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[EQUATION] Let [MATH] be an infinite ultraproduct of members of [MATH] . For any [MATH] and [MATH] , if [MATH] , then there is [MATH] such that [MATH] As [MATH] and [MATH] is infinite, we get [MATH] is infinite. On the other hand, if [MATH] , then by the definition one-dimensional asymptotic class, there must be some [... |
Lemma 9 Let [MATH] be a one-dimensional asymptotic class, [MATH] be a finite set of algebraic formulas of the form [MATH] [MATH] could be empty) and [MATH] any finite set of formulas of the form [MATH] (the length of [MATH] can vary and [MATH] is non-empty). Then there are [MATH] and [MATH] such that the following hold... |
For any [MATH] with [MATH] , there exists [MATH] with [MATH] and [MATH] such that for any [MATH] and [MATH] , we have [MATH] covers [MATH] and avoids [MATH] in [MATH] |
In particular, [MATH] should imply the equation ( ) and the inequality ( ), which are defined throughout the proof. Proof. By Definition , for each [MATH] there are finitely many [MATH] and [MATH] , such that for any [MATH] and [MATH] |
[EQUATION] Take [MATH] . Let [EQUATION] We claim that there is some [MATH] such that for any [MATH] and [MATH] , we have [MATH] . Otherwise, there are [MATH] and [MATH] such that the following holds: |
[MATH] [MATH] for each [MATH] [MATH] for each [MATH] Therefore, [EQUATION] By the definition of one-dimensional asymptotic class, there is some [MATH] such that |
[EQUATION] Since [MATH] , there is clearly a contradiction. Assume [MATH] . Fix any [MATH] with [MATH] , for [MATH] , define inductively the following sets: [MATH] and [MATH] |
[MATH] [MATH] Suppose [MATH] are defined. There are two cases. If [MATH] and [MATH] , define [MATH] [MATH] [MATH] If [MATH] , define |
[MATH] [MATH] Choose an element [MATH] in [MATH] such that [MATH] has the maximal cardinality among [MATH] [MATH] and [MATH] The construction stops either when [MATH] is empty, that is [MATH] covers [MATH] for any [MATH] , or when [MATH] and [MATH] for some [MATH] and [MATH] |
Let [MATH] be a maximal sequence of the construction. Define [MATH] if [MATH] and [MATH] Claim 10 There is [MATH] such that if [MATH] and [MATH] , then [MATH] is always defined. |
Proof. Suppose [MATH] and [MATH] . We first estimate the size of [MATH] in terms of [MATH] when the latter is not empty during the construction of [MATH] |
Suppose all [MATH] have no more than [MATH] -many solutions over any parameter [MATH] [MATH] can be empty). Let [MATH] and [MATH] . Then |
[MATH] Therefore, [EQUATION] By construction, [MATH] . As [MATH] is maximal among [MATH] , we get [EQUATION] Let [MATH] , then [EQUATION] |
As [MATH] , for each [MATH] we have [MATH] . And by the definition of [MATH] , for any [MATH] , if [MATH] , then [MATH] . Hence, [MATH] . On the other hand, |
[EQUATION] Hence, [EQUATION] Let [MATH] . Define [EQUATION] Then there is some [MATH] such that whenever [MATH] , we have [EQUATION] |
In particular, we have [EQUATION] Therefore, when [MATH] , we have [MATH] , and hence, [EQUATION] Consequently, [EQUATION] There is some [MATH] such that whenever [MATH] , we have [MATH] . Fix some [MATH] with [MATH] and let |
[EQUATION] be a maximal sequence. We claim that for each [MATH] , if [MATH] , then [MATH] . Otherwise, [MATH] is in the sequence. By the argument above, [MATH] By calculation, we have |
[EQUATION] Hence, [MATH] . We conclude [MATH] . Therefore, [MATH] and by induction, for each [MATH] , we have [MATH] . Now we can see that [MATH] |
Consider the set [MATH] . By inequality ( ), [EQUATION] By inequality ( ) and [MATH] , we get [EQUATION] Hence [MATH] . As [MATH] is the end term of a maximal sequence, it can only be the case that [MATH] and [MATH] |
Therefore, if [MATH] and [MATH] , then [MATH] exists and [EQUATION] where [MATH] Take any [MATH] with [MATH] , let [MATH] as defined in Claim 10 and for [MATH] , define [MATH] if [MATH] or [MATH] and [MATH] . By construction we have [MATH] covers [MATH] and avoids [MATH] in [MATH] for any [MATH] and [MATH] |
