the arithmetic of cuts in models of arithmeticweb.mat.bham.ac.uk/r.w.kaye/papers/arithcuts1.pdf ·...

The Arithmetic of Cuts in Models of Arithmetic

Richard Kaye

29th May 2012

Abstract

We present a number of results on the structure of initial segmentsmodels of Peano arithmetic with the arithmetic operations of addition,subtraction, multiplication, logarithm and a notion of derivative on cuts.Each of these operations is defined in two dual ways, often with quitedifferent results, and we attempt to systemise the issues and show howvarious calculations may be carried out.

The main issues arise because the notion of ‘infinitesimal’ is relative tothe object being considered and not absolute, and this work constitutes acase study. It also presents some important background on using cuts asnumbers, often assumed but not studied in detail in models of arithmetic.It also provides an analysis of the kinds of duality that play a part in theseconsiderations.

1 Introduction and preliminaries

Dedekind’s construction of the real numbers shows that there is a arithmeticstructure with addition and multiplication (and more) on the set of proper cutsof Q. For a nonstandard model of arithmetic, the set of its cuts is the mostimportant structural feature and a great deal of work in nonstandard models(for example Friedman [2], Paris and Kirby [7]) involves their cuts. It is commonknowledge that these cuts also have arithmetic structure, and some rudimentaryform of this structure is sometmes used to name cuts, but I am unaware of anypaper where this structure is worked out fully.

In fact, the arithmetic structure of such cuts is an ‘elementary’ but surpris-ingly tricky topic in models of Peano arithmetic: there are a number of possibledefinitions of addition and multiplication (for example) on initial segments ofsuch models, and one doesn’t get a simple algebraic structure such as an orderedfield or even another nonstandard model of Peano arithmetic. When workingon a related problem and trying to apply such operations the author quicklydiscovered that calculations using these operations are rather more tricky thanwas expected. This paper contains an attempt to systemise the issues, presentthe main unifying ideas in a helpful way, and show how the various calculationsmay be carried out.

To motivate the work presented here more thoroughly, it has already beennoted that initial segments and cuts are key to understanding many structuralproperties of models of arithmetic. Therefore a model should be understood inconjunction with its Dedekind completion. Indeed, in many ways this is essentialin work with uncountable models. Friedman showed that nonstandard models

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are isomorphic to many of their initial segements, and the different ways thishappens can give important structural information. Paris and Kirby showedthat indicators give rise to initial segments, and conversely that the existence ofinitial segments with particular properties can tell one about first order formulastrue in the model.

Another application, one that may be considered part of an on-going researchprogramme, concerns using initial segments to count external sets. In otherwords, it tries to explain the importance of initial segments by mapping externalsubsets of the model to initial segments. If M � PA and X ⊆ M is a boundedsubset, but otherwise arbitrary, we may approximate X by internal (M -finite)sets x ∈M such that (the extension of) x is a subset of X. How closely X canbe approximated in this way becomes of central interest. We define the lowercardinality of X, cardX as the supremum of all cardinalities cardx ∈M takenover such x ⊆ X. Then cardX is in general a cut of M . Similarly we mayapproximate X by internal sets x′ ∈ M such that x′ is a superset of X. Theupper cardinality of X, cardX as the infimum of all cardx′ ∈M taken over suchx′ ⊇ X. In many cases these cuts agree and cardX = cardX, and the structuralproperties of X are related to the size of this cut. This is what is happening ina number of important results, including the characterisation of closed normalsubgroups of AutM when M is countable and recursively saturated [4, 5], theexternal normal subgroup structure of internal groups such as Sym(n) [1] andforthcoming work by Kaye and Reading [8, 6].

In all such cases, it is essential to understand the arithmetic structure of suchinitial segments. For example, Reading has considered the size of transversalsand the index [G:H] of externally defined subgroups H 6 G of an internalgroup g ∈ M . Of course one expects [G:H] = card(G)/ card(H) and thereforeone needs the appropriate notion of division on initial segments. The currentpaper is intended to be supplementary to the more exciting work on applicationsof the cardinality-as-cut idea, and we have found that a detailed understandingof this arithmetic is helpful to understand the strengths and limitations of thisidea. For further work in these directions, the reader is directed to Reading’sthesis at Birmingham university and planned future papers by him and thepresent author.

To sum up the findings of the present paper: arithmetic of cuts behavesformally in a similar way to that of arithmetic of cuts in the rationals. However,it is more complicated in that there are usually two different limiting processesto calculate each operation, corresponding to upper and lower cardinality above,and in many cases these operations give different answers. Often the distinctionbetween these two operations can be understood in terms of the derivative ∂I ofa cut, or one of its variations ∂

nI, to be defined below. This is worked through

in some detail for the additive structure below. The derivative bahaves in somerespects like a derivative in classical analysis, with some further complications.In the main, these complications arise from the fact that unlike Q or R where thenotion of ‘infinitesimal’ dx (and in particular the size of such values) is ratheruniform across the whole structure, but the notion of ‘infinitesimal’ di ∈ ∂Ivaries with the cut I under consideration.

The rest of this section will contain some preliminaries, setting the sceneand stating a few convenient but important conventions that will aid our workhere considerably. Some of these conditions are slightly different to the normalones found in the literature, so some attention is required.

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Throughout, M will be a nonstandard model of Peano Arithmetic (PA). Forbackground on models of PA see Kaye [3]. Very little general theory is neededhere, however. The most important result we need is the principle of overspill.

It is important for the arithmetic structure of M and its cuts that we considernegative elements and ‘negative cuts’. (Definitions will be given shortly.)

Definition 1.1. The structure Z = Z(M) is the full ring associated with M . Itis defined as 0 together with two copies of {x ∈M : x > 0}, one copy for positiveelements and one for negative elements, and with +, ·, < defined appropriately.Z is a discretely ordered ring.

Lower case letters a, b, c, . . . , x, y, z, . . . will always range over elements of Zand upper case letters A,B,C, . . . ,X, Y, Z, . . . range over subsets of Z.

Definition 1.2. An initial segment I ⊆e Z is a subset of M which is downward-closed:

∀x ∈ Z ∀y ∈ I (x < y → x ∈ I).

We identify the elements a ∈ Z with the initial segment {x ∈ Z : x 6 a}. Wedenote the set of initial segements of Z by Dcl(M), Z(M) or simply Z, andrefer to it as the Dedekind completion of M or Z.

Note that we make no assumption that an initial segment is proper nor thatit is closed under any function such as successor. We use −∞ to denote theinitial segment ∅ and ∞ to denote Z. Note too the essential convention thata number a is identified with the set of elements of Z less than or equal toit. This is necessary for the formulas below to work in all cases, and to avoidconsiderable extra complexity.

Variables I, J,K, . . . range over initial segments of Z, so I, J,K, . . . mightbe, but need not be, elements of Z itself.

Definition 1.3. The ordering relation 6 on Z is ⊆. We write I < J to meanI is a subset of J and I, J are not equal.

Proposition 1.4. The ordering 6 on Z is a linear order on Z. As ordered sets,we have M ⊆ Z ⊆ Z, i.e. the ordering 6 on Z extends that on Z.

The proof is a straightforward check using the definitions. (The same appliesto a number of other propositions given in this paper without proofs, and in allsuch cases the reader should be able to supply necessary proofs easily enough.)

To aid calculations and to make this material manageable, it is essential atall times to bear in mind various dualities. Chief amongst these is the dualitybetween 6 and >, and between ‘initial segments’ and ‘end segments’. We mightsay a subset I is an end segment of Z if I is closed upwards under >. Thedual of an initial segment is an end segment which is almost always its settheoretic complement. To reduce the amount of terminology we shall redefine‘complement’ slightly to make the identification exact.

