Saharon Shelah- Quite Complete Real Closed Fields

8/3/2019 Saharon Shelah- Quite Complete Real Closed Fields

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QUITE COMPLETE REAL CLOSED FIELDS

SAHARON SHELAH

Abstract. We prove that any ordered field can be extended to onefor which every decreasing sequence of bounded closed intervals, of anylength, has a nonempty intersection; equivalently, there are no Dedekindcuts with equal cofinality from both sides. Here we strengthes the resultsfrom the published version.

1. Introduction

Laszlo Csirmaz raised the question of the existence of nonarchimedeanordered fields with the following completeness property: any decreasing se-quence of closed bounded intervals, of any ordinal length, has nonemptyintersection. We will refer to such fields as symmetrically complete for rea-sons indicated below.

Theorem 1.1. Let K be an arbitrary ordered field. Then there is a sym-metrically complete real closed field containing K.

The construction shows that there is even a symmetric-closure in a

natural sense, and that the cardinality may be taken to be at most 2|K|++1.

I thank the referee for rewriting the paper.In September 2005 (after the paper appeared), lecturing on it in the

Rutgers logic seminar without details, Cherlin asked where d(K) wasused in the argument. Checking the proof, it appears that this was used (inthe published version) but we can improve the result: the completion take 2 + 1 steps instead of a bound depending on K.

Note that by [2], consistently with ZFC is (i.e., after forcing extension)there an 1-saturated ultrapower of the field of the real (hence a real closedfield) which is Scott complete. Also compared to the published version weexpand 5 dealing with relatives of the symmetric closure; note also that

if K is an order field which is symmetrically closed or just has no cut ofcofinality (cf(K), cf(K)) then K is real closed.

Our problem translate to considering cuts of K and their pair of cofinal-ities. Our strategy is:

Key words and phrases. Real closed fields, ordered fields, cuts, completeness, cofinality.First typed: March 2000.

Research supported by the United States-Israel Binational Science Foundation. Publica-tion 757.

1


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2 SAHARON SHELAH

(a) we consider some properties of cuts (of a real closed field) beingDedekind, being Scott, being positive, being additive, being multi-

plicative,(b) we define dependence relation on the set of cuts ofK, which satisfies

the Steinitz assumptions,(c) realizing a maximal independent family of cuts with the right pairs

of cofinality, we get a one step symmetric closure.It is fine: we can show existence and uniqueness. But will iterating

this atomic closure eventually terminate?(d) For a field K we define a similar chain inside K, its minimal length

being h(K)(e) we define the depth d(K) of K(f) we show that h(K), and d(K) are quite closed and show that our

iterated closure from (c) do increase the cardinality much(g) we finally show that after 2 steps the iterated closure from (c)

termenate.

2. Real Closed Fields

Any ordered field embeds in a real closed field, and in fact has a uniquereal closure. We will find it convenient to work mainly with real closedfields throughout. Accordingly, we will need various properties of real closedfields. We assume some familiarity with quantifier elimination, real closure,and the like, and we use the following consequence ofo-minimality. (Readersunfamiliar with o-minimality in general may simply remain in the context

of real closed fields, or in geometrical language, semialgebraic geometry.)LABELmonotone Fact 2.1. (1) LetK be a real closed field, and let f be a parametrically

definable function of one variable defined over K. Then f is piece-wise monotonic, with each piece either constant or strictly mono-tonic; that is we can find finitely many intervals (allowing and as end points), f is constant strictly more or strictly decreasingon each interval. Moreover this holds uniformly and definably in de-

finable families, with a bound on the number of pieces required, andwith each piece an internal whose endpoints are definable from thedefining parameters for the function.

LABEL 2.1.1

Notation 2.2. K, L are ordered fields usuall real closed. If K L, X L,then K(X) is the subfield of L generated by K X. K+ = {c K : c > 0}.

(2A) Cuts.

