BETTER NONSTANDARD UNIVERSES

WITH APPLICATIONS

R. JIN

Rutgers UniversityNew Brunswick, NJ 08903 USAandCollege of CharlestonCharleston, SC 29424 USAemail: rjin@math.rutgers.edu

1. Introduction

There are various reasons why some nonstandard universes are consideredto be better than others. Di�erent people may have di�erent opinions andmay choose di�erent nonstandard universes to work within for di�erentpurposes. We think it is reasonable to consider that a nonstandard uni-verse which possesses stronger power for deriving results in either standardor nonstandard mathematics, or which supplies more convenient tools sothat, in practice, some complicated derivation procedures admit signi�cantsimpli�cations, is better than the one which doesn't.

In the early time of nonstandard analysis what people needed from anonstandard universe is basically the existence of an in�nitesimal. Sincethe introduction of �-saturation (�-saturation was �rst singled out in [23]),nonstandard analysis has experienced great prosperity. Given an in�niteregular cardinal �, a nonstandard universe V is said to be �-saturated if ev-ery family of less then � internal sets in V with �nite intersection propertyhas non-empty intersection. @1-saturation is also called countable satura-tion. Countable saturation is one of the most commonly used propertiesin nonstandard analysis. Countable saturation gives us great conveniencein handling countable sequences of internal sets or countable constructionswith each step internal. For example, countable saturation laid down afoundation for the invention of Loeb measure construction by P. Loeb in[22], and guarantees the completeness of the nonstandard hull of a metricspace. Part of the reason why countable saturation o�ers so much was made

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clear by C. W. Henson and H. J. Keisler in [10], where they showed that:second-order nonstandard arithmetic with countable saturation implies ex-actly the same sentences of standard second-order arithmetic as are impliedby the system of standard third-order arithmetic.1 Bearing those reasonsin mind we consider that a nonstandard universe with countable saturationis better than the one without it.

Could we �nd even a better property for nonstandard universes? Bythat we mean to �nd a saturation-like property essentially stronger andevidently more useful than countable saturation for dealing with countablesequences of internal objects.

In x2 we introduce a property called the isomorphism property suggestedby C. W. Henson in [7]. Henson's isomorphism property is elegant, strongerthan countable saturation and very useful. In [8] Henson pointed out thatthe nonstandard universe satisfying the isomorphism property could be anultrapower2 of the standard superstructure by a result of S. Shelah [30].In both [7] and [8] Henson studied nonstandard hulls of Banach spaces ina nonstandard universe satisfying the isomorphism property. The readershould see there that the isomorphism property makes the subject simpleand clear. With three more sophisticated applications we illustrate more ofthe strength of the isomorphism property.

In x3 we push the issue further towards the direction of x2. We introducea property called the special model axiom suggested by D. Ross in [26] anda property called full saturation. The special model axiom is even strongerthan the isomorphism property and full saturation is the strongest amongthem. Ross mentioned in [26] that the nonstandard universe satisfying thespecial model axiom could be an ultralimit of the standard superstructure.But it couldn't be an ultrapower without extra set theoretic assumptionsbeyond ZFC. The existence of a fully saturated nonstandard universe iseven questionable. In fact, the existence of a fully saturated nonstandarduniverse is undecidable in ZFC. In x3 we present one application of thespecial model axiom and one application of full saturation. With thosetwo applications the reader should have a taste of the power of those twoproperties.

Besides the reasons we mentioned above, there might be some otherreasons why some nonstandard universes are better than others. For ex-ample, one may consider that a nonstandard universe is better if the hy-

1The complete result in [10] is: Let k > 2; n 2 N and 1 6 m 6 1. Then �PA(k) +

�k�1m �CA+i+

n -saturation and PA(k+n+1) +�k+nm �CA have the same consequences

in the language of PA(k).2By an ultrapower or an ultralimit in this chapter we always mean a nonstandard

universe of the standard superstructure obtained by a bounded ultrapower or a boundedultralimit construction, respectively.

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perreal �eld in it bears a stronger resemblance to the standard real �eld.In x4 we introduce such a property called the �-Bolzano-Weierstrass prop-erty for some uncountable regular cardinal �. By a result of H. J. Keislerand J. H. Schmerl in [21] one can construct a countably saturated nonstan-dard universe satisfying the �-Bolzano-Weierstrass property for some � e.g.� = (2@0)+. But the �-Bolzano-Weierstrass property is inconsistent withthe isomorphism property. This shows that the better-ness of nonstandarduniverses in x4 takes a di�erent direction from the better-ness in the senseof xx2,3.

Finally, we would like to mention, but not to include the details, onemore kind of nonstandard universes called Minimal Nonstandard Universe.A minimal nonstandard universe is an ultrapower of the standard super-structure modulo a selective ultra�lter on a countable set. Note that suchnonstandard universes are countably saturated. Some interesting applica-tions of minimal nonstandard universes to measure theory were obtainedby C. W. Henson and B. Wattenberg in [11] and by M. Benedikt in [2] and[3]. But the existence of a selective ultra�lter is undecidable in ZFC. Forfurther study on this subject the reader should consult [11],[2] and [3].

In the end of each section we include some exercises. Star symbols � areadded to indicate hard-ness.

Remark The de�nitions of the isomorphism property and the specialmodel axiom could be trivially generalized to the �-isomorphism propertyand the �-special model axiom for any in�nite regular cardinal �. In fact,it was the �-forms of those properties which were originally de�ned. In thischapter we need only the forms with � = @1 for all applications. Hence wedrop the cardinal � in the de�nitions for simplicity. Similar to the gener-alization of countable saturation to �-saturation the reader should be ableto generalize those properties, without any di�culties, to the �-forms forfuture applications.

We assume the reader has a basic training in model theory. A one-semester graduate level model theory course should be more than enough.Our notation in model theory are consistent with that in [5]. By a nonstan-dard universe in this chapter we mean always a triple (V(N );V(� N); �),where V(N ) is the superstructure on N , the set of all standard naturalnumbers as urelements, V(�N) is the superstructure on �N , and � is a non-standard extension from V(N ) to V(�N) de�ned in the �rst chapter by C.W. Henson in this volume. We call V(N ) the standard superstructure. Ourrestriction on the set of urelements for the standard superstructure to be Ninstead of an arbitrary set S is just for simplicity. Note that the standardsuperstructure V(N ) contains a copy of the standard real �eld. In order toavoid writing the whole expression (V(N );V(� N); �) we always denote V

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for a nonstandard universe. We write �; �; ; : : : for ordinal numbers anddenote ! for the �rst in�nite ordinal. For any set S we denote card(S) forthe (external) cardinality of S. For a model A with base set A we oftenwrite card(A) for the cardinality of A. We reserve the notion jAj for theinternal cardinality of A when A is an internal set. For each nonstandarduniverse V we denote �V for the cardinality of the family of all internal setsin V. Note that �V =

Sn2! card(

�Vn(N)). By a language we always meana �rst-order language.

