Abstract and Applied Analysis

Volume 2014 (2014), Article ID 184071, 11 pages

http://dx.doi.org/10.1155/2014/184071

Research Article

Isomorphic Universality and the Number of Pairwise Nonisomorphic Models in the Class of Banach Spaces

Mirna Džamonja

School of Mathematics, University of East Anglia, Norwich NR4 7TJ, UK

Received 7 December 2013; Revised 22 March 2014; Accepted 30 March 2014; Published 20 May 2014

Academic Editor: S. A. Mohiuddine

Copyright © 2014 Mirna Džamonja. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

We develop the framework of natural spaces to study isomorphic embeddings of Banach spaces. We then use it to show that a sufficient failure of the generalized continuum hypothesis implies that the universality number of Banach spaces of a given density under a certain kind of positive embedding (very positive embedding) is high. An example of a very positive embedding is a positive onto embedding between and for 0-dimensional and such that the following requirement holds for all and in : if , then there are constants and with and .

1. Introduction

In this paper we join a recent trend that has seen a combination of model theory and set theory address questions coming from analysis and topology. Examples are the spectacular proof by Malliaris and Shelah of the in [1] or more directly connected to this paper, work by Shelah and Usvyatsov on the isometric universality of Banach spaces [2], which will be mentioned in more detail below. We are specifically interested in the isomorphic embeddings of Banach spaces, in particular in the universality number of this class. This is the minimal number of the Banach spaces of a given density which isomorphically embed all the other spaces of the same density, allowing, depending on the context, the embeddings or the spaces to have extra properties. In many contexts, for example, when working just with plain isomorphic embeddings, it suffices to work with spaces of the form and even to assume that is 0-dimensional and so of the form where is a Boolean algebra and its Stone space. This is because every Banach space of a given density embeds isometrically into one of the form with the same density. In the main body of this paper we concentrate on the spaces of the form .

The topic of universality of Banach spaces has already received a significant input from set theory, notably in the work of Brech and Koszmider [3, 4], which will be reviewed below. The new element we bring is the study of the topic not only by set-theoretic methods such as forcing, but also by the methods coming from classification theory in model theory and pcf theory. These ideas were explored in the context of isometric embeddings in [2], but the methods applicable to isometries do not at all apply in the context of isomorphic embeddings; hence we have needed to construct a new framework. Using this framework we are able to provide a template (Theorem 16(1)) of results which state that, for an uncountable cardinal number under certain cardinal arithmetic assumptions, there is no universal Banach space of density , under certain kinds of isomorphic embeddings. The general kind of embeddings considered in the template is called very positive embeddings and it includes many natural examples of embeddings. One is presented in the following theorem which easily follows from Theorem 16(1) (see Section 5.3).

Theorem 1. Suppose that and are two regular cardinals with

and that .

Then the minimal number of spaces of the form of density needed to embed all Banach spaces of the form of density by an embeddings satisfying the following conditions (i)–(iii) is , where

(i) is positive; that is, ;(ii) is onto;(iii)if , , and , then there are constants ,   such that and .

In particular there is no surjectively universal space of density , under the embeddings satisfying (i)–(iii).

It would of course be desirable to weaken the condition of very positivity. In recent work, Shelah [5] introduces a model-theoretic property called the olive property which generalizes the model-theoretic tools that can be used to apply the method of invariants which were used in the context of linear orders in [6] and which are necessary for [2] and for this work. The restriction to very positivity in our work basically comes from the connection of the invariants to linear orders. Therefore it is a promising direction for future work to find a Banach space isomorphism context where one could use the olive property in place of the order.

An additional consideration of the paper is the subject of the number of pairwise nonisomorphic Banach spaces of a given density. For example, the celebrated Kaplansky theorem [7] shows that if and admit a bijective isomorphism which also preserves pointwise order, then and are homeomorphic. Coupled with model-theoretic results which show that for every uncountable there are pairwise nonisomorphic Boolean algebras of size and with the Stone representation theorem, this gives that there are pairwise non-order-isomorphic Banach spaces of density . It follows from our work, as we show in Theorem 16(2), that the same is true under the weaker assumption of very positive embeddings (that the assumption of very positivity is strictly weaker than the assumption of preserving order follows from our examples in Section 5). A very general model-theoretic study of the properties which lead to a large number of pairwise nonisomorphic models in metric structures is undertaken by Shelah and Usvyatsov in [8] and in the future it may yield more general results about nonisomorphic Banach spaces.