LABEL 2.2 Definition 2.3. (1) A cutin a real closed field K is a pair C = (C, C+)

with K the disjoint union ofC and C+, and C < C+. The cut is aDedekind cut if both sides are nonempty, and C has no maximum,while C+ has no minimum. For L K let CL = (C L, C+ L),be a cut of L.


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QUITE COMPLETE REAL CLOSED FIELDS 3

(2) The cofinality of a cut C is the pair (, ) with the cofinality ofC and the coinitiality of C+ (i.e., the cofinality to the left). If

the cut is not a Dedekind cut, then one includes 0 and 1 as possiblevalues for these invariants.

(3) A cut of cofinality (, ) is symmetric if = .(4) A real closed field is symmetrically complete if it has no symmetric

cuts.(5) A cut is positive if C K+ is nonempty.

LABEL 2.2.1Conclusion 2.4. (1) If L is a real closed field extending K and a, b

L \ K realizes the same cut of K then a, b realizes the same typesover K in L.

(2) IfC is a non-Dedekind cut ofK then for some , cf(C) {(, 0), (0, ), (, 1), (1, )}so = cf(K)

Proof: By 2.1. monotone

We will need to consider some more specialized properties of cuts.LABEL 2.3A

Definition 2.5. Let K be a real closed field, C a cut in K.

(1) The cut C is a Scott cut if it is a Dedekind cut, and for all r > 0 inK, there are elements a C, b C+ with b a < r .

(2) The cut C is additive if C is closed under addition and containssome positive element.

(3) The cut C is multiplicative ifC K+ is closed under multiplicationand contains 2.

(4) Cadd is the cut with left side {r K : r + C C}.

(5) Cmlt is the cut with left side {r K : r (C

K+) C

}.Observe that LABEL 2.3.1

Observation 2.6. (1) Scott cuts are symmetric, in fact both cofinalitiesare equal to cf(K).

(2) If C is a positive Dedekind cut which is not a Scott cut, then Caddis an additive cut, or note: Cadd = {c : c 0} is impossible as notscott

(3) if C is an additive cut which is not a multiplicative cut, then Cmlt isa multiplicative cut.

LABEL 2.3.13Definition 2.7. (1) If K L are ordered fields, then a cut C in K is

said to be realized, or filled, by an element a of L if the cut inducedby a on K is the cut C.

(2) IfC1, C2 K and C1 < C2 but no a K satisfies C1 < a < C2 thenthe cut ofK defined (or induce or canonically extends) by (C1, C2) is({a K : a < c for some c C1}, {b K : c b for some c C2}),e.g., (C1, C2) may be a cut of a subfield of K.

LABEL scottLemma 2.8. LetK be a real closed field. Then there is a real closed field Lextending K in which every Scott cut has a unique realization, and no otherDedekind cuts are filled.


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4 SAHARON SHELAH

This is called the Scott completion of K, and is strictly analogous tothe classical Dedekind completion. The statement found in [?] is worded

differently, without referring directly to cuts, though the relevant cuts areintroduced in the course of the proof. The result is also given in greatergenerality there.

LABELmultiplicative Lemma 2.9. Let K be a real closed field, C a multiplicative cut in K, and

L the real closure of K(x), where x realizes the cut C. Then for any y Lrealizing the same cut, we have x1/n < y < xn for some n.

Proof. Let OK be {a K : |a| C}, and let OL be the convex closure in

L of OK. Then these are valuation rings, corresponding to valuations on Kand L which will be called vK and vL respectively.

The value group K of vK is a divisible ordered abelian group, and the

value group of the restriction of vL to K(X) is K Z

with = vL(x)negative, and infinitesimal relative to K. The value group of vL is thedivisible hull of K Z.

Now if y L induces the same cut C on K, then vL(y) = qvL(x) for somepositive rational q. Hence u = y/xq is a unit of OL, and thus u, u

1 < x

for all positive rational . So xq < y < xq+ and the claim follows. LABEL additive

Lemma 2.10. Let K L be real closed fields, and C an additive cut inL. Let C and Cmlt be the cuts induced on K by C and Cmlt respectively.Suppose that Cmlt = (C

)mlt, and that x, y L are two realizations of thecut C, with x C and y C+. Then y/x induces the cut Cmlt on K.

