aa r X i v : . [ m a t h . GN ] J u l ORDER ISOMORPHISMS BETWEEN BASES OF TOPOLOGIES
JAVIER CABELLO S ´ANCHEZ
Abstract.
In this paper we will study the representations of isomorphismsbetween bases of topological spaces. It turns out that the perfect setting for thisstudy is that of regular open subsets of complete metric spaces, but we haveachieved some results about arbitrary bases in complete metric spaces and alsoabout regular open subsets of Hausdorff regular topological spaces. Introduction
This paper has a twofold goal. In the first part, we shall restrict ourselves to thestudy of complete metric spaces and order preserving bijections between arbitrarybases of their topologies. Namely, we will show that given a couple of completemetric spaces, say X and Y , every order preserving bijection between bases oftheir topologies induces a homeomorphism between dense G δ subspaces X ⊂ X and Y ⊂ Y . Later, we restrict ourselves to the bases of regular open sets onthe wider class of Hausdorff, regular topological spaces and show that whenever X ⊂ X is dense, the lattices R ( X ) and R ( X ) of regular open subsets are naturallyisomorphic and analyse some consequences of this. The mix of both parts givesan explicit representation of every isomorphism between lattices of regular opensets in complete metric spaces that may be considered as the main result in thispaper.1.1. Plan of the paper.
Apart from this Introduction, the present paper hasSection 2, where we prove the main results of the paper, and Section 3, thatcontains some remarks and applications of the main results.1.2.
Notations and conventions.
We will denote the interior of A ⊂ X asint X A , unless the space X is clear by the context, in which case we will just writeint A . The same way, A X or A will denote the closure of A in X .We will denote by R ( X ) the lattice of regular open subsets of X and B X will bea base of the topology of X , please recall that an open subset U of some topologicalspace X is regular if and only if U = int U . e-mail address: [email protected]: Lattices; complete metric spaces; locally compact spaces; open regular sets; partialordered sets.Mathematics Subject Classification: 54E50, 54H12. We say that T : B X → B Y is an isomorphism when it is a bijection that preservesinclusion, i.e., when T ( U ) ⊂ T ( V ) is equivalent to U ⊂ V .2. The main result
In this Section we will prove our main result, Theorem 2.14. Actually, it is justa consequence of Theorem 2.8 and Proposition 2.13, but as both results are a littlemore general than Theorem 2.14 we have decided to separate them. We have splitthe proof in several intermediate minor results.
Lemma 2.1.
Let ( X, d X ) and ( Y, d Y ) be complete metric spaces or locally com-pact topological spaces and B X , B Y , bases of their topologies. Suppose there is anisomorphism T : B X → B Y . Then, there exist dense subspaces X ⊂ X and Y ⊂ Y and a homeomorphism τ : X → Y such that τ ( x ) ∈ T ( U ) if and only if x ∈ U ∩ X .Proof. It is the same as in [3, Lemma 2]. (cid:3)
Remark . In the conditions of Lemma 2.1, we will denote R ( x ) = \ x ∈ U ∈B X T ( U ) . What the proof of [3, Lemma 2] shows is that the subset X , defined as the points x ∈ X for which R ( x ) is a singleton, is dense in X .The following Theorem is just a translation of the Th´eor`eme fondamental inLavrentieff’s [5].
Theorem 2.3 (Lavrentieff, [5]) . If there exists a bicontinuous, univocal and recip-rocal correspondence between two given sets (inside an m -dimensional space), itis possible to determine another correspondence with the same nature between thepoints of two G δ sets containing the given sets, the second correspondence agreeingwith the first in the points of the two given sets. A more general statement of Lavrentieff’s Theorem can be found in [8, Theorem24.9]:
Theorem 2.4 (Lavrentieff) . If X and Y are complete metric spaces and h is ahomeomorphism of A ⊂ X onto B ⊂ Y , then h can be extended to a homeomor-phism h ∗ of A ∗ onto B ∗ , where A ∗ and B ∗ are G δ -sets in X and Y , respectively,and A ⊂ A ∗ ⊂ A , B ⊂ B ∗ ⊂ B . As for the following Theorem, the author has been unable to find Alexandroff’swork [1], but Hausdorff references the result in [4] as follows:
Theorem 2.5 (Alexandroff, [1, 4]) . Every G δ subset in a complete space is home-omorphic to a complete space. RDER ISOMORPHISMS BETWEEN BASES OF TOPOLOGIES 3
Combining Theorems 2.4 and 2.5 with Lemma 2.1 we obtain:
Proposition 2.6.