We should finish this introduction by mentioning that for uncountable the successor of a regular it is not difficult to construct specific models of set theory in which there are no isomorphically universal Banach spaces of density ; for example, the classical Cohen model will do (paper [9] addresses this and finer versions of it). The point of Theorem 16 is that it is not a result which is true just in some specifically constructed model; it is a result which holds as soon as certain cardinal arithmetic assumptions are fulfilled. Another remark is that on the basis of what is known in the literature and what we obtain here, no known result differentiates between the universality number of Banach spaces of a given density under isometries or under isomorphisms. Furthermore, it is not known how to differentiate them from the universality number of Boolean algebras.

2. Background

For a quasi-ordered class , the universality number is defined as the smallest size of such that for every there is such that . In Banach space theory we find many examples of classes whose universality numbers have been studied, with respect to isomorphic, isometric, and other kinds of embeddings. A classical result by Banach [10, page 185] shows that is isometrically universal for all separable Banach spaces.

For the nonseparable case, the situation is more complex. The cardinal arithmetic assumption GCH automatically gives one universal model for each uncountable density, as explained below. Specific models of the failure of GCH were studied by Brech and Koszmider [3] who considered Banach spaces of density the continuum and proved that in the Cohen model for many Cohen reals there is an isomorphically universal Banach space of density . In [9] there are negative universality results in Cohen and Cohen-like models; for example, the isomorphic universality number for Banach spaces of density is in the Cohen model and moreover one Cohen real adds a w.c.g. Banach space of density which does not embed into any Banach space with a dense set of size in the ground model. In [4], which studies w.c.g. Banach spaces, it is stated (page 1268), without proof, that Koszmider and Thompson noted that a version of the proof from [3] gives a model where there is no isomorphically universal Banach space of density . Let us briefly explain the known positive universality results in the context of GCH. Throughout, stands for an infinite cardinal.

By combining the Stone duality theorem, the fact that any Banach space is isometric to a subspace of and that has a totally disconnected continuous preimage, Brech and Koszmider proved the following.

Fact 1 (see [3], Fact 1.1). The universality number of the class of Boolean algebras of size is greater or equal to the universality number of the class of Banach spaces of density with isometric embeddings, which is greater or equal than the universality number of the class of Banach spaces of density with isomorphic embeddings.

The class of spaces of the form for a Boolean algebra of size is isometrically universal for the class of Banach spaces of density , and in particular its universality number with either isometric or isomorphic embeddings is the same as the universality number of the whole class of Banach spaces of density .

Fact 1 is only interesting in the context of uncountable , since for we have a universal Boolean algebra as well as an isometrically universal Banach space, as explained above. On the other hand, it is known from the classical model theory (see [11] for saturated and special models) that in the presence of GCH there is a universal Boolean algebra at every uncountable cardinal, so the questions of universality for the above classes are interesting in the context of the failure of the relevant instances of GCH. Negative universality results for Boolean algebras are known to hold when GCH fails sufficiently by the work of Kojman and Shelah [6] and in Cohen-like extensions by the work of Shelah (see [6] for a proof). Shelah and Usvyatsov proved in [2] that, in the models where the negative universality results that were obtained for Boolean algebras in [6] hold, the same negative universality results hold for Banach spaces under isometric embeddings. The smallest cardinal at which these results can apply is . For example, if is a regular cardinal greater than but smaller than (so ), there is no universal under isometries Banach space of density .

Conjecture 2. The universality number of the class of Banach spaces of density with isomorphic embeddings is the same as the universality number of the class of Boolean algebras of size .