Proof. If a K and ax y, then a (Cmlt)+, by definition, working in L.

On the other hand if a K and ax < y, then a [(C)mlt], which byhypothesis is (Cmlt)

. LABELdedekind Lemma 2.11. Let K L be real closed fields, and C a positive Dedekind

cut in L which is not additive. Let C and Cadd be the cuts induced on Kby C and Cadd respectively. Suppose that C

add = (C

)add. Suppose thatx, y L are two realizations of the cut C, with x C and y C+. Theny x induces the cut Cadd on K.

Proof. If a K and a + x y, then a (Cadd)+, by definition, working in

L.On the other hand if a K and a + x < y, then a [(C)add]

, which by

hypothesis is (Cadd).

(2B) Independent cuts.We will rely heavily on the following notion of independence.

Definition 2.12. Let K be a real closed field, and C a set of cuts in K.We say that the cuts in C are dependent if for every real closed field Lcontaining realizations aC (C C) of the cuts over K, the set {aC : C C}is algebraically dependent over K.


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The following merely rephrases the definition.LABEL 2.9

Lemma 2.13. Let K be a real closed field and C a set of cuts over K.

(1) the following are equivalent.(A) C is independent(B) For each set C0 C, and each ordered field L containing K,

if aC L is a realization of the cut C for each C C0, thenthe real closure of K(aC : C C0) does not realize any cuts inC \ C0.

(2) For C0 and L as in clause (B) above, every C C \ C0 define a(unique) cut C of L (see Definition 2.3(2)) and {C : C C \ C0} is 2.2an independent set of cuts of L

(3) Assume K : is an increasing sequence of real closed fields,C a set of cuts of K0, and C C < C define a cut of K.

Then each C C define a cut of K which we callC[] and ifC is anindependent set of cuts of K0 then {C

[] : C C} is an independentset of cuts of K.

Note that this dependence relation satisfies the Steinitz axioms for a de-pendence relation. We will make use of it to realize certain sets of types ina controlled and canonical way.

Lemma 2.14. Let K be a real closed field, and C a set of cuts over K.Then there is a real closed field L generated over K (as a real closed field)by a set of realizations of some independent family of cuts included in C,in which all of the cuts C are realized. Furthermore, such an extension is

unique up to isomorphism over K.Proof. Clearly we must take L to be the real closure of K(aC : C C0),where C0 is some maximal independent subset of C; and equally clearly, thisworks.

It remains to check the uniqueness. This comes down to the following:for any real closed field L extending K, and for any choice of independentcuts C1, . . . , C n in K which are realized by elements a1, . . . , an ofL, the realclosure of the field K(a1, . . . , an) is uniquely determined by the cuts. One

proceeds by induction on n. The real closure K of K(an) is determined bythe cut Cn by 2.4; and as none of the other cuts are realized in it, they 2.2.1extend canonically to cuts C1, . . . , C

n1 over K, which are independent over

K. At this point induction applies. Lemma 2.15. LetK be a real closed field, and C a set of Dedekind cuts inK. Suppose that C is a Dedekind cut of cofinality (, ) which is dependenton C, and let C0 be the set {C

C : cof(C) = (, ) or (, )}. Then C isdependent on C0, and in particular C0 is nonempty.

Proof. It is enough to prove this for the case that C is independent. If thisfails, we may replace the base field K by the real closure K over K of a setof realizations of C0. Then since none of the cuts in C \ C0 are realized, and


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6 SAHARON SHELAH

C is not realized, these cuts extend canonically to cuts over K, and hencewe may suppose C0 = . We may also suppose C is finite, and after a second

extension of K we may even assume that C consists of a single cut C0. Thisis the essential case.

