Generalized Analytic Sets

B. Zilber Kemerovo University

July, 1996

1 Introduction

We aim here to develop a model theory of analytic functions on C and, what is of specific interest, on algebraically closed fields with non-Archimedian valuation. There is a natural interest in considering fields in languages L_{F} which are just sets of names for entire functions from a family F, containing + and · with the natural interpretation in the structure. A canonical example

F = {exp, +, ·} = exp

on complex numbers is already very non-trivial. The corresponding structure C_{exp}, of course, interprets the integers. Nevertheless the model theory for such structures may be quite interesting. In particular, J.Mycielski [M] asks whether C_{exp} has an automorphism different from the complex conjugation. The question is unanswered but in more general situation the answer is ’no’. A known counterexample by Fatou and Bieberbach provides two entire functions f_{1} and f_{2} of two complex variables such that the image of C^{2} under

⟨f_{1}, f_{2}⟩ : C^{2} → C^{2}

is an open set which has a complement containing an open subset. As a matter of fact f_{1} and f_{2} are both obtained by compositions of +, · and a two-variable entire function φ. One can rather easily see that the structure C_{φ}, where the index corresponds to the language {φ, +, ·}, has only continuous automorphisms, i.e. the identity and the conjugation.

One can show that the problem of J.Mycielski is closely connected to the following problem:

Does C_{exp} have the property of quasi-minimality: any definable (with parameters) subset of the structure is either countable or co-countable?

Again for the Fatou-Bieberbach example C_{φ} the answer is negative. Thus in general the approach has a small chance to lead to an interesting model theory of entire functions.

Nevertheless Algebraic Geometry gives a hint for such a theory. Actually, a modeltheoretician can view Algebraic Geometry as Model Theory of positively definable subsets of some structures. The projective n-space P^{n}(K) over an algebraically closed field K with a structure of Zariski closed relations on it is a canonical structure of this sort. Also, as is well known, some branches of Analytic Theory play a very important role in Algebraic Geometry.

A structure we consider here has P^{1}(K) as the universum and a language of the form L̄_{F} where F is as above while the interpretation is slightly different as explained below. K is an algebraically closed field with a valuation and complete with respect to the valuation. For such fields there is a good analytic theory developed, including the case of non-Archimedean valuation. The latter case is treated in the framework of so called Rigid Analytic Geometry (see [BGR], [GP]). Before giving a precise explanation of the interpretation of L̄_{F} let us change, for convenience, the notation P^{1}(K) to K̄ assuming also K̄ = K ∩ {∞}, K̄^{n} will then denote the n−th cartesian power. Now, a symbol F ∈ L̄_{F} attached to an entire function f : K^{n} → K from F is interpreted as a subset F of K̄^{n+1} which is

(i) the closure of the graph graph(f) of the function in Zariski topology, if f is a polynomial function, and (ii) F ∩K^{n+1} = graph(f), F \K^{n+1} = {⟨x_{1}, …, x_{n+1}⟩ ∈ K̄^{n+1} : x_{1} = ∞∨…∨x_{n+1} = ∞} otherwise. This sort of subsets F of K̄^{n+1} is not analytic in general, but it satisfies the property that F has a subset F′ which is analytic in K̄^{n+1} and F \ F′ is analytic in K̄^{n+1} \ F′. (In (i) F′ = ∅ and in (ii) F′ = F \ K^{n+1}.) This can be naturally generalized by using the inductive definition: A set S ⊆ K̄^{n+1} is generalized analytic in K̄^{n+1} if S is analytic in K̄^{n+1} or there is S′ ⊆ S which is generalized analytic in K̄^{n+1} and S \ S′ is analytic in K̄^{n+1} \ S′. The main result of the paper is Theorem A. Class of generalized analytic subsets of K̄^{n} for all n is closed under cartesian products and the positive logical operations: ∨, ∧, ∃ and ∀. Moreover, the closure of a subclass of generalized analytic sets containing all singletons under cartesian products and ∨, ∧, ∃ is closed under ∀. This is true both for the case of C and for the non-Archimede- an case.

The second main result is important in the case of non-Archimedean fields while for C it follows immediately from the fact that generalized analytic subsets are closed in the real topology.

Theorem B. For any F the structure (K̄, L̄_{F}) is atomic compact in the sense of J.Mycielski [W], or in other words, for any n the topology on K̄^{n} whose base of closed sets consists of generalized analytic subsets is compact.

Notice that K̄^{n} is not compact in any non-Archimedean valuation topology but is ’compact’ in Grothendieck topology [BGR] which deals with admissible open sets and coverings only.

Theorem B is important in developing the approach of [Z], since it shows the existence of a specializations π such that the classical analytic functions are at least continuous on their domains in the π-topology. Yet we were not able to show that the functions are analytic in the sense of [Z]. Theorem A has an interesting geometric consequence showing that the analytic theory is not very far from the algebraic theory of fields. We introduce the notion of analytic closure ancl which generalizes the notion of algebraic notion in fields. In particular, if f(x, ȳ) is an analytic function definable in (K̄, L̄_{F}) then all the solutions of the equation f(x, ā) = 0 are in the ancl(ā). We use also a condition of ’selfsufficiency’ of the language L̄_{F} which is satisfied for a suitable extension of any given language.

