The millennium question over the reals, the complex numbers and - - PowerPoint PPT Presentation

Technion, Spring semester 2013 238900-13

The millennium question over the reals, the complex numbers and other general structures.

Research seminar 238900-13 Johann A. Makowsky∗

∗ Faculty of Computer Science,

Technion - Israel Institute of Technology, Haifa, Israel janos@cs.technion.ac.il

Graph polynomial project: http://www.cs.technion.ac.il/∼janos/RESEARCH/gp-homepage.html

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Lecture 1: Summary

• Introducing the topic.
• Register machines for arbitrary rings R.
• Basic observations concerning unit cost over R.
• Defining decidability DECR over R.
• Defining PR and NPR over R.
• Proving computability of NPR ⊂ DECR over the real and complex num-

bers using quantifier elimination (QE).

• What is QE?

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Lecture 2: Summary

We shall consider two cases where PR = NPR. 0n one case we add the FORTRAN-function: sin(x), in the oder case we disregard multiplication and the order.

• Adding sin(x) to R.

R = R, +, ×, <, sin x, 0, 1. We define Psin, NPsin and DECsin in the natural way.

• Disregarding multiplication and order and test only for equality:

R = R, +, 0, 1 We define Plin, NPlin and DEClin in the natural way. Theorem:(Klaus Meer) Psin = NPsin and Plin = NPlin.

Klaus Meer,

• A note on a P = NP for a restricted class of real machines,

Journal of Complexity 8 (1992), 451-453

• Real number models under various sets of operations,

Journal of Complexity 9 (1993), 366-372 File:lec-2 3

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Lecture 2: Summary (contd)

We shall also discuss adding other FORTRAN functions: exp, log, sin

• Adding exp(x) to R.

R = R, +, ×, <, exp(x), 0, 1. We define Pexp and NPexp in the natural way.

• Adding two functions:

F1 = {exp(x), sin(x)} or F2 = {exp(x), log(x)} to R. R = R, +, ×, <, exp(x), log(x), sin(x)0, 1. We define PF1, PF2, NPF1 and NPF2 in the natural way. Theorem:(Mihai Prunescu) Pexp = NPexp and PF1 = NPF1. Note: PF2=NPF2 remains open.

Mihai Prunescu,

• P = NP for a the reals with Various Analytic Functions.

Journal of Complexity 17 (2001), 17-26 File:lec-2 4

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Exercise: Computability of Z in R

We want to decide whether the sets Z ⊂ R and Q ⊂ R are computable.

• The decision problems for the sets Z and Q are computable in BSS over

the field R, but there is no bound on the length of the computation.

• Are the problems (R, Z) and (R, Q) in NPR?
• The problems (R, Z) and (R, Q) are in Psin. In fact they are computable

in constant time. We use the fact that sin(k · π) = 0 for k ∈ Z.

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The set AMeer

We look at the set AMeer = A = {t ∈ [0, 2π] : ∃k ∈ N (k · t 2π ∈ N)} . We note that

• t ∈ A iff

t 2π ∈ Q ∩ [0, 1], hence A is countable and dense in [0, 2π].

• The problem (R, [0, 2π]) is in Psin.

Here we use a constant a = 2π.

• We study the problem (R, AMeer), respectively ([0, 2π], AMeer).

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([0, 2π], AMeer) is in NPsin

• Input x ∈ R always has size 1.
• We show that using a single guess ([0, 2π], AMeer) can be solved

in constant time.

• 1. GUESS k ∈ R.
• 2. TEST k ≥ 0 and then sin(kπ) = 0.
• 3. Two yes show that k ∈ N.
• 4. TEST sin(k·x

2 ) = 0.

• 5. Yes shows that x ∈ A.

Q.E.D.