So at this point we have a realization a of C0 over the real closed fieldK, and a realization b of C over K, with b algebraic, and hence definable,over a, relative to K. Thus b is the value at a of a K-definable function(being the -th root for some i), not locally constant near a, and by Fact2.1 it follows that there is an interval b with endpoints in K and a definablemonotonefunction which is order isomorphism or anti-isomorphism from the intervalI0 to an interval I1 including a, with the cuts corresponding. So the cutshave some (or inverted) cofinalities. This contradicts the supposition thatC0 has become empty, and proves the claim.

For our purposes, the following case is the main one. We combine ourprevious lemma with the uniqueness statement.

LABELsymmetry Proposition 2.16. Let K be a real closed field, and C a maximal indepen-

dent set of symmetric cuts in K. Let L be an ordered field containing Ktogether with realizations aC of each C C. Then the real closure K

ofK(aC : C C) realizes the symmetric cuts of K and no others. Further-more, the result of this construction is unique up to isomorphism. Moreoverif L is a real closed field extending K which realizes every cut in C then K

can be embedded into L over k.

Proof: Clear (the no other by 2.18 below)scottsymmetric

Evidently, this construction deserves a name.LABEL hull

Definition 2.17. (1) Let K be a real closed field. A symmetric hull ofK is a real closed field generated over K by a set of realizations of amaximal independent set of symmetric cuts.

(2) We say that K = K : is an associated symmetric -chain

over K when:(a) K0 = K,(b) K is an order field ,(c) K is increasing continuous with (d) K+1 is a symmetric hull of K for <

(3) In (2) we replace

-chain by chain if K

is symmetricallycomplete but < K+1 = K

While this is unique up to isomorphism, there is certainly no reason toexpect it to be symmetrically complete, and the construction will need tobe iterated. The considerations of the next section will help to bound thelength of the iteration.

LABELscottsymmetric Lemma 2.18. (1) For regular < there is a real closed field with an

(, )-cut


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(2) Let K be a real closed field, and L its symmetric hull. Then everyScott cut in K has a unique realization in L.

(3) Assume L is the real closure of K {aC : C C}, C an independentset of cuts of K and aC realizing the cut C in L for C C.(a) If every C C is a Dedekind cut then every cut of K realized

in L is a dedkind cut(b) if one non-Dedkind cut of L is realized in L then evey non-

Dedekind cut of K is realized in L.

Proof. (1) First we choose Ki a real closed field Ki increasing continuouswith i , K0 = K and for i < the element ai Ki+1 \ Kiis above all members of Ki. Second we choose a real closed fieldKi increasing continuous with i such that K0 = K, and fori < , bi K

i+1i \Ki is above aj for j < and below any b K

i such

that (j < )(j < b), in Ki. Lasly in K, ({aj : j < }, {bi : i < })determine a cut i.e., ({a : a < aj for some j < }, {b : bi < b fori < }), in K, is such a cut.

(2) Recall that every Scott cut is symmetric. One can form the sym-metric hull of K by first taking its Scott completion K1, realizingonly the Scott cuts (uniquely), and then taking the symmetric hullof K1, by part (3) we are done.

(3) Easy.

3. Height and Depth

Definition 3.1. Let K be a real closed field.

(1) The height of K, h(K), is the least ordinal for which we can find acontinuous increasing sequence Ki (i ) of real closed fields withK0 countable, K = K, and Ki+1 generated over Ki, as a real closedfield, by a set of realizations of cuts which is independent. [[we couldhave chosen K0 as the algebraic members of K]]

(2) Let h+(K) be max(|h(K)|+, 1)

Remark:

(1) 1 is the first uncountable cardinal(2) h+(K) is the first uncountable cardinal strictly greater than h(K)),

so regular.Observe that the height ofK is an ordinal of cardinality at most |K| (or is

undefined, you can let it be , a case which by 3.3 do not occurs). We need 3.3to understand the relationship of the height of K with its order-theoreticstructure, which for our purposes is controlled by the following parameter.