Theorem C. The analytic closure in (K̄, L̄_{F}) induces a pregeometry on K̄, i.e. it satisfies

(i) A ⊆ ancl(A), (ii) ancl(A) = ancl(ancl(A)), (iii) b ∈ ancl(A ∪ {c}) \ ancl(A) implies c ∈ ancl(A ∪ {b}).

The material of the paper is dependent on the techniques of Rigid Analytic Geometry (which generalizes the classical analytic theory in the context we are interested in) and the reader who wants to follow the proofs has to consult [BGR] which we refer to. But for those who are interested in the case of the complex numbers only there is an easier way: ignoring the word ’admissible ’ one comes to the simpliest case of the Grothendieck topology - the usual real topology on C. Then some of the statements reduce to triviality while others transform to classical facts and techniques of complex analytic theory.

2 Generalized analytic sets. Basic properties

In what follows we are heavily dependent on the terminology and facts of Rigid Analytic Geometry as represented in [BGR]. The references point to this book if the source is not mentioned explicitly. We start by giving a very brief review of main notions of Rigid Analytic Geometry.

The basic objects are affinoid varieties equipped with algebras of analytic functions, which are analogues of ’nice’ domains in C^{n}. An analytic variety is constructed by pasting of affinoid varieties in a certain way. In the article we always deal with analytic varieties pasted from countably many affinoid varieties and we assume this everywhere. The notions of an admissible open set and an admissible open covering are given by a definition starting with an assumption that an embedding of an affinoid variety into an analytic variety which agrees on algebras of analytic functions is admissible open and that any finite covering of an affinoid variety by affinoid varieties is admissible . The definition in [BGR], section 9.1.1 assumes (we use the enumeration of the book):

(b)

(i) if U and V are admissible open sets, then U ∩ V is; (ii) for U an admissible open set {U} is an admissible covering of U; (iii) if U is an admissible open set, {U_{i} : i ∈ I} an admissible covering of U and {V_{ij} : j ∈ J_{i}} admissible coverings of U_{i} for all i ∈ I, then {V_{ij} : j ∈ J_{i}, i ∈ I} is an admissible covering of U; (iv) if U and V are admissible open sets with V ⊆ U and {U_{i} : i ∈ I} an admissible covering of U, then {V ∩ U_{i} : i ∈ I is admissible covering of V.

Also the following properties are satisfied by any analytic variety X (9.1.2 and 9.3.1/4): (G_{0}) ∅ and X are admissible open subsets of X. (G_{1}) Let U ⊆ X be an admissible open set and let V ⊆ U. Assume that there exists an admissible covering {U_{i} : i ∈ I} of U such that V ∩ U_{i} is admissible open in X for all i ∈ I. Then V is admissible open in X. (G_{2}) Let {U_{i} : i ∈ I} be a covering of an admissible open set U ⊆ X such that U_{i} is admissible open in X for all i ∈ I. Assume further that {U_{i} : i ∈ I} has a refinement which is admissible . Then {U_{i} : i ∈ I} is an admissible covering of U.

Now we are ready to introduce the main notion of the present paper.

Definition Let U be an admissible open subset of an analytic variety X and S ⊆ U. The analytic rank of S in U ark_{U}(S) is a natural mumber or ∞ and is defined by induction as follows: ark_{U}(S) = 0 iff S = ∅; ark_{U}(S) ≤ n + 1 iff there is S′ ⊆ S with ark_{U}(S′) ≤ n, U \ S′ is admissible open in X and S^{0} = S \ S′ is an analytic subset of U \ S′; ark_{U}(S) ≤ ∞ for all S; ark_{U}(S) = min_{n}(ark_{U}(S) ≤ n) where n ranges in N ∪ {∞}. A subset S ⊆ U is said to be g.analytic in U (generalized analytic) if ark_{U}(S) = n for some natural n.

We omit the subscript U when it doesn’t lead to ambiguity. The notation X, S^{0} and S′ will be used throughout the paper as in the definition.

Actually the definition does not need the assumption that U \S′ is admissible open in X as the following shows (remember that by definition X has an admissible covering by affinoid varieties)

Lemma 2.1 If S is a g.analytic subset of an admissible open U such that there is an admissible covering of U by affinoid varieties then U \S is admissible open . Moreover U \ S has an admissible covering by affinoid varieties.

Proof Let {X_{i} : i ∈ I} be an admissible open covering of U by affinoid varieties. By the properties G_{1} and G_{2} it is enough to prove the statement for S ∩ X_{i} as a subset of X_{i} for any i ∈ I. So we assume U = X is an affinoid variety. First we consider the case S is analytic in X. We may assume S = {x ∈ X : f_{1}(x) = 0& … &f_{k}(x) = 0} for some functions f_{1}, …, f_{k} analytic on X. Then 9.1.4/5 gives the required. By induction we may assume now that U = X \ S′ is admissible open and has an admissible covering by affinoid varieties. Since S^{0} is an analytic subset of U, by the above proved V = U \ S^{0} = X \ S is admissible open in U. Thus V is admissible open in X. Let {X_{i} : i ∈ I} be an admissible covering of U by affinoid varieties. Then {X_{i} ∩ V : i ∈ I} is an admissible covering of V by (b)(iv) of the definition. Also by the above proved each X_{i} ∩ V has an admissible covering {X_{ij} : j ∈ J_{i}} by affinoid varieties. By (b)(iii) {X_{ij} : i ∈ I, j ∈ J_{i}} is an admissible covering of V by affinoid varieties.□

Remark The admissible coverings in Lemma can be assumed countable because of our assumtion above and the fact that 9.1.4/5 provides a countable covering.