It follows from the above discussion that Conjecture 2 would improve Fact 1(1) and it would imply the negative universality results of Shelah and Usvyatsov. For all we know at this point Conjecture 2 could be a theorem of ZFC; that is, it is not known to fail at any even consistently. A particular case of Conjecture 2 is the following Conjecture 3, which summarizes the most interesting case from the point of view of Banach space theory.

Conjecture 3. The universality number of the class of Banach spaces of density with isomorphic embeddings is the same as the universality number of the class of Banach spaces of density with isometric embeddings.

There is a considerable amount of study of other kinds of embeddings of Banach spaces, but isometries and isomorphisms and our work will fit into that area. Sticking to the spaces of the form , among the classically studied isomorphic embeddings are those that preserve multiplication or the ones that preserve the pointwise order of functions. It is known for either one of them (Gelfand and Kolmogorov [12] for the former and Kaplansky [7] for the latter) that if they are onto, they actually characterize the topological structure of the space; that is, if is an onto embeddings which either preserves multiplication or the pointwise order, then and are homeomorphic. We will show that, in moving from the order preserving onto assumption just a small bit, we no longer have the preservation of the homeomorphic structure, but under the assumption that GCH fails sufficiently, we do have a large number of pairwise nonisomorphic spaces and a large universality number.

We now finish the introduction by giving some background information for the readers less familiar with Banach space theory.

Definition 4. A Banach space is a normed vector space complete in the metric induced by the norm. A linear embeddings between Banach spaces is an isometry if for every , we have , where we use to denote . A linear embeddings between Banach spaces is an isomorphism if there is a constant such that, for every , we have .

Remark 5. Every isometry is an isomorphism. An isomorphism is in particular an injective continuous function, and in fact, a linear map is an isomorphism if and only if both and are linear and continuous.

For an isomorphism, we define .

Throughout the paper letters and will be used for Boolean algebras, , for infinite cardinals, and and for compact spaces. The space is the space of all continuous real-valued functions on with the topology given by the supremum norm . We will write St for the Stone space of a Boolean algebra , which is defined as the space of all ultrafilters on with the topology generated by sets as a clopen basis. Let us note that Fact 1 implies.

Observation 1. The universality number of Banach spaces of density , under any kind of embeddings, is either 1 or .

This is so because if for any , we had that were a universal family of Banach spaces of density , then we could assume that each for some Boolean algebras of size . Therefore we could find a single algebra of size such that all embed into it (simply by freely generating an algebra by a disjoint union of all ) and hence would be a single universal Banach space of density .

3. Natural Spaces of Functions

Our methods will involve a combination of model theory, set theory, and Banach space theory. In this section we introduce a simple model-theoretic structure which will be used to achieve that mixture of methods.

Suppose that is a Boolean algebra. We will associate to it a structure whose role is to represent the space , where is the Stone space of , . The idea is as follows. We are interested in the set of all simple functions with rational coefficients defined on , so functions of the type , where each is rational, , and denotes the basic clopen set in determined by . Every element of is the limit of a sequence of such functions, since the limits of such sequences form exactly the class of Lebesgue integrable functions, which of course includes . Let us then consider the vector space freely generated by over , call it (this vector space figures in [3] with the notation and is considered in a different context). Hence every simple function on with rational coefficients corresponds uniquely to an element of , via an identification of each with . Using coordinatwise addition and scalar multiplication, the product becomes a vector space. Any function in can be identified with an element of this vector space, namely, a sequence of simple rational functions whose limit is , and hence can be identified with a subset of .

To encapsulate this discussion we will work with vector spaces with rational coefficients and with two distinguished unary predicates , satisfying . With our motivation in mind, we will call them function spaces. If such a space is the space of sequences of simple rational functions over a Stone space and , correspond, respectively, to the set of such sequences which converge or converge to 0, then we call a natural space and we denote it by . In spaces of the form for an element of we define as the norm in of the limit of . If is an embeddings between and , we will say that is a constant of the embeddings if for every of , we have that . Not every embeddings has such a constant, but we will only work with the ones which do.

We will mostly be interested in a specific case of the representation of continuous functions as limits of simple functions, given by the following observation.

Lemma 6. Suppose that is the Stone space of a Boolean algebra and let be a function in with for some . Then there is a sequence of simple functions, where each is of the form , with each rational in and , such that .