In order to be coherent we would like to give equivalent forms of count-able saturation and �-saturation in next exercise, in terms of internallypresented structures. First we de�ne internally presented structures.

De�nition 1.1 Given a nonstandard universe V and given a language L.An L-structure A is called internally presented in V i� the base set andevery L-relation or L-function in A are internal in V.

Note that an internally presented structure itself may not be internalwhen L contains in�nitely many symbols.

1.1. EXERCISES

Exercise 1.2 Show that(1) a nonstandard universe V is countably saturated i� for any countable

language L, for every internally presented L-structure A, and for any set of�rst-order L-formulas �(x) with one free variable x consistent with Th(A),the set �(x) is realizable (or satis�able) in A.

(2) for any in�nite regular cardinal � a nonstandard universe V is �-saturated i� for any language L with card(L) < �, for every internallypresented L-structure A, and for any set of �rst-order L-formulas �(x)with one free variable x consistent with Th(A), the set �(x) is realizable inA.

2. The Isomorphism Property

De�nition 2.1 A nonstandard universe V is said to satisfy the isomor-phism property i� any two elementarily equivalent, internally presented L-structures for some countable language L, are isomorphic.

Remark Given two structures of some language, people might check theelementary equivalence between them by playing an Ehrenfeucht-Fraiss�egame or constructing a set of partial isomorphisms with back-forth property[1]. So if one want to show that two internally presented structures of somecountable language in a nonstandard universe satisfying the isomorphismproperty are isomorphic, he might need only to show that he could win the

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related Ehrenfeucht-Fraiss�e game or construct a set of partial isomorphismswith back-forth property.

From now on we will write IP for the isomorphism property.

Proposition 2.2 (C. W. Henson [7]) If V satis�es IP , then every in�niteinternal set has cardinality �V .

Proof: It su�ces to prove that any two in�nite internal sets have samecardinality. Let A and B be two internal sets. Then A and B are twoelementarily equivalent structures of the empty language. By IP there is abijection between them. 2

Proposition 2.3 (C. W. Henson [7]) If V satis�es IP , then V is countablysaturated.

Proposition 2.3 is a trivial corollary of Lemma 2.6. See the remark rightafter Lemma 2.6.

Proposition 2.4 (C. W. Henson [7] and S. Shelah [30]) Suppose � is anin�nite cardinal such that �i! = �.3 Then there is an ultra�lter F on �such that the ultrapower of the standard superstructure modulo F satis�esIP .

The �rst two applications need an equivalent form of IP in terms of therealizability of a \second-order type" due to S. Shelah and the author [16].

Let L be a countable language and X be a new n-ary relation symbolnot in L. Suppose �(X) is a set of �rst-order L[fXg-sentences and A is anL-structure with base set A. We say that �(X) is realizable in A i� thereexists R � An such that

(A; R) j= '(R)

for every '(X) 2 �(X).

De�nition 2.5 A nonstandard universe V is said to satisfy the resplen-dency property i� for any countable language L, for any n-ary new relationsymbol X not in L, for any internally presented L-structure A, and for anyset of �rst-order L [ fXg-sentences �(X), if �(X) [ Th(A) is consistent,then �(X) is realizable in A.

Lemma 2.6 (R. Jin and S. Shelah [16]) A nonstandard universe V satis�esIP i� V satis�es the resplendency property.

Remark Comparing with Exercise 1.2, it is easy to see that the resplen-dency property is a natural generalization of countable saturation. To get

3Note that i0 = @0, in+1 = 2in and i! =S

n<!in. Note also that the cardinality

of the standard superstructure is i!.

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countable saturation simply view the new relation symbol X in the de�ni-tion as an individual variable. This justi�es why we call IP a saturation-likeproperty.

2.1. UNLIMITED LOEB MEASURE SPACES

Given an internal ��nite additive measure space (;A; �) such that A con-tains every singleton set, �() > n for every n 2 N , and �(fxg) < 1

nfor

every x 2 and n 2 N . Following the Loeb construction we can extend thestandard �nitely additive measure �� on A to a countably additive measureL� on the �-algebra generated by A. But this measure is not complete. Inorder to make the measure complete, we have to throw in the subsets of allL�-measure zero sets. We need to de�ne a �-algebra of all \measurable"sets for this. For each S � let the outer measure of S be

��(S) = inff��(A) : A 2 A ^ S � Ag

and the inner measure of S be

��(S) = supf��(A) : A 2 A^A � Sg:

If L�() were �nite, then it would be immediate that a set S is measurablei� its outer measure and inner measure coincide. But now we have L�() =1 (this is what the word \unlimited" means). By a conventional methodwe de�ne the �-algebra B of all \measurable" sets as following:

B = fS � : (8A 2 A) (��(A) <1! ��(S \A) = ��(S \A))g:

Note that B is a �-algebra, A � B, and every subset of an L�-measure zeroset is in B.

When trying to extend L� to a complete measure on B one encounters aproblem. For an S 2 B should one let L�(S) = ��(S) or let L�(S) =

��(S)?There would have no problem if every S 2 B had same outer measureand inner measure. But this may not be the case. So one has at least twodi�erent ways to extend L� to a complete countably additive measure onB if there are some sets in B having di�erent outer measure and innermeasure.

Theorem 2.7 (R. Jin and S. Shelah [16]) Suppose V satis�es IP . Thenevery unlimited, non-atomic Loeb measure space has a measurable subsetwith in�nite outer measure and zero inner measure.

Proof: Given an unlimited, non-atomic Loeb measure space (;B; L�)generated by an internal space (;A; �) as above. We need to �nd a setS � such that for any A 2 A, if �(A) is �nite, then ��(A \ S) = 0, andif S � A, then L�(A) =1.

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We will use the resplendency property in the proof. The main idea isthe following: First we write a set of sentences �(X) to express that X isthe desired set, i.e. a measurable set with in�nite outer measure and zeroinner measure. Then we prove that �(X) is consistent with our internallypresented structure. By the resplendency property one could �nd a setrealizing �(X) in our structure. Therefore, the desired set exists.