Definition 1.5. The complement of an initial segment I is

Ic = ∪{z ∈ Z : ∀w ∈ I z > w}.

The complement of an end segment E is

Ec = {z ∈ Z : ∀w ∈ I z 6 w}.

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In other words the complement of an initial segment I is its normal set-theoretic complement but with max I added in case this maximum exists, i.e. incase I = a = max I for some a ∈ Z. The complement of an end segment is dualto this.

Given an initial segment I ⊆ M we usually use lower case i, i1, i2, . . . torange over elements of I, i.e. elements of Z less than or equal to I. We usei′, i′1, i

′2, . . . to range over elements of Ic, i.e. elements of Z greater than or equal

to I.We use X \Y for set difference, and this is not to be confused with X−Y as

defined below. We usually make no cardinality assumption on M , and often PAcan be weakened to one of its subtheories—but this will not be made explicithere.

Definition 1.6. A cut of M is a (not necessarily proper) I ∈ Z \ Z, i.e., is aninitial segment that is not an element of Z.

Definition 1.7. If X ⊆ Z we set

inf X = {x ∈ Z : ∀y ∈ X x 6 y}

and

supX = {x ∈ Z : ∃y ∈ X x 6 y}.

Thus inf Z = −∞, inf ∅ =∞, inf M = 0 and supZ = supM =∞, sup∅ =−∞. Also inf a = sup a = a for all a ∈M .

2 Additive structure

The goal of this section is to define and understand the binary operation ofaddition on initial segments, in particular on cuts. We start though with theunary operation of −.

Definition 2.1. If I ∈ Z we define

−I = {x : I 6 −x} = {−x : x ∈ Ic}.

Proposition 2.2. The unary − operation has the following properties.

(a) −− I = I for each I ∈ Z.

(b) The − operation applied to a ∈ Z gives −a. In particular −0 = 0.

(c) The − operation applied to Z and ∅ gives ∅ and Z respectively, justifyingthe notation −∞ for ∅.

Proposition 2.3. For I, J ∈ Z, we have

I 6 J ⇔ −J 6 −I.

Definition 2.4. If I ∈ Z we define

|I| = I if I > 0, |I| = −I otherwise.

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We now define addition on Z. The various conventions chosen earlier havebeen set up specifically to make this definition as simple as possible and so thatit extends the usual addition on Z.

Definition 2.5. For I, J ⊆e M we let

I + J = {i + j : i 6 I and j 6 J}.

It is immediate that I + J is an initial segment if both I, J are, and that itgives the correct answer for a + b, 0 + I, etc.

Note that if I is empty or J is empty there is no pair of elements (i, j) ∈ Z×Zsuch that i 6 I and j 6 J . (This is the only reasonable logical convention here,though does need some care.) Thus −∞+ I = −∞ for all I, including the caseI = Z. Hence we have the simple but important result that −∞ +∞ = −∞.(The apparent asymmetry between ∞ and −∞ here will be resolved shortly.)The other basic properties of + are straightforward.

Proposition 2.6. For I, J,K ∈ Z, we have:

(a) I + (J + K) = (I + J) + K;

(b) I + J = J + I;

(c) I 6 J implies I + K 6 J + K;

(d) I + 0 = I.

(e) I +−∞ = −∞ and J +∞ =∞ for J > −∞.

It is not true in general that I < J implies I+K < J +K. Counterexamplesare easy to find, including but not restricted to K = ±∞. See Proposition 2.30below.

The dual of + is the following operation.

Definition 2.7. For I, J ⊆e M , let

I ⊕ J = −((−I) + (−J)).

The reader can check that I ⊕ J = inf{i′ + j′ : i′ > I, j′ > J}. Once again,this addition agrees with addition on a, b ∈ Z.

We will investigate soon when it is the case that I+J = I⊕J . One inclusionis, however, always true.

Proposition 2.8. For I, J ∈ Z, we have I + J 6 I ⊕ J .

By duality, ⊕ has all of the basic properties enjoyed by +.

Proposition 2.9. For I, J,K ∈ Z, we have:

(a) I ⊕ (J ⊕K) = (I ⊕ J)⊕K;

(b) I ⊕ J = J ⊕ I;

(c) I 6 J implies I ⊕K 6 J ⊕K;

(d) I ⊕ 0 = I.

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(e) I ⊕∞ =∞ and J ⊕−∞ = −∞ for J <∞.

Proof. Straightforward. For example, I ⊕ J = −((−I) + (−J)) = −((−J) +(−I)) = J ⊕ I.

The next proposition records a particular case where addition of cuts is easy.

Proposition 2.10. Let I, J ∈ Z be nonnegative and closed under +. ThenI + J = I ⊕ J = max(I, J).

The additive structure of Z is understood best in terms of another operationon cuts, defined next.

Definition 2.11. Given I ∈ Z we define the derivative of I as

∂I = {x ∈ Z : ∀y ∈ Z (y 6 I ⇒ x + y 6 I)} ∈ Z.

We use di, di1 etc. as variables ranging over ∂I, and di′ for variables overelements of ∂Ic.

Thus ∂I is an initial segment of Z. We will see shortly that it measures thesize of the ‘boundary’ between I and Ic.

Note that x ∈ ∂I can be defined by any of the equivalent statements

∀y ∈ Z (y 6 I ⇒ x + y 6 I),

∀y ∈ Z (y I ⇒ x + y > I)

or

∀y ∈ Z (y > I ⇒ x + y > I).

Example 2.12. ∂a = 0 for each a ∈ Z. Also, ∂∞ = ∂(−∞) =∞.

Proposition 2.13. Given I ∈ Z, ∂I is non-negative and closed under addition.In fact ∂I = I whenever I > 0 is closed under addition.

It is immediate from the definition that I + ∂I = I for all I ∈ Z. It is nottrue in general that I ⊕ ∂I = I.

The notion of ∂I is the ‘fuzziness’ or ‘uncertainty’ we know I by. For eachk ∈ ∂I and i 6 I 6 i′ we have i + k 6 I 6 i′ − k. Moreover,

Proposition 2.14. For I ∈ Z and di′ > ∂I there is i ∈ I with i 6 I ∂I there is i 6 I with i + k′ 66 I.

Example 2.15. Given a non-negative cut K ∈ Z closed under addition andx ∈M we may define the cuts

x + K = sup{x + k : k 6 K}

and

x−K = sup{x− k′ : k′ > K}.

These have derivative equal to K itself.

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Definition 2.16. For nonnegative cut K ∈ Z closed under +, I is a downwardsK-limit if I = a−K for some a ∈M . I is a upwards K-limit if I = a + K forsome a ∈M .

Proposition 2.17. Let K ∈ Z be a nonnegative cut. Then no cut I is both anupwards K-limit and a downwards K-limit.

Proof. Suppose x, y ∈M with

I = sup{x + k : k ∈ K} = inf{y − k : k ∈ K}.

Then, for all k ∈ K, x + k < y − k so by overspill there is k′ > K withx + k′ < y − k′, hence I < x + k′ < y − k′ < I.

Example 2.18. To find cuts with a given derivative K that are neither of theform a + K nor a − K for some a, we may use a tree argument. We assumefor simplicity that M is countable and show that there are continuum manycuts with derivative K but only countably many upwards and downwards K-limits. One inductively takes ui, vi ∈ M with ui + K < vi and shows usingProposition 2.17 that there is always w with ui + K < w, w + K < vi. Thensetting ui+1, vi+1 to be either ui, w or w, vi we continue and all the limits ofthese sequences can be arranged to all have derivative K.