Let X and Y be complete metric spaces and T : B X → B Y an inclusion preserving bijection. Then, there exist a complete metric space Z and dense G δ subspaces X ⊂ X, Y ⊂ Y such that Z, X and Y are mutuallyhomeomorphic. Of course, if Z is an in Proposition 2.6 then every dense G δ subset Z ′ ⊂ Z fulfilsthe same, so it is clear that there is no minimal Z whatsoever. In spite of this, itis very easy to determine some maximal Z : Theorem 2.7.
The greatest possible space Z in the preceding Proposition is (home-omorphic to) ( X , d Z ) , where X is the subset given in Lemma 2.1 and (1) d Z ( x, x ′ ) = max { d X ( x, x ′ ) , d Y ( τ ( x ) , τ ( x ′ )) } . A more explicit, though less clear, way to state Theorem 2.7 is the following:
Theorem 2.8.
The metric d Z makes X complete and, moreover, if Z ′ embeds inboth X and Y respectively via φ ′ X and φ ′ Y in such a way that φ ′ X ( z ) ∈ U if andonly if φ ′ Y ( z ) ∈ T ( U ) , then φ ′ X embeds Z ′ in X .Proof. For the first part, take a d Z -Cauchy sequence ( x n ) in X and let y n = τ ( x n )for every n . It is clear that both ( x n ) and ( y n ) are d X -Cauchy and d Y -Cauchy,respectively, so let x = lim( x n ) ∈ X, y = lim( y n ) ∈ Y , these limits exist because X and Y are complete. It is clear that any sequence (˜ x n ) ⊂ X converges to x ifand only if y = lim( τ (˜ x n )). This readily implies that R ( x ) = { y } , so x ∈ X andthis means that X is complete with d Z .Now we must see that every metric space Z ′ that embeds in both X and Y is embeddable in X , whenever the embeddings respect the isomorphism betweenthe bases. For this, as X is endowed with the restriction of the topology of X and Z ′ is homeomorphic to φ ′ X ( Z ′ ) ⊂ X , the only we need is φ ′ X ( Z ′ ) ⊂ X . So,suppose x ∈ φ ′ X ( Z ′ ) \ X and let z ∈ ( φ ′ X ) − ( x ). As Z ′ also embeds in Y , thereexists y = φ ′ Y ( z ) ∈ φ ′ Y ( Z ′ ) \ Y , too, with the property that x ∈ U if and only if y ∈ T ( U ). By the very definition of X and Y this means that x ∈ X , y ∈ Y and τ ( x ) = y . (cid:3) Now, we approach Proposition 2.13, the main result about regular topologicalspaces. For this, the following elementary results will come in handy.
Lemma 2.9.
Let Z be a topological space and A ⊂ Z . A is a regular open subsetof Z if and only if for every open V ⊂ Z, V ⊂ A implies V ⊂ A .Proof. It is obvious. (cid:3)
Lemma 2.10.
Let X be a topological space. Whenever Y ⊂ X is dense and U ⊂ X is open, one has U X = U ∩ Y X . JAVIER CABELLO S ´ANCHEZ
Proof.
Let x ∈ U X . This is equivalent to the fact that every open neighbourhood V of x has nonempty intersection with U . So, V ∩ U is a nonempty open subset of X and the density of Y implies that V ∩ U ∩ Y is also nonempty, so x ∈ U ∩ Y X and we have U X ⊂ U ∩ Y X . The other inclusion is trivial. (cid:3) Lemma 2.11.
Let X be a topological space and U, V ∈ R ( X ) such that U V .Then, there is ∅ 6 = W ∈ R ( X ) such that W ∩ U = ∅ and W ⊂ V .Proof. Actually, U \ V is regular because U and X \ V are regular and (cid:0) U \ V (cid:1) = U ∩ (cid:0) X \ V (cid:1) . This set is nonempty because U ⊂ V , along with the monotonicityof the interior operator, would imply U = int X U ⊂ int X (cid:0) V (cid:1) = V. (cid:3) Remark . If in Lemma 2.11 X is regular and Hausdorff, then V can be takenas any open subset that contains U strictly. Proposition 2.13.
