realcomplete and proximally fine spaces with monotone neighborhood bases at all their points. If X and Y have homeomorphic Samuel compactifications, then X and Y are uniformly homeomorphic. Proof. By the previous corollary there exists a proximal homeomorphism between γcX = cX and γcY = cY , thus between X and Y . Since both X, Y are proximally fine, the proximal homeomorphism is a uniform homeomorphism. 2 The special case of Theorem 3.21 of uniform spaces having monotone uniform bases is weaker than the corresponding Theorems 3.12 and 3.1. The procedure above works if we know that a trace of the neighborhood filter of x ∈ X on a disjoint collection {Gn } of open sets with x ∈ Gn \ Gn cannot be a free ultrafilter. There is another approach using extension of maps. In this case we have to use the special homeomorphism λ : sX → sY generated sX by the isomorphism L : U (X) → U (Y ). By [8], both L and λ are related by the property λ(cozr f ) = sY
cozr (Lf )
for every f ∈ U (X), f ≥ 0.
Theorem 3.22. If Y is a locally fine space then λ(X) ⊂ γcY . Proof. Assume that for some x ∈ X we have ξ = λ(x) ∈ sY \ γcY . Since ξ is not Cauchy in cY , there is the situation described before Lemma 3.18. We have a uniformly discrete sequence {cozr gn } with ξ ∈ cozr gn \ cozr gn . Take fn = L−1 (gn ) and denote by ϕn the L-image of the constant function on X
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having value n. The function on cozr gn coinciding with ϕn on cozr gn is uniformly continuous and can be extended to a uniformly continuous function h on Y (both assertions follow from the fact that Y is locally fine). Now L−1 (h) coincides with n on cozr fn , which contradicts the facts that it is bounded on a neighborhood of x and x ∈ cozr (fn ). 2 Corollary 3.23. If X, Y are locally fine and uniformly realcomplete and have isomorphic lattices U (X), U (Y ), they are proximally homeomorphic. The following special case is formulated as a theorem since it belongs to Banach–Stone-like results. Theorem 3.24. Let X, Y be locally fine, proximally fine and uniformly realcomplete. If the lattices U (X) and U (Y ) are isomorphic then X, Y are uniformly homeomorphic. Realcompact topological space endowed with its fine uniformity is uniformly realcomplete and, trivially, locally fine and proximally fine. So, Shirota theorem for topological spaces follows from the previous assertion. Corollary 3.25. If X, Y are realcompact topological spaces having isomorphic lattices C(X), C(Y ), then X and Y are homeomorphic. Complete uniform spaces need not be uniformly realcomplete. To formulate the previous results for complete spaces instead of uniformly realcomplete spaces, we need a uniform version of another Shirota theorem (a completely regular space is realcompact provided it has a complete uniformity and no closed discrete subspace of Ulam measurable cardinality). There are several results in that direction. A space X is called star-finite if it has a base of star-finite uniform covers. A space X has inversion property (or is inversion-closed) if 1/f ∈ U (X) provided f ∈ U (X), f (x) = 0 for every x ∈ X. Ulam measurable cardinal is the first uncountable two-valued measurable cardinal. The next results are proved in [10, Thm. VII.18] (item 1), [15] (item 2) and [16, 2.8] (item 3). In fact, the last quoted paper contains a result that X is complete iff cX is complete and every minimal cX-Cauchy filter has countable intersection property. Proposition 3.26. Let X be a complete uniform space having no uniformly discrete subspace of Ulam measurable cardinality. Then cX is complete in either of the following cases: 1. X is locally fine; 2. X is star-finite; 3. X has inversion property. Theorem 3.27. Let X, Y be locally fine, proximally fine and complete having no uniformly discrete subspace of Ulam measurable cardinality. If the lattices U (X) and U (Y ) are isomorphic then X, Y are uniformly homeomorphic. We conjecture that a similar result holds if one assumes that X is inversion-closed instead of being locally fine. For star-finite spaces it may be a different situation. The assumption on non-measurability of covering numbers of X, Y is substantial. If we take for X a uniformly discrete space of measurable cardinality and Y = γcX, then both X, Y are complete, zerodimensional and locally fine, sX = sY = βX. The lattices U (X) and U (Y ) are isomorphic when we assign to f ∈ U (X) its uniformly continuous extension onto γcX. Thus λ : sX → sY is identity and λ(X) Y . Moreover, X and Y are not homeomorphic. Notice that X is not uniformly realcomplete.