Example 2.19. Given a nonnegative cut K closed under +, and a, b ∈M , wecan easily calculate

(a + K) + (b + K) = (a + K)⊕ (b + K) = (a + b) + K

(a + K) + (b−K) = (a + b)−K

(a + K)⊕ (b−K) = (a + b) + K

(a−K) + (b−K) = (a−K)⊕ (b−K) = (a + b)−K

In particular, (a + K) + (b−K) < (a + K)⊕ (b−K).

Example 2.20. Example 2.19 is not the only way one can get I + J < I ⊕ J .We can give other examples by a tree argument similar to that in Example2.18. Suppose M is countable, and x, K are given. Inductively suppose we haveui, vi, ri, si with

ui + K < vi

ri + K < si

ui + ri + K < x

ui + ri + K < x

vi + si > x + K.

Then we can split the contruction into two, making a binary tree as follows.Choose ui < w < vi and ri < t < si with w + t = x. If we can perform thischoice with ui + K < w, w + K < vi and ri + K < t, t + K < si we then takeui+1, vi+1, ri+1, si+1 to be either ui, w, t, si or w, vi, ri, t and continue, arraningthat the limits of the ui, vis is a cut I with derivative K and the limits of theri, sis is a cut J with derivative K and I + J = x−K < x + K = I ⊕ J .

There is always ui < w < vi and ri < t < si with w + t = x becauseui + ri < x < vi + si. For any such w we have either ui +K < w or w +K < vi

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as ui + K < vi. Suppose that ui + K < w, other cases being similar, but thatvi − w ∈ K. As ri + K < t or t + K < si and vi + si > x + K we must havet + K < si. Suppose that t− ri ∈ K. Then for each k ∈ K we have

∃w′, t′ (w′ + t′ = x ∧ ui + k < w′ < vi − k ∧ ri + k < t′ < si − k)

for we may just take w′ = w − k, t′ = t + k. By overspill there is w′, t′ withw′ + t′ = x∧ ui + k′ < w′ < vi − k′ ∧ ri + k′ < t′ < si − k′ for some k′ > K andwe are done.

The derivative operation on Z is a kind of valuation, as the next propositionshows.

Proposition 2.21. For cuts I, J ,

∂(−I) = ∂I

and

∂(I + J) = ∂(I ⊕ J) = ∂I + ∂J = ∂I ⊕ ∂J = max(∂I, ∂J).

Proof. The first is an easy check.For the second, note that ∂I and ∂J are nonnegative and closed under + so

Proposition 2.10 applies. Without loss of generality assume K = ∂J > ∂I. Ifi ∈ I, j ∈ J and k ∈ K then (i+j)+k = i+(j+k) ∈ I+J showing ∂(I+J) > ∂J .Similarly if i′ > I, j′ > J and k ∈ K then i′ + j′ − k = i′ + (j′ − k) > I ⊕ J so∂(I ⊕ J) > ∂J .

Conversely, if k′ > K let i ∈ I with i′ = i+k′ > I and j ∈ J with j′ = j+k′ >J . Then for i1 ∈ I and j1 ∈ J , i′+j′ = (i+k′)+(j+k′) = (i+j)+2k′ 6 i1 +j1implies i+k′ 6 i1 or j+k′ 6 j1, which are both impossible. Thus i′+j′ > I+J ,showing I + J is not closed under addition by 2k′, so k′ > ∂(I + J). Similarly,i′ + j′ − 2k′ ∈ I + J 6 I ⊕ J but i′ + j′ > I ⊕ J so I ⊕ J is not closed underaddition by 2k′ and k′ > ∂(I ⊕ J).

Proposition 2.22. If I, J are cuts and ∂I < ∂J then I + J = I ⊕ J .

Proof. If I, J are as given there are i0 6 I, k0 6 ∂J with i′0 = i0 +k0 > I. Then

I ⊕ J = inf{i′ + j′ : i′ > I, j′ > J}= inf{i0 + j′ : j′ > J}

since the difference between sufficiently small i′ > I and i0 is less than k0 6 ∂J .Thus I ⊕ J = i0 + J . Similarly

I + J = sup{i + j : i ∈ I, j ∈ J}= sup{i0 + j : j ∈ J}

so I + J = i0 + J also.

In some cases, I⊕∂I can be taken as a canonical cut greater than or equal toI with approximately the same amount of information as I, but which is nevera downwards ∂I-limit. The following explains exactly when I < I ⊕ ∂I.

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Proposition 2.23. For a cut I ∈ Z,

I = I + ∂I 6 I ⊕ ∂I.

Also, I = I ⊕ ∂I unless some (equivalently all) x ∈ (I ⊕ ∂I) \ I has I = x− ∂Iand I ⊕ ∂I = x + ∂I.

Proof. Let K = ∂I. It is easy to see that I = I + K, as i 6 i + k I for each k ∈ K, and given i′ > I, k′ > K we have x + k < i′ + k′

as x < (i′ − k) + k′. This shows that I 6 x − K < x + K 6 I ⊕ K. To seethat the nonstrict inequalities here are in fact equalities, suppose at least one isstrict. Then there are I < x < x + K < y K with x + 2k′ 6 y. Now let i ∈ I withi′ = i + k′ > i. Then i < I < x < y < I ⊕K < i′ + k′ with i′ + k′ − i = 2k′

which is impossible.

Proposition 2.24. Given cuts I, J , we have

I + J 6 I ⊕ J 6 (I + J)⊕ ∂(I + J) = (I + J)⊕max(∂I, ∂J)

with at least one of the inequalities here being equality.

Proof. Let K = max(∂I, ∂J) = ∂(I + J). Suppose x > I + J , k′ > K. Thenthere are i 6 I and j 6 J with i′ = i + [k′/2] > I and j′ = j + [k′/2] > J . Soi′ + j′ = i + j + k′ 6 x + k′ hence I ⊕ J 6 (I + J)⊕ ∂(I + J) and so

I + J 6 I ⊕ J 6 (I + J)⊕ ∂(I + J).

Either all here are equal or I +J = a−K, (I +J)⊕∂(I +J) = a+K, in whichcase I ⊕ J (having derivative K) must be equal to one of these.

This describes the cases when + and ⊕ differ.

Proposition 2.25. Suppose I, J are cuts. Then I + J 6 I ⊕ J with inequalityholding if and only if I = a− J , J = a− I, for some a ∈ Z.

Proof. Most of the work has already been done. I +J 6 I ⊕J is true generally,and I + J J (sincea − i = j ∈ J gives a = i + j 6 I + J) so i = a + x ∈ a − J , and conversely if−x > J then a + x 6 I (for a + x > I implies a = (a + x) +−x > I ⊕ J) hencea− J 6 I. J = a− I is similar.

For the converse, assuming I = a − J or J = a − I we may calculateI + J = a− J + J = a− ∂J and I ⊕ J = (a− J)⊕ J = a + ∂J .

It may be easier to remember the conclusion to the previous result using thederivative. If I + J < I ⊕ J then ∂I = ∂J and I = a− ∂I, J = a+ ∂J for somea, or vice versa.

Corollary 2.26. If I, J ∈ Z have I + J < I ⊕ J then

I ⊕ J = (I + J)⊕ ∂(I + J) = (I ⊕ J)⊕ ∂(I ⊕ J).

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Proof. If K = ∂(I + J) = ∂(I ⊕ J) then (I ⊕ J)⊕K = (I ⊕ J) + K = (I ⊕ J)since (I ⊕ J) is not a downwards K-limit.

It is interesting to see what happens when I + J = I ⊕ J and only one partof the Proposition 2.25 holds.

Proposition 2.27. Let I, J,K be cuts with K = max(∂I, ∂J) and I+J = I⊕J .