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Real-analytic functions

S.G. Krantz and H.R. Parks, A Primer for Real-Analytic Functions, Birkh¨ auser, 2002 (2nd edition)

A function f : R → R is real-analytic on an open set U ⊆ R if for all x ∈ U

• f has derivatives of all orders at x, and
• for every a ∈ U there is a neighborhood a ∈ V ⊂ U such that for all x ∈ V

f agrees with its Taylor series, i.e., f(x) =

• n

f(n)(a) n! (x − a)n Examples: polynomials, 1

x, sin, cos, exp, log are real analytic.

The function f(x) = x ∈ Q 1 x ∈ R − Q is not real-analytic.

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Properties of real-analytic functions

Proposition:(Classical)

• The set of real-analytic functions on U is closed under

scalar multiplication, pointwise addition multiplication and composition.

• The reciprocal of an analytic function that is nowhere zero is analytic.
• The inverse of an invertible analytic function whose derivative is nowhere

zero is analytic.

• Assume rn are distinct zeroes of f and lim rn = r is in a connected

component Dr the domain D of f. Then f(x) = 0 for all x ∈ Dr. Hence, if f is not constant on Dr, it has at most countable many zeroes in Dr.

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([0, 2π], AMeer) is not in DECsin hence not Psin.

There is no deterministic program using sin(x) which always terminates and decides ([0, 2π], AMeer).

• We proceed by contradiction.
• We do not allow division, and discuss later what effect divsion has.
• Assume we have a program which decides ([0, 2π], AMeer). It may have a

fixed number of constants, c1, . . . , cs ∈ R.

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Evaluating paths in the computation tree

Let γ be a path in the (unwound) computation tree of a program over R and sin.

• γ evaluates a term Tγ(x, c1, . . . , cs).
• Tγ(x, c1, . . . , cs) represents a a real fγ(x) function which is real-analytic.
• Since the program always terminates, there are at most countably many

paths γ.

• We can replace functions fγ(x) which are identically 0 by constant as-

signments.

• Let So there at most countably many values x ∈ R for which fγ(x) = 0.

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Heading for the contradiction

We now look at the set B = {t ∈ [0, 2π] : ∃γ (fγ(t) = 0)} We note:

• B is countable.
• Let t0 ∈ [0, 2π] − B. So for all γ we have fγ(t0) = 0.
• Since each path γ is finite there is some open set U(t0) ⊆ R such that
• ur program gives the saem answer for all inputs x ∈ U(t0), i.e.,

U(t0) ⊆ A or U(t0) ⊆ [0, 2π] − A.

• But this is impossible since A is countable and dense in [0, 2π].

Q.E.D.

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Meer’s Theorem

We have shown Theorem:(K. Meer 1992) Let F be any set of real-analytic functions such that (R, Z) ∈ NPF. Then PF = NPF. Problem: What limitations does the requirement (R, Z) ∈ NPF impose?

• For a set D ⊆ R the problem is in NPF iff it is existentially first order de-

finable using addition, multiplication, order, constants and unary function symbols for functions from F.

• For F = {exp(x)} we have (R, Z) ∈ NPF.

This is so, because every set D ⊆ R with (R, D) ∈ NPexp has only finitely many connected components.

• L. van den Dries and C. Miller, The field of reals with restricted analytic functions and

exponentiation, Israel Journal of Mathematics (1994). File:lec-2 13

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The role of division

Our proof shows really: Theorem: (Meer-Prunescu) Let D ⊆ R and R − D be dense in [a, b] ⊂ R. Then ([a, b], D) ∈ DECF for any set F of real-analytic functions.

• This allows to include division, although division is not a total function.
• To obtain Then PF = NPF we still need to show that ([a, b], D) ∈ NPF.
• To show that Pexp = NPexp Prunescu uses a different approach.

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Semi-algebraic sets

Let D ⊂ Rn.

• D is semi-algebraic if it is the solution set of a quantifierfree first order formula over

the ordered field of R.

• A function f : Rn → R is semi-algebraic, if its graph is a semi-algebraic set.
• A function f is essentially non-semi-algebraic if for no open set U ⊆ Rn the function f|U

is semi-algebraic.

