An Introduction to the Fine Structure of the Constructible Hierarchy

J.E.Fenstad, P.G.Hinman (eds.), Generalized Recursion Theory @ North-Holland Pu bl. Comp. 19 74 ~

AN INTRODUCTION TO THE FINE STRUCTURE OF THE CONSTRUCTIBLE HIERARCHY (Results of Ronald Jensen) Keith J. DEVLIN University of Oslo

$0. Introduction We shall work in Zermelo-Fraenkel set theory (including the axiom of choice) throughout, and denote this theory by ZFC. We shall adopt the usual, well-known, notations and conventions of contemporary set theory (e.g. an ordinal is defined to be the set of all smaller ordinals, cardinals are initial ordinals, etc.). The paper is entirely self-contained, but some familiarity with the usual definition of the constructible universe, L, in terms of definability, and the proof that L is a model of ZFC + GCH + V = L, will be helpful. The exposition is based, with permission, very strongly on a set of notes written by Ronald Jensen and entitled “The Fine Structure of the Constructible Hierarchy”. Except where otherwise stated, the results are entirely those of Professor Jensen. It is convenient at this point for us to express our appreciation of several dluminating discussions with Professor Jensen on his work in general.

’

Since we wrote this paper, a slightly revised version of these notes has been p u b lished as a research paper. See R.B. Jensen, “The Fine Structure of the Constructible Hierarchy”, Annals of Mathematical Logic, Vol. 4 [ 19721, p. 229. The present paper represents a lengthy discourse on an expansion of the earlier parts of Jensen’s paper, and it is hoped that the somewhat more leisurely pace we adopt here (as opposed to Jensen’s paper) will be of benefit to those not predominantly interested in the set theoretical consequences of the Fine Structure Theory. For those who are so inclined, the notation we use is almost identical to that of Jensen, so this paper should provide a good introduction to Jensen’s. 123

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Previously, Jensen worked, as did most other people, with the usual “constructible hierarchy”. Thus, one defines, inductively, sets La, a E OR, by setting Lo = 0, LA = U,
+

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125

structure of the Jensen hierarchy. A corresponding theory may also be developed for the L-hierarchy, the only difference being that some awkward complications arise because of the above mentioned defects in this definition.

3 1. Preliminaries We shall be concerned with first-order structures of the form M = ( M ,E,A >, where M is a non-empty set and A C M . In general, we shall write ( M ,A ) for ( M ,E, A ). The (first-order) language for such structures consists of the following: (i) variables u j , j E w (generally denoted by u, w , x , y , z , etc.) (Vbl.). (ii) predicates =, E, A . (iii) bounded quantifiers (Vui E uj), (3ui E uj), i,j E w , i # j . (iv) unbounded quantifiers ( h i ) , ( 3 u j ) , i E w . (v) connectives A, v, 1, +,-. Finite strings of variables (or of elements ofM) are denoted by v , X, etc. We w r i t e u E X f o r u l E X A ... A u , E X , w h e r e w e h a v e ~ = a ~...,, a,.Similarly for 3 v , V v , etc. The notions of primitive formula (PFml), formula (Fml), free variable, statement, and satisfaction are assumed known. A formula of this language is Co (orn,) if it contains no unbounded quantifiers. Let n 2 I , and let Q, denote V if n is even and 3 if n is odd. A formula is C,(ll,) if it is of the form 3 x 1 V x 2 3 x 3... Q , x ~ ~ ~ ( V X , ~...XQn+l~,cp) ~ V X ~ where cp is C,. A formula in wluch the predicate A does not occur is called an E-fomulu. kMdenotes the satisfaction relation for M . Thus, kM is the set of all (p, (2)) such that p is a formula of the above language and 2; E M and p holds z in M at the point (2). We generally write k Mcp[~]for (cp, (2)) E kM.‘t=Mn denotes the set of all (cp, (2)) E bM such that cp is En. Let N C M . A set R C M is E f;l(N) (IIF(N)) iff there is a En (n,) formula cp(u,v) and elements a E N such that for all x EM, R(x) q [ x ,u ] . The set of all such R is also denoted by X r ( N ) (II,”(N>>. setXz(N) = nn,M(N). Write C for XF(@ and E,(M) for C p ( M ) . Similarly for II, A. If cp is a formula, cpM denotes the relation { x EM1 cp[x]}. Similarly, and more generally, define cpM [a]for u E M as {x EM1 b~ ~ [ xa]}. , Let F be a class of structures of the form M = ( M , A). A relation R is

-

u,,;x,M(N),A,M(N)=x,M(N)

bM

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uniformly E n ( M) for M E F iff there is a Z, formula q(u, v ) and elements a E n { M I M E F } s u c h t h a t w h e n e v e r M E F , R nM=cpM[a].

52. Rudimentary functions A function f : Vn -+ V is rudimentary (rud) iff it is generated by the following schemata: (i) f(xl, ..., x,) = x i , I Ii In. (ii) f ( x l , ..., x , ) = x i - xi, 1 Ii, j i n. (iii) f(xl, ..., x , ) = { x i ,xi}, 1 Ii,j In. (iv) f ( x l , ...,x , 1= h (81 ( X I ,...,x , ), ...,gk(x 1 , ...,x, 11, where g l , ...,gk, h are rud. (v) f(xl, ...,x , ) = U y E x lh ( y , x 2 , ...,x , ) , where h is rud. For example, the following functions are clearly rud: f ( x )= u x i f(x) = xi u xi (= {Xi ,Xi})

f@)= f@)=

u

{XI

( x ) = { t q } , t X l , ( X 2 , . . . , x , >)I.