Moreover, we can assume that for every , is the sum of at most functions of the form .

Proof. By multiplying by a constant if necessary, we can assume that . Functions of the form with each real, contain the constant function 1, form an algebra, and separate the points of ; hence by the Stone-Weierstrass theorem, they form a dense subset of . Notice that every function can be, by changing the coefficients and the sets if necessary, represented in the form where all s are pairwise disjoint, so we can without loss of generality work only with such functions. Given and with s disjoint, we can find for rational numbers with ; hence the function is approximated within by , showing that also functions with rational coefficients and disjoint s are dense. Now given a function in with and , let be a function with rational coefficients and disjoint s satisfying , recalling that the in is the supremum norm. Define now

and consider the function . We claim that . By the assumption that s are disjoint, for any , there is at most one such that . If or , then . If and , then as . If and , then as .

To finish the proof, we observe that given a sequence of functions of the above form which converges to a function , we can define by induction on , , and if the number of coefficients in is , and otherwise. Hence is the sequence with possible repetitions of each element finitely many times and so .

Definition 7. Suppose that is a sequence in and suppose that was obtained by first replacing each with an equivalent function with disjoint s and then replacing the coefficients by using the procedure described in the proof of Lemma 6. We say that is a top-up of .

Corollary 8. Suppose that and are Boolean algebras, and let denote the linear subspace of spanned by the functions whose rational coefficients are in . Then for every in , there is with .

Proof. Let ; hence can be written as where and are both continuous and positive. Therefore, by the closure of under linear combinations, it suffices to prove the corollary in the case of . Let , and we now apply Lemma 6.

The point of these definitions is the connection between the embeddability in the class of spaces of the form St and the class of function spaces. Namely, we have the following.

Theorem 9. Suppose that and are Boolean algebras, and let denote the linear subspace of spanned by the family of sequences of simple functions whose coefficients are rationals in and which satisfy that each has at most many elements. Then if there is an isomorphic embeddings from to , then there is an isomorphic embeddings from to satisfying that for every in , if , then .

Proof. Let be an isomorphic embeddings, so . We intend to define an isomorphic embeddings from to . By linearity it is sufficient to work with the basis of . Let us use the notation for the projection on the th coordinate. First we define the action of on those which have the property that there is at most one such that is not the identity zero function, and then is a function of the form for some . If there is no such with , then we let be the element of whose all projections are zero. Otherwise, let be such that and consider , which is well defined. We have no reason to believe that is a simple function with rational coefficients. However, there is a function which is a simple function with rational coefficients and whose distance to in is less than . We define to be the unique element of such that the only which is not identically zero is and .

Now suppose that is such that for exactly one , is not identity zero, and for some and some rational in . For , let be the element of whose th projection is and all other projections are identity zero. Hence we have already defined , and we let .

Finally suppose that is any element of . Therefore for every , we have already defined , where is on the th coordinate. Let . Hence we have defined a linear embeddings of to . We extend this embeddings to by linearity. We need to check that this embeddings preserves and , and again it suffices to concentrate on the basis . So suppose that is in , and let be its image under . We will show that by showing that it is a Cauchy sequence. Let ; we will consider . Let and . We have

which goes to 0 as .

At the end suppose that is in , and let be its image under . By the definition of , we have that (this was the point of requiring the members of to be sequences whose th element has coefficients). Since , we have , so in conclusion, .

4. Invariants for the Natural Spaces and Very Positive Embeddings

We will now adapt the Kojman-Shelah method of invariants [6], to the natural spaces and a specific kind of isomorphic embeddings between Banach spaces, which we call very positive embeddings (see Definition 12). From this point on we assume that is a regular uncountable cardinal.

Definition 10. (1) Suppose that is a model of size . A filtration of is a continuous increasing sequence of elementary submodels of , each of size .

(2) For a regular cardinal , we use the notation for .

(3) A club guessing sequence on is a sequence such that each is a club in , and for every club , there is such that .

Observation 2. Suppose that and there is a club guessing sequence . Then there is a club guessing sequence such that, for all ,

where is the increasing enumeration of , for each .