• Let F be a set of real-analytic functions. f is semi-analytic if it is the solution set of a

quantifierfree first order formula over the ordered field of R with function symbols for functions from F.

• f is sub-analytic if it is the solution set of a first order formula over the ordered field
• f R with function symbols for functions from F.
• A real-analytic function is tame on U if it has no analytical singularities on the boundaries
• f U.
• A total real-analytic function f is always tame.

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Prunescu’s Theorem

Theorem: (M. Prunescu 2001) Let F be a set of real-analytic tame functions containing at least one function which is essentially non semi-algebraic. Then PF = NPF. Comments:

• F0 = {exp} is tame.

F1 = {exp, log} is not tame. F2 = {exp, log |(1,∞)} is tame.

• AP runescu = {(x, y, z) ∈ R3 : y > 0 ∧ z = y · exp(x

y)}

• AP runescu ∈ NPFi for i = 0, 1, 2.
• AP runescu ∈ PF0.

AP runescu ∈ PF1 in constant time. AP runescu ∈ PF2 but AP runescu ∈ DECF2.

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Still to be done (as projects)

• Meer’s Theorem: Plin = NPlin.
• Note that: P<

lin = NP< lin is still open.

• However, it is known that NP<

lin ⊆ DEC< lin.

This is shown using quantifier elimination.

• Prunescu’s Theorem for tame functions.

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Lecture 3-4: Summary

• The structure of the set of accepted inputs of deterministic machines.
• The structure of the set of accepted inputs of non-deterministic ma-

chines.

• Simulating non-determinsitic machines by determinsitic machines is quan-

tifier elimination QE.

• For arbitrary finite vocabularies τ and a τ-structure M PM = NPM implies

Th(M) has QE.

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Lecture 3-4: Summary (continued)

• A short course in QE, I: The theory of R as an infinite set.
• A short course in QE, II: R as a dense linear order.
• A short course in QE, III: R as an ordered additive abelian group.
• A short course in QE, IV: R as an ordered real closed field.
• A short course in QE, V: Z as an ordered abelian group.

Pressburger arithmetic.

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R as an infinite set (an exercise)

Equality programs:

• We get an input (x1, . . . , xm).
• We have tests xi = xj?
• We can reassign values xi := xj;
• We can shift the registers.
• There ar no alegebraic operations or order comparisons.

What are the decision problems is this very restricted formalisms?

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R as an densly ordered set (an exercise)

Equality programs:

• We get an input (x1, . . . , xm).
• We have tests xi = xj? and xi ≤ xj?
• We can reassign values xi := xj;
• We can shift the registers.
• There ar no alegebraic operations or order comparisons.

What are the decision problems is this very restricted formalisms?

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Two easy quantifier elimination theorems

The following first order theories have QE:

• For τ= = ∅, i.e., equality only Th=(N), Th=(Z), Th=(Q) and Th=(R).
• For τ< = {<}, i.e., equality and order only Th<(Q) and Th<(R).

The following first order theories do not have QE:

• For τ< = {<}, i.e., equality and order only Th<(N) and Th<(Z).

Let’s try to prove this from almost nothing!

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The general scheme for proving QE, I

Let T be a set of τ-sentences. Given a formula φ(¯ x) with distinct free variables ¯ x = (x1, . . . , xm). We want to find a formula ψ(¯ x) with the saem free variables but without quantifiers such that T | = ∀¯ [x]φ(¯ x) ↔ ψ(¯ x)

• First put φ(¯

x) into Prenex Normal Form PNF with Q1y1Q2y2 . . . ∃ynB(¯ y, ¯ x) where B is in Disjunctive Normal Form DNF or Q1y1Q2y2 . . . ∀ynB(¯ y, ¯ x) where B is in Conjuctive Normal Form CNF.

• We want to eliminate the quantifier which binds yn.