We will need the following technical Lemma specific for the non-Archimedean case.

Lemma 2.2 Let S_{1}, S_{2} and S_{1} ∩ S_{2} all be g.analytic in an analytic variety X. Then the two sets V_{1} = X \S_{1} and V_{2} = X \S_{2} form an admissible covering of X \(S_{1} ∩S_{2}).

Proof Let {X_{i} : i ∈ I} be an admissible covering of X by affinoid varieties. Clearly, for all i ∈ I X_{i} = (X_{i} ∩ V_{1}) ∪ (X_{i} ∩ V_{2}), since X_{i} ∩ S_{1} ∩ S_{2} = ∅. If X_{i} ∩ V_{1}, X_{i} ∩ V_{2} is an admissible covering of X_{i} for all i ∈ I then {X_{i} ∩ V_{k} : i ∈ I, k = 1, 2} is an admissible covering of X \(S_{1} ∩S_{2}), by (b)(iii). This covering refines V_{1}, V_{2}. Thus the latter is an admissible covering. So we come to consider each X_{i} and thus reduce the proof to the case: X is an affinoid variety and S_{1} ∩ S_{2} = ∅. We procede now by induction on ark(S_{1})+ark(S_{2}). In the case S_{1} and S_{2} are analytic the statement is proved in 9.1.4/5. So, we assume ark(S_{1}) > 1. By induction hypothesis X \ S′_{1} and X \ S_{2} form an admissible covering of X. Let {Y_{j} : j ∈ J} and {Z_{l} : l ∈ L} be admissible coverings by affinoid varieties of X \ S′_{1} and X \ S_{2} respectively. Then, again by (b)(iii), the covering {Y_{j}, Z_{l} : j ∈ J, l ∈ L} of X is admissible . By definition an admissible covering of an affinoid variety must contain a finite subcovering (which is always admissible by definition). Let

{Y_{1}, …, Y_{k}, Z_{1}, …, Z_{l}}

be the subcovering. For all j ∈ J Y_{j} ⊆ X \ S′_{1}, hence Y_{j} ∩ S_{1} = Y_{j} ∩ S_{1}^{0} is analytic in Y_{j} and by induction hypothesis Y_{j} \ (Y_{j} ∩ S_{1}), Y_{j} \ (Y_{j} ∩ S_{2}) is an admissible covering of Y_{j}. By (b)(iii) we get the admissibility of the covering

{Y_{1} \ (Y_{1} ∩ S_{1}), …, Y_{k} \ (Y_{k} ∩ S_{1}), Y_{1} \ (Y_{1} ∩ S_{2}), …, Y_{k} \ (Y_{k} ∩ S_{2}), Z_{1}, …, Z_{l}}.

But the latter is a refinement of V_{1}, V_{2}, since

Y_{1} \ (Y_{1} ∩ S_{1}), …, Y_{k} \ (Y_{k} ∩ S_{1}) ⊆ V_{1},
Y_{1} \ (Y_{1} ∩ S_{2}), …, Y_{k} \ (Y_{k} ∩ S_{2}), Z_{1}, …, Z_{l} ⊆ V_{2}.

Hence V_{1}, V_{2} is an admissible covering of X. □

Lemma 2.3 Let X be an analytic variety and {X_{i} : i ∈ I} be an admissible finite covering of X. Then S ⊆ X is g.analytic in X iff S ∩ X_{i} is g.analytic in X_{i} for any i ∈ I. Moreover,

ark_{X}(S) = max_{i∈I}ark_{X_{i}}(S ∩ X_{i}).

Proof Immediately from the definition by induction on ark(S).□

Proposition 1 Let X be an analytic variety

(i) If ark_{X}(S_{1}) ≤ n and ark_{X\S_{1}}(S_{2} \ S_{1}) ≤ m then ark_{X}(S_{1} ∪ S_{2}) ≤ n + m. (ii) If ark_{X}(S_{1}) ≤ n_{1} and ark_{X}(S_{2}) ≤ n_{2} then ark_{X}(S_{1} ∪ S_{2}) ≤ max(n_{1}, n_{2}) and ark_{X}(S_{1} ∩ S_{2}) ≤ n_{1} · n_{2}.

Proof of (i). Suppose the assumption holds. By induction on n we assume that ark_{X}(S′_{1} ∪ S_{2}) ≤ n + m − 1 and by definition S_{1}^{0} \ S_{2} is analytic in X \ (S′_{1} ∪ S_{2}). Hence by definition ark_{X}(S_{1} ∪ S_{2}) ≤ n + m. Proof of (ii). Suppose the assumption holds. Then S_{1}^{0} ∪ S_{2}^{0} \ (S′_{1} ∪ S′_{2}) is analytic in X \ (S′_{1} ∪ S′_{2}). By induction hypothesis ark_{X}(S′_{1} ∪ S′_{2}) ≤ max(n_{1} − 1, n_{2} − 1). Now ark_{X}(S_{1} ∪ S_{2}) ≤ max(n_{1}, n_{2}) follows immediately from (i). To analyze the intersection we use the induction on n_{1} · n_{2}. Notice that

S_{1} ∩ S_{2} = (S_{1}^{0} ∩ S_{2}) ∪ (S′_{1} ∩ S_{2}).