We �rst form an internally presented structure A of a �nite languageLA. Let

A = ( [A [ �R; ;A; �R;2; �;\;r;+; �; <; 0; 1);

where ;A and �R are unary relations, 2 is a membership relation between and A, � is the measure function from A to �R, \ and r are Booleanoperators on A, and (�R; +; �; <; 0; 1) is the hyperreal �eld in V. Noticethat we abuse the notation here. Rigorously, we should use � 2; �\; �+;etc. We often omit � when the meaning is obvious. Let X be a new unaryrelation symbol not in LA. We now express X as a desired set by a set ofL [ fXg-sentences �(X). Let

�(X) = f�(X)g [ f n(X) : n 2 Ng [ f�n(X) : n 2 Ng;

where�(X) =: 8x(X(x)! (x));

n(X) =: 8U(A(U) ^ �(U) < n!9V (A(V ) ^ 8x(X(x) ^ x 2 U ! x 2 V ) ^ �(V ) < 1

n));

�n(X) =: 8U(A(U) ^ 8x(X(x)! x 2 U)! �(U) > n):

Note that in A, the elements n and 1nare de�nable. The sentence �(X)

says that X is a subset of , the sentence n(X) says that if U 2 A hasmeasure < n, then the outer measure of U \ X is < 1

n, and �n(X) says

that if U 2 A and X � U , then U has measure > n. So if the set �(X) isrealized by some set S in A, then S is clearly the desired set.

By the resplendency property we need only to prove that �(X) is con-sistent with Th(A).

Claim �(X) [ Th(A) is consistent.Proof of Claim: By Downward L�owenheim-Skolem theorem we can

�nd a countable LA-structure A0 such that A0 j= Th(A). Let 0[A0[R0

be the base set of A0. It su�ces to prove that �(X) is realizable in A0.So we want to �nd a set S0 � 0 such that S0 realizes �(X) in A0. LetfAn : n 2 Ng be an enumeration of the set

fA 2 A0 : 9n 2 N(�(A) < n)g:

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We choose xn inductively such that

xn 2 0 r ((n�1[

k=0

Ak) [ fxk : k < ng):

The set 0 r ((Sn�1k=0 Ak) [ fxk : k < ng) is non-empty because it has

unlimited measure. Let S0 = fxn : n 2 Ng. It is easy to check that (A0; S0)realizes �(X). 2

Remark: The result in the theorem was �rst proved by Henson [9] in anonstandard universe called a polyenlargement. Then Ross in [26] provedthe result by assuming the special model axiom. It is still open if one couldsettle the result by countable saturation or �-saturation for any in�niteregular cardinal �.

2.2. THE EXISTENCE OF BAD CUTS

Let H be a hyper�nite integer and let H = f0; 1; : : : ;H � 1g. H is called ahyper�nite time line in [20]. Let L� be the Loeb probability measure on Hgenerated by the internal normalized uniform counting measure �. Througha standard part map x 7! �( x

H) the hyper�nite time lineH together with L�

could be closely associated with the standard unit interval [0; 1] togetherwith Lebesgue measure on it. What about the order topology? Could onede�ne a topology on H which resembles the usual order topology on [0; 1]?Note that the natural order topology onH is discrete, hence not interesting.H. J. Keisler and S. C. Leth in [20] gave the de�nition of a U -topology onH for each cut U � H.

An in�nite initial segment U of �N is called a cut (or additive cut insome literature) if a + b 2 U for any a; b 2 U . Given a cut U � H, a setO � H is U -open i� for any a 2 O there exists a d 2 H r U such that[a� d; a+ d] � O, where [a� d; a+ d] = fx 2 H : a� d 6 x 6 a+ dg. AllU -open sets form a U -topology. Given a point x 2 H, by the U -monad of xwe mean the set fy 2 H : jy � xj 2 Ug. Note that no U -topology could beHausdor� because it could not separate two points within a U -monad. Butif we view each U -monad as a \point" and consider the natural order ofthose monads, then the resulting \order" topology is just the U -topology.

With a U -topology we can de�ne U -nowhere dense sets and U -meagersets (recall that a meager set is a countable union of nowhere dense sets).

De�nition 2.8 (H. J. Keisler and S. C. Leth [20]) A cut U is called a goodcut if there exists, in H, a U -meager set of Loeb measure one.

Recall that the standard unit interval has a meager set of Lebesguemeasure one. So a good cut U will make H together with Loeb measure

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and the U -topology much like [0; 1] together with Lebesgue measure andthe order topology.

Keisler and Leth proved in [20] that almost all cuts are good. They werealso able to construct a bad cut in some nonstandard universe under anassumption beyond ZFC in set theory, e.g. 2@0 < 2@1 . In next theorem weconstruct a bad cut in every nonstandard universe satisfying IP in ZFC.

Theorem 2.9 ([14]) If V satis�es IP , then every hyper�nite time line hasbad cuts.

In the proof we need an equivalent form of bad-ness from [20]. Givena cut U in �N , an internal increasing function f : f0; 1; : : : ; Lfg 7!

�N forsome Lf 2

�N is called a U -crossing sequence i� the set U\ff(n) : n 6 Lfgis upper unbounded in U .

Lemma 2.10 (H. J. Keisler and S. C. Leth [20], Proposition 4.5) For anycut U � H the following are equivalent.

(1) There is a U -meager set of Loeb measure one.(2) There is a U -crossing sequence f such that

X

n<Lf

f(n)

f(n+ 1)< 1:

Proof of Theorem 2.9: By Lemma 2.10 it su�ces to construct a cutU � H such that for every U -crossing sequence f the sum

X

n<Lf

f(n)

f(n+ 1)

is not �nite. The main idea of the proof is similar to the idea in the proofof Theorem 2.7. Let F be the set of all internal increasing functions f fromf0; 1; : : : ; Lfg for some Lf 2 H, to H. Then F is internal. We form aninternally presented structure A. Let

A = (H [F [ �R;H;F ; �R; R; S;+; �; <; 0; 1);

where H;F and �R are unary relations, R is a ternary relation such thatha; b; fi 2 R i� f 2 F , a 2 dom(f) and f(a) = b, S is a function from F to�R such that for any f 2 F

S(f) =X

n<Lf

f(n)

f(n+ 1);

and (�R; +; �; <; 0; 1) is the hyperreal �eld in V. Let L be the language ofA. Note that the following �rst-order L-sentences are true in A.

�n =: 9x(H(x) ^ x > n ^ 8y(H(y)! y 6 x))

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for each n 2 N , and

� =: 8f8x8y(F(f) ^H(x) ^H(y) ^ x < y !9g(F(g) ^ range(g) = range(f) \ [x; y]));

where the formula range(g) = range(f) \ [x; y] is an abbreviation of the�rst-order L-formula

8z(9uR(u; z; g)$ x 6 z ^ z 6 y ^ 9uR(u; z; f)):

Let X 62 L be a unary predicate symbol. We de�ne �(X) to be the set ofL [ fXg-sentences which contains exactly the following:

'1(X) =: 8x(X(x)!H(x))

'2(X) =: 8x8y(x 6 y ^H(x) ^X(y)! X(x))

'3(X) =: 8x8y(X(x) ^X(y)! X(x+ y))

n =: 8f(F(f) ^ 8x(X(x)!9y9z(R(y; z; f) ^X(z) ^ x 6 z))! S(f) > n)

for each n 2 N .Note that the sentences '1(X), '2(X) and '3(X) say that X is a cut

in H. The sentences n(X) for n 2 N say that if f is a crossing sequenceof X, then the internal sum S(f) is not �nite. So �(X) describes that Xis a bad cut by Lemma 2.10. So if �(X) is realizable in A, then H mustcontain a bad cut. By Lemma 2.6 it su�ces to show that �(X) [ Th(A) isconsistent.