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The general scheme for proving QE, II

We continue:

• In the case ∃ynB(¯

y, ¯ x) B =

i Bi and

∃ynB(¯ y, ¯ x) is equivalent to

• i

∃ynBi . Furthermore Bi is a conjunction of atomic or negated atomic formulas.

• In the case ∀ynB(¯

y, ¯ x) we distribute ∀ over conjunctions.

• To do an induction we are with eliminating

∃y

• j

Ci,j(y) where Bi =

j Cj.

• The case ∀yB(y) is analogous.

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Lecture 5

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What is know about PR = NPR for R an (ordered) ring or field.

• We explain the table from [BCSS], page 111.
• We use the project from 2002 by D. Guez and E. Watted.

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Lecture 6

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P= NP on various less expressive structures.

We want to explore P = NP for reducts of subrings of the real and complex numbers. In particular we look at the structures (with possibly additional constants for elements)

• Infinite sets with equality only. R, N, Z.
• Infinite linear orders:

Dense orders: Q, <Q, R, <R Discrete orders: N, <N with first element. Z, <Z without first element.

• Infinite abelian groups:

R, +R, Q, +Q, Z, +Z, R − {0}, ×R, Q − {0}, ×Q, Z − {0}, ×Z,

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Elementary equivalent structures, I

Two τ-structures A, B are elementarily equivalent if for all τ-sentences φ of FOL(τ) we have A | = φ iff B | = φ. We write A ≡ B. A theory T is a set of formulas without free variables. T ⊆ FOL(τ) is complete if for any two τ-structures A, B which satisfy T we have A ≡ B. Using the path decomposition theorem we have Proposition 1 Let A, B be two τ-structures with A ≡ B. Then we have

PA = NPA iff PB = NPB

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Infinite sets with equality and order.

Corollary 2 (i) For any two infinite sets A, B we have

PA = NPA iff PB = NPB

if order is absent. (ii) For the densly ordered structures R, <R Q, <Q we have

PR,<R = NPR,<R iff PQ,<Q = NPQ,<Q

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Problems in NPA on infinite sets A with/without order.

We define the problems: HALF: Input: a1, . . . , a2n ∈ A. Problem: Is there a set J ⊆ [2n] of size n such that

• i,j∈J,i=j

ai = aj OHALF: Input: a1, . . . , a2n, ∈ A. Problem: Is there a set J ⊆ [2n] of size n such that (

• i,j∈J,i=j

ai = aj) ∧ (

• i∈J

ai ≤ a2n) Clearly, HALF and OHALF are in NP over the appropriate structures. Are they in P over the same structure?

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PA = NPA on infinite sets A with order.

NSUC: Input: a1 < a2 ∈ A. Problem: Is there b ∈ A such that a1 < b < a2 Again this is in NPZ,<, and the input size is 2. But in Z, <Z the formula ∃y(x < y < z) is not equivalent to a quantifierfree formula.

We have seen this before, using the fact that quantifierfree formulas are preserved under substructures.

Theorem 3 (i) NSUC is not computable, hence NPZ,< is not computable over Z, <Z. (ii) PZ,< = NPZ,< and hence, PA,< = NPA,< for any A, <A ≡ Z, <Z.

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Cases I could not resolve

• Infinite sets with equality only.

Let A = {a1, . . . , a2n} HALF asks whether |A| ≥ n. EVEN asks whether |A| is even. Are they computable with equality only? I guess NO.

• What about dense orders?

Are HALF or EVEN computable over dense linear orders? I guess NO.

• What about Z, <Z, sucZ, 0Z?

Here sucZ(a) = a +Z 1 is the successor relation on Z and 0 is a constant. We note that the complete theories Th(A) of an infinite set, Th(Q, <) and Th(Z, <Z, sucZ, 0) admit quantifier elimination.