Definition 3.2. Let K be a real closed field. The depth of K, denotedd(K), is the least regular cardinal greater than the length of every strictlyincreasing sequence in K.


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Observe that the depth is uncountable. The following estimate is straight-forward, and what we will really need is the estimate in the other direction,

which will be given momentarily.LABEL 3.3

Lemma 3.3. Let K be a real closed field. Then h(K) d(K).

Proof. One builds a continuous strictly increasing tower K of real closedsubfields of K starting with any countable subfield of K. If is limit, wedefine K =

x, so we arrive at acontradiction.

LABEL 3.4Proposition 3.4. Let K be a real closed field. Then d(K) h+(K).

Proof. Let > h(K) be regular and uncountable, and let K ( h(K))be a continuous increasing chain of real closed fields, with K0 countable,Kh(K) = K, and Ki+1 generated over Ki, as a real closed field, by a set ofrealizations of independent cuts.

For h(K) and X K, let K,X be the real closure of K(X) inside

K. We recast our claim as follows to allow an inductive argument. For X K with |X| < , and any h(K), we have d(K,X) .

Now this claim is trivial for = 0 as K0 is countable, and the claim passessmoothly through limit ordinals up to h(K), so we need only consider thepassage from to = + 1. So K is K,S with S a set of realizations ofindependent cuts over K, and similarly K,X is K,XS.

Consider the claim in the following form:

d(K,XS0) for S0 S

In this form, it is clear if|S0| < , as it is included in the inductive hypothesisfor , and the case |S0| reduces at once to the case |S0| = . So we nowassume that S0 is a set of realizations of independent cuts of K.

By 2.13(2),(3) we can find a subset S1 of S0 of cardinality 0 + |X| such2.9that:

(a) if s S0 \ S1 then the cut C which s induce on K is not realizedin the real closure K of K(X0 S1)

(b) the cuts which the s S0 \ S1 induce on K form an independent

family then after moving S1 into X, we may suppose that S0 is a setof realizations of cuts which are independent over K,X .


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Let {s : < } list S0 \ S1. For , let L = K,X{s: let Bi denote the cut induced on L by ai. With held fixed,and with i varying, as d(L) we find that the cuts Bi stabilize for largeenough i < (and furthermore, ai / L). Accordingly, for each we may

select j < above j0

such that the cuts B

i coincide for all i j.Now fix a limit ordinal < such that for all < we have j < . Wemay also require that ai L for i < . Then (B

) =


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10 SAHARON SHELAH

(2) If K has no symmetric cut except possibly Scott cuts, and L is asymmetric closure of K (equivalently the Scott completion of K)

then L is symmetrically complete.

Proof: Striaght.

4. Proof of the Theorem

We now consider the following construction. Given a real closed field K,we form a continuous increasing chain K by setting K0 = K, taking K+1to be the symmetric hull of K in the sense of Definition 2.17, and takinghullunions at limit ordinals.

If at some stage K is symmetrically complete, that is K = K+1, thenwe have the desired symmetrically complete extension ofK, and furthermore

our extension is prime in a natural sense. We claim in fact:LABELtermination Proposition 4.1. (1) Assume K is a real closed field, = 2,

=max(h+(K), 2) and K ( + 1) is an associated continuoussymmetric -chain , then(a) if cf(K) = then K is symmetrically complete so K+1 = K.(b) K+1 is symmetrically closed

(2) Also

(i) |K| 2h+(K)+1 , and |K+1| 2

h+(K)+1

(ii) ifK is a symmetrically complete extension of K thenK+1 canbe embedded into K over K.

(iii) K is unbounded inK+1 (and no non-Dedekind cut ofK is real-

ized in K and no non-symmetric Dedekind cut of K is realizedin K.

(iv) Any two real closed fields extending K which are symmetricallycomplete and embeddable into each other, are isomorphism overK (so we can say K is the closure)

(v) For some unique there is an associated continuous chainover K of length .