Let A0 be a countable elementary submodel of A. Then Th(A0) =Th(A). If we show that �(X) is realizable in A0, then it is clear thatTh(A) [ �(X) is consistent.

Claim �(X) is realizable in A0.Proof of Claim: Let A0 = H0 [ F 0 [ R0 be the base set of A0 and let

F 0 = ffi : i 2 Ng. We now inductively construct an increasing sequencehai : i 2 Ni and a decreasing sequence hbi : i 2 Ni in H

0 such that for eachi 2 N

(a) ai < bi,(b) 2ai < ai+1,(c) bi=ai is not �nite in R

0 ,(d) if f 2 F 0 such that

range(f) = range(fi) \ fx 2 H0 : ai 6 x 6 big;

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if S(f) is �nite in R0 and if Lf is not �nite, then there is a k 2 f0; 1; : : : ; Lf�1g\H0 such that f(k) 6 ai+1 and f(k+1) > bi+1 (i.e., f has a jump acrossthe interval (ai+1; bi+1)).

We show �rst that the claim follows from the construction. Let

U = fx 2 H0 : (9i 2 N) (x 6 ai)g:

Then '1(U) and '2(U) are trivially true in (A0; U). The sentence '3(U) istrue in (A0; U) by condition (b). Given any fi 2 F

0 such that fi is a crossingsequence of U . To show that n(U) is true in (A0; U) for any n 2 N weneed only to show that S(fi) is not �nite. Suppose S(fi) is �nite. By thefact that � is true in A0 there exists a g 2 F 0 such that

range(g) = range(fi) \ fx 2 H0 : ai 6 x 6 big:

Then S(g) is also �nite because S(g) 6 S(fi). Since fi is a crossing sequenceof U , ai 2 U and bi 62 U , then g is also a crossing sequence of U . Hence Lgis not �nite (since no �nite sequence could be a crossing sequence of anycut). By condition (d) we know that g has a jump from ai+1 to bi+1, i.e.g(k) 6 ai+1 and g(k + 1) > bi+1 for some k 2 dom(g). So g can't be acrossing sequence of U , a contradiction.

We now do the inductive construction. Choose any a1 and b1 in H0 such

that b1=a1 is not �nite (for example, a1 = 1 and b1 = H). Suppose we havefound hai : i < ki and hbi : i < ki for some k > 1 such that they satisfy theconditions (a)|(d). We need to �nd ak and bk. Let g 2 F

0 be such that

range(g) = range(fk�1) \ fx 2 H0 : ak�1 6 x 6 bk�1g:

Case 1: S(g) is not �nite or Lg is �nite. Simply let a0k = ak�1 andb0k = bk�1.

Case 2: S(g) is �nite and Lg is not �nite. Let m 2 N be such thatS(g) < m. Since g is an element in A0 and A0 � A, then there is a t in A0

such that

t = minfg(n)

g(n+ 1): n 2 H0 ^ n < Lgg:

Let n0 2 H0 and n0 < Lg be such that t = g(n0)=g(n0 + 1). Then

tLg 6X

n<Lg

g(n)

g(n+ 1)= S(g) < m:

So we have g(n0+1)=g(n0) > Lg=m. Now let a0k = g(n0) and b0k = g(n0+1).

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Clearly, we have that b0k=a0k is not �nite. Let ak = 2a0k and bk = b0k � 1.

Then it is easy to see that bk=ak is still not �nite. Now it is obvious thatthe sequences

hai : i < k + 1i and hbi : i < k + 1i

satisfy conditions (a)|(d). 2

2.3. DEFINABILITY IN CONSTRAINT QUERY LANGUAGES

Although the following application is from database theory, we will presentit in a purely model theoretic way. No knowledge on database theory isassumed. The application is due to M. Benedikt et al. [4].

Let L be the language of ordered �elds and let L0 = fX1; : : : ;Xng,where Xi is an ni-ary relation symbol not in L. Let R be the standard real�eld.

By an L0-�nite expansion of R we mean an expansion of R to an L[L0-structure Re = (R; R1; : : : ; Rn) such that the interpretation e(Xi) = Ri ofevery symbol Xi in L

0 is a �nite relation in R.

For an L0-�nite expansion Re let the carrier of Re be the set

Ce = fr 2 R : 9Xi 2 L0 9(a1; : : : ; ani) 2 R

ni

(r 2 fa1; : : : ; anig ^ (a1; : : : ; ani) 2 e(Xi))g:

Clearly, Ce is �nite.

Two L[L0-sentences � and are equivalent over R i� for any L0-�niteexpansion Re,

Re j= � i� Re j= :

Given an L0-�nite expansion Re and an order-preserving injection

� : Ce 7! R;

let R�(e) be another L0-�nite expansion such that C�(e) = �[Ce] and

�(e)(Xi) = f(�(a1); : : : ;�(ani)) : (a1; : : : ; ani) 2 e(Xi)g:

It is easy to see that Re �L0 [ f<g and R�(e) �L

0 [ f<g, the reducts of Re

and R�(e) on L0 [ f<g, are isomorphic.

De�nition 2.11 An L [ L0-sentence � is order-invariant in R i� for anyL0-�nite expansion Re and for any order-preserving injection � : Ce 7! R

Re j= � i� R�(e) j= �:

13

Theorem 2.12 (M. Benedikt et al. [4]) For any order-invariant L [ L0-sentence � there exists an L0 [ f<g-sentence such that � and areequivalent over R.4

The proof of this theorem is long. The main idea of the proof is like thefollowing: By assuming the contrary we �rst �nd two L0-��nite expansions�Re and

�Re0 so that they agree on all L0 [ f<g-sentences but disagree on�, i.e

�Re j= ' i� �Re0 j= '

for every L0 [ f<g-sentence ' and

�Re j= � i� �Re0 j= :�:

Then we show, by the order-invariance of �, that �Re and�Re0 could be

chosen so that Ce and Ce0 are subsets of an L-indiscernible sequence in�R.

Finally, one could derive a contradiction by showing that �Re and �Re0

agree on �.

Proof of Theorem 2.12 Suppose the theorem is not true and let � bea witness. Let f'n : n 2 Ng be an enumeration of all L0 [ f<g-sentences.

Claim 1 For any m 2 N there are two L0-�nite expansions Re andRe0 such that they agree on every 'n for n < m and disagree on �.