Theorem 11 Let [MATH] be a one-dimensional asymptotic class in a countable language [MATH] . Let [MATH] be an infinite ultraproduct of members among [MATH] . Then exact pseudofinite [MATH] -expansions of [MATH] exist. |
Proof. Let [MATH] be a list of all formulas in [MATH] such that [MATH] is in one variable and [MATH] is a tuple of variables. For [MATH] , let [MATH] |
Let [MATH] be a list of all formulas such that [MATH] is algebraic ( [MATH] can be empty). Let [MATH] By Lemma , there are [MATH] such that for any [MATH] with [MATH] there exists [MATH] with [MATH] such that [MATH] covers [MATH] and avoids [MATH] in [MATH] for all [MATH] and [MATH] |
For any [MATH] , let [MATH] (set [MATH] ). Define [MATH] if [MATH] ; otherwise let [MATH] Claim 12 [MATH] is an exact pseudofinite [MATH] -expansion of [MATH] |
Proof. We only need to show that [MATH] is an [MATH] -expansion of [MATH] . We verify the conditions one by one. 1. [MATH] clear. |
2. [MATH] is an [MATH] -independent subset: Suppose, towards a contradiction, that there are [MATH] which are not [MATH] -independent. We may assume that any proper subset of [MATH] is an [MATH] -independent set. Suppose for [MATH] , each [MATH] . Let [MATH] be an ordering of [MATH] . Define |
[EQUATION] Let [MATH] be the collections of all the orderings of [MATH] . Since [MATH] is finite and [MATH] , we have exactly one [MATH] . We may assume that [MATH] . Suppose [MATH] . By assumption, [MATH] . Since [MATH] satisfies the exchange property, we have [MATH] . Let [MATH] witness algebraicity (i.e., [MATH] is ... |
Let [MATH] . Since [MATH] is infinite, [MATH] . For any [MATH] , we have [MATH] , hence [MATH] avoids [MATH] . As [MATH] in [MATH] , by construction, the set [MATH] avoids [MATH] , we get |
[EQUATION] for any [MATH] . We conclude [MATH] , contradiction. 3. Density/coheir property: As [MATH] is pseudofinite, it is [MATH] -saturated. Therefore, we only need to show that for any [MATH] , if [MATH] is non-algebraic, then there is [MATH] such that [MATH] . We may assume that [MATH] |
Let [MATH] . Then [MATH] . Note that [MATH] for any [MATH] . Therefore [MATH] covers [MATH] in [MATH] for any [MATH] Suppose [MATH] for [MATH] . Let |
[EQUATION] As [MATH] is non-algebraic, [MATH] For any [MATH] , since [MATH] covers [MATH] in [MATH] and [MATH] , there is some [MATH] such that [MATH] . For [MATH] , choose [MATH] randomly. Let [MATH] . Then [MATH] and [MATH] , i.e., [MATH] |
4. Extension Property: Suppose [MATH] is finite dimensional. Let [MATH] be a base of [MATH] . Suppose [MATH] for each [MATH] . Let [MATH] . Let |
[EQUATION] and define [MATH] . By essentially the same argument as [MATH] -independence of [MATH] , we have [EQUATION] By the fact that [MATH] is pseudofinite, hence [MATH] -saturated, we only need to show that for any [MATH] , if [MATH] is non-algebraic, then there is [MATH] such that [MATH] . We may assume that [MATH... |
Consider the size of [MATH] . We have [EQUATION] where as above [MATH] [MATH] and [MATH] with [MATH] is the largest number of solutions of [MATH] over parameters for [MATH] |
Let [MATH] and [MATH] Note that there is some [MATH] such that for all [MATH] we have [MATH] . Hence [EQUATION] where [MATH] is defined as the equation ( ). By the inequality ( ), we have |
[EQUATION] Therefore, [EQUATION] for all [MATH] As [MATH] , there must be some [EQUATION] for all [MATH] . Choose [MATH] at random for [MATH] . Set [MATH] , then [MATH] and [MATH] .∎ |
Corollary 13 Let [MATH] be a one-dimensional asymptotic class in a language [MATH] and [MATH] be an infinite ultraproduct of members of [MATH] . Suppose [MATH] of [MATH] is non-trivial. Then the exact pseudofinite [MATH] -expansion [MATH] is a pseudofinite structure whose theory is supersimple of [MATH] -rank [MATH] |