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Infinite abelian groups

After Prunescu, JSL 2002

Theorem 4 (Bourgarde; Hemmerling and Gassner; Prunescu) Let G = (A, +A, 0A) be an infinite abelian group. Then PG = NPG. This generalizes K. Meer’s Theorem Plin = NPlin for the additive group of R stated in Lecture 1. For the ordered abelian group of the reals Rovs we have Theorem 5 (H. Fournier and P. Koiran 2001)

Povs = NPovs iff P/poly = NP/poly,

• resp. P0
• vs = NP0
• vs iff P = NP, for the parameterfree case.

We shall read Prunescu’s proof carefully to see why it does not work in the ordered case.

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The Nullsack problem

We call Nullsack the following problem NSG ⊆ G∞: NSG = {(x1, . . . , xn) : n ∈ N and ∃J = ∅, J ⊆ [n] with

• j∈J

xj = 0}

• NSG ∈ NPG with boolean guesses and parameterfree.
• NSG is computable in exponential time deterministically.

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Fundamental Theorems on abelian groups

• Every finitely generated abelian group G is isomorphic to

Zn ⊕ Zq1 ⊕ . . . ⊕ Zqt where n is its rank and qs, s ≤ t are powers of primes.

• G is torsion-free if t = 0.
• F.W. Levi (1942):

An abelian group is orderable iff it is torsion-free.

• O. H¨
• lder:

Every archimedian abelian ordered group is an ordered subgroup of the reals R, +R, <R, 0R, .

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Pr¨ ufer’s theorems on abelian gropus

• Every finite abelian group is isomorphic to the direct sum of cyclic gropus
• f prime order.
• Pr¨

ufer: An abelian group of bounded exponent is isomorphic to a direct sum of cyclic groups.

• Let G be infinite abelian group of bounded exponent. There is a prime

number p, such that there infinitely many elements on G of order p.

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p-elementary abelian groups

• Let Hp =

i∈N Zp, where the elements are infinite sequences of elements

• f Zp where all but finitely many are 0.
• Hp =

i∈N Zp is an infinite dimensional, countable vector space over the

field GF(p).

• Let H = {Z, Hp : p a prime}.
• the groups Hp are not finitely generated.

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Prunescu’s Theorem: For G an abelian group PG = NPG. Step 1

The following was proved by K. Meer for the additive group of R and by B. Poizat for H2. We first generalize this to groups in H: Proposition 6 (M. Prunescu) Let H ∈ H. Then NSH ∈ PH, hence PH = NPH.

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Proof of NSH ∈ PH, I

• Let m, n ∈ N and m, n ≥ 1, and ¯

a ∈ {0, 1}n, ¯ b1, . . . ,¯ bm ∈ Zn.

• The system

¯ a · ¯ x = 0 ¯ b1 · ¯ x = 0 . . . ¯ bm · ¯ x = 0 (1) has infinitely many solutions in Hn, provided that no ¯ bi is a multiple of ¯ a, and in case that H = Hp, no inequation reduces to 0 = 0 modulo p.

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Proof of NSH ∈ PH, II

• Assume NSH can be solved deterministically in polynomial time p(n).
• Choose n such that 2n − 1 > p(n).
• Using that the system (1) has infinitely many solutions, we construct

Y, N ∈ Hn such that Y ∈ NSH, N ∈ NSH, but both traverse the same set

• f < p(n) non-trivial tests negatively.
• Thus we reach a contradiction.

Q.E.D. What happens here in the case of ordered abelian groups?

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Strategy for the proof of NSG ∈ PG, for any abelian group G.

We proceed as follows:

• We look at ultraproducts of G.
• We show that every (non-trivial) ultrapower G∗ of G is

elementarily equivalent to G.

• Lemma 7

We show that every (non-trivial) ultrapower G∗ of G contains both G and some H ∈ H, such that G ∩ H = {0}.

• Then we show that NSG∗ ∈ PG∗.
• Using that G∗ ≡ G we conclude that NSG ∈ PG.

