Juliette Kennedy and Saharon Shelah- On embedding models of arithmetic of cardinality aleph-1 into...

8/3/2019 Juliette Kennedy and Saharon Shelah- On embedding models of arithmetic of cardinality aleph-1 into reduced powers

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On embedding models of arithmetic of

cardinality 1 into reduced powers

Juliette Kennedy

Department of Mathematics

University of Helsinki

Helsinki, Finland

Saharon Shelah

Institute of Mathematics

Hebrew University

Jerusalem, Israel

October 6, 2003

Abstract

In the early 1970s S.Tennenbaum proved that all countable modelsofPA+1Th(N) are embeddable into the reduced product N

/F,where F is the cofinite filter. In this paper we show that ifM is a

model ofPA

+ 1 Th(N), and |M| = 1, then M is embeddableinto N/D, where D is any regular filter on .

1 Preliminaries

Let LA, the language of arithmetic, be the first order language with non-logical symbols +, , 0, 1, . N denotes the standard LA structure. We shallbe concerned with the following theories: The theory 1-T h(N), defined asthe set of all universal formulas true in the standard LA structure N. Hence-forth we refer to theories satisfying 1-T h(N) as Diophantine correct. We will

also refer to the theory P A, which consists of the following axioms (x < yabbreviates x y x = y):

1.) x , y, z((x + y) + z = x + (y + z))

Research partially supported by grant 40734 of the Academy of Finland.Research partially supported by the United States-Israel Binational Science Founda-

tion. Publication number [728]

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2.) x, y(x + y = y + x)

3.) x , y, z((x y) z = x (y z))4.) x, y(x y = y x)5.) x , y, z(x (y + z) = x y + x z)6.) x((x + 0 = x) (x 0 = 0))7.) x(x 1 = x)8.) x , y, z((x < y y < z) x < z)9.) x x x10.) x, y(x < y x = y y < x)11.) x,y,z(x < y x + z < y + z)12.) x,y,z(0 < z x < y x z < y z)

13.) x, y(x < y z(x + z = y)14.) 0 < 1 x(x > 0 x 1)15.) x(x 0)

Thus the theory P A is the theory of nonnegative parts of discretelyordered rings. For interesting examples of these models see [2].

2 The Countable Case

We begin by presenting the two embedding theorems of Stanley Tennenbaum,

which represent countable models of P A by means of sequences of realnumbers. We note that Theorem 1 follows from the 1-saturation of thestructure N/F ([1]), however we present Tennenbaums construction as itconstructs the embeddings directly:

We consider first the reduced power (of LA structures) N/F, where Fis the cofinite filter in the boolean algebra of subsets of N. Let A be thestandard LA structure with domain all nonnegative real algebraic numbers.We also consider the reduced power A/F.

Iff is a function from N to N, let [f] denote the equivalence class off inN/F. We use a similar notation for A/F. When no confusion is possible,

we will use f and [f] interchangeably.

Theorem 1 (Tennenbaum) LetM be a countable Diophantine correct modelof P A. Then M can be embedded inN/F.

Proof. Let m1, m2, . . . be the distinct elements of M. Let P1, P2, . . . beall polynomial equations over N in the variables x1, x2, . . . such that M |=

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Pi(x1/m1, x2/m2, . . .). Each system of equations P1 Pn has a solu-

tion in M. Thus, by Diophantine correctness, there is a sequence of naturalnumbers v1(n), v2(n), . . . for which

N |= (P1 Pn)(x1/v1(n), x2/v2(n), . . .).

Note that if the variable xi does not appear in P1 Pn, then the choiceof vi(n) is completely arbitrary. Our embedding h : M N/F is givenby:

mi [n.vi(n)].

In the figure below, the i-th row is the solution in integers to P1 Pn,

and the i-th column is h(mi).

m1 m2 . . . mn . . .P1 v1(1) v2(1) . . . vn(1) . . .P2 v1(2) v2(2) . . . vn(2) . . ....

......

...Pn v1(n) v2(n) . . . vn(n) . . ....

......

...

Note that if me is the element 0M of M, then the polynomial equation

xe = 0 appears as one of the Ps. It follows that, for n sufficiently large,ve(n) = 0. Thus h(0

M) is the equivalence class of the zero function. Similarly,h maps every standard integer ofM to the class of the corresponding constantfunction.

We show that h is a homomorphism. Suppose M |= mi + mj = mk. Thenthe polynomial xi + xj = xk must be one of the Ps, say Pr. Ifn r, then byconstruction vi(n) + vj(n) = vk(n). Hence N

/F |= h(mi) + h(mj) = h(mk),as required. A similar argument works for multiplication. Suppose M |=mi mj. By an axiom ofP A, for some k, M |= mi + mk = mj. Thus, aswe have shown, h(mi) + h(mk) = h(mj). It follows from the definition of therelation in N that h(m

i) h(m

j).