Remark: Let [MATH] be an infinite ultraproduct of a one-dimensional asymptotic class. We can also make the [MATH] -expansion [MATH] satisfying |
[EQUATION] that is the pseudofinite coarse dimension of [MATH] with respect to [MATH] is zero. This is because by Lemma we know that [MATH] where [MATH] depends only on [MATH] and [MATH] . If we redefine |
[EQUATION] we see that additionally [MATH] Note that for generic element [MATH] , we have [MATH] while [MATH] for any element [MATH] . In a following project, together with other collaborators, we found this fact generalises to all definable sets. That is, the coarse dimension of a definable set equals to the coefficie... |
Groups in H-structures This section deals with definable groups in [MATH] -structures when the base theory is supersimple of [MATH] -rank one. We ask whether there are any new definable groups in [MATH] -structures. As we said before, in |
the authors have partially solved the question by showing that in stable theories the connected component of an [MATH] -definable group in an [MATH] -structure is isomorphic to some [MATH] -definable group. We record their results here. |
Fact 14 , Proposition 6.5] Let [MATH] be a group in a language [MATH] with [MATH] and assume that [MATH] is an [MATH] -saturated [MATH] -structure. Let [MATH] be finite and let [MATH] be an [MATH] -definable subgroup defined over [MATH] . Then [MATH] is [MATH] -definable over [MATH] |
Fact 15 , Proposition 6.6] Let [MATH] be a stable structure of [MATH] -rank one in a language [MATH] and let [MATH] be a subset of [MATH] such that [MATH] is an [MATH] -saturated [MATH] -structure. Let [MATH] be countable and let [MATH] be an [MATH] -definable group over [MATH] . Let [MATH] be the connected component o... |
In this section, we will show that in supersimple theories, all [MATH] -definable groups in [MATH] -structures are definably isomorphic to [MATH] -definable groups. |
We first introduce some basic notions and facts about [MATH] -structures developed in , as well as some results about groups in simple theories that we will use later. |
Let [MATH] be an [MATH] -structure. To simplify the notation, we write with subscript/superscript [MATH] for notions in [MATH] and no subscript/superscript for [MATH] . We also write [MATH] -independent to denote forking independence in [MATH] [MATH] -independent for [MATH] respectively), and [MATH] -generic for generi... |
Definition 16 Let [MATH] be a subset of an [MATH] -structure [MATH] . We say that [MATH] is [MATH] -independent if [MATH] Remark: Note that this is not the same as being [MATH] -independent in the sense of forking in [MATH] |
Definition 17 Let [MATH] be a tuple in an [MATH] -structure [MATH] and let [MATH] be [MATH] -independent. Define the [MATH] -basis of [MATH] over [MATH] , denoted by [MATH] , as the smallest tuple [MATH] in [MATH] such that [MATH] |
By , Proposition 3.9] [MATH] -bases exist and are unique up to permutation. Here is a useful observation: Lemma 18 Let [MATH] be an [MATH] -structure and [MATH] be a tuple. Suppose a subset [MATH] is [MATH] -independent and [MATH] . Then [MATH] |
Proof. Suppose not, then [MATH] . There is a finite tuple [MATH] such that [MATH] . Denote the dimension of the underlying geometric theory as [MATH] . Let [MATH] be a finite tuple such that [MATH] . Let [MATH] be a tuple containing both [MATH] and [MATH] . Then [MATH] . By the choice of [MATH] , we have |
[EQUATION] By assumption, [MATH] . Therefore, [EQUATION] We conclude that [MATH] Since [MATH] is [MATH] -independent, we also have [MATH] . By additivity of [MATH] , we have |
[EQUATION] a contradiction. Definition 19 Let [MATH] be a structure. A set [MATH] is hyper-definable over [MATH] if there is a type-definable set [MATH] for some [MATH] and a type-definable equivalence relation [MATH] on [MATH] both defined over [MATH] such that [MATH] |