• If I⊕J is an upward K-limit then there are i ∈ I and j ∈ J with I 6 i+K,J 6 j + K and equality holding in at least one of these two cases.

• If I + J is a downward K-limit then there are i′ > I and j′ > J withI 6 i′ −K, J 6 j′ −K and equality holding in at least one of these twocases.

Proof. For the first, if I +J = I⊕J = x+K for some x then x = i+ j for somei ∈ I, j ∈ J , and for each k′ > K i + j + k′ > I ⊕ J . Therefore for each k′ > Kthere are i′ > I and j′ > J with i′ + j′ = i + j + k′, or (i′ − i) + (j′ − j) = k′.Thus both i′ − i 6 k′ and j′ − j 6 k′, and hence I 6 i + K, J 6 j + K, for ifi1 ∈ I = i+ k′ with k′ > K then by the above there is i′ > I with i′ − i 6 ii − iso i1 ∈ I is greater than i′ > I, a contradiction. Similarly for j. At least one ofthe inequalities I 6 i+K, J 6 j+K must be an equality as K is the derivativeof either I or J or both.

The other case is similar.

For a given positive K closed under addition there are many cuts I with∂I = K. Cuts of the form a + K or a − K are examples. So are J + K andJ ⊕ K when ∂J 6 K. Others can be obtained from a tree argument as inExample 2.18. It is natural to ask, given I and K = ∂I, whether and how Imay be ‘simplified’ to J +K or J ⊕K with ∂J < K. The following two resultsshow that this can only be done in particular circumstances, and if so we mayassume without loss that J = j ∈ Z is not a cut.

Proposition 2.28. If J+∂I = I with J 6 I then either I = J or else I = j+∂Ifor some j 6 J .

Proof. Let K = ∂I and assume that I is not j + K for all j 6 J . Then givenarbitrary j0 6 J there is i > j0 + K in I and hence j1 ∈ J and k1 ∈ Kwith j1 + k1 > i. It follows that j1 − j0 > K and hence ∂J > K. ThereforeI = J + ∂I 6 J + ∂J = J .

Proposition 2.29. If J⊕∂I = I with J 6 I then either I = J or else I = i+∂Iand J = i− ∂I for all i 6 I.

Proof. Let K = ∂I and assume that J i+k for all j′ > J , k′ > K > k. Thus i− j′ 6 k′−kfor all such j′, k, k′ showing that i − j′ 6 K for all j′ > J . It follows thatJ = i−K and I = (i−K)⊕K = i + K.

Finally, for this section, we show how to use ⊕ and ∂ to determine whenI + K = J + K for I 6 J .

Proposition 2.30. Given I 6 J and K in Z we have I + K < J + K if andonly if one of the following holds.

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(a) I ⊕ ∂(K) < J .

(b) For some j 6 J and k ∈ K, I + ∂(K) = j − ∂(K), I ⊕ ∂(K) = j + ∂(K)and K = k + ∂(K).

Proof. Suppose I 6 J and K are given.If I ⊕ ∂(K) < J we show that I + K < J + K. Let j 6 J and i′ > I,

k′ > ∂(K) be such that i′ + k′ < j. Since k′ > ∂K there is k 6 K withk + k′ > K. Now i′ + k′ + k < j + k 6 J + K but i′ > I and k′ + k > K soI + K < J + K.

Similarly, if j 6 J and k ∈ K with I+∂(K) = j−∂(K), I⊕∂(K) = j+∂(K)and K = k + ∂(K) then I +K = I + k + ∂(K) = k + j− ∂(K) does not containj + k, so I + K < J + K.

Conversely, suppose I ⊕ ∂(K) > J and I + K < J + K. We shall show theexistence of j, k as given in (b). The proof will give a little more informationconcerning this exceptional case.

We may assume ∂I 6 ∂K, since if ∂I > ∂K we would have I ⊕ ∂K = Isince each for each i′ > I there is c′ with ∂I > c′ > ∂(K) and i′ − c′ > I. Thismeans that i′ = (i′ − c′) + c′ is the sum of an element of Ic and an element of∂(K)c, so this would show that J 6 I ⊕ ∂(K) 6 I 6 J .

Next, we show there is k ∈ K with K = k + ∂(K). Take arbitrary j ∈ Jand k ∈ K. We attempt to show that there is some i ∈ I and k1 ∈ K withi + k1 = j + k. Suppose first that k + ∂(K) < K. Then there is c′ > ∂K withk + c′ 6 K. Since c′ > ∂I there is i ∈ I J . Therefore j + k J , then for each j 6 J and k 6 K there is i ∈ I and c ∈ ∂(K)with i + c = j so j + k = i + c + k ∈ I + K as c + k ∈ K so I + K > J + K. Sowe may assume I + ∂(K) < J and in particular I + ∂(K) < I ⊕ ∂(K).

So we have I + ∂(K) < J 6 I ⊕ ∂(K) and ∂I 6 ∂(K). Therefore thereis j ∈ J \ I + ∂(K) and for any such j we have I + ∂(K) = j − ∂(K) 6 j 6j + ∂(K) = I ⊕ ∂(K) and J 6 j + ∂(K). This is what we were required toprove.

3 Subtraction and duality

The unary subtraction operation −I has already been defined. It is natural toextend this to a binary subtraction operation as usual. In fact, as well as I − Jwe also have a dual operation to this.

Definition 3.1. Given cuts I, J ∈ Z with I > J , define

• I − J = I + (−J) = {i− j′ : i 6 I, j′ > J},

• I J = I ⊕ (−J) = −(J − I) = inf{i′ − j : i′ > I, j 6 J}.

The following collects together properties of − and that follow from pre-vious results about addition.

Proposition 3.2. Let I, J ∈ Z. Then:

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(a) I − 0 = I 0 = I;

(b) I − J 6 I J ;

(c) ∂(I − J) = ∂(I J) = max(∂I, ∂J);

(d) I − J = I J whenever ∂I 6= ∂J ;

(e) I − J < I J iff ∂I = ∂J and I − J = a − ∂I, I J = a + ∂I for some(equivalently all) I − J < a < I J .

Proof. By Propositions 2.6, 2.8, 2.21, 2.25 and 2.22.

Remark 3.3. It might be argued that the definitions of I − J and I J arenot the ‘correct’ ones as they give rather naıve upper and lower bounds for thedifference, and do not give the finer details one expects. In contrast to thesedefinitions, the following might be proposed for I − J .

inf{sup{i′ − j′ : j′ > J} : i′ > I}sup{inf{i′ − j′ : i′ > I} : j′ > J}sup{inf{i− j : j 6 J} : i 6 I}inf{sup{i− j : i 6 I} : j 6 J}.

However, it is easy to check that each of these alternative definitions give a valuebetween I − J and I J , so by Proposition 3.2(d) are equal to both I − J andI J when ∂I 6= ∂J . Also, if ∂I = ∂J then each of the alternative definitionshas the same derivative, as can be checked, and hence by Proposition 3.2(e) areall equal to one of the two values I − J and I J . Therefore it seems thatthe extra complexity of the alternative definitions is not required in the generaltheory.

Note that Proposition 3.2(c) above highlights one aspect in which the deriva-tive is not like a valuation function. As ∂(I−J) = ∂(IJ) = max(∂I, ∂J), thederivative operation is rather well-behaved on the difference even when ∂I = ∂J .(In valuation theory, the valuation of a difference of two points with the samevalue, can be anything below this value.) On the other hand, it is preciselywhen ∂I = ∂J that we have difficulty evaluating I − J and I J .

One special case for I − J and I J is when I = J . This is given next.

Proposition 3.4. For I ∈ Z we have I − I = −∂I and I I = ∂I.