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4. Problems In the preceding section we found some general cases when a lattice isomorphism L : U (X) → U (Y ) entails a proximal homeomorphism X → Y . To get a uniform homeomorphism X → Y , the spaces X and Y must have, in some sense, the same position in their proximity classes. The case when both spaces are the coarsest members of their proximity classes is trivial, since then both spaces are precompact, thus compact because of completeness. In all our cases we took the other easiest situation, namely when both spaces are the finest members of their proximity classes, i.e., they are proximally fine. What about other situations? A corresponding transfer to uniform spaces of Shirota theorem for topological spaces (if X, Y are realcompact spaces and C(X), C(Y ) are lattice isomorphic, then X, Y are homeomorphic) would be the result for R-generated uniform spaces. We were able to prove special cases only and the next question is open. Problem 1. Is it true that if X, Y are complete R-generated spaces with lattice isomorphic U (X), U (Y ) then X and Y are uniformly (or proximally) homeomorphic? We have a Banach–Stone-like theorems for products of complete metrizable spaces and for products of complete spaces having monotone bases that are uniformly zero-dimensional if metrizable. We conjecture that the first mentioned case can be included in the second one after we remove the condition on zero-dimensionality for metrizable spaces. Problem 2. Is it true that if X, Y are products of complete spaces having monotone bases, with lattice isomorphic U (X), U (Y ) then X and Y are uniformly (or proximally) homeomorphic? As we already mentioned after Theorem 3.27 we do not know what is the situation for inversion-closed spaces. Problem 3. Is it true that if X, Y are complete inversion-closed spaces with lattice isomorphic U (X), U (Y ) then X and Y are uniformly (or proximally) homeomorphic? Acknowledgements The first author acknowledges the support of the grant GACR P201/12/0290. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]
S. Banach, Théorie des opérations linéaires, Monogr. Mat., vol. 1, 1932, Warszawa. F. Cabello Sánchez, J. Cabello Sánchez, Lattices of uniformly continuous functions, Topol. Appl. 160 (2013) 50–55. E. Čech, On bicompact spaces, Ann. Math. 38 (1937) 823–844. V.A. Efremovič, Geometry of proximity (Russian), Mat. Sb. 31 (1952) 189–200. R. Engelking, General Topology, Heldermann Verlag, Berlin, 1989. I. Gelfand, A.N. Kolmogoroff, On rings of continuous functions on a topological space, Dokl. Akad. Nauk SSSR 22 (1939) 11–15. M. Hušek, Factorization of Mappings (Products of Proximally Fine Spaces), Seminar Uniform Spaces 1973–1974, Math. Inst. Academy, Prague, 1974, pp. 173–190. M. Hušek, Lattices of uniformly continuous functions determine sublattices of bounded functions, Topol. Appl. 182 (2015) 71–76. M. Hušek, A. Pulgarín, Banach–Stone-like theorems for lattices of uniformly continuous functions, Quaest. Math. 35 (2012) 417–430. Isbell, Uniform Spaces, Math. Surv., vol. 12, Amer. Math. Soc., Rhode Island, 1964. I. Kaplansky, Lattices of continuous functions, Bull. Am. Math. Soc. 53 (1947) 617–623. A.N. Milgram, Multiplicative semigroups of continuous functions, Duke Math. J. 16 (1949) 377–383. J. Pelant, Locally fine uniformities and normal covers, Czechoslov. Math. J. 37 (1987) 181–187. I. Ramm, A.S. Švarc, Geometry of proximity, uniform geometry and topology (Russian), Mat. Sb. 33 (1953) 157–180.
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[15] G.D. Reynolds, M.D. Rice, Completeness and covering properties of uniform spaces, Q. J. Appl. Math. Oxf. 29 (1978) 367–374. [16] M.D. Rice, Subcategories of uniform spaces, Trans. Am. Math. Soc. 201 (1975) 305–314. [17] T. Shirota, A generalization of a theorem of I. Kaplansky, Osaka Math. J. 4 (1952) 121–132. [18] M.H. Stone, Applications of the theory of Boolean rings to general topology, Trans. Am. Math. Soc. 41 (1937) 375–481. [19] V. Vilhelm, C. Vitner, Continuity in metric spaces (Czech), Čas. Pěst. Mat. 77 (1952) 147–173.