Let X be a topological space and Y ⊂ X a dense subset. Then T : R ( X ) → R ( Y ) , defined as T ( U ) = U ∩ Y , is a lattice isomorphism with inverse S ( V ) = int V .Proof. We need to show that T and S are mutually inverse.Let U ∈ R ( X ), the first we need to show is that T is well-defined, i.e., that T ( U ) = U ∩ Y is regular in Y .Let V ⊂ X an open subset such that V ∩ Y ⊂ U ∩ Y Y . Then, as the closure in X preserves inclusions, we have V X = V ∩ Y X ⊂ U ∩ Y Y X ⊂ U X , where the first equality holds because of Lemma 2.10. Taking interiors in X alsopreserves inclusions, so we obtain V ⊂ int X (cid:16) V X (cid:17) ⊂ int X (cid:16) U X (cid:17) = U, which readily implies that V ∩ Y ⊂ U ∩ Y and we obtain V ∩ Y ∈ R ( Y ) fromLemma 2.9. It is clear that S ( V ) ∈ R ( X ) for every V ∈ R ( Y ), so both maps arewell-defined.Furthermore, Lemma 2.10 implies that, for any regular U ⊂ X, S ◦ T ( U ) = S ( U ∩ Y ) = int X (cid:16) U ∩ Y X (cid:17) = int X (cid:16) U X (cid:17) = U. As for the composition
T ◦ S , we have
T ◦ S ( V ) = T (cid:16) int X (cid:16) V X (cid:17)(cid:17) = int X (cid:16) V X (cid:17) ∩ Y RDER ISOMORPHISMS BETWEEN BASES OF TOPOLOGIES 5 for any V ∈ R ( Y ). Let W ⊂ X be an open subset for which V = W ∩ Y , the verydefinition of inherited topology implies that there exists such W . The previousequalities can be rewitten as T ◦ S ( W ∩ Y ) = T (cid:16) int X (cid:16) W ∩ Y X (cid:17)(cid:17) = T (cid:16) int X (cid:16) W X (cid:17)(cid:17) = int X (cid:16) W X (cid:17) ∩ Y, so we need W ∩ Y = int X (cid:16) W X (cid:17) ∩ Y . It is clear that W ∩ Y ⊆ int X (cid:16) W X (cid:17) ∩ Y ,so what we need is int X (cid:16) W X (cid:17) ∩ Y ⊆ W ∩ Y . Both subsets are regular in Y , soif this inclusion does not hold, there would exist an open H ⊂ YH = ∅ , H ⊂ int X (cid:16) W X (cid:17) ∩ Y, H ∩ ( W ∩ Y ) = ∅ , so, by Lemma 2.11 there is an open G ⊂ X such that H = G ∩ Y and so(2) G ∩ Y = ∅ , G ∩ Y ( ∗ ) ⊂ int X (cid:16) W X (cid:17) ∩ Y, ( G ∩ Y ) ∩ ( W ∩ Y ) = ∅ , and this is absurd. Indeed, the inclusion marked with ( ∗ ) implies that we maysubstitute G by G ∩ int X (cid:16) W X (cid:17) , so both equalities in (2) hold for some open G ⊂ int X (cid:16) W X (cid:17) . As Y is dense and G and W are open, the last equality impliesthat G ∩ W = ∅ . Of course, this implies G ∩ int X (cid:16) W X (cid:17) = ∅ , which means G = ∅ and we are done. (cid:3) Now we are in conditions to state our main result:
Theorem 2.14.
Let X , Y and Z be complete metric spaces, φ X : Z ֒ → X and φ Y : Z ֒ → Y be continuous, dense, embeddings and X = φ X ( Z ) . Then, T : R ( X ) → R ( Y ) given by the composition U U ∩ X φ − X ( U ∩ X ) φ Y ( φ − X ( U ∩ X )) int (cid:16) φ Y ( φ − X ( U ∩ X )) (cid:17) is an isomorphism between the lattices of open regular subsets of X and Y andevery isomorphism arises this way. The “every isomorphism arises this way” part is due to Theorem 2.8, while the“the composition is an isomorphism” part is consequence of Proposition 2.13.3.