To see that h is one to one, suppose that mi = mj. Since in models ofP A the order relation is total, we may assume that mi < mj . Again by theaxioms ofP A, we can choose mk such that mi + mk + 1 = mj. As we haveshown, N/F |= h(mi) + h(mk) + h(1) = h(mj). Since h(1) is the class ofthe constant function 1, it follows that h(mi) = h(mj). 2

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Corollary 2 LetM be a countable model of the 1-T h(N). ThenM can be

embedded in N.

Proof. The models of the 1-T h(N) are precisely the substructures of modelsof T h(N). Thus, M extends to a model of P A, which can be embedded in

N as in Theorem 1. 2

Before turning to the theorem for the non-Diophantine correct case, weobserve first that the given embedding depends upon a particular choiceof enumeration m1, m2, . . . of M, since different enumerations will in generalproduce different polynomials. We also note that different choices of solutionyield different embeddings. Also, as we shall see below, we need not restrict

ourselves to Diophantine formulas: we can carry out the construction for LAformulas of any complexity which hold in M.

We state the non-Diophantine correct case of the theorem:

Theorem 3 (Tennenbaum) Let M be a countable model of P A. Then Mcan be embedded inA/F.

Proof. Given an enumeration m1, m2, . . . of M, we form conjunctions ofpolynomial equations Pn exactly as before. We wish to produce solutionsof P1 Pn in the nonnegative algebraic reals for each n. We proceed asfollows: The model M can be embedded in a real closed field F by a stan-dard construction. (Embed M in an ordered integral domain, then form the(ordered) quotient field, and then the real closure.) Choose k so large thatx1, . . . , xk are all the variables that occur in the conjunction P1 Pn.The sentence x1 . . . xk(P1 Pn x1 0 x2 0 xk 0) is true inM, hence in F. It is a theorem of Tarski that the theory of real closed fields iscomplete. Thus, this same sentence must be true in the field of real algebraicnumbers. This means we can choose nonnegative algebraic real numbersv1(n), v2(n) . . . satisfying the conjunction P1 Pn. Let h : M A/Fbe given by

mi [n.vi(n)].

The proof that h is a homomorphism, and furthermore an embedding, pro-ceeds exactly as before, once we note that the equivalence classes all consistof nonnegative sequences of real algebraic numbers. 2

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Remark 4 Under any of the embeddings given above, if M |= P A then

nonstandard elements of M are mapped to equivalence classes of functionstending to infinity. Why? If f is a function in the image of M, and fdoes not tend to infinity, then choose an integer k such that f is less than kinfinitely often. Since M |= P A, either [f] [k] or [k] [f]. The secondalternative contradicts the definition of in N. Hence [f] [k], i.e., [f] isstandard.

Remark 5 LetF be a countable ordered field. ThenF is embedded inR/F,where R is the field of real algebraic numbers. The proof is mutatis mutandisthe same as in Theorem 3, except that due to the presence of negative elements

we must demonstrate differently that the mapping obtained is one to one. Butthis must be the case, since every homomorphism of fields has this property.

Remark 6 For any pair of LA structures A and B satisfying P A, if A iscountable and if A satisfies the 1-T h(B) then there is an embedding of Ainto B/F. In particular, if M is a model of P A, then every countableextension of M satisfying the 1-T h(M) can be embedded in M

/F.

Remark 7 Given Theorem 1, one can ask, what is a necessary and sufficientcondition for a function to belong to a model of arithmetic inside N? For apartial solution to this question, see [3].

3 The Uncountable Case

We now show that some of the restrictions of Theorem 1 can be to someextent relaxed, i.e. we will prove Theorem 1 for models M of cardinality 1and with an arbitrary regular filter D in place of the cofinite filter. We notethat for filters D on for which B/D is 1-saturated, where B is the twoelement Boolean algebra, this follows from the result of Shelah in [5], thatthe reduced power N/D is 1-saturated.

Our strategy is similar to the strategy of the proof of Theorem 1, inthat we give an inductive proof on larger and larger initial segments of theelementary diagram of M. However we must now consider formulas whosevariables are taken from a set of1 variables. This requires representing eachordinal < in terms of finite sets un, which sets determine the variableshandled at each stage of the construction. The other technicality we require

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is the use of the following function g(x, y), which bounds the size of the

formulas handled at each stage of the induction.Let h(n, m) = the total number of non-equivalent Diophantine formulas

(x1, . . . xm) of length n. Define

g(n, n) = h(n, 0)

g(n, m 1) = 2 + h(g(n, m), m) (g(n, m) + 3), for m n.