Fact 20 , Lemma 2.8, Corollary 3.14, Proposition 6.2] Let [MATH] be an [MATH] -structure. 1. Let [MATH] be [MATH] -independent tuples such that [MATH] . Then [MATH] |
2. Let [MATH] be a subset of [MATH] , then [MATH] 3. Suppose [MATH] is superrosy of thorn-rank one and [MATH] is [MATH] -saturated. Let [MATH] be an [MATH] -definable group over some finite [MATH] -independent set [MATH] . Let [MATH] be a generic element of the group. Then [MATH] |
Fact 21 , Proposition 5.6] Let [MATH] be a [MATH] -saturated [MATH] -structure and [MATH] be [MATH] -closed and [MATH] . Suppose [MATH] is supersimple of SU-rank one and [MATH] . Then [MATH] if and only if none of the following holds: |
[MATH] [MATH] [MATH] Fact 22 , Lemma 4.4.8] Let [MATH] be a type-definable/hyper-definable group in a simple theory. Let [MATH] be a non-empty type-definable/hyper-definable subset of [MATH] . Suppose for independent [MATH] we have [MATH] , and put [MATH] . Then [MATH] is a type-definable/hyper-definable subgroup of [M... |
Fact 23 , Theorem 4.7.1] We fix an ambient simple theory. Let [MATH] be a partial type and [MATH] be a partial type-definable function defined on pairs of independent realizations of [MATH] , both over [MATH] such that |
1. Generic independence: for independent realizations [MATH] of [MATH] the product [MATH] realizes [MATH] and is independent from [MATH] and from [MATH] |
2. Generic associativity: for three independent realizations [MATH] of [MATH] , we have [MATH] 3. Generic surjectivity: for any independent [MATH] realizing [MATH] , there are [MATH] and [MATH] independent from [MATH] and from [MATH] , with [MATH] and [MATH] |
Then there are a hyper-definable group [MATH] and a hyper-definable bijection from [MATH] to the generic types of [MATH] , such that generically [MATH] is mapped to the group multiplication. [MATH] is unique up to definable isomorphism. |
We proceed by some lemmas, most of which are about the properties of generic elements of definable groups in [MATH] -structures. |
In the following we will assume [MATH] is an cardinal with [MATH] Lemma 24 Let [MATH] be a [MATH] -saturated [MATH] -structure such that [MATH] is supersimple of [MATH] -rank one. Let [MATH] be an [MATH] -(type-)definable group over some set [MATH] with [MATH] and [MATH] . Let [MATH] be [MATH] -independent and [MATH] -... |
Proof. By Fact 20 (3), [MATH] . That is [MATH] and [MATH] By assumption, [MATH] . Hence, [MATH] . Thus, [MATH] . Together with [MATH] , we get [MATH] . Hence, [MATH] . Again, as [MATH] , we have [MATH] . Since [MATH] , we conclude that [MATH] . Therefore, [MATH] |
As [MATH] , we have [EQUATION] Take [MATH] with [MATH] . As [MATH] , we still have [MATH] Therefore, [MATH] and [MATH] are [MATH] -independent tuples of the same [MATH] -type. By Fact 20 (1), [MATH] . As [MATH] is in the [MATH] -definable closure of [MATH] , we get [MATH] . Hence, [MATH] as we have claimed. |
The proof of [MATH] is similar. Lemma 25 Let [MATH] be a [MATH] -saturated model of [MATH] . Let [MATH] be an [MATH] -type-definable group over [MATH] with [MATH] and [MATH] . Then there are a partial [MATH] -type [MATH] and a partial [MATH] -type [MATH] over [MATH] such that: |
1. [MATH] is the set of all [MATH] -generics in [MATH] 2. For any complete [MATH] -type [MATH] over [MATH] with [MATH] , there is a complete [MATH] -type [MATH] over [MATH] such that [MATH] |
3. Let [MATH] be three realizations of [MATH] over [MATH] . Then there are [MATH] such that [MATH] realise [MATH] [MATH] and [MATH] . In addition, if [MATH] are [MATH] -independent, then [MATH] are [MATH] -independent. |
Proof. Suppose [MATH] is defined by a partial type [MATH] . Let [MATH] be the partial [MATH] -type over [MATH] which contains [MATH] and is closed under implication such that for all [MATH] [MATH] if and only if [MATH] is [MATH] -generic in [MATH] . Let [MATH] be the restriction of [MATH] in the language [MATH] |