Proof of Claim 1: Otherwise � will be equivalent to some Booleancombination of those 'n's for n < m. 2(Claim 1)

Let V be a nonstandard universe satisfying IP .By Claim 1 and Transfer Principle we can �nd two L0-��nite expansions

�Re and �Re0 such that they agree on every 'n for n < m and disagreeon �. By countable saturation one can let m be 1. So we have now twoL0-��nite expansions �Re and

�Re0 such that they agree on f'n : n 2 Ngand disagree on �. Let jCej = He and let jCe0 j = He0 .

Again by Transfer Principle and countable saturation one can �nd anL-indiscernible increasing internal sequence hri : i 2

�Ni in �R.

Claim 2 We can assume that Ce and Ce0 are the sets fri : i < Hegand fri : i < He0g, respectively.

Proof of Claim 2: For �Re there is an internal order-preserving bi-jection � from Ce to fri : i < Heg. By Transfer Principle and the order-invariance of � one has that �Re and

�R�(e) agree on �.�Re and

�R�(e)

agree also on f'n : n 2 Ng because the reducts of �Re and �R�(e) onL0 [ f<g are isomorphic. Same for �Re0 . 2(Claim 2)

4The original theorem in [4] is more general with R replaced by any in�nite o-minimalstructure.

14

Claim 3 There is an order-preserving bijection I from Ce to Ce0 suchthat

(a1; : : : ; ani) 2 e(Xi) i� (I(a1); : : : ; I(ani)) 2 e0(Xi)

for i = 1; : : : ; n.Proof of Claim 3: Note that \x 2 Ce" could be written as an L0-

formula. Because the submodel of �Re �L0 [ f<g generated by Ce and the

submodel of �Re0 �L0 [ f<g generated by Ce0 are elementarily equivalent,

by IP , there is an isomorphism I between them. (Here is the place we useIP .) 2(Claim 3)

Let LC be the language L[fci : i < Heg, where ci is a constant symbolnot in L. Note that �RC = (�R; ri)i<He and �RC0 = (�R; I(ri))i<He aretwo LC-structures by interpreting ci as ri or as I(ri), respectively.

Claim 4 �RC and �RC0 are elementarily equivalent.Proof of Claim 4: It is a straightforward consequence of indiscernibility

of ri's. 2(Claim 4)

Let LmC be the language LC [ fd1 : : : ; dmg and let Lm be the languageL [ fd1 : : : ; dmg, where di's are new constant symbols.

We need a property, called o-minimality, of real �elds in next claim. LetF be either a standard real �eld or a hyperreal �eld. for any L-formula�(x; �y) and any �r in F the set de�ned by �(x; �r) in F is a �nite unionof intervals. Let those intervals be maximal and let E�r

� be the set of allendpoints of those intervals. Clearly, E�r

� is �nite and every point in E�r� is

de�nable by some L-formula with parameters from �r.

Claim 5 Suppose �u = (u1; : : : ; um) and �v = (v1; : : : ; vm) are in�Rm

such that (�RC ; �u) and (�RC0 ; �v) are elementarily equivalent in LmC . Then(1) for every um+1 2

�R there is a vm+1 2�R such that (�RC ; �u; um+1)

and (�RC0 ; �v; vm+1) are elementarily equivalent in Lm+1C ,

(2) for every vm+1 2�R there is a um+1 2

�R such that (�RC ; �u; um+1)and (�RC0 ; �v; vm+1) are elementarily equivalent in Lm+1

C .Proof of Claim 5: By symmetry we need only to prove (1). Given um+1,

let �(x) be the set all LmC -formulas realized in (�RC ; �u) by um+1. If �(x)were countable, then vm+1 could be chosen easily by countable saturation.Unfortunately, �(x) is not countable. To overcome this di�culty we useo-minimality. We want to choose a countable set �(x) � �(x) so that forany �(x) 2 �(x) there is a �(x) 2 �(x) such that

(�RC ; �u) j= 8x(�(x)! �(x)):

Given any Lm-formula �(x; �y), where �y = fy1; : : : ; yng. Let

l� = max(fa : 9�r 2 Cne (a 2 E

�r� ^ a 6 um+1)g [ f�1g)

15

and let

r� = min(fa : 9�r 2 Cne (a 2 E

�r� ^ a > um+1)g [ f+1g):

Note that (1) max and min above exist because the sets are hyper�nite, (2)l� 6 um+1 6 r�, (3) if l� < um+1 < r�, then for any �r 2 Cn

e the formula�(x; �r) has a constant truth value in (l�; r�), (4) l� and r� are de�nablefrom some �r 2 Cn

e .Case 1: um+1 = l� or um+1 = r� for some Lm-formula �. Then um+1

is de�nable by some LmC -formula �(x). Let �(x) = f�(x)g.Case 2: l� < um+1 < r� for any Lm-formula �. Let ��(x) be an L

mC -

formula saying that l� < x < r� and let

�(x) = f�� : � is an Lm-formula.g:

Then �(x) is countable because there are only countably many Lm-formulas.Obviously �(x) is �nitely realizable in (�RC0 ; �v). By countable saturation�(x) is realized in (�RC0 ; �v) by some vm+1 2

�R.It is easy now to check that (�RC ; �u; um+1) and (�RC0 ; �v; vm+1) are

elementarily equivalent in Lm+1C . 2(Claim 5)

Note that in Claim 5 um+1 2 Ce i� vm+1 = I(um+1) 2 Ce0 .

Claim 6 Suppose (�RC ; �u) and (�RC0 ; �v) are elementarily equivalent.Suppose (�x) is a quanti�er-free L [ L0-formula. Then

�Re j= (�u) i� �Re0 j= (�v):

Proof of Claim 6: It su�ces to assume that is an atomic formula.Case 1: contains no symbol from L0. Then it is trivial by the ele-

mentary equivalence.Case 2: contains symbols from L0. Then =: Xi(��(�x)) for an Xi 2

L0 and an ni-tuple of L-terms �� . Note that L0 contains no function symbols.