Proof. If k 6 −∂I then −k > ∂I so there is i ∈ I with i′ = i − k > I hencek = i− i′ 6 I − I. Conversely if −k 6 I − I then −k = i− i′ for some i 6 I 6 i′

hence k 6 ∂I. The case for I I is dual, and I I = −(I − I) = ∂I.

We also have the following easy identities.

Proposition 3.5. For all I, J,K ∈ Z

(a) I − (J ⊕K) = (I − J)−K

(b) I (J + K) = (I J)K

(c) (I + J)−K = I + (J −K)

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(d) (I ⊕ J)K = I ⊕ (J K)

Proof. For the first, note that I−(J⊕K) = I−−(−J+−K) = I+(−J+−K) =(I − J)−K. The others are similar.

If I⊕∂I was a canonical cut greater than equal to I, with similar properties,the corresponding (dual) downward operation is I − ∂I.

Proposition 3.6. Let I be a cut. Then I ∂I = I. Also I − ∂I 6 I withequality except in the case when for some (equivalently for all) I − ∂I < a < Iwe have I − ∂I = a− ∂I, I = a + ∂I.

Proof. This is the dual of Proposition 2.23. For example, I ∂I = −(∂I − I) =−(∂(−I) + (−I)) = − − I = I and I − ∂I = −(∂I I) = −(∂(−I) ⊕ (−I)) 6−− I = I.

Proposition 3.7. Given I 6 J and K in Z we have I⊕K < J ⊕K if and onlyif one of the following holds.

(a) J − ∂K > I, or

(b) for some i′ > I and k′ > K, J ∂K = i + ∂K, J − ∂K = i − ∂K andK = k − ∂K.

Proof. Dual to Proposition 2.30.

Proposition 3.8. Let I, J ∈ Z. Then

(I J)⊕ J = I ⊕ ∂J

and

(I − J) + J = I − ∂J.

Proof. Let K = ∂J . Then

(I J)⊕ J = inf{(i′ − j) + j′ : i′ > I, j ∈ J 6 j′}= inf{i′ + k′ : i′ > I, k′ > K}= I ⊕K.

Also,

(I − J) + J = sup{(i− j′) + j : i 6 I, j 6 J 6 j′}= inf{i− k′ : i 6 I, k′ > K}= I −K

as required.

Proposition 3.9. Suppose I,K ∈ Z are cuts with ∂I < ∂K. Then

I − ∂K = I ∂K < I < I + ∂K = I ⊕ ∂K

and I⊕∂K = x+∂K, I−∂K = x−∂K for any x with (I−∂K) < x 6 (I⊕∂K).

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Proof. For simplicity suppose K = ∂K. Obviously I 6 I + K 6 I ⊕ K andI−K 6 IK. Let k ∈ K and i ∈ I with i′ = i+k > I. Then i = i′−k > IKshowing I K < I. For these i, i′, k, if i1 ∈ I, k1 ∈ K then i+ k > i1 so i1 + k1is at most i + (k + k1) showing I + K = i + K, and since I + K is an upwardsK-limit, I ⊕ K = I + K. Also, if k′1 > K then i − k′1 ∈ I − K showingi−K = I −K = I K as ∂(I −K) = ∂(I K) = K.

We can also write down duals of Propositions 2.28 and 2.29.

Proposition 3.10. If J − ∂I = I with J > I then I = J of J = j + ∂I,I = j − ∂I for all j ∈ J \ I.

Proof. From the assumptions and duality we have −J ⊕ ∂(−I) = −I with−I > −J and the result follows by Proposition 2.29.

Proposition 3.11. If J ∂I = I, J > I then I = J or I = j − ∂I some j ∈ J .

Proof. From the assumptions we have −J + ∂(−I) = −I with −I > −J . Nowapply Proposition 2.28.

The following proposition evaluates all terms of the form ‘(I minus J) plus J ’with addition and subtraction being either the normal +,− or the dual form,⊕,.

Proposition 3.12. Let I, J ∈ Z and let K = max(∂I, ∂J).

• If K = ∂J > ∂I we have I − J = I J and (I J)⊕ J = I ⊕K = i + Kfor some i ∈ I, (I − J) + J = I −K = i−K for some i ∈ I.

• If K = ∂I = ∂J then I − J = I J except when I − J = x − K andI J = x + K for some (equivalently all) x ∈ (I J) \ (I − J). Also(I J) ⊕ J = I ⊕ K, which equals I unless I is a downwards K-limit;(I J)⊕ J = I −K, which equals I unless I is an upwards K-limit; and(I J) + J , (I − J)⊕ J are both equal to one of these two.

• If ∂I > ∂J then I − J = I J and (I J)⊕ J = (I J) + J = I.

Proof. If K = ∂J > ∂I Proposition 3.8 applies. Let I ⊕K = I + K = i + Kand I K = I − K = i − K with i ∈ I, k ∈ K and i′ = i + k > I. ThenI − J = sup{i1 − j′1 : i1 ∈ I, j1 > J} = sup{i − j′1 : j1 > J} since each i1 ∈ Iis bounded above by i + k for some k ∈ K = ∂J . so I − J = i − J . SimilarlyI J = i′ − J . But as i′ = i + k, k ∈ ∂J , these are equal.

If K = ∂I = ∂J , and suppose either IJ is not an upwards K-limit or I−Jis not an downwards K-limit. Then there are IJ > x > x−K > y > I−J forsome x, y, and as K is closed under addition there is k′ > K with 2k′ < x− y.Now take i I J > x andi− j′ < I −J < y. But (i′− j)− (i− j′) = 2k′ < x− y which is a contradiction.A previous calculation (Proposition 3.8) shows that (I J) ⊕ J = I ⊕K andProposition 2.23 identifies when this equals I; similarly (IJ)⊕J = I−K, byProposition 3.8, and Proposition 3.9 identifying when this is equal to I. Finallyand (I J) + J , (I − J)⊕ J are between these two with the same derivative, soeach must be equal to one of these two.

If K = ∂I < ∂J there is j ∈ J and k ∈ K with jk = j′ > J . It is easy tocheck from this that I − J = sup{i − j′ : i ∈ I}, I J = inf{i′ − j : i′ > I},

14

and that these are equal because j′ = k + j and I is closed under addition byk. Thus (I J)⊕ J = I ⊕ ∂J = I and (I − J)− J = I − ∂J = I.

Example 3.13. Let K be a proper cut closed under addition and a > b ∈ M .Then (a+K)−(b+K) = sup{(a+k)−(b+k′) : k < K < k′} = sup{(a−b)−k′ :K < k′} = (a− b)−K. Other calculations can be carried out in the same way,we list them here.

(a + K)− (b + K) = (a− b)−K

(a + K) (b + K) = (a− b) + K

(a + K)− (b−K) = (a + K) (b−K) = (a− b) + K

(a−K)− (b + K) = (a−K) (b + K) = (a− b)−K

(a−K)− (b−K) = (a− b)−K

(a−K) (b−K) = (a− b) + K

However, as in Example 2.18, these are not the only ways to get cuts I, J withgiven derivative K and I J > I − J .

Proposition 3.14. Given I ∈ Z and I < x < y ∈ M , the following areequivalent: x + I = y + I; x− I = y − I; and y − x ∈ ∂I.

Proof. We show x+I = y+I iff y−x ∈ ∂I. The other is similar. If x+I = y+Ithen for all i ∈ I there is i1 ∈ I with y + i = x+ i1 so i+ (y− x) ∈ I and hencey−x ∈ ∂I as i was arbitrary. Conversely if y−x ∈ ∂I then for all i ∈ I there isi1 = i+(y−x) ∈ I such that x+ i1 = x+(y−x)+ i = y+ i, so x+I 6 y+I.