Proof. First of all notice that by passing to subsets if necessary we can without loss of generality assume that each has order type . Given , let consist of the points of of cofinality , and let be the closure of in . Since , we have that is unbounded in , so it is clear that is a club of , and since we have , we obtain that the resulting sequence is a club guessing sequence on . It also follows that , so the increasing enumeration as claimed exists.

The main definition we need is the definition of the invariant. Let us suppose that is regular and that is a club guessing sequence with an increasing enumeration of , for each and satisfying the requirement (4). This sequence will be fixed throughout. The existence of such a sequence will be discussed at the end of the section, but for the moment let us say that Shelah (see Theorem 15) proved that such a sequence exist in many circumstances, notably for any regular .

Definition 11. Suppose that is a Boolean algebra of size , a filtration of , that and that . An ordinal is an element of the invariant if and only if there is such that, for every in , we have

We will be interested in the kind of embeddings between Banach spaces which will allow us to define appropriate which preserve the invariants; see the Preservation Lemma 13. We have succeeded to do this in the case of a special kind of positive embeddings, as defined in the following definition.

Definition 12. We say that an isomorphic embeddings is very positive if the following requirements hold:

(i) (positivity),(ii)for every with , there is with and ,(iii)if , , and , then there is definable from with .

We do not know if very positive embeddings was studied in the literature, but clearly, one kind of embeddings that is very positive is an order preserving onto embeddings. In this case we have Kaplansky’s theorem [7] mentioned above, which shows that in the presence of such an embeddings from onto we have that and are homeomorphic. We show in the example in Section 5 that the analogue is not true for very positive embeddings. In particular the question of the number of pairwise nonisomorphic by very positive embeddings spaces of the form does not reduce to the well-studied and understood question of the number of pairwise nonisomorphic Boolean algebras of a given cardinality (which for any infinite is always equal to , see Shelah’s [13]).

Let us now make a further assumption on :

Lemma 13 (Preservation Lemma). Let and be Boolean algebras of size and suppose that is a very positive embeddings. Let and be any filtrations of and , respectively, and let denote the linear subspace of spanned by the set of sequences of functions whose rational coefficients are in and that satisfy that the th coordinate has nonzero coefficients.

If is an isomorphic embeddings satisfying that for every , then there is a club of such that, for every with and for every with and , we have that

Proof. We may assume that the underlying set of and is the ordinal . Let us define a model with the universe two disjoint copies of the -sequences of the simple functions on with rational coefficients, interpreted as the elements of and , all the symbols of and with interpretations induced from these models, and the symbols , , and . By the assumption (6), there is a club of such that, for every of cofinality not , we have that restricted to the sequences whose ordinal coefficients are is an elementary submodel of and that it has universe corresponding to . Let us denote the latter model by .

Suppose now that . Choose with and . Let . By the choice of , we have that . Suppose first that and let demonstrate this. Let , which is well defined as . Notice that the requirement (5) will hold if we replace by any top-up (see Definition 7) as , and hence for all , we have . Since the topping up procedure is definable in , we may assume that for all and . By Lemma 6 applied within , we can assume that . By the choice of , we have that and similarly . By the fact that and since is an isomorphism, we have that . We would like to use to witness that , so let us try. We have already established that and . It remains to check the property (5) of .

Suppose for a contradiction that there is such that but that . Applying (ii) we can find with and . By Corollary 8, we can assume that there is with , and hence . Translating the properties guaranteed by (ii) into the terms of and applying the elementarity of , we can assume that . Now we apply (iii) to find , definable from and satisfying . Being definable from , has an approximation with definable from ; hence . By topping up if necessary as in Lemma 6 and in the above paragraph, we can assume that every element in satisfies ; therefore contradicts the choice of .

Now let us prove the other direction of the desired equality. Let as exemplified by some . As in the previous paragraphs, we can assume that , and hence by (ii) we can assume that for some which is not 0 we have and by the same argument as above we can assume that there is such that . Now we claim that exemplifies that . Suppose for a contradiction that and . Let . As before, we can assume that and each . So by the positivity we have , by the choice of we have that , and by elementarity we have . It follows that contradicts .