Claim.

ark_{X\(S′_{1}∩S_{2})}(S_{1}^{0} ∩ S_{2}) ≤ n_{2}.

By induction hypothesis S′_{1} ∩ S_{2} is g.analytic and by Lemma 2.2 X \ (S′_{1} ∩ S_{2}) has an admissible covering X \ S′_{1}, X \ S_{2}. Both S_{1}^{0} ∩ (X \ S′_{1}) and S_{2} ∩ (X \ S′_{1}) are g.analytic subsets of X \ S′_{1} of ranks not greater than 1 and n_{2} correspondingly, by definition. Hence

ark_{X\S′_{1}}(S_{1}^{0} ∩ S_{2}) ≤ n_{2}.

From the other hand (X \ S_{2}) ∩ (S_{1}^{0} ∩ S_{2}) = ∅. By Lemma 2.3 the claim follows. Now use the induction hypothesis to count ark(S′_{1} ∩ S_{2}) ≤ (n_{1} − 1) · n_{2} and by (i) of Proposition we have the required. □

Proposition 2 If S_{1} ⊆ K̄^{m_{1}}, S_{2} ⊆ K̄^{m_{2}} are g.analytic then S_{1} × S_{2} is g.analytic in K̄^{m_{1}+m_{2}}.

Proof Notice that

S_{1} × S_{2} = S_{1} × K̄^{m_{2}} ∩ K̄^{m_{1}} × S_{2}.

So we need to prove that both members of the union on the right are g.analytic. By induction on ark(S_{1}) we assume S′_{1} × K̄^{m_{2}} is g.analytic . By definition S_{1}^{0} × K̄^{m_{2}} is analytic in K̄^{m_{1}+m_{2}} \ S′_{1} × K̄^{m_{2}}. Hence S_{1} × K̄^{m_{2}} is g.analytic in K̄^{m_{1}+m_{2}}. Symmetrically the second member is too.□

Proposition 3 If S ⊆ K̄^{m+1}, ark(S) ≤ n and pr : K̄^{m+1} → K̄^{m} the projection (forgetting the first coordinate) then ark(pr(S)) ≤ n.

Proof This is based on Proper Mapping Theorem of Kiehl (see 9.6.3/3 and [K]) which states that under a proper morphism σ : M → N of analytic varieties the image σ(S) of an analytic subset S ⊆ M is analytic in N. From 9.6.2 we know that pr is proper (see the discussion after the proof of 9.6.2/4) and also that for any admissible open U ⊆ K̄^{m} the restriction pr_{U} : pr^{−1}(U) → U is proper (9.6.2/3). Let S = S^{0} ∪ S′, ark(S′) ≤ n − 1. Notice that

pr(S) = pr([K̄ × (K̄^{m} \ pr(S′))] ∩ S^{0}) ˙∪pr(S′).

By induction ark(pr(S′)) ≤ n − 1 and by Lemma 2.1 K̄^{m} \ pr(S′) is admissible open and thus the Proper Mapping Theorem is applicable to

pr : K̄ × (K̄^{m} \ pr(S′)) → K̄^{m} \ pr(S′)

by the remark above. Since S^{0} is analytic in K̄^{m+1} \ S′ the subset [K̄×(K̄^{m}\pr(S′))]∩S^{0} is analytic in the smaller admissible open set K̄×(K̄^{m}\pr(S′)). Hence pr([K̄ × (K̄^{m} \ pr(S′))] ∩ S^{0}) is analytic in K̄^{m} \ pr(S′). □

Lemma 2.4 For U as in Lemma 2.1 any finite covering of U by sets of the form U \ S_{j} (j = 1, …, k) with S_{j} g.analytic in U is admissible .

Proof Again we assume U = X is affinoid . Let {V_{j} : j = 1, …, k} be a finite covering of X, V_{j} = X \ S_{j}. We use induction on

a = Σ_{j}ark(S_{j}).

If ark(S_{j}) ≤ 1 for all j then the covering is admissible by 9.1.4/5. Thus we can start by assuming ark(S_{1}) > 1. Denote U = X \ S′_{1}. Let {X_{i} : i ∈ I} be an admissible covering of U by affinoid varieties. Then {X_{i} ∩ V_{j} : j = 1, …, k} is a finite covering of X_{i} for each i ∈ I and is admissible by induction hypothesis since

X_{i} ∩ V_{1} = X_{i} \ (X_{i} ∩ S_{1}) = X_{i} \ (X_{i} ∩ S_{1}^{0}),

ark(X_{i} ∩ S_{1}^{0}) ≤ 1 and ark(X_{i} ∩ S_{j}) ≤ ark(S_{j}) for all j. Hence by (b)(iii) the covering

{X_{i} ∩ V_{j} : i ∈ I, j = 1, …, k}

of U is admissible . Then {U ∩ V_{j} : j = 1, …, k} is an admissible covering of U, by G_{2}, since it has an admissible refinement. By induction hypothesis {U, V_{2}, …, V_{k}} is an admissible covering of X. Hence {U ∩ V_{1}, …, U ∩ V_{k}, V_{2}, …, V_{k}} is an admissible covering of X, which is a refinement of {V_{j} : j = 1, …, k}, hence the latter is admissible . □

Lemma 2.5 Let S_{1}, S_{2} be g.analytic subsets of X, and S_{2} an analytic subset of X \ S_{1}. Then S_{2} is analytic in X.