Using what we have learnt about addition and subtraction we shall nowsolve the following: given I, J ∈ Z we wish to find all possible X ∈ Z withX +J = I. In the case ∂J > ∂I there will be no solutions X. So we will assumeK = ∂I > ∂J .

Under these assumptions we have (I − J) + J = I − ∂J , by Proposition 3.8.If ∂J < ∂I or if I is not an upwards K-limit then I − ∂J = I and we have asolution X = I − J to X + J = I.

So assume now that I = i+K and ∂J = ∂I = K. Then I −K = i−K. Wetry X = I J . Here, (I J) + J 6 (I J)⊕ J = I so (I J) + J = I unlessadditionally J = j′ −K for some j′ > J , by Proposition 2.25.

This proves one direction of the following.

Proposition 3.15. Suppose I, J ∈ Z. Then there is a solution (X = I − J orIJ) to X +J = I if and only if ∂J 6 ∂I and the following holds: it is not thecase that ∂I = ∂J , I is an upwards ∂I-limit and J is a downwards ∂J-limit.

Proof. One direction has been shown. For the other, suppose X + J = I,K = ∂I = ∂J , I = i + K and j = j′ −K. Then for some k′ > K and x 6 Xwe must have x + j′ − k′ = i. But then by overspill there is k′′ > K withk′− k′′ > K so x+ j′− (k′− k′′) = i+ k′′ > i+K contradicting X +J = I.

Proposition 2.30 says something about the uniqueness about such X. IfX1 +J = I and X2 +J = I with X1 6 X2, then for ∂J < ∂I we have X1 = X2.This is because Proposition 2.30 implies X1 +∂J > X2. But ∂J < ∂X1 = ∂I soX1 + ∂J = X1 > X2 hence X1 = X2. On the other hand, if ∂I = ∂J , X1 6 X2

15

and X1 + J = I, X2 + J = I, Proposition 3.14 says that X1, X2 must be ‘close’in the sense that for all X1 6 x1 < x2 6 X2 we have x2 − x1 6 ∂J = ∂I,but leaves open the possibility that they may not be identical. Proposition 2.30gives the exact result.

Proposition 3.16. Suppose I, J ∈ Z and ∂J < ∂I. Then:

(a) if ∂I = ∂J the solution to X + J = I given by Proposition 3.15 when itexists is unique;

(b) if ∂I > ∂J and X+J = I then any X ′ such that ∂X ′ = ∂I and X ′⊕∂J >X, X ⊕ ∂J > X ′ is also a solution.

Proof. For (b) just use Proposition 2.30, noting that as ∂J < ∂I = ∂X = ∂X ′

we have X + ∂J = X ⊕ ∂J and similarly for X ′.

4 Rationals and the logarithm

Additional structure on Z, in particular for multiplication and division, becomescomplicated by the fact that our elements of Z are integers, and so we oftenneed to take integer parts. The theory goes more smoothly when we pass tothe field of fractions Q = Q(M) of Z(M) and initial segments I ⊆e Q. Themain difference is that the order on Q is dense, rather than discrete, and thischanges the formal details of some definitions though the spirit of the definitionsremains the same. In particular for each r ∈ Q there is an initial segment with nomaximum element corresponding to r, i.e. the initial segment {p ∈ Q : p < r}.

Definition 4.1. We let Q denote the set of all initial segments of Q withoutmaximum element and identify each r ∈ Q with {p ∈ Q : p < r} ∈ Q. Wecontinue to use I, J,K as ranging over initial segments, the context indicatingwhat these are initial segments of.

Complementation of initial segments and end segments is defined in Q ina symmetric way by Ic = {q ∈ Q : ∃r 6∈ I (q > r)} for initial segments I andEc = {q ∈ Q : ∃r 6∈ E (q < r)} for end segments E. The order relation 6 on Qis set inclusion and for X ⊆ Q we define

inf X =⋃{x ∈ Q : ∀y ∈ X x < y}

and

supX =⋂{x ∈ Q : ∃y ∈ X x < y}.

We can also consider the ‘localisation’ of Q to a single denominator n. Theresult is similar to Z.

Definition 4.2. For positive n ∈ Z we let Qn = {z/n : z ∈ Z} ⊆ Q andQn the set of initial segments of Qn, possibly with maximum element. ThusZ = Q1 and Z = Q1. Because elements of Qn may have maximum elements,complementation is defined in a way similar to that in Z.

To each I ∈ Qn we associate IQ = sup I = inf(Ic) in Q, the inf and suptaking place in Q and each r ∈ Q being identified with its set of predecessors,as mentioned above. Similarly, if E is an end segment of Qn we define EQ =(supEc)c = (inf E)c.

16

The set Q has its own additive structure, with relations and operations+,⊕,−,, ∂,< and 6 on Q all defined in an formally identical way to theseoperations of Z using the particular operations of inf, sup and complementationin Q. Similarly, +,⊕,−,, ∂,<,6 are all defined on Qn for each positive n ∈Z. The additive structure of Qn is obviously isomorphic to that of Z via themap z 7→ z/n. The relationship between each Qn and Q is given by the nextproposition.

Proposition 4.3. For each positive n ∈ Z, The map I 7→ IQ from Qn to Q isa embedding preserving all additive structure from +,⊕,−,, ∂,<,6.

Proof. This is just axiom checking, and easy. The map is easily seen to be 1–1.Many statements are obvious, such as: I 6 J implies IQ 6 JQ and I + J = Kimplies IQ + JQ = KQ. That (−I)Q = −(IQ) follows from the definition of−I as {−n : n ∈ Ic} and the fact that IcQ = IQ

c. Hence I ⊕ J = K implies−I + −J = −K which implies −IQ + −JQ = −KQ and hence IQ ⊕ JQ = KQ.The definition of ∂ does not cause difficulties either.

Because of this, when passing from Qn to Q, we shall identify each I ∈ Qn

with IQ. Thus our additive theory up to this point can be applied to each Qn

and to Q as well.In the other direction, each I ∈ Q determines two (possibly different) initial

segments In = infQn Ic and In = supQnI in each Qn, and this of course will

occasionally require some care.It will be helpful to have a definition of logarithms on Q. Any base could be

taken, but for simplicity we will use base e.

Definition 4.4. We denote by Q+

the set of I ∈ Q with I > 0. Our base modelM � PA can describe bounds on logarithms up to rational numbers in Q, and

we define the logarithm operation log : Q+ → Q by

log I = inf{q ∈ Q : ∀p 6 I (p > 0→ q > loge p)}.

It is easy to check that this logarithm can also be written as

log I = sup{q ∈ Q : ∃p 6 I (p > 0 ∧ q 6 loge p)}.

(Recall: by the definition of inf and sup in Q these cannot contain maximumelements.)

Proposition 4.5. The map log : Q+ → Q is 1–1 and onto.

We denote the inverse log−1 by exp.

Proposition 4.6. Given I > 1 is in Q,

(a) I is closed under addition iff log I is closed under successor; and

(b) I is closed under multiplication iff log I is closed under addition.

The log and exp functions allow us to extend our additive theory to other op-erations, including multiplication. Multiplication and division is covered brieflyin the next section. We conclude this section with a discussion of derivative.

17

Definition 4.7. The derivative ∂ : Q→ Q is defined in the same way as on Z,

as ∂I = {x ∈ Q : ∀y ∈ I x + y ∈ I}. We also denote ∂ by ∂1

and define

∂n+1

I = exp(∂n(log I))

for all I ∈ Q for which ∂n(log I) is defined.

Here, ∂1(log I) is defined for I > 0, ∂

2(log I) for I > 1, and so on, so ∂

2I is

defined for all I > 0. Also, since ∂I > 0 for all I, we have ∂2I > 1 for all I > 0.