The next task is to construct lots of Boolean algebras with different invariants for and then to us the Preservation Lemma to show that no fixed can embed them all.

Lemma 14 (Construction Lemma). Suppose that . Then the club guessing sequence can be chosen so that for any which is a closed set of limit ordinals, there is a Boolean algebra , a filtration of and a club of such that for every there is with and .

The proof of this lemma is presented in Section 6. The following theorem of Shelah will be used in the proof of the Construction Lemma as well as in the proof of Theorem 16.

Theorem 15 (Shelah, [14, Claim 1.4]). Let be two regular cardinals with . Then there is a stationary set and sequences , such that

(i) and ;(ii)for every club of , there is with ;(iii) and ;(iv)if is a nonaccumulation point of , then ;(v)the nonaccumulation points of every are successor ordinals.

Claim 1.4. in [14] does not state property (c) explicitly, but it follows from the first line of the proof of that Claim.

Now we present the main theorem of the paper.

Theorem 16. Suppose that and are two regular cardinals with and that .

Then

(1)the minimal number of spaces of the form of density needed to embed all Banach spaces of the form of density very positively is . In particular there is no very-positively universal space of density ;(2)there are at least pairwise non-very positively isomorphic Banach spaces of density .

Proof. Fix sequences and as guaranteed by Theorem 15. Notice that satisfies that, with being the increasing enumeration of , we have that . Hence letting for and an arbitrary club of of order type satisfying for , the sequence can be used in the context of the Preservation Lemma 13. It can also be used in the context of the Construction Lemma 14. Let us therefore find the Boolean algebras as described in the statement of the Construction Lemma. Notice that there are many different choices for .

Suppose for a contradiction that there is a family for some for some algebras of size which is very positively universal for all for Boolean algebras of size . Notice that the assumptions we have made on imply that , so the size of each is . Let be the family of all subsets of that appear as invariants of elements of ; hence the size of is (since we have assumed ), and in particular there is a closed set of limit ordinals such that . Let . Suppose that is a very positive embeddings of into some , and let be an embeddings of into satisfying that for every , we have , which exists by Theorem 9. Let be a club of as guaranteed by the Preservation Lemma, let be a club of as guaranteed by the Construction Lemma, let , and suppose that is such that . Then by the choice of there is in whose invariant is , but then also has invariant , by the choice of , and hence we have a contradiction with the choice of .

Consider the family . By the argument in 1 for every , the set that embeds very positively into has size , so clearly every is very positively isomorphic with many . Hence we can choose pairwise non-very positively isomorphic elements of by a simple induction.

5. Examples

5.1. Cardinal Arithmetic

An example of circumstances when Theorem 16 applies is when

5.2. Very Positive Embeddings onto Do Not Give Rise to Homeomorphism

We give an example of two 0-dimensional spaces and which are not homeomorphic; yet they admit a very-positive isomorphism onto. The example itself was constructed by Plebanek in [15, Example 5.3], when considering positive onto isomorphisms.

Let consist of two disjoint convergent sequences with and with , and let consist of a single convergent sequence with . Define by letting for all

Plebanek shows that is a positive isomorphism onto and moreover he calculates the inverse which is given by

We will show that is a very positive embeddings. Considering property (ii) of Definition 12, suppose that with ; we need to find with and . Since is onto, there is such that and since is an isomorphism and , we also have .

For the property (iii), we will have an existential proof of the existence of the as required, given and as in the assumptions. First let us deal with the case that . Let be the family of all non-negative functions in for which there is exactly one point with non-zero value, and on that point the value is equal to that of . Each element of is clearly definable from . We claim that some can be chosen to demonstrate (iii). Namely, since we do not have , we cannot have by positivity. Hence there is some value with , and by the continuity of the functions and , there must be some such . Then letting and for gives a function in , and we have and . Let us now deal with the general case.