Claim: Item 2 holds. If not, then there exists [MATH] -type [MATH] over [MATH] extending [MATH] such that [MATH] is inconsistent. By compactness, there is some [MATH] such that [MATH] . As [MATH] is closed under implication, [MATH] , hence also [MATH] , which contradicts that [MATH] |
Now we prove item 3. Write [MATH] [MATH] and [MATH] , where [MATH] [MATH] [MATH] [MATH] and [MATH] [MATH] (We remark that [MATH] can be empty.) As [MATH] and [MATH] has [MATH] -rank 1, we get [MATH] are [MATH] -independent. By the axioms of of [MATH] and [MATH] -saturation, there are [MATH] in [MATH] such that [MATH] a... |
[EQUATION] Let [MATH] be such that [EQUATION] Define [MATH] [MATH] and [MATH] Since [MATH] and [MATH] , we get [MATH] . Therefore, [MATH] . Hence, [MATH] |
We only need to show that [MATH] and [MATH] satisfy [MATH] . Let [MATH] . By item 2, there is a complete [MATH] -type [MATH] over [MATH] extending [MATH] . Let [MATH] be a realization of [MATH] . By Fact 20 (3), [MATH] . Therefore, both [MATH] and [MATH] are [MATH] -independent and |
[EQUATION] By Fact 20 (1), [MATH] . Hence [MATH] . Similarly, [MATH] and [MATH] are realizations of [MATH] In addition, if [MATH] are [MATH] -independent, then [MATH] and [MATH] are such that [MATH] [MATH] and [MATH] [MATH] . As [MATH] and [MATH] , we get |
[EQUATION] Therefore, [MATH] , whence [MATH] . Together with [MATH] we get [MATH] . The other [MATH] -independences among [MATH] are similar. Hence, [MATH] are [MATH] -independent. |
Lemma 26 Let [MATH] be two languages. Let [MATH] be an [MATH] -structure. Suppose [MATH] is [MATH] -hyper-definable and [MATH] is [MATH] -type-definable in [MATH] such that there is an [MATH] -isomorphism from [MATH] to [MATH] , then [MATH] is [MATH] -type-interpretable. |
Proof. Suppose [MATH] is [MATH] -type-definable, [MATH] where [MATH] and [MATH] are [MATH] -type-definable and [MATH] is [MATH] -type-definable which induces an isomorphism between [MATH] and [MATH] |
As [MATH] is the graph of a function from [MATH] to [MATH] , we have: [EQUATION] By compactness, there are some [MATH] such that |
[EQUATION] is an [MATH] -definable graph of a partial function. Let [MATH] be the [MATH] -definable equivalence relation given by [MATH] if and only if there is some [MATH] such that both [MATH] and [MATH] hold. We claim that |
[EQUATION] Let [MATH] . Suppose [MATH] holds. As [MATH] is an isomorphism between [MATH] and [MATH] , there is some [MATH] with [MATH] and [MATH] . Therefore, both [MATH] and [MATH] hold and so does [MATH] . On the other hand, if [MATH] holds, then there is [MATH] with [MATH] and [MATH] . Let [MATH] such that [MATH] an... |
As [MATH] is defined by [MATH] , by compactness, there is some [MATH] such that on [MATH] we have [EQUATION] Thus, [MATH] is [MATH] -definable and it agrees with [MATH] on [MATH] . We have |
[EQUATION] By compactness, there are [MATH] such that [MATH] is an equivalence relation on [MATH] . Therefore, [MATH] is [MATH] -definable. |
We first consider [MATH] -(type-)definable subgroups of [MATH] -(type-)definable groups. We generalize Fact 14 to supersimple theories. |
Theorem 27 Let [MATH] be non-trivial of [MATH] -rank one and let [MATH] be [MATH] -saturated. Suppose [MATH] is an [MATH] -(type-)definable group and [MATH] is an [MATH] -(type-)definable subgroup of [MATH] , both defined over some set [MATH] with [MATH] . Then [MATH] is [MATH] -(type-) definable ovear [MATH] |
Proof. Suppose [MATH] . Let [MATH] and [MATH] be defined as in Lemma 25 with [MATH] . Suppose [MATH] is defined by the partial [MATH] -type [MATH] . As [MATH] is closed under implication, [MATH] . Therefore, [MATH] |
By Fact 22 [MATH] . We will show that [MATH] also satisfies the conditions of Fact 22 in [MATH] Let [MATH] . Since [MATH] , we have [MATH] . Take two [MATH] -independent realizations [MATH] of [MATH] . By Lemma 25 , there are [MATH] both realising [MATH] such that [MATH] and [MATH] . Therefore, [MATH] is also generic i... |