Proof By Lemma 2.4 {X \ S_{1}, X \ S_{2}} is an admissible covering of X. By 2.1 there is an admissible covering {X_{i} : i ∈ I} of X \ S_{1} by affinoid varieties. Since S_{2} is analytic in the set by further refinement we may assume S_{2} ∩ X_{i} is a zero-set of a system of equations for each i ∈ I. For any admissible covering {X_{j} : j ∈ J} of X \S_{2} by affinoid varieties we have S_{2} ∩ X_{j} is a zero-set of a system of equations since it is empty for all j ∈ J. Thus the admissible covering {X_{j} : j ∈ I ∪ J} of X witnesses that S_{2} is analytic in X. □

3 Dimension and irreducibility

Recall that for an affinoid variety X dimension is defined as Krull dimension of the algebra A of analytic functions on X. It is also important to have in mind that a rational subdomain X′ of an irreducible X (i.e. such that algebra A is an integral domain) is of the same dimension as X (see 7.3.2, in particular 7.3.2/8(ii)). Definition For a g.analytic S the dimension of S dm(S) is defined by induction on ark(S) as follows: dm(∅) = −1; dm(S) = max{dim(S^{0}), dm(S′)}, where dim(S^{0}) is the dimension of S^{0} as an analytic subset of X \ S′. Thus dm is a natural generalization of the analytic dimension. We will also need some theory of irreducibility.

Definition Let M be an affinoid variety and X ⊆ M admissible open . We call X irreducible in M if either X is a rational subdomains of M which is algebraically irreducible, i.e. the algebra A of analytic functions on X is without zero divisors or X is connected in Grothendieck topology and there is an admissible covering {X_{j} : j ∈ J} of X by rational subdomains of M which are algebraically irreducible. Notice that by the definition M itself is irreducible iff the algebra of M is an integral domain.

Observe the following

Lemma 3.1 Assume M is an affinoid variety

(i) If Y is a rational algebraically irreducible subdomain of M and Y′ a rational subdomain of Y then Y′ is irreducible in M; (ii) If X ⊆ M is admissible open irreducible in M and S a proper g.analytic subset of X then X \ S is irreducible in M; (iii) If S ⊆ X, S is g.analytic in X, X admissible open irreducible and dm(S) = dim(M), then S = X.

Proof of (i). If algebra A of Y is without zero-divisors then algebra A′ of Y′ is of the form A′ = A⟨f, g^{−1}⟩ for f = (f_{1}, … f_{n}), g elements of A. It is shown in 6.1.4 that A⟨f, g^{−1}⟩ is a completion of A[g^{−1}] under appropriate semi-norm. Since A[g^{−1}] is an integral domain, A⟨f, g^{−1}⟩ is too. Also Y′ is a rational subdomains of M by transitivity (7.2.4/2) and this finishes the proof. Proof of (ii). First we consider the case when S is a zero-set of an analytic function in X and X is a rational algebraically irreducible subdomain of M. 9.1.4/5 states that X \ S is a union of assending chain of rational subdomains of X and (i) finishes the proof. Now assume S = S_{1} ∩ … S_{k} with X as above and each S_{m} is a zero-set of an analytic function on X. Then

X \ S = (X \ S_{1}) ∪ … ∪ (X \ S_{k})

and we get an admissible covering of X \ S taking the union of admissible covers of X \ S_{m}’s. Since X is irreducible X ≠ S_{m_{1}} ∪ S_{m_{2}} and hence (X \ S_{m_{1}}) ∩ (X \ S_{m_{2}}) ≠ ∅. This implies the connectedness of X \ S and finally the irreducibility of the set. Now suppose S is an analytic subset of an open irreducible X ⊆ M. Let {X_{j} : j ∈ J} be the covering as in definition. We may assume that S ∩ X_{j} is a zero set of finitely many analytic functions on X_{j}. Then by above proved X_{j} \ S is irreducible. Thus X_{j} \ S is connected and has an admissible covering {Y_{ji} : i ∈ I_{j}} for each j ∈ J such that Y_{ji} are rational subdomains of M. Now {Y_{ji} : i ∈ I_{j}, j ∈ J} is an admissible covering of X \ S. The latter is connected since otherwise

⋃{X_{j} \ S : j ∈ J} = V ˙∪U

for two disjoint open subsets U and V forming an admissible covering. Then by the irreducibility of X_{j} \S either (X_{j} \S)∩V = ∅ or (X_{j} \S)∩U = ∅ for each j ∈ J. Since X is connected there are j_{1} and j_{2} in J such that (X_{j_{1}} \S)∩V ≠ ∅, (X_{j_{2}} \S)∩U ≠ ∅ and X_{j_{1}} ∩ X_{j_{2}} ≠ ∅. The latter intersection then must be a subset of a proper analytic subset S ∩ X_{j_{1}} but this is impossible since X_{j_{1}} ∩ X_{j_{2}} is a rational subdomain of X_{j_{1}} by 7.2.3/7.