Since , it follows that . Suppose first that ; hence certainly . Let be such that (so we can take if and if ). We are going to define a function for by letting , for and otherwise. Since is continuous and , we have, that for large enough , , and hence and . Since and both and are continuous, we must have that, for large enough , , and therefore for large enough we have . So, some will work to exemplify (iii).

Suppose now that . If , by continuity is eventually . If , we can choose as in the previous paragraph, so we are done by a similar argument. Otherwise we have that so by looking at and , we obtain . Also we have that is eventually . Choose such that and define by letting and for , , we see that for large enough , we have that .

If then eventually , which is a contradiction and so . If then we conclude from the definition of that, for every , and similarly , therefore , and we can use the very first argument.

5.3. Specific Very Positive Embeddings

Theorem 17. Theorem 16(1) implies Theorem 1.

Proof. We just need to show that the assumptions of Theorem 1 imply those of Theorem 16(1). The cardinal arithmetic assumption and the requirement of positivity are the same in both theorems, so we proceed to show that any as in the assumptions of Theorem 1 satisfies the requirements (ii) and (iii) of the definition of very positivity. Requirement (ii) follows easily by the surjectivity of . Finally (iii), letting , be as given by (iii) in Theorem 1 and suffices for (iii) of the very positivity.

5.4. Not Every Positive onto Embeddings Satisfies the Requirements of Theorem 1

Note that it is a consequence of the assumptions of Theorem 1 that if , then for every , .

6. Proof of the Construction Lemma

We present a proof of Lemma 14. Let and sequences , be as in the statement of Theorem 15, while is such that for . For all the definitions of invariants we use here, the value of the invariant is the same with respect to as it is with respect to , so we will not make a difference between the two. We start with a construction lemma for a certain family of linear orders, as obtained by Kojman and Shelah in [6]. Let us give their definition of the invariants of linear orders.

Definition 18. Suppose that is a linear order with the universe and is a filtration of . Then for every such that the universe of is , we define

Lemma 3.7 in [6] proves that, under the assumptions we have stated, for every closed set of limit ordinals in , there is a linear order with universe and a filtration of such that for every with we have (Lemma 3.7 in [6] also states the assumption , but this assumption is not used in the proof of the lemma, only in the proof of the final result).

The idea of our proof is to transform the Kojman-Shelah construction first into a construction of a family of Boolean algebras of size and then to use these Boolean algebras to define natural spaces of functions with appropriate invariants.

Definition 19. Suppose that is a Boolean algebra with the set of generators and is a filtration of , while is such that is generated by . We define

where means that for any element of we have if and only if .

Definition 20. Suppose that is a linear order with universe . We define a Boolean algebra as being generated by freely except for the equations

Since the equations in (13) are finitely consistent with the axioms of a Boolean algebra, it follows from the compactness theorem that the algebra is well defined. Now we will see a translation between the calculation of the invariants of the linear orders and the associated Boolean algebras.

Sublemma . Let be a linear order on , and let be the algebra associated to as per Definition 20. Let and be any filtrations of and , respectively. Then there is a club such that for every one has

and moreover, for any , this is exemplified by if and only if is exemplified by .

Proof. Let be the club of such that the universe of is , is an elementary submodel of , and is generated by and is an elementary submodel of . Suppose that .

First suppose that as exemplified by . Let , we need to prove   and  .

Case 1 ( ). Hence by the choice of , we have and , . Therefore .

Suppose that is in and satisfies . By the Disjunctive Normal Form for Boolean algebras, we can assume that for some and . It suffices to prove that for every we have . Fix an and without loss of generality assume that , as otherwise the conclusion is trivial.

Let , for . For simplicity assume that both of these sets are nonempty, as otherwise the proof is easier. Let be the -minimal element of ; hence . Let be the -maximal element of ; hence . In conclusion, . Since we have assumed that , we cannot have , equivalently . Hence we have . Similarly, since we can conclude that . Finally, if we had , then we would obtain , in contradiction with , and therefore .

Suppose now that . Therefore . On the other hand, , and hence we must have , which, taking into account , gives that , and hence , a contradiction. Hence we have . By the choice of , we have , and hence and in particular , as required. Since the roles of and in this proof were symmetric, we can prove in the same way that for any in which satisfies we also have .