Applications and remarks
In this Section, we are going to show how Proposition 2.13 gives in an easy waysome properties of β N and conclude with a couple of examples that show that thehypotheses imposed in the main results are necessary. But first, we need to dealwith an error in some outstanding work. In [2] F. Cabello and the author showedthat some results in [6] were not properly proved. Later, in [3] the same authors JAVIER CABELLO S ´ANCHEZ proved that, even when the proof of [6, Theorem 6] was incorrect, the result wastrue. Now, we are going to explain what the error in [6] was:
Definition 3.1 (Definition 2) . A distributive lattice with smallest element 0 satis-fying Wallman’s disjunction property is an
R-lattice if there exists a binary relation ≫ in L which satisfies: • If h ≥ f and f ≫ g , then h ≫ g . • If f ≫ g and f ≫ g , then f ∧ f ≫ g ∧ g . • If f ≫ g , then there exists h such that f ≫ h ≫ g . • For every f = 0 there exist g and g = 0 such that g ≫ f ≫ g . • If g ≫ f ≫ g , then there exists h such that h ∨ f = g and h ∧ g = 0 . Immediately after Definition 2 we find this:
Theorem 3.2 (Theorem 1) . A distributive lattice with smallest element 0 is anR-lattice if and only if it is isomorphic to a sublattice of the lattice of all regularopen sets on a locally compact space X . This sublattice is an open base and itselements have compact closures. The open regular set in X associated to f ∈ L is denoted by U ( f ). With thisnotation, the next statement is: Theorem 3.3 (Theorem 2) . Let L be an R-lattice. Then there exists uniquelya locally compact space X which satisfies the property in Theorem 1 and where U ( f ) ⊃ U ( g ) if and only if f ≫ g . Our Proposition 2.13 contradicts the uniqueness of X in the statement of Theo-rem 2 and we may actually explicit a lattice isomorphism between R ( X ) and R ( Y )for different compact metric spaces X and Y . Namely, we just need to take thesimplest compactifications of R and the composition of the lattice isomorphismspredicted by Proposition 2.13: Example 3.4.
Let X = R ∪ {−∞ , ∞} and Y = R ∪ { N } . Then, T : R ( X ) → R ( Y ) , defined by T ( U ) = U if U ∩ {−∞ , ∞} = ∅ U ∩ R if U ∩ {−∞ , ∞} = {∞} U ∩ R if U ∩ {−∞ , ∞} = {−∞} ( U ∩ R ) ∪ { N } if {−∞ , ∞} ⊂ U is a lattice isomorphism whose inverse is given by S ( V ) = (cid:26) V if N V ( V ∩ R ) ∪ {−∞ , ∞} if N ∈ V RDER ISOMORPHISMS BETWEEN BASES OF TOPOLOGIES 7
It seems that the problem here is that the definition of R-lattice, Definition 2,does not include the relation ≫ , but in Theorem 2 and its consequences the authorconsiders ≫ as a unique, fixed, relation given by ( L , ≤ ). It is clear that the abovespaces generate, say, different ≫ X and ≫ Y in the isomorphic lattices R ( X ) and R ( Y ). This leads to the error already noted in [2], Section 5.Actually, with the definition of R-lattice given in [6], in seems that the originalpurpose of the definition is lost. Indeed, the relation ≫ may be taken as ≥ in quitea few lattices. This leads to a topology where every regular open set is clopen, inSection 3.1 we will see an example of a far from trivial topological space wherethis is true. Given a lattice ( L , ≥ ), the relation between each possible ≫ and theunique locally compact topological space given by Theorem 2 probably deserves acloser look.Anyway, if we include ≫ in the definition, then [6, Theorem 2] is true. So letus put everything in order. Definition 3.5 (Shirota) . Let ( L , ≤ ) be a distributive lattice with min L = 0 and ≫ be a relation in L . The triple ( L , ≤ , ≫ ) is an R-lattice if the following holds: (1) For every a = b ∈ L , there exists h ∈ L such that either a ∧ h = 0 and b ∧ h = 0 or the other way round. (Una forma de Wallman’s disjunctionProperty). (2) If h ≥ f and f ≫ g , then h ≫ g . (3) If f ≫ g and f ≫ g , then f ∧ f ≫ g ∧ g . (4) If f ≫ g , then there exists h such that f ≫ h ≫ g . (5) For every f = 0 there exist g and g = 0 such that g ≫ f ≫ g . (6) If g ≫ f ≫ g , then there exists h such that h ∨ f = g and h ∧ g = 0 . With Definition 3.5 everything works and this result remains valid, but it doesnot lead to the consequences stated there as Theorems 3 to 6.