[EQUATION] As [MATH] and group operations are [MATH] -definable, we have [EQUATION] Therefore, [MATH] , whence [MATH] . By Fact 22 we get an [MATH] -type-definable group [MATH] such that [MATH] contains all [MATH] -generics in [MATH] |
Clearly, [MATH] . Let [MATH] be an [MATH] -generic element in [MATH] . By Fact 20 (3), we have [MATH] . Since [MATH] is also [MATH] -generic in [MATH] , we get [MATH] By Lemma 25 there is an [MATH] satisfying [MATH] such that [MATH] . As [MATH] is [MATH] -generic in [MATH] [MATH] By Fact 20 (1), [MATH] . Hence, [MATH] ... |
Now we consider general [MATH] -(type-)definable groups. The following is a generalization of Fact 15 Theorem 28 Let [MATH] be supersimple of [MATH] -rank one and [MATH] be [MATH] -saturated. Let [MATH] be an [MATH] -(type-)definable group over a set [MATH] of size less than [MATH] . Then [MATH] is [MATH] -definably is... |
Proof. Suppose [MATH] is type-definable. Let [MATH] and [MATH] be defined as in Lemma 25 . In the following, we will extend [MATH] -generically and [MATH] -type-definably the group operation [MATH] of [MATH] to [MATH] on [MATH] |
Let [MATH] be the partial [MATH] -type over [MATH] such that [MATH] are [MATH] -independent and [MATH] -generic in [MATH] over [MATH] if and only if [MATH] for any [MATH] . For [MATH] , we have [MATH] by Lemma 24 . That is [MATH] for some [MATH] -definable function [MATH] over [MATH] . Let [MATH] be the [MATH] -formula... |
[EQUATION] Then we can see that [EQUATION] By compactness, there are [MATH] such that [EQUATION] Let [MATH] [MATH] be two pairs of realizations of [MATH] such that [MATH] . Note that [MATH] is an [MATH] -generic element in [MATH] . By Fact 20 (3), [MATH] . Similarly, [MATH] . Applying Fact 20 (1), we get [MATH] . There... |
[EQUATION] By compactness, there is some finite set of [MATH] formulas [MATH] such that the [MATH] -type of any pair [MATH] determines [MATH] or [MATH] for any [MATH] . Hence, there are [MATH] -formulas [MATH] such that |
[EQUATION] and for any [MATH] , we have [EQUATION] Let [MATH] be the partial [MATH] -type over [MATH] such that [MATH] if and only if [MATH] are [MATH] -independent over [MATH] . By Lemma 25 , for [MATH] , there are [MATH] realizing [MATH] such that [MATH] and [MATH] . Note that [MATH] . Hence, |
[EQUATION] for some [MATH] . As [MATH] , we also have [EQUATION] Define [MATH] . As [MATH] and [MATH] , we also have [MATH] . Note that [MATH] is defined by [MATH] if and only if [MATH] . Hence, [MATH] is an [MATH] -type-definable function from [MATH] to [MATH] and [MATH] agrees with [MATH] on [MATH] |
We now verify all the conditions of the group chunk theorem (Fact 23 ) in order to obtain an [MATH] -hyper-definable group out of the generically given group operation. |
Lemma 29 The [MATH] -type-definable function [MATH] satisfies all the conditions in Fact 23 Proof. Generic independence: Let [MATH] be [MATH] -independent realizations of [MATH] and [MATH] . Then there are [MATH] -independent and [MATH] -generic elements [MATH] over [MATH] such that [MATH] . Let [MATH] . Since [MATH] i... |
Generic associativity: Let [MATH] be [MATH] -independent realizations of [MATH] . By Lemma 25 , there are [MATH] -generic and [MATH] -independent realizations [MATH] such that |
[EQUATION] Now we have [EQUATION] Since [MATH] we get [MATH] Generic surjectivity: for any [MATH] -independent realizations [MATH] of [MATH] , there are [MATH] -independent realizations [MATH] of [MATH] such that [MATH] . Let [MATH] . Then [MATH] is [MATH] -independent from [MATH] and from [MATH] . By Lemma 24 [MATH] .... |
By Fact 23 , there are an [MATH] -hyper-definable group [MATH] over [MATH] , and an [MATH] -type-definable embedding [MATH] over [MATH] such that [MATH] contains all [MATH] -generics of [MATH] |
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