If �Re j= (�u), then ��(�u) 2 e(Xi). Hence ��(�u) 2 Cnie . Since (�RC ; �u) and

(�RC0 ; �v) are elementarily equivalent, then I(��(�u)) = ��(�v). Note that I isan isomorphism from the submodel of �Re �L

0[f<g generated by Ce, to thesubmodel of �Re0 �L

0[f<g generated by Ce0 . Then I(��(�u)) = ��(�v) 2 e0(Xi).Hence �Re0 j= (�v). 2(Claim 6)

Claim 7 �Re j= � i� �Re0 j= �.Proof of Claim 7: Without loss of generality we assume that � has the

form 8�xn9�xn�1 : : : (�xn; : : : ; �x1), where is a quanti�er-free L[L0-formula.Suppose �Re j= �. By a back-and-forth argument using Claim 5 we are ableto �nd �un; : : : ; �u1 and �vn; : : : ; �v1 such that (�RC ; �un; : : : ; �u1) is elementarilyequivalent to (�RC0 ; �vn; : : : ; �v1), and

(�Re; �un; : : : ; �u1) j= (�un; : : : ; �u1):

16

By Claim 6 we have

(�Re0 ; �vn; : : : ; �v1) j= (�vn; : : : ; �v1):

This shows that for any �vn there exists �vn�1 such that for any �vn�2 thereexists �vn�3...... such that

(�Re0 ; �vn; : : : ; �v1) j= (�vn; : : : ; �v1):

So it is clear that �Re0 j= �. 2

2.4. EXERCISES

Exercise 2.13 Let F = (F ; +; �; <; 0; 1) be an ordered �eld. An upperbounded initial segment I of F is called a regular gap i� it has no leastupper bound and for any positive � 2 F there exists an r 2 I such thatr + � 62 I. F is called Scott complete i� F has no regular gap. Show thefollowing:

(1) Suppose F is a sub�eld of F 0. If an initial segment I in F has aleast upper bound in F 0 r F , then I is a regular gap in F .

(2) If V satis�es IP , then the hyperreal �eld in V is not Scott complete.Hint: Write a set of formulas �(X) to express that X is a regular gap.

Exercise 2.14 Show that IP implies the existence of an external set S ��N such that S \ f0; 1; : : : ;Hg is internal for every H 2 �N . This meansthat the nonstandard universe satisfying IP can't be classless.

Exercise 2.15 Show that IP implies the existence of a bijection f betweentwo �in�nite sets A and B such that for any ��nite sets a � A and b � Bthe restrictions f �a and f�1 �b are internal.

�Exercise 2.16 Let IP0 be IP restricted only on �nite languages in De�-nition 2.1. Prove that IP is equivalent to IP0 plus countable saturation. See[13] for help.

��Exercise 2.17 Find a nonstandard universe V satisfying IP such that

cf(�N) 6= coin(�N r N);

where cf(�N) is the smallest cardinality of some set S � �N co�nal in�N and coin(�N r N) is the smallest cardinality of some set S � �N r N

coinitial in �N r N . See [12] for help.

Exercise 2.18 (M. Benedikt et al. [4]) Let L be any language including <and L0 be same as in x2.3. Let A be any in�nite totally-ordered L-structure(not necessarily o-minimal). By an L0-bounded quanti�er sentence we meanan L [ L0-sentence built up from atomic formulas in L [ L0 via the usual

17

logical connectives and the quanti�cations 8�x 2 Xi and 9�x 2 Xj for someXi;Xj 2 L

0.Show that for any L0-bounded quanti�er sentence �, if � is order-invariant

in A, then there is an L0[f<g-sentence such that � and are equivalentover A.

3. The Special Model Axiom and Full Saturation

Given a language L, an L-structure A is called a special model if there isa sequence hA� : � < card(A)i of L-structures such that

(1) for any � < � < card(A), A� is an elementary submodel of A� ,(2) A =

S�<card(A)A�, and

(3) for any � < card(A), A�+1 is (card(�))+-saturated.

The sequence hA� : � < card(A)i is called a specializing chain for A.

De�nition 3.1 A nonstandard universe V is said to satisfy the specialmodel axiom i� any internally presented structure of some countable lan-guage is a special model.

We will write SMA for the special model axiom.

Proposition 3.2 (D. Ross [26]) Suppose V satis�es SMA. Then every in-�nite internal set in V has cardinality �V .

Proof: It su�ces to show that any two in�nite internal sets have samecardinality. Given two in�nite internal sets C and D, we want to derive acontradiction by assuming that card(C) = � < card(D) = �. We form aninternally presented structure

A = (C [D;C;D);

where C;D are unary relations. Then one has card(A) = �. Suppose hA� :� < �i is a specializing chain for A. Let C� [ D� be the base set of A�.Note that C� � C for every � < �. Since A�+1 is �+-saturated, thencard(C�+1) > �+. This contradicts the fact C�+1 � C. So card(D) 6card(C). By symmetry we have card(C) = card(D). 2

Proposition 3.3 (D. Ross [26]) If V satis�es SMA, then V satis�es IP .

Proof: It is proved in [5] that any two elementarily equivalent specialmodels with same cardinality are isomorphic. 2

Proposition 3.4 For any strong limit5 cardinal � with cf(�) > @0 thereis an ultralimit V of the standard superstructure such that V satis�es SMAand �V = �.

5A cardinal � is called a strong limit i� for every � < � one has 2� < �.

18

Consult [5] for the proof.

De�nition 3.5 A nonstandard universe V is fully saturated i� every inter-nally presented L-structure A for some countable language L is a saturatedmodel, i.e. A is card(A)-saturated.

Proposition 3.6 If V is fully saturated, then V satis�es SMA.

Proof: A saturated model is trivially a special model. 2

Proposition 3.7 For any cardinal � such that � > i! and �<� = � thereexists a fully saturated nonstandard universe V such that �V = �.

Consult [5] for the proof.

Next we give one application of SMA and one application of full satu-ration.

3.1. COMPACTNESS OF LOEB PROBABILITY SPACES

Given any probability space (;B; P ), a family C � B is called compact i�for any D � C, D has f.i.p., i.e. every �nite subfamily of D has non-emptyintersection, implies

TD 6= �. A family C � B is called inner-regular i� for

any B 2 BP (B) = supfP (C) : C 2 C ^ C � Bg:

De�nition 3.8 (D. Ross [27]) A probability space (;B; P ) is called com-pact i� there is a compact, inner-regular family C � B.

A Radon space is an example of a compact space. Ross showed in [27]that a compact probability space could be topologized so that the resultingtopological measure space is Radon. In [27] a question whether a Loebprobability space is compact, is posed. The following theorem is one ofmany results in [17] concerning the compactness of Loeb probability spaces.

Theorem 3.9 (R. Jin and S. Shelah [17]) Assume CH (Continuum Hy-pothesis). Suppose V satis�es SMA and cf(�V) = @1. Then every non-atomic Loeb probability space is compact.

We need more notation in the proof. For any set S we write 2S for theset of all functions from S to f0; 1g. Let 2<N be the set

Sn2N 2

n, where n

could be viewed as a set f0; : : : ; n� 1g. A set t � 2<N is called a tree i� forany s; s0 2 2<N, s0 � s and s 2 t imply s0 2 t. By a branch of t we meana maximal totally ordered subset of t with order �. We denote T for thetrees without maximal nodes and [T ] for all branches of T . Let's consider2N as a Cantor space with the usual probability measure �, i.e.