Finally consider the general case. By induction on ark(S) we assume that X′ = X \S′ is an irreducible analytic variety. Then by the above proved X′ \ S^{0} is irreducible open too. But the latter is equal to X \ S. Proof of (iii). First consider the case when S is analytic in X. By definition there is an admissible covering {X_{j} : j ∈ J} of X which witnesses the irreducibility of X in M. We may assume that S ∩ X_{j} is given by an ideal α_{j} of the algebra A_{j} of X_{j}. For some j

dim(S ∩ X_{j}) = dim(X_{j})

and hence the Krull dimension of A_{j} is equal to that of A_{j}/α_{j}. Since A_{j} is prime, α_{j} = 0 and hence X_{j} ⊆ S. If X_{i}∩X_{j} ≠ ∅ then the intersection is a rational subdomain of both X_{j_{1}} and X_{j_{2}} and has dimension equal to that of X_{j_{1}} and X_{j_{2}}. Hence X_{i} ⊆ S too. By connectedness it follows S = X. Consider now the general case. We may assume by induction on ark(S) that dm(S′) < dim(X) and thus S^{0} is an analytic subset of X \ S′ of equal dimension. Now the required follows from (ii) and the above proved.□

Lemma 3.1 has the following useful

Corollary 1 If S is a proper g.analytic subset of K^{n} then dm(S) < n. In particular, for n = 1 such an S must be countable.

Proof Remember that K^{n} has an admissible countable covering by ascending algebraically irreducible affinoid varieties and apply (iv) of Lemma. The countability of S for n = 1 follows from the fact that the number of zeroes of an analytic function on an affinoid variety is finite.□

4 Main Theorems

Proposition 4 [Theorem B] Let {S_{i} : i ∈ I} be a family of g.analytic subsets of K̄^{m} such that

⋂_{i∈I} S_{i} = ∅.

Then there is finite I_{0} ⊆ I with

⋂_{i∈I_{0}} S_{i} = ∅.

Proof Since K̄^{m} has a finite admissible covering by affinoid varieties we may assume that {S_{i} : i ∈ I} is a family of g.analytic subsets of an affinoid variety X. We consider the lexicographic ordering of the set of pairs

{⟨dm(S_{i}), ark(S_{i})⟩ : i ∈ I}

and proceed by induction on

⟨d, a⟩ = min{⟨dm(S_{i}), ark(S_{i})⟩ : i ∈ I}.

Let S_{i} = S be the member of the family where the minimum is reached. For ⟨d, a⟩ ≤ ⟨0, 1⟩ the statement is evident since then S is finite or empty. Consider now the general situation. If there is no finite I_{1} ⊆ I such that

S′ ∩ ⋂_{i∈I_{1}} S_{i} = ∅

then the new family obtained by replacing S with S′ and leaving all the rest untouched is of smaller ⟨d, a⟩ and satisfies the assumptions of the theorem while the statement of the theorem is false. This contradicts the induction hypothesis. So we may assume there is finite I_{1} as above and thus assuming for any j ∈ I

S_{j} ⊆ ⋂_{i∈I_{1}} S_{i}

we have

S ∩ S_{j} = S^{0} ∩ S_{j}.

We may again assume S_{i} ⊆ S and thus S_{i} ⊆ S^{0} for any i ∈ I. Remember that S^{0} is an analytic subset of U = X \ S′ and by minimality dim(S^{0}) = dm(S_{i}) for all i ∈ I. We also assume 1 ∈ I and S_{i} ⊆ S_{1} for all i ∈ I. Claim For any admissible covering {X_{j} : j ∈ J} of U there is a finite J_{0} ⊆ J such that

S_{1} ⊆ {X_{j} : j ∈ J_{0}}.

Proof {U, (X \ S_{1})} is a covering of X by the assumptions. It is admissible by Lemma 2.4. Hence {X_{j} : j ∈ J} ∪ {U} is an admissible covering of X. By qcompactness of X there is a finite subcovering {X_{j} : j ∈ J_{0}} ∪ {U} and the J_{0} satisfies the required.

By the Claim and Lemma 2.1 we may assume that all X_{j} are affinoid varieties and for some j ∈ J_{0} the family {S_{i} ∩ X_{j} : i ∈ I} has empty intersection while every finite subfamily has a common point. To simplify the notation we also put S_{i} = S_{i} ∩ X_{j} and S^{0} = S^{0} ∩ X_{j}. Moreover, we assume S^{0} is a zero-set of an ideal of the algebra of X_{j} and thus S^{0} is an affinoid variety itself. Let M_{1}, … M_{k} be the irreducible components of S^{0}, (i.e. the analytic subsets corresponding to the decomposition of the algebra of S^{0} to prime ideals). Again we may restrict the family {S_{i} : i ∈ I} to the components and thus come to considering the cases S_{i} ⊆ M_{m} for some m ≤ k and all i ∈ I. If dim(M_{m}) < a we are done by the induction hypothesis. It remains to consider the case dm(S_{i}) = dim(M_{m}). Applying Lemma 3.1(iii) one sees S_{i} = M_{m} which finishes the proof of the proposition.

Proposition 5 For a g.analitic S(x, y) ⊆ K̄^{n+1} the set defined by ∀yS(x, y) is g.analytic. Moreover, the set can be obtained from g.analytic set defined by S(x, y) by finetely many operations of the form ∃y(y = a&S(x, y) (substitution of a constant a) and finite intersections.

Proof For a finite A ⊆ K̄ denote

T_{A} = {x ∈ K̄^{n} : ∀a ∈ AS(x, a)}.