The proof of Proposition 4.1 occupies the remainder of this section.termination

LABEL induced Lemma 4.2. Suppose that K is a real closed field, and that K : () is a continuous chain of iterated symmetric hulls of any length. Let

x K \ K with > 0 arbitrary. Then the cut induced on K by x issymmetric.

Proof. Let be minimal such that the cut in question is filled in K+1. Thenthe cut induced on K by x is the canonical extension of the cut induced onK by x, and is symmetric by Proposition 2.16. symmetry

We now begin the proof by contradiction of Proposition 4.1(1). We as-terminationsume therefore that the chain is strictly increasing at every step up to K,and that there is a symmetric cut C over K.


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Remark: If we use = max{h+(K), 2} then 2 is regular and greaterthan h(K); in particular d(K) by 3.4. Furthermore, as > h(K), we 3.4

can view the chain K as a continuation of a chain Ki (i h(K)) of the sortoccurring in the definition ofh(K), with Kh(K) = K0; then the concatenatedchain gives a construction of K of length at most h(K) + < , and henceh(K) < for all < , and in particular d(K) for all < by 3.4. 3.4

For < , let C denote the cut induced on K by C.LABELcutsymmetryLemma 4.3. (1) For any < , the cut C is symmetric.

(2) For every < ,C is bounded in C and C+ is bounded in C

+

from below.

Proof:

(1) Suppose C is not symmetric. Then the cut C is not realized in

K, by Lemma 4.2. Hence the cut C is the canonical extension of inducedC to K, contradicting its supposed symmetry.

(2) Assume C is unbounded from below in C; so necessarily cf (C )

and [, ) cf(C ) = cf(C ).

For every [, ), as K+1 is a symmetric hule of K, by part(1) it follows that some a K+1 realizes the cut C hence by theprevious sentence, a C

+. So for every limit < , E {a : [, )} is a subset of C

+ unbounded from below, hence

cf(C) = (cf(C ), cf(C

+ )) = (cf(C

, cf()). As 0 = 1 are regulars

< for some (, )cf(C ) = cf() hence C is not symmetric,contradicting part (1).

In particular, the cut C is realized in K+1, and thus we have the fol-lowing.

Corollary 4.4. For any limit ordinal , the two-sided cofinality of Cis cof().

After these preliminaries, we divide the analysis of the supposed cut C intoa number of cases, each of which leads to a contradiction. Let C : < be an increasing sequence of members of K+ cofinal in it.

(Case I) C is a Scott cut

If C is a Scott cut and cf(K) = 2, there is nothing to be proved, soassume cf(K) = 2.By 4.3(2), we can find a , a

+ : < , such that a

C

, a+ C+ both cutsymmetry

realizes the cut C. For some club E of consisting of limit ordinals we have < E a , a

+ K. As C is a Scott cut necessarily a

a

+ : <

is a decreasing sequence of negative members of K with no negative lowerbound, so 1/(a+ a

) : < is increasing cofinal in K, so cf(K) = .

But K is cofinal in K hence cf(K) = = 2, contradicting what we haveassumed in the beginning of the case.


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12 SAHARON SHELAH

(Case II) C is a multiplicative cutLet < have uncountable cofinality (recall 2).The cut C is realized in K+1 by some element a. As C is multiplicative,

either all positive rational powers of a lie in C, or all positive rationalpowers of a lie in C+.

On the other hand, K+1 may be constructed in two stages as follows.First realize all the cuts in a maximal independent set of symmetric cutsin K, with the exception of the cut C, getting a field K

; then take the

real closure of K(a), where a fills the caonical extension of the cut C toK. As seen in Lemma 2.9, there are only two cuts which may possibly bemultiplicativeinduced by C on K+1, and each has countable cofinality from one side, and

uncountable cofinality from the other.So C+1 is not symmetric, and this is a contradiction.