∂2I is the multiplicative analogue of ∂I.

Proposition 4.8. For I > 0 in Q, ∂2I = {x ∈ Q : ∀y ∈ I x · y ∈ I}.

In other words, ∂2I is the limit of i′/i where i′ > I > i.

Corollary 4.9. For I > 0 in Q, ∂(log I) = log(∂2I).

Proof. This is almost directly from the definition of ∂2: ∂

2I = exp(∂(log I))

hence ∂(log I) = log(∂2I).

Note that if I > 1 is closed under multiplication then ∂2I = I so ∂(log I) =

log I. Other cases for calculating ∂ log I will be given in the next section.

Example 4.10. Given q ∈ Q and K ∈ Q with ∞ > K > 0 and working in Qwe can define

q + K = sup{q + k : k < K} = inf{q + k′ : k′ > K}q −K = sup{q − k′ : k′ > K} = inf{q − k : k < K}qK = sup{qk : k < K} = inf{qk′ : k′ > K}q/K = sup{q/k′ : k′ > K} = inf{q/k : k < K}.

The first two of these just use additive structure of Q and ∂(q+K) = ∂(q−K) =

K, log(q + K) = sup{log(q + k) : k < K}, etc., and ∂2(q + K) = sup{(q +

k′)/(q + k) : k < K < k′}, with little extra information available on thesein general. On the other hand, one can calculate log(qK) = log(q) + log(K),

log(q/K) = log(q)− log(K), and ∂2(qK) = ∂

2log(q/K) = ∂

2K.

Example 4.11. For 0 < K < 1 in Q one has ∂2K = inf{k′/k : k′ > K > k} so

1 6 ∂2K 6 1/K. In general 1 6 ∂

2K 6 max(K, 1/K) for K > 0.

5 Multiplicative structure and beyond

Unsurprisingly, the structure of cuts in Z, Qn or Q with addition and multipli-cation combined is rather complicated, and most calculations seem to requirelooking at a large number of cases. Where possible, problems should be solvedin the muliplicative domain by using logarithms to reduce to additive problems,but there are limitations to this approach. This section concludes the paperwith an incomplete overview of the multiplicative theory.

As for addition and subtraction, there are two obvious definitions of multi-plication, depending on whether one looks from below or from above. And asfor Dedekind’s construction of the reals, the definitions are made slightly morecomplicated because we need to consider the cases of positive and negative cutsseparately.

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Definition 5.1. Let I, J ∈ Z where I > 0 and J > 0. Then we define

I · J = sup{ij : 0 6 i ∈ I, 0 6 j ∈ J}

and

I � J = inf{i′j′ : i′ > I, j′ > J}.

We extend this to other X,Y ∈ Z using the unary subtraction operation asfollows. Assuming that I > 0 and J > 0, we let

I · (−J) = (−I) · J = −(I � J),

I � (−J) = (−I)� J = −(I · J),

(−I) · (−J) = (I � J)

and

(−I)� (−J) = (I · J).

Our guiding principle here is that X · Y is always a supremum and X � Yis always a (larger) infimum.

Proposition 5.2. For all X,Y ∈ Z, X · Y 6 X � Y .

In the sequel we will mainly consider structural properties of positive I, J .This is for simplicity: results on other cuts can be read off using the abovedefinitions and use of the absolute value operation. For example, it is easy tocheck that ∂(aI) = |a|∂I.

Example 4.10 defines 1/I for positive I ∈ Q. We may also define 1/(−I) =−(1/I). This enables us to define two division operations on cuts I, J .

Definition 5.3. For I, J ∈ Q, (or via the usual identifications in Qn, Z) wedefine I/J and I � J by

I/J = I · (1/J) and I � J = I � (1/J).

Note that division yields cuts in Q. These may be approximated in Qn, Zwith possible ambiguity and/or loss of precision.

A number of straightforward basic results hold for multiplication and divisionby translation between the multiplicative and additive structure using the logand exp functions.

Proposition 5.4. Let I > 0 be in Q. Then ∂2I is closed under multiplication.

Proposition 5.5. Let I, J > 0 be in Q. Then I · J = exp(log I + log J) andI � J = exp(log I ⊕ log J).

Proposition 5.6. Let I, J > 0 be in Q. Then I · J = I � J unless ∂2I = ∂

2J

and for K = ∂2I and any a ∈ I � J \ I · J we have I · J = a/K, I � J = aK.

As expected, examples where · and � differ do occur: let K ∈ Z be a positivecut closed under multiplication. Then one can calculate aK · b/K = (ab)/Kwhereas aK � b/K = (ab)K.

Probably the most interesting results here concern combinations of the fam-ilies of additive and multiplicative operations.

The next result is analogous to the familiar result d/dx(log y) = (1/y)(dy/dx).We will see that it is not in general particularly exact.

19

Proposition 5.7. Let I ∈ Q be such that I > 0. Then

∂I · 1

I6 ∂ log I 6 ∂I � 1

I.

Proof. Let k ∈ ∂I · (1/I) and i 6 I. To show log(i) + k 6 log I we may assumek 6 di/i′ where di ∈ ∂I and i′ > I. Note too that 2di = di + di 6 I sodi/i′ < 1/2. Then

log(i) + k 6 log(i) + di/i′ 6 log(i + idi/i′)

which holds by the familiar alternating power series for log(1 + x) (formalisedand proved in PA) showing log(1 + x) 6 x for 0 < x < 1/2 and di/i′ < 1/2.But then i/i′ 6 1 and so i + idi/i′ 6 i + di 6 I, so log(i) + k 6 log I.

For the other inequality, suppose k′ > ∂I � I. Then there are di′ > ∂I andi ∈ I with i + di′ > I and k > di′/i. So

log(i) + k > log(i) + di′/i > log(i + di′) > log I

as required.

Corollary 5.8. For I > 0 we have

∂I · 1

I6 log ∂

2I 6 ∂I � 1

I.

Example 5.9. Let N ⊆e Z be a positive cut containing N and closed undermultiplication. Let a > N in Z. We look at the cuts I = aN and J = a/N .

For I = aN we can calculate ∂2I = inf{n′/n : n′ > N > n} = N , since N

is closed under multiplication. Also ∂I = aN , ∂I/I = sup{an/an′ : n′ > N >n} = 1/N, and ∂(I) � I = inf{an′/an : n′ > N > n} = N , again by closureunder multiplication. Also log(aN) = log a + logN so ∂ log(aN) = logN sincelogN is closed under addition.

Similarly for J = a/N , ∂2J = inf{(a/n)/(a/n′) : n′ > N > n} = N , ∂(J) =

a/N , 1/(a/N) = N/a, ∂(J)/J = sup{(a/n′) · (n/a) : n′ > N > n} = 1/Nand ∂(J) � J = inf{(a/n) · (n′/a) : n′ > N > n} = N . Also log(a/N) =inf{log(a/n) : n ∈ N} = inf{log(a) − log(n) : 0 < n ∈ N} = log(a) − logN , so∂ log(a/N) = logN .

The derivatives of products I · J , and of powers of I are interesting.

Proposition 5.10. Suppose I, J > 0. Then either ∂(I · J) = ∂(I � J) or

∂(I · J) = a/K and ∂(I · J) = aK for K = ∂2I = ∂

2J and I · J < a 0 and K = ∂

2I = ∂

2J . Then

a direct calculation (using the closure of K under multiplication) shows that∂(a/K) = a/K and ∂(aK) = aK.

The following is analogous to the product rule for differentiation.

Proposition 5.11. For I, J > 0 in Q,

max{I · ∂J, ∂I · J} 6 ∂(I · J) 6 ∂(I � J) 6 max{I � ∂J, ∂I � J}.