Theorem 3.6 (Theorem 2) . Let ( L , ≤ , ≫ ) be an R-lattice. Then there existsuniquely a locally compact space X which satisfies the property in Theorem 1 andwhere U ( f ) ⊃ U ( g ) if and only if f ≫ g . The Stone- ˇCech compactification of N . We will analyse the isomorphismgiven in Proposition 2.13 when Y = N and X = β N , the Stone- ˇCech compactifica-tion of N .This is not going to lead to new results, but it seems to be interesting inspite of this. These are very different spaces, so it could be surprising the fact thatthey share the same lattice of regular open subsets. In any case, as N is discrete,every V ⊂ N is regular and this, along with Proposition 2.13, implies that R ( β N ) = { int( U ) : U ⊂ N } . As β N is regular, every open W ⊂ β N is the union of its regular open subsets,and these regular open subsets are determined by the integers they contain, so W JAVIER CABELLO S ´ANCHEZ is determined by a collection A W of subsets of N . Of course, if W contains S ( J )for some J ⊂ N and I ⊂ J then S ( I ) ⊂ W , too. This means that A W is closedfor inclusions. Furthermore, as J is open in β N , see [7, p. 144, Subsection 3.9], if S ( I ) ⊂ W and S ( J ) ⊂ W then S ( I ) ∪ S ( J ) = S ( I ∪ J ), so S ( I ∪ J ) ⊂ W and A W is closed for pairwise supremum. Summing up, A W is an ideal of the lattice P ( N )for every proper open subset ∅ ( W ( β N –and every S ( J ) ∈ R ( β N ) is closed, so“clopen” and “regular open” are equivalent in β N .It is also clear that every ideal A in P ( N ) defines an open W A ∈ β N as ∪{S ( I ) : I ∈ A} , that these identifications are mutually inverse and that W ⊂ V if and onlyif A W ⊂ A V , so each maximal ideal in P ( N ) defines a maximal open proper subsetof β N . As β N is Hausdorff, these maximal open subsets are exactly β N \ { x } forsome x , so each point is dually defined by a maximal ideal. In other words, everypoint in β N is associated to an ultrafilter in P ( N ).As our final comment in this Just for fun
Remark, we have that β N is the onlyHausdorff compactification of N that fulfils: ♠ If J, I ⊂ N are disjoint, then their closures in the compactification aredisjoint, too,although this is just a particular case of a result by ˇCech, see [7, p. 25-26].3.2. The hypotheses are minimal.
In some sense, Theorem 2.14 is optimal.Here we see that there is no way to extend it if we omit any of the hypotheses.
Remark . Consider any infinite set Z endowed with the cofinite topology τ cof .It is clear that every pair of nonempty open subsets of Z meet, so every nonemptyopen subset is dense in Z and this implies that the only regular open subsets of Z are Z and ∅ . Of course the same applies to any uncountable set endowed with thecocountable topology τ con , so ( R , τ cof ) and ( R , τ con ) have the same regular opensubsets. Nevertheless, there is no way to identify homeomorphically any coupleof dense subsets of R with each topology. In order to avoid this pathologicalbehaviour we needed to consider just regular Hausdorff spaces since these spacesare the only reasonable spaces for which the regular subsets comprise a base ofthe topology. In other words, Theorem 2.14 will not extend to general topologicalspaces. Remark . There is no “non-complete metric spaces” result, in part because I and Q have the obvious isomorphism between their bases of regular open subsetsand they are, nevertheless, disjoint subsets in R . This means that when trying togeneralise Theorem 2.14 the problem may come not only from the lack of separationof the topologies as in Remark 3.7 but also from the, so to say, lack of points inthe spaces even when they are metric. RDER ISOMORPHISMS BETWEEN BASES OF TOPOLOGIES 9 Acknowledgements
Supported in part by DGICYT project MTM2016-76958-C2-1-P (Spain) andJunta de Extremadura programs GR · · References [1] P. S. Alexandroff. Sur les ensembles de la premi`ere classe et les ensembles abstraits.
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