(Case III) C is an additive cut

Let b : < be increasing cofinal in C K+ and let b

+ : < be

decreasing unbounded from below in C+. Clearly < b+/b (Cmlt)

+

and easily b+/b : < is a decreasing sequence of members of (Cmlt)

+

unbounded from below in it(Quotation ??)

For some club E of we have: if E then b : < is increasing

sequence in C

(K)+ cofinal in it and b+ : < is a decreasing sequencein C+ K unbounded from below in it. So b

+ /b

: < is a decreasing

sequence of members of (Cmlt)+ unbounded from below in it. [[Also it is

a decreasing sequence of members of (CK)+mlt. Hence (Cmlt)

+ K =(CK)

+mlt i.e., (C)mlt = (Cmlt), recalling 2.5(5).]]2.3A

According to the property ofb+/b : < the cofinality of (Cmlt) from

the right is .Now if the cofinality of Cmlt from the left is also , then by 2.6(3) we2.3.1

contradict Case II. On the other hand if the cofinality of Cmlt from the leftis less than , then from some point downward this cofinality stabilizes; butthen, choosing large and of some other cofinality (again, since 2 thisis possible), we contradict Lemma 4.3.cutsymmetry

(Case IV) C is a positive Dedekind cut, but not a Scott cut

One argues as in the preceding case, considering Cadd and using Lemma2.11, which leads to a symmetric additive cut and thus a contradiction todedekindthe previous case. In details choose b : < , b

+ : < as in case III,

so b+ b : < is a decreasing sequence in C

+add unbounded from below

in it


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(Quatation ?)

[[As in (case III) for a club of < , (Cadd)+ K = (CK)

+add hence

(C)mlt = (Cmlt)]]. If the cofinality of Cadd from below is also , recall thatCadd is an additive cut by 2.6(2) contradiction to case III. 2.3.1

As no cases remain, Proposition 4.1(1) is proved, and thus the construc- terminationtion of a symmetrically complete extension terminates.

As for clause (i) of 4.1(2), to estimate the cardinality of the result- terminationing symmetrically complete extension, recall that it has height at most = max(h+(K), 2) max(|K|+, 2) and hence cardinality at most 2

.

Moreover, similarly for any < , |K| 2h+(K)+1 hence

|K| = |


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(1) We say that a real closed field is - complete if no Dedekind cut ofK has cofinality

(2) We say that L is a -hull of K when there is a maximal subset C ofcut(K) := {C : C a Dedekind cut of K with cofinality } whichis independent and aC L for C C realizing C such that L is thereal closure of K {aC : C C }

(2A) We say that L is a weak -hull of K when there is a subset C ofcut(K) which is independent and aC L for C C realizing Csuch that L is the real closure of K {aC : C C}

(3) We say that K : is an associated -chain over K

when: K is a real closed field, increasing continuous with , K0 = Kand K+1 is a -hull of K for <

(3A) We say that K : is a weak associated -chain over

K when: K

is increasingcontinuous sequence of real closed fields,K = K0 and K+1 a weak -hull ofK for <

. We may omit meaning {(, ) : , are regular infinite cardinals}. We may omitover K

(4) We say that K : is an associated -chain over K when:

it is an associated -chian over K, < K+1 = K andK is complete

(5) Let d(K) be the minimal regular cardinal (so infinite such that:for every non-Scott Dedekind cut C of K both cofinalities of C are< ).

LABEL 5.2Theorem 5.2. Let K be a real closed field and be as in Definition 5.1.