�(ff 2 2N : f(n) = 0g) =1

2

19

for every n 2 N . Note that every closed subset of 2N could be written as[T ] for some tree T � 2<N. By CH we can �x an enumeration ff : < @1gof 2N. For each � < @1 and m 2 N we choose a tree T�;m such that

[T�;m] \ ff : < �g = � and �([T�;m]) >m

m+ 1:

This can be done because �(ff : < �g) = 0.Given a probability space (;B; P ) and a sequence of measurable sets

hAn : n 2 Ni. The sequence hAn : n 2 Ni is called independent i� for anym 2 N and for any h 2 2m

P (m�1\

n=0

Ah(n)n ) =

m�1Y

n=0

P (Ah(n)n );

where A0n = An and A1

n = rAn.

Proof of Theorem 3.9: Given a non-atomic Loeb probability space(;B; L�) generated by an internal probability space (;A; �). Choose anindependent sequence hAn : n 2 Ni in A such that L�(An) =

12 for each

n 2 N . For any tree T � 2<N let

AT =\

n2N

[

h22n\T

n�1\

i=0

Ah(i)i :

It is easy to check that L�(AT ) = �([T ]). Note that AT is a countableintersection of internal sets. We are now ready to construct an inner-regular,compact family C. Note that one needs only to deal with the inner-regularityfor all sets in A.

Let A be the internally presented structure same as the one in the proofof Theorem 2.7, i.e.

A = ( [A [ �R; ;A; �R;2; �;\;r;+; �; <; 0; 1):

By SMA there is a specializing chain hA� : � < �Vi for A. Let f�� :� < @1g be an increasing sequence of regular cardinals co�nal in �V . LetB� =

S�<��

A�. Suppose the base set of B� is � [ A� [ R� . We can

choose an enumeration fa� : � < �Vg of A such that for every � < @1

fa� : � < ��g � A�:

Without loss of generality we assume An 2 A0 for every n 2 N . For any� < �V let

g(�) = minf� < @1 : � < ��g:

20

For each a� 2 A and each m 2 N we choose b�;m � a� \ATg(�);m such that

b�;m 2 Ag(�)+1 and L�(b�;m) = L�(a� \ATg(�);m). Note that b�;m exists by

countable saturation of Bg(�)+1. Now let

C = fb�;m : � < �V ^m 2 Ng [ fAn : n 2 Ng:

Claim C is compact and inner-regular.Proof of Claim: Clearly, C is inner-regular. Given anyD � C with f.i.p.,

we want to show thatTD 6= �. Without loss of generality we assume that

D is maximal. For each n 2 N one has either A0n 2 D or A1

n 2 D. So there

is a function h0 2 2N such that Ah0(n)n 2 D for every n 2 N . Let

� =[fg(�) : 9m(b�;m 2 D)g:

Case 1: � < @1. Then D � A�+1 and card(D) 6 �� . Since B�+1 is(��)

+-saturated, thenTD 6= �.

Case 2: � = @1. For each b�;m 2 D and for every n 2 N we have

(n�1\

i=0

Ah0(i)i ) \ b�;m 6= �:

But that means

(n�1\

i=0

Ah0(i)i ) \ATg(�);m 6= �:

Note that

ATg(�);m =\

n2N

[

h22n\Tg(�);m

n�1\

i=0

Ah(i)i :

By a careful check one can see that h0 � n 2 Tg(�);m for every n 2 N . Soh0 2 [Tg(�);m]. But h

0 = f for some < @1 (recall that we have a �xed

enumeration of 2N). So when g(�) > one has h0 62 [Tg(�);m] because

[Tg(�);m] \ ff 0 : 0 < g(�)g = �:

This contradicts that � = @1. 2

Remark In Theorem 3.9 the assumptions CH and cf(�V) = @1 couldn'tbe eliminated. If one replaces cf(�V) = @1 by cf(�V) = (2@0)+ (with orwithout CH), then the result will be just opposite. If one replaces CH by:CH (we can even weaken the condition cf(�V) = @1 to cf(�V) = � for anyuncountable regular � 6 2@0), then the compactness of Loeb probabilityspaces is undecidable in ZFC.

21

3.2. AUTOMORPHISMS OF LOEB MEASURE ALGEBRAS

Let (;B; L�) be a Loeb probability space generated by the internal nor-malized uniform counting measure (;A; �). We denote �B for the Loebalgebra, i.e. the Boolean algebra B modulo the ideal of L�-measure zerosets. For each element B 2 B we denote �B 2 �B for the equivalence classcontaining B. Note that each �B 2 �B contains an internal set in A.

De�nition 3.10 An automorphism of �B is a bijection � from �B to �B suchthat � is a Boolean algebra homomorphism and preserves the measure, i.e.for any A;B 2 B, �( �A) = �B implies L�(A) = L�(B).

De�nition 3.11 A bijection T : 7! is called a point-automorphismi� both T and T�1 are measurable and for any B 2 B one has L�(B) =L�(T [B]).

It is easy to see that a point-automorphism induces, in a natural way,an automorphism of �B.

Theorem 3.12 (D. Ross [25]) Suppose V is fully saturated. Suppose(;B; L�) is a Loeb probability space generated by an internal normalizeduniform counting measure space (;A; �). Then every automorphism � of�B is induced by a point-automorphism T .

Proof: Let A = fA� : � < �Vg. We construct two sequences hB� : � <�Vi and hC� : � < �Vi such that

(1) fB� : � < �Vg = fC� : � < �Vg = A,(2) for any � < �V , �( �B�) = �C�,(3) for any �0 < �1 < : : : < �n�1 < �V and for any h 2 2n one has

jn�1\

i=0

Bh(i)�i

j = jn�1\

i=0

Ch(i)�i

j;

where A0 = A, A1 = r A and j � j means internal cardinality.

Claim The theorem follows from the construction.Proof of Claim: For any x 2 there is an � such that fxg = B�

by (1). It is easy to see by (3) that C� is also a singleton fyg for somey 2 . Let T (x) = y. Then it is easy to check again by (3) that T is awell-de�ned bijection. Also it is not hard to check that T [B�] = C�. Soone has jAj = jT [A]j for every A 2 A. This implies that T and T�1 aremeasurable and preserve the measure. So T is a point-automorphism. By(2) and (3) one can easily see that � is induced by T . 2(Claim)

We now construct B� and C� by induction. Let

A = ( [A [ �R; ;A; �R;2; �;\;r;+; �; <; 0; 1)

22

be the internally presented structure same as in the proof of Theorem 2.7.Suppose we have found fB� : � < �g and fC� : � < �g such that (2) and(3) are true up to stage �.