This is evidently a g.analytic set of the form assumed in the formulation of Proposition. Suppose b ∈ T_{A} and ¬∀yS(b, y). Consider by induction on k ≤ ark(T_{A}) sets

T_{A}^{(k+1)} = (T_{A}^{(k)})′, where T_{A}^{(0)} = T_{A}.

By the theory above the sets of the form U_{k} = K̄^{n} \ T_{A}^{(k)} for 1 < k ≤ ark(T_{A}) are admissible open and there is an admissible covering of U_{k} by affinoid varieties {X_{k,j} : j ∈ J_{k}}. Also T_{A}^{(k−1)} ∩ U_{k} are analytic in U_{k} and we can choose X_{k,j} so that the T_{A}^{(k−1)} ∩ X_{k,j} is given as a zero-set of an ideal of analytic functions on X_{k,j}. Then by algebraic means T_{A}^{(k−1)} ∩ X_{k,j} can be decomposed into a finite union of irreducible analytic subvarieties. Call these subvarieties for all k and j, depending also on the covering {X_{k,j} : j ∈ J}, components. Now, the components which are not subsets of T = {x ∈ K̄^{n} : ∀yS(x, y)} call superfluous. Define d to be the minimal number such that there is finite A and an admissible covers {X_{k,j} : j ∈ J} for which all the superfluous components are of dimension less than d. We are going to see that d = 0 hence for the choice of A which realizes the minimum all components are not supefluous. This immediately implies T ∩ X_{k,j} = T_{A} ∩ X_{k,j} for all k, j and thus T = T_{A}. Suppose d > 0. Then in each superfluous component choose a b which does not belong to T. The countable set consisting of all chosen b denote B. Since for any b ∈ B S(b, K̄) ≠ K̄, the set is countable by Corollary 1. Hence

K̄ \ ⋃_{b∈B} S(b, K̄) ≠ ∅.

Let a be an element from the set. Then T (A ∪ {a}) does not intersect B and thus it intersects the superfluous components of T_{A} on lower dimensions. This contradicts the minimality of d. □

Summing up Propositions 1, 3 and 5 one gets Theorem A. Proposition 4 is Theorem B.

5 Analytic closure

In this section F is the set of names for g.analytic relations on K̄ and (K̄, L̄_{F}) is the structure with the g.analytic relations corresponding to F being the basic ones.

Definition An n-ary relation T on K̄ (equivalently, subset of K̄^{n}) is called Cdefinable-closed in (K̄, L̄_{F}) (C ⊆ K̄) if T is obtained from the basic relations of L̄_{F} by positive logical operations and substitution of constants for the names of elements from C. T is called C-definable-open in (K̄, L̄_{F}) if T is the complement to a C-definable-closed in (K̄, L̄_{F}). T is called C-definable-analytic in (K̄, L̄_{F}) if there are S ⊆ K̄^{n} C-definable-closed and U ⊆ K̄^{n} C-definable-open in (K̄, L̄_{F}) such that T = S ∩U and T is analytic in U.

Definition For C ⊆ K̄, a ∈ K̄ in the structure (K̄, L̄_{F}) it is said that a is in analytic closure of C or a ∈ ancl(C) iff there is an (n + m + 1)−ary ∅-definable-analytic relation S and a string c̄ from C length n such that the number of realizations x of ∃ȳS(x, ȳ, c̄) is countable (length(ȳ)=m) and ⟨a, b̄, c̄⟩ ∈ S for some b̄ ∈ K̄^{m}.

Below in this section we consider for simplicity the case K = C but the final results are true for the non-Archimedean case too.

We need an assumption for the language L̄_{F}. First we have to introduce a notation. Recall that dim_{s}S for an analytic set S and its point s denotes the local dimension of the set in the point s. Given the projection pr_{k} : K̄^{n} → K̄^{n−k}, forgetting the first k coordinates, let

m_{k}(S) = min_{s∈S}{dim_{s}(pr_{k}^{−1}(pr_{k}(s)) ∩ S)}

and define the set

lrg_{k}(S) = {s ∈ S : dim_{s}(pr_{k}^{−1}(pr_{k}(s)) ∩ S) > m_{k}(S)}

( the set of the points in S with large local dimension in the fiber).

Lemma 5.1 For any S ⊆ U analytic and smooth in an open U ⊆ K̄^{n} and any k < n the set lrg_{k}(S) is analytic in U.

Proof is given in Appendix.

Recall that under notation above the subset sng(S) ⊆ S of all the singular points of S is analytic in U of dimension less than that of S. See e.g. [C],I,s.5.2.

Definition The language L̄_{F} is said to be sufficient if for any ∅-definable-analytic set S ⊆ K̄^{n} and k < n the sets sng(S) and lrg_{k}(S) are ∅-definable-analytic .

Lemma 5.2 For any sufficient L̄_{F} if there are (n + m + 1)−ary ∅-definable-analytic S and ⟨c̄, d̄⟩ ∈ S (length(c̄) = n, length(d̄) = m + 1) then there is (n + m + 1)-ary ∅-definable-analytic S^{∗} ⊆ S such that

(i) ⟨c̄, d̄⟩ ∈ S^{∗}, (ii) S^{∗} is smooth, (iii) lrg_{n}(S^{∗}) = ∅.