Case 2 ( , so by the choice of ). We have , so , and similarly . We also have and similarly for ; hence we need to prove that . As in Case 1, it suffices to show that, for every with , we have (the equality cannot occur), and vice versa. Let us start with the forward direction. As before, from , we conclude . Also, if , then we have , contradicting that . Hence .

If , then , so , and hence , as required. So assume that . Hence and so , a contradiction. This finishes the proof of the forward direction, and the other direction follows from the symmetry of the roles of and in the proof.

Now suppose that as exemplified by . Let ; we need to prove . If , then , hence by the assumption, and hence by the definition of . The other direction follows by symmetry.

The next step is going from the invariants of Boolean algebras to the invariants of natural spaces.

Sublemma . Let be one of the algebras described in the above, and let be its filtration. Then there is a club of such that, for every with , one has that

and moreover, for any , this is exemplified by if and only if is exemplified by . Here, the invariant on the left refers to the invariant in the natural space and the invariant on the right to the invariant in the algebra . The notation is used for the sequence in .

Proof. Let be a model consisting of , , two disjoint copies of the -sequences of the simple functions on with rational coefficients, interpreted as the elements of and all the symbols of with induced interpretations induced from these models. Recall the assumption that for all , we have and notice that it implies that there is a club of such that, for every of cofinality , the model is -saturated in ; that is, it realizes all the types with countably many parameters in which are realized in . Let be any filtration of , let be a club witnessing Sublemma , and let be such that .

Suppose as exemplified by but with and the limit of satisfies . By topping up if necessary (see Definition 7), we may assume that each , and by throwing away unnecessary elements of , we may assume that every . We can then assume that for each there are pairwise disjoint and such that . Since , there has to be a with a nonempty intersection with . By applying the Disjunctive Normal Form, we can assume that for some and . Therefore there is such that . Then we have that , and hence there has to be such that . It follows that . Let . From the choice of , using Sublemma , we have that for , if and only if . We go through a case analysis like in the proof of Sublemma . If , then we have , so , a contradiction. If , then , a contradiction. Therefore, .

Claim 1. Suppose that as exemplified by some . Without loss of generality, we can assume that for some .

Proof of the Claim. First let us notice that if , then for , we have , so we can without loss of generality assume that . Similarly we can assume that , and then by applying a similar logic, we can also assume that for all and that for all . Each is a simple function with rational coefficients defined on (without loss of generality) disjoint basic clopen sets of the form where and . Let enumerate all the relevant . For each and , let be the truth value of “.” Consider the following sentence with parameters , and the elements of ; there is such that

(i)for all if , we have ;(ii)for all and , we have if and only if .

This sentence is true as exemplified by , so by the choice of , it is true in ; say as exemplified by . Let us note that in , we have or , and let us assume that , as the other case is symmetric. We claim that exemplifies that . If not, we can find with and . By the triangle inequality it follows that

We have that for every both and are equal to if and if . The possible difference is on , where the former function is equal to and the latter to . We now claim that is constant on .

Clearly, it suffices to show that each is constant on , and by the choice of , it suffices to show that, for each and each , is constant on . Let for some . If , then so is constantly 0 on . In addition, we have , so is constantly 1 on . If , then by the choice of , so , and hence is constantly 1 on . In addition, so is constantly 0 on , and the statement is proved.

Let be such that is constantly on . Say as the other case is symmetric. Hence on is , in particular we have

and hence is a function which contradicts that exemplifies that . A contradiction and hence exemplifies that as required.

Now suppose that and assume without loss of generality by Claim 1 that for some . We need to prove that as exemplified by . But indeed, if for some and , we have that , then there is in with and then clearly the sequence is not in and is below , a contradiction.

With this, putting together all the sublemmas and the Kojman-Shelah result, we finish the proof of the Construction Lemma.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The author thanks EPSRC for the Grants EP/G068720 and EP/I00498 which supported this research and the University of Wroclaw in Poland for their invitation in October 2010, when some of the preliminary results were presented. The author especially thanks Grzegorz Plebanek for the many productive conversations during the development of this paper.

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