Theorem 8 LetM be a model ofP A of cardinality1 which is Diophantinecorrect and letD be a regular filter on . ThenM can be embedded inN/D.

Proof. Let M = {a : < 1}. Let {An}n be a family witnessing theregularity of D. We define, for each < 1 a function f NN. Ourembedding is then a [f]. We need first a lemma:

Lemma 9 There exists a family of sets un, with < 1, and n N, suchthat for each n,

(i) |un| < n + 1

(ii) un un+1

(iii)

n un = + 1

(iv) un un = u

n (+ 1)

(v) limn|un|n+1

= 0

Suppose we have the lemma and suppose we have defined f for all < .We choose f NN componentwise, i.e. we choose f(n) for each n separatelyso that f(n) satisfies the following condition:

(),n: If = (..., x,...)un is a Diophantine formula, such that the length

of is g(n, |un|), then

M |= (..., a,...)un N |= (..., f(n),...)un .

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(A formula is said to be Diophantine if it has the form

x0, . . . , xn1(t1(x0, . . . , xn1) = t2(x0, . . . , xn1)),

where t1 and t2 are LA-terms.) Now suppose = 0, n > 0. By (ii) and (iii),u0n = {0} for all n. We claim that ()0,n holds for each n. To see this, fix nand let

= {(x) | M |= (a0), where is Diophantine with || g(n, 1)}.

This is up to equivalence a finite set of formulas. Also, M |= x

{(x) | } and therefore by Diophantine correctness N |= x

{(x) | }.

If k witnesses this formula, set f0(n) = k. Clearly now ()0,n holds.Now assume (),n holds for all < and for each n. We choose f(n)

for each n as follows. Fix n < and let

= {(x0, x1, . . . xk) | M |= (a,...,a,...)un\{}),

where is Diophantine and || g(n, |un|)},

for k = |un| 1 (the case that |un| 1 reduces to the previous case). This

is again a finite set of formulas, up to equivalence. Now

M |=

{(a, aun\{})| }

and thereforeM |= x

{(x, aun\{})| }.

Let (x1, . . . , xk) x

{(x, x1, . . . , xk)| }. Note that

|| 2 + h(g(n, |un|), |un|) (g(n, |u

n|) + 3) = g(n, |u

n| 1).

Now let = max (un \ {}). Then un = u

n \ {}, by (iv) of the lemma.

Since (),n holds for each n and we know that M |= (aun) then weknow by the induction hypothesis that N |= (f(n)un). Thus N |=x

{(x,...,f(n),...)un)| }. Let k witness this formula and setf(n) = k. Then (),n holds with a [f(n)].

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This mapping is an embedding of M into N/D. To see this, suppose

M |= (a, a, a) where is the formula x0 + x1 = x2. Let n = n0 belarge enough so that || g(n0, |un0|), where is chosen so that u

n0

contains , , . Then for all n n0, (),n holds and N |= f(n) + f(n) = f(n), forall n n0. Let An1 be an element of the chosen regular family of D suchthat An1 {0, . . . , n0} = . Then since f(n) + f(n) = f(n), for n n0,this holds also for n An1. The proof that multiplication is preserved isthe same so we omit it. Finally, as in the countable case, we note that ourmapping is one-to-one and hence an embedding.

We now prove the lemma, by induction on . Let u0n = {0} for all n < .

Case 1. is a successor ordinal, i.e. = +1. Let n0 be such that n n0implies |u

n|n

< 12

. then we set

un =

{} n < n0un {} n n0.

Then (i), (ii) and (iii) are trivial. Proof of (iv): Suppose un = un {}.

Case 1.1. = . Then

un = un

= u

n + 1= un (+ 1).

Case 1.2. un. Then by the induction hypothesis

un = un (+ 1)

= un (+ 1),

since un implies .

Proof of (v):

limn

|un|

n lim

n

|un| + 1

n= 0,

by the induction hypothesis.

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Case 2. is a limit ordinal > 0. Let n be an increasing cofinal -sequence

converging to , for all < 1. Let us choose natural numbers n0, n1,... suchthat i u

i+1ni and n ni+1 implies ni |u

i+1n | < n. Now we let

un = uin {}, if ni n < ni+1.

To prove (iv), let un. We wish to show that un = u

n (+ 1). Suppose

ni n < ni+1. Then un = uin {}. But then u

n implies u

in {}.

If = , then as before, un = un (+ 1). So suppose u

in . Then

un = ui (+ 1) = un (+ 1).

Finally, we prove (iii): Let un, a limit. Let ni be such that un =uini {}. We have u

ini

= ui+1ni (i + 1). Therefore u

i+1ni {} = u

n+1.

To prove (v), let > 0 and choose i so that ni > 2. We observe that ni nimplies

limn

|un|

n= lim

n

|uin |

n+

1

n

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