5.1

(1) There is a -hull L of K, see Definition 5.1(1)5.1(2) L in (1) is unique up to isomorphism over K, and K is cofinal in it(3) For every ordinal there is an associated -chain over K, see

Definition 5.1(2) and it is unique up to isomorphism over K5.1(3A) If K :

is a weak associated -chain over this K it iscofinal in all its members in particular in K

(4) If = cf() < d(K) moreover for some , K has a non-Scott Ded-kind cut of cofinality (, ) and ()[(, ) ] then there is noassociated -chain over K, moreover no -complete extension of K

(5) 0 d(K) d(K)

(6) If d(K) = 0 then K is isomorphic to the field of the reals

(7) If K :

is a weak associated

-chain over K thend(K) Max{||+, d(K)},Check

(7A) In (7) also |K | |K| + || +

{|K| : (, ) and < d(K)}(8) Assume satisfies = cf(K), = cf() > 0 and for every regular

< d(K) + for some regluar < we have (, ) / , then thereis an associated -chain over K of length (compare with 4)

(9) If ( regular ) ( regular ) [(, ) / ] then for every real closedfield K there is an associated -chain over it


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(10 If there is an associated -chain K : over K then

(a) K is a real closed field, -complete extending K

(b) [universally]] if K is a real closed field extending K then Kcan be embeded into K over K

(c) [uniqueness] if L is a -complete real closed field extending Kwhich can be embeded over K into L for every-complete, -complete real closed field extending K then L is isomorphism toK over K.

(d) in (c), |L| 2


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16 SAHARON SHELAH

induce C on K, then CK is a cut of K of cofinality thesame as C, hence a -cut of K hence is realized in K+1 by

the construction , say by a, contradiction to CK induce Con K. Hence

()2 for no < does CK induce C on K. This means that forevery < some a K\K realizes C, so a C

a C+

so as we can replace C by |{b : b C+}, {b : b C}|without loss of generaliy for arbitrarily large < , a C

,so remain ??

()3 a C K realizes the cut CK

()4 cf(C) = so let cf(C) = (, )

Case 1: If = let be regular < such that (, ) / . Let{b : < be a decreasing sequence in C

+ unbounded from belowin it so without loss of generaliy(*) {b : < } k()

Now for some club E of we have(*) if E then > () and {a : < } is an unbounded subset

of C K hence cf(CK) = (cf(), )Choose E, such that cf() = . But then a realizes a {(cf(), )}-cut i.e., {(, )}-cut i.e., {(, )}-cut which is a non -cut by thechoice of , contradiction to ()1.

Case 2: = We repeat the proof of 4.1 after 4.3, in (case I) using = cf(K)termination

cutsymmetry (9) For every field K define n by induction on n <

0

= |K| (or d(K))

n = Min{ regular and if < n then for some < n+1 we have(, ) / }.

Now K, (

{n : n < })+ satisfies the condition in (8).

(10) Easy. [Detials?]What about cf(K), we have not changed it in all our completion.

It doesnt make much differnece as ??.

LABEL 10.4Claim 5.3. (1) For K and regular ( 0) there is L such that

(a) K L(b) cf(L) =

(c) if K L

and cf(L

) = then we can embed L into L

over K(d) if > 0 then in clause (c) we can add: there is an embeddingf of L into L over K such that Rang(f) is unbounded in L

(2) For K and = 0 there are L1, L, such that:(a) K L(b) cf(L) = 0(c) cfK L then for some {1, 2} there is an embedding f of L

into L over K such that Range() is unbounded in L

(3) If cf(K) > 0 then in part (2) we cannot have L1 = L2,


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(4) If cf(K) = 0 then in part (2) we can use L1 = K = L2 similarly inpart (1).

We have concentrated on real closed fields. This is justified by

LABEL 5.6Claim 5.4. (1) Assume K is an ordered field, = cf(K) and K has no

{(, )}-cut. Then K is real closed.(2) In Theorem 5.2 if we add (cf(K), cf(K)) / and deal with ordered 5.2

fields, it still holds.LABEL 5.7

Claim 5.5. (1) If F is an ordered field of cardinality then there is F

such that(a) F is a real closed field F of cardinality (b) F extends F

(c) cf(F) = 0(d) if C is a Dedekind cut of F of cofinality (1, 2) then 1 = 2(2) If I is a linear order of cardinality then we can find F such that

(a) F is a real closed field of cardinality (b) I is embeddable into F (and if it is (F+0 ,

Documents

Saharon Shelah- Quite Complete Real Closed Fields