Proof To satisfy (i) and (ii) take for S_{1} an ∅-definable-analytic subset of S containing ⟨c̄, d̄⟩ of the smallest possible dimension. Then sng(S_{1}) does not contain ⟨c̄, d̄⟩ and is analytic in some ∅-definable-open U_{1} (the same as for S_{1}). Put now S_{2} = S_{1} \sng(S_{1}) and U_{2} = U_{1} \ sng(S_{1}). Evidently S_{2} is analytic in the open set U_{2} and both are ∅-definable. To get (iii) consider two cases. Case 1. If ⟨c̄, d̄⟩ ∉ lrg_{n}(S_{2}) then analogously to the construction of S_{2} put S^{∗} = S_{2} \ lrg_{n}(S_{2}) and we get the required. Case 2. If ⟨c̄, d̄⟩ ∈ lrg_{n}(S_{2}) then put S_{3} = lrg_{n}(S_{2}) and by the definition of lrg one has m_{n}(S_{2}) < m_{n}(S_{3}). If the same case holds for S_{3} we repeat the step on. Since m_{n}(S_{l}) < dim(S) for any analytic subset S_{l} of S, we necessarily come to case 1 and get the required.□

Lemma 5.3 Let

S = ⋃_{i∈I} S_{i},

S and S_{i} be analytic in an open U, S_{i} ∩ S_{j} = ∅ for i ≠ j, dim S = d, dim S_{i} = r for all i ∈ I. Then d = r iff I is countable.

Proof If d = r then each S_{i} is a union of analytic components of S and since the number of components is countable, I is countable. Suppose r < d and I is countable (and is just N) towards the contradiction. We may assume that S is irreducible and, moreover, S is just a ball in C^{d}. Then one can find a descending sequence of closed balls B_{i} ⊆ S such that B_{i} ∩ S_{j} = ∅ for all j ≤ i. their common point is in S and not in S_{i} for any i ∈ I. The contradiction.□

We assume below that we work in a structure with a sufficient language.

Proposition 6 In the definition of the analytic closure S can be assumed such that for any c̄′ ∃x∃ȳS(x, ȳ, c̄′) implies that the number of realizations x of ∃ȳS(x, ȳ, c̄′) is countable.

Proof By Lemma 5.2 we can assume that for a natural r

dim S(K̄, K̄^{m}, c̄′) = r

for any c̄′ such that ∃ȳS(x, ȳ, c̄′). For the same reasons we assume that for a natural s

dim S(a′, K̄^{m}, c̄′) = s

for any a′, c̄′ such that ∃ȳS(a′, ȳ, c̄′). Counting the dimensions in case c̄′ = c̄ one sees that r = s. It follows that the number of solutions of ∃ȳS(x, ȳ, c̄′) is countable. □

Lemma 5.4 For any C ⊆ K̄

(i) C ⊆ ancl(C), (ii) ancl(C) = ancl(ancl(C)).

Proof (i) follows from the fact that the diagonal of K̄^{2} is analytic . (ii) follows from Proposition 6 and the fact that a countable union of countable sets is countable.□

Lemma 5.5 ancl satisfies the reciprocity principle a ∈ ancl(C ∪ {d}) \ ancl(C) implies d ∈ ancl(C ∪ {c}).

Proof Let a ∈ ancl(d, C). This is witnessed by some ∅-definable-analytic S and the fact that ⟨a, b̄, c̄, d⟩ ∈ S while ∃ȳS(x, ȳ, c̄, d) has finitely many solutions. By Proposition 6 we may assume

dim S(K̄, K̄^{m}, c̄, d′) = r = dim S(a′, K̄^{m}, c̄, d′)

whenever ∃ȳS(a′, ȳ, c̄, d′). Then counting the dimensions we have dim S(K̄, K̄^{m}, c̄, K̄) is equal either to r or to r + 1. In the first case the analytic set of m + 2-tuples S(K̄, K̄^{m}, c̄, K̄) can be decomposed into nonintersecting analytic subsets of the same dimension of the form

{⟨x, ȳ, d′⟩ : S(x, ȳ, c̄, d′) holds }

for d′ satisfying ∃x∃ȳS(x, ȳ, c̄, d′). It follows that the number of those d′ is countable. Hence d ∈ ancl(C) in this case. In the second case first consider the subcase when dim S(a, K̄^{m}, c̄, K̄) = r + 1. Then again by the proposition for any a′ such that the set is non-empty we assume dim S(a′, K̄^{m}, c̄, K̄) = r + 1. Then the number of those a′ is countable, and we have a ∈ ancl(C). If dim S(a, K̄^{m}, c̄, K̄) = r then analogously we see that the number of d′ for which ∃ȳS(a, ȳ, c̄, d′) holds is countable. Hence d ∈ ancl(a, C). □

Thus we proved Theorem C Under assumption of sufficiency of L̄_{F} ancl determines a pregeometry on K̄, i.e. for any C ⊆ K̄ the following holds:

(i) C ⊆ ancl(C), (ii) ancl(C) = ancl(ancl(C)), (iii) a ∈ ancl(C ∪ {d}) \ ancl(C) implies c ∈ ancl(C ∪ {c}).

REFERENCES

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[M] J.Mycielski, Remarks on infinite systems of equations, Algebra Universalis, v.21 (1985), 307-309

[Z] B.Zilber, Quasi-Riemannian manifolds, To appear in Proc. Logic Coll., Kiehl-1993