Case 1: � is even. We pick B� �rst. Let

= minf� : A� 62 fB� : � < �gg

and let B� = A . This step guarantees (1). Let �( �B�) = �A for someA 2 A. We de�ne a set of formulas ��(x) with only one free variable x,which expresses that x is a candidate for C�. The set ��(x) contains exactlythe following:

(a) A(x), i.e. x is an internal subset of ,(b) �(x�A) < 1

mfor every m 2 N , i.e. the symmetric di�erence of x

and A will have Loeb measure zero,(c)

�((n�1\

i=0

Bh(i)�i

) \Bj�) = �((

n�1\

i=0

Ch(i)�i

) \ xj)

for any �0 < �1 < : : : < �n�1 < �, for any h 2 2n and for any j = 0; 1.Note that for any internal sets A;B � one has �(A) = �(B) i� jAj = jBj.

Since card(��(x)) = card(�) < �V and ��(x) is clearly �nitely realiz-able, then, by full saturation, ��(x) is realized by some C 2 A. Let C� = C.Clearly, (2) and (3) are true up to stage �+ 1.

Case 2: � is odd. We pick C� �rst and then B� by symmetry.Finally, (1) is true because of the way we choose B� when � is even,

and C� when � is odd. 2

3.3. EXERCISES

Exercise 3.13 (D. Ross [26]) Show that SMA implies that

cf(�N) = coin(�N r N) = cf(�V):

By comparing with Exercise 2.17 this exercise witnesses that IP does notimply SMA.

Exercise 3.14 Let SMA0 be SMA restricted only on �nite languages inDe�nition 3.1. Show that SMA is equivalent to SMA0 plus countable satu-ration.

Exercise 3.15 Assuming CH. Suppose V is an ultrapower of the standardsuperstructure modulo an ultra�lter on a countable set. Show that everyLoeb probability space generated by an internal normalized uniform countingmeasure on a hyper�nite set in V is compact.

23

�Exercise 3.16 Show that Theorem 3.12 is still true if V satis�es SMAinstead of full saturation (see [15] for hint).

Exercise 3.17 (D. Ross [25]) Let V be fully saturated. Let (;B; L�) be aLoeb probability space generated by an internal normalized uniform count-ing measure on a hyper�nite set . Let �B be the Loeb algebra. Show thefollowing:

(1) There exist automorphisms � of �B, which are not induced by anyinternal point-automorphisms,

(2) Let �C � �B be a subalgebra such that card( �C) < �V . Then for anyautomorphism � of �C there exists an internal point-automorphism T suchthat � is induced by T .

�Exercise 3.18 If V satis�es SMA but is not fully saturated, then for anyLoeb probability space (;B; L�) as in Exercise 3.17 there exists a subal-gebra �C � �B with card( �C) < �V and there exists an automorphism � of�C such that � is not induced by any internal point-automorphism. (Thisexercise and Exercise 3.16 are related, in fact, to a question posed in [26].This exercise witnesses that SMA does not imply full saturation.) Hint:Show �rst that �V is a singular strong limit cardinal. Then construct asubalgebra of cardinality cf(�V) together with an automorphism by a diag-onal method so that the automorphism avoids to be induced by any internalpoint-automorphisms (see [15] for more help).

4. The �-Bolzano-Weierstrass Property

Let F = (F ; +; �; <; 0; 1) be an ordered �eld. Let � be an uncountableregular cardinal. By a bounded �-sequence in F we mean a sequence ha� :� < �i in F such that fa� : � < �g � [�r; r] for some positive r 2 F .A �-sequence ha� : � < �i in F is said to converge in F i� there is anr 2 F such that for any positive � 2 F there is an � < � such thatfa� : � < � < �g � [r � �; r + �].

De�nition 4.1 Given a nonstandard universe V and let �R be the hyperreal�eld in V. Let � be an uncountable regular cardinal less than or equal tocard(�R). �R satis�es the �-Bolzano-Weierstrass property i� every bounded�-sequence in �R has a convergent �-subsequence in �R. V satis�es the �-Bolzano-Weierstrass property i� �R does.

Clearly, the �-Bolzano-Weierstrass property is a natural generalizationof the Bolzano-Weierstrass property for the standard real �eld.

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Proposition 4.2 (H. J. Keisler and J. H. Schmerl [21]) There exist count-ably saturated nonstandard universes satisfying the �-Bolzano-Weierstrassproperty for � = (2@0)+.6

De�nition 4.3 A nonstandard universe V is called �-Archimedean i�card(�N) = � and card(f0; 1; : : : ;Hg) < � for every H 2 �N .

Proposition 4.4 (J. Cowles and R. LaGrange [6]) If V satis�es the �-Bolzano-Weierstrass property, then V is �-Archimedean.

Proof: Suppose V is not �-Archimedean.Case 1: card(� N) < �. Since the set of all � rational numbers has

cardinality card(�N) < �, then there are no �-convergent sequences in �R.But there are �-sequences in �R.

Case 2: card(�N) > �. Because V is not �-Archimedean, there existsan H 2 �N and an S � f0; 1; : : : ;Hg such that card(S) = �. Clearly, theset S could be ordered as a �-sequence. That sequence is bounded and hasno convergent �-subsequence because it is discrete. 2

Remark Suppose U is a regular ultra�lter on any �. Then the ultrapowerof the standard superstructure modulo U could never be �-Archimedean forany �. So the existence of an �-Archimedean ultrapower would imply theexistence of some non-regular ultra�lters, which may imply the consistencyof some large cardinals. By assuming the consistency of a measurable car-dinal it is consistent that there exists a �-Archimedean ultrapower for some� [18]. It is still open whether such kind of ultrapowers could satisfy the�-Bolzano-Weierstrass property.

Proposition 4.5 (J. Cowles and R. LaGrange [6]) If V satis�es the �-Bolzano-Weierstrass property, then the hyperreal �eld in V is Scott com-plete. (See Exercise 2.13 for the de�nition of Scott completeness.)

Proof: Suppose �R is not Scott complete. Then there is an upper boundedregular gap I in �R. It is easy to see that the co�nality of I is same as theco�nality of �N . But V is �-Archimedean. So there exists an increasing�-sequence co�nal in I. Clearly, the sequence is bounded and has no con-vergent �-subsequence. 2

Remarks: (1) Since IP implies the hyperreal �eld is not Scott complete,the �-Bolzano-Weierstrass property is inconsistent with IP . (2) The readerwho is interested in doing research on this subject should consult the papers[6], [18], [19], [21], [28], [29], [31], [32] and [33].

6The exact result in [21] is that: Suppose � and � are uncountable cardinals, � isregular, � < �, and �� < � whenever � < � and � < �. Then there exists a �-saturatednonstandard universe satisfying the �-Bolzano-Weierstrass property.

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4.1. EXERCISES

Exercise 4.6 (1) Let F be an ordered �eld. Show that the unit interval[0; 1] in F is compact i� F is isomorphic to the standard real �eld.

(2) Suppose @1 = card(�N) in V (V may not be countably saturated).Show that V satis�es the @1-Bolzano-Weierstrass property i� the unit in-terval of the hyperreal �eld is Lindel�of.

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