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Short course on ultraproducts

More details on the blackboard. (i) Filters and ultrafilters. (ii) Non-principal ultrafilters. (iii) Ultraproducts and ultrapowers. (iv) Proof of Lemma 7.

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Filters and ultrafilters

Let I be an infinite set. Intuitively, a filter F is a collection of large sets of I. A filter F is a family of subsets of I such that

• 1. The empty set ∅ is not an element of F.
• 2. If A and B are subsets of I, A is a subset of B, and A is an element of

F, then B is also an element of F.

• 3. If A and B are elements of I, then so is the intersection of A and B.

F is an ultrafilter if additionally

• 4. If A is a subset of I, then either A or I − A is an element of F.

Properties 1 and 3 imply that A and I − A cannot both be elements of F. File:lec-6 44

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Examples of filters and ultrafilters, I.

Let I = N.

• Let A ⊆ N. FA = {B ⊆ N : A ⊆ B}.

FA is an ultrafilter.

• A filter F on a set I is principal if it is of the form FA = {B ⊆ N : A ⊆ B}.

for some subset A ⊆ I. Every principal filter is an ultrafilter.

• F1 be the family of co-finite sets.

F1 is a non-principal filter but not an ultrafilter. Neither the set of even numbers nor the set of odd numbers is in F1.

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Examples of filters and ultrafilters, II

Let I = R and

• F2 be the set of uncountable sets.
• F3 be the set of co-countable sets.
• F4 be the set of dense subsets of R.

Discuss the filter properties of Fi for i = 2, 3, 4. Discuss ultrafilters on finite sets I.

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Ultraproducts and ultrapowers

Let I be an index set and F be a non-principal ultrafilter on I. Let Ai : i ∈ I a family of τ-structures.

• A =

i∈I Ai is the cartesian product of these structures.

• For ¯

a,¯ b we define ¯ a ∼F ¯ b iff {i ∈ I : ai = bi} ∈ F.

• The ultraproduct

i∈I Ai/F is the quotient structure A/ ∼F.

• In case all the structures Ai are the same

we speak of the ultrapower

I A/F.

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The Ultrafilter Theorem

Theorem 8 Every filter F over a set I is contained in some ultrafilter U over I. Proof: We use the well-ordering theorem, that every set can be well-ordered. Let Uα : α ≤ β be a well-ordering of the powerset of I. We put F0 = F. For each 0 < α < β we check whether Uα ∈ Fα or I −Uα ∈ Fα. If yes, we put Fα+ = Fα. If no, we put Fα+ = Fα[Uα], which is the smallest filter containing Fα and Uα. If δ < β is a limit ordinal, we put Fδ =

α<δ Fα.

It is now easy to check that Fβ is an ultrafilter. Q.E.D.

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• Los’ Theorem

Theorem 9 (Fundamental Theorem of Ultraproducts) Let

i∈I Ai/F an ultraproduct of τ-structures Ai : i ∈ I,

and φ(x1, . . . , xn) ∈ FOL(τ) be a first order formula. Let ¯ aj : j ∈ [n].

• i∈I

Ai/F | = φ(¯ a1, . . . ,¯ an) iff {i ∈ I : Ai | = φ(ai,1, . . . , ai,n)} ∈ F The proof is by induction.

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What we need for Prunescu’s Theorem

• Let A be a τ-structure. Then A ≡

I A/F.

In particular, this holds for A = G, an abelian group.

• Let G be an abelian group,and G∗ =

I A/F an ultrapower of G.

Then G∗ contains a group G′ which is an isomorphic copy of G.

We map a ∈ G into the constant sequence ai = a.

• Then G∗ contains an isomorphic copy of H

for some H ∈ H with H ∩ G = {0}.

Let ¯ a consist of infinitely many different coordinates ai. If the ai’s have anbounded order, we take the H to be the subgroup of G∗ generated by ¯ a, which isomorphic to Z and 0 is the only element also in G′. If all the ai’s have order bounded by m, we use Pr¨ ufer’s Theorem. File:lec-6 50