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Proof. ∂(I · J) 6 ∂(I � J) holds by the previous result.For the first inequality, assume that I ·∂J 6 ∂I ·J and suppose i ∈ I, j ∈ J ,

di ∈ ∂I and j1 ∈ J . We must show ij + j1di ∈ I · J . Without loss we mayassume j = j1 for if j < j1 simply replace j by j1 and prove a stronger result.Then ij + jdi = (i + di)j ∈ I · J as di ∈ ∂I.

For the other inequality, suppose I � ∂J 6 ∂I � J and choose di′ > ∂I,j′ > J . So j′di′ is a ‘typical’ element in (∂I�J)c and it will suffice to find i ∈ Iand j ∈ J with ij + j′di′ > I � J . Choose i0 6 I 6 i′0 with i0 + di′ = i′0. Nowfrom I�∂J 6 ∂I�J there are i′ > I and dj′ > ∂J with i′dj′ 6 j′di′. Withoutloss we may assume i′ 6 i′0 and we let i = i0 − i′0 + i′ so i 6 I 6 i′ = i + di′.Now choose j ∈ J with j′ = j + dj′ > J . Hence

ij + jdi′ + i′dj′ = (i + di′)(j + dj′) = i′j′ > I � J.

Thus one or both of jdi′, i′dj′ is greater than or equal to ∂(I ·J). But jdi′ 6 j′di′

and i′dj′ 6 j′di′ so j′di′ > ∂(I · J), as required.

Note that I · ∂J and ∂I · J are closed under addition since ∂I, ∂J are, so

max{I · ∂J, ∂I · J} = I · ∂J + ∂I · J = I · ∂J ⊕ ∂I · J

Similarly for max{I � ∂J, ∂I � J}.

Example 5.12. If N > N is closed under addition and a > N , setting I = aNand J = a/N , then I, J and hence I · J , I � J are closed under addition andwe have ∂(I · J) = I · J = a2/N , and ∂(I � J) = I � J = a2N .

Observe that, in calculating powers of cuts by standard natural numberpowers, In = I · I · · · · · I or In = I � I � · · · � I it doesn’t matter whether onetakes · or � since two lower bounds i, j 6 I can be replaced by max(i, j), andsimilarly i′, j′ > I can be replaced by min(i′, j′). In other words, for I > 0 andstandard n ∈ N, In = sup{in : 0 I}. Hence we deducefrom the previous proposition that,

Corollary 5.13. For I > 0 in Q and n ∈ N, In−1 · ∂I 6 ∂(In) 6 In−1 � ∂I.

At present I am unable to give an example of a cut I > 1 such that In−1 ·∂Iand In−1 � ∂I are different.

For the derivative of 1/I, observe first the following.

Proposition 5.14. For I > 0 closed under + we have ∂(1/I) = 1/I.

Proof. For i 6 I 6 i′,

1

i′+

1

i′=

1

i′/2

so ∂(1/I) > 1/I. But ∂(1/I) 6 1/I holds by definition, hence the result.

We wonder if, more generally, it is true that ∂(1/I) = ∂I/I2. The followinggets close to proving this.

Proposition 5.15. For I > 0 we have

∂I · 1

I26 ∂

1

I6 ∂I � 1

I2.

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Proof. For di ∈ ∂I and i′ > I,

di · 1

i′2+

1

i′6

1

i′ − di

hence ∂I · (1/I2) 6 ∂(1/I). For the other inequality, given di′ > ∂I and i 6 I,we want i′ > I with 1/i′ + di′/i2 > 1/I. Choose (without loss of generality,increasing i if necessary) i′ = i + di′/2 > I so

1

i′+

di′

i2>

1

i′ − (i′2/i2)di′/2>

1

i′ − di′/2=

1

i>

1

I,

which follows from the series expansion 1/(x− h) = 1/x + h/x2 + h2/x3 + · · · ,as required.

We conclude with some results concerning the interesting cut exp(1/N) =inf{ n√

e : n ∈ N}.

Proposition 5.16. Suppose I ∈ Q and I > 0. Suppose also that ∂2I >

exp(1/N). Then I is closed under addition.

Proof. By assumption ∂ log I > 1/N. Let 0 1/n for some n ∈ N, so ndl > 1. Then log I > log i + ndl, since log Iis closed under repeated addition by dl. But log i + ndl > log(i + indl) hencei + indl 6 I and so 2i 6 I.

Corollary 5.17. If I ∈ Q, I > 0 and ∂2I > exp(1/N) then ∂I · Ik = ∂I � Ik

for all k ∈ N, and in particular ∂(Ik+1) = ∂I · Ik.

Proof. By the previous proposition I is closed under addition so I = ∂I and theresult is a triviality.

Example 5.18. Let I be the cut exp(1/N) in Q. We can check that ∂2I =

exp ∂(1/N) = exp(1/N) = I. Also, given n′ > N we have 2e1/n′> 2 > e1/2 and

hence I is not closed under +.We can also show that ∂ exp(1/N) = 1/N. In one direction, suppose that

0 < x 6 1/n′ for some n′ > N and n′′ > N. Then we must find an upper boundfor exp(1/n′′)+x. But in PA one can prove that for sufficiently large z > y > 0,exp(1/(y/3)) > exp(1/y) + 1/z, since by expanding exp(1/y) + 1/z = 1 + 1/y +1/2y2 + · · ·+ 1/z which is at most 1 + 3/y. Thus exp(1/n′′) + x 6 exp(1/n′′) +1/n′ 6 exp(1/(n′′/3)) 6 exp(1/N as required, showing ∂ exp(1/N) > 1/N. Theother direction is easier, as given x = 1/n (n ∈ N) we have 2n/2 = 2 > exp(1/N)so for a 6 exp(1/N) we have a + 2nx > exp(1/N) > a hence for some i < nwe have a + (i + 1)x > exp(1/N) > a + ix showing x > ∂ exp(1/N). Note alsothat 1/N = 1/N2 exp(1/N), which is the answer for ∂ exp(1/N) expected fromelementary calculus.

References

[1] John Allsup and Richard Kaye. Normal subgroups of nonstandard symmetricand alternating groups. Arch. Math. Logic, 46(2):107–121, 2007.

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[2] Harvey Friedman. Countable models of set theories. In Cambridge SummerSchool in Mathematical Logic (Cambridge, 1971), pages 539–573. LectureNotes in Math., Vol. 337. Springer, Berlin, 1973.

[3] Richard Kaye. Models of Peano arithmetic. The Clarendon Press OxfordUniversity Press, New York, 1991. Oxford Science Publications.

[4] Richard Kaye. A Galois correspondence for countable recursively saturatedmodels of Peano arithmetic. In Automorphisms of first-order structures,pages 293–312. Oxford Univ. Press, New York, 1994.

[5] Richard Kaye. Infinitary definitions of equivalence relations in models of PA.Ann. Pure Appl. Logic, 89(1):37–43, 1997. Logic Colloquium ’94 (Clermont-Ferrand).

[6] Richard Kaye and Alan Reading. Counting external sets in a models ofarithmetic. In preparation, 2012.

[7] L. A. S. Kirby and J. B. Paris. Initial segments of models of Peano’s ax-ioms. In Set theory and hierarchy theory, V (Proc. Third Conf., Bierutowice,1976), pages 211–226. Lecture Notes in Math., Vol. 619. Springer, Berlin,1977.

[8] Alan Reading. Unpublished M.Phil thesis, University of Birmingham, 2012.

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the arithmetic of cuts in models of arithmeticweb.mat.bham.ac.uk/r.w.kaye/papers/arithcuts1.pdf ·...

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