And i f f ( y , x ) is rud, so i s g ( y , x ) = ( f ( z , x ) l zE y ) (= U z E y { ( f ( z , x ) , z ) } ) . We say that R C Vn is rudimentary (rud) iff there is a rud function r : Vn + V such that R = {(x) I r(x) # 8). For example, 6 is rud, since .Y~x+-+I-xf!J. We list some basic properties of rudimentary functions and relations. (1)

Iff, R are rud, so is

Proof. By definition, there is a rud r such that R ( x ) ++ r ( x )#

0. Then

= UyEr(x)f(X).

(2)

Let xR be the characteristic function o f R . R is rud if xR is rud.

Proof. By (1).

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121

R is rud iff 1 R is rud.

Proof. By (2), since xqR(x)= 1 - xR(x). (4)

Let fi : V" -+ V be rud, i = 1, ..., m. Let R j C V n be rud and mutually disjoint, i = 1, ..., m , and such that U E I R j= V n .Define f : V n + V by f ( x ) =fi(x) iffRi(x). Thenfis rud.

By (l), (5)

fi is rud. But f(x)= uEl fi(X).

IfR(y, x) is rud, so isf(y,x) = y n { z l R ( z , x ) ) .

Then h is rud. But f ( y , x )= U Z E , , h ( z , x ) .

(6)

SupposeR(y,x) is rud and (Vx)(3!y)R(y,x).Set the unique z € y such that R ( z , x ) ,if such a z exists.

8,

otherwise.

Then f is rud.

Proof.f(y,x)= U [ y n { z I R ( z , x ) ) ] .

(7)

-

IfR(y,x) is rud, so is (32 € y ) R ( z , x ) .

Proof. Take r rud so that R ( y , x )

r ( y , x )f

8. Then

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128

(8)

IfRi(x) are rud, i = 1, ..., rn, then so are m

rn

n Ri,

U Ri,

i=1

i=l

(9)

(Trivial).

Let (-)o,(-)l, denote the inverse functions to (-, -). Then (-)*, (-)1 are rud. More generally, let (-);, ...,(-):-1 denote the inverse functions to ( x l , ...,xn).Then (-)", ..., (-):-1 are rud.

Proof. the uniquez E Ux such that ( 3 u l , u z E U x ) ( x = ( u l , u 2 ) Au1 = z ) (XI0 =

6,

if no such z exists.

etc.

(10) The function the unique z E f ( X J > = X(Y> =

6,

u u x such that ( z , y )E x

if no such z exists.

is rud. (By definition.) (1 1) dom and ran are rud.

Proof. dom (x) = { z E U Ux I (3w E U Ux)((w, z ) E x)} ran(x)= { Z E U U X ~ ( ~ ~ E U U X ) ( ( Z , W ) E X ) ) . (12) f(x,y)

= x Xy =

( 13) f(x, y ) = x

UuExUvEy( ( u ,u ) } is rud.

r y = x n (ran(x) X y ) is rud.

( 14) f(x, y ) = x"y = ran (x f y ) is rud.

(15) f(x) = xpl is rud.

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Proof. Set h(z) = ((z)l,(z)o). Then h is rud. But clearly, f(x) = x-’ = h”(x n (ran (x) X dom (x))). Recalling our preliminary discussion ( S O ) , we observe that though rud functions increase rank, they only do so by a finite amount. More precisely, by induction on the rud definition * of a given rud function f, we see that there is a p € w such that for all x 1 , ...,x,, rank (f(x ...,x,)) < max{rank(xl), ..., rank(x,)} + p . ~

* Note : In future, we shall often refer to “the rud definition o f f ” ,

or simply “the definition off ”. We mean an arbitrary such definition, the actual choice being irrelevant, and hence assumed made once and for all. We now prove that the rud functions do in fact encompass all of the “simply definable” functions we spoke about in SO. First, let us call a function f : V n + V simple iff whenever cp(z, y) is a Co €-formula, there is a Co €-formula J/ such that k cp(f(x),y) ++ $(x,Y). A useful characterisation of this concept is given by the following:

Proposition.A function f : Vn -+ V is simple iff (i) the predicate x € f(y)is Z ;: and ; so is (VX ~ f ( y ) ) ~ ( x ) . (ii) wheneverA(x) is ,z Proof. (+) By definition. (+) Let f satisfy (i) and (ii), and le‘t cp(x,y) be a To €-formula. An easy induction on the length of cp shows that (p(f(x),y) is equivalent to a Lo Eformula; so f is simple. Using this proposition, and easy induction on the definition off yields:

Lemma 1 . Iff is rud then f is simple.

Now, since there are C x functions which increase rank by an infinite amount, it is clear that the converse to the above lemma is false. However, we do have: Lemma 2. R C Vn is XriffR is rud

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Proof. (+) Let R be Zx.By ( 3 ) ,(7), and (8) above, an easy induction on the Cx definition of R shows that R is rud. (+) Let R be rud. Then xR is rud. SOby Lemma 1, xR is simple. Using our above proposition, an easy induction on the rud definition of xR shows that xR and hence R , is Ex. ~

We require some generalisations of these concepts. Let A C V. We say that a function f is rud in A iff f is generated by the schemata for rud functions and the function xA . Let p E V. We say that a function f is rud in parameter p iff f is generated by the schemata for rud functions and the constant functions h(;r) = p . Lemma 3. I f f is nid in A C V, there are rud functions gl,...,gn such that f is expressible (in a uniform way with respect to the rud definition o f f ) as a composition of gl, ...,g, and the function h ( x ) = A n x.

Proof. By induction on the (nid) definition o f f

A set X is said to be rud closed if for all rud functions f , f “ X n C X . A structure M = ( M ,A ) is said to be rud closed if for all functions f which are r u d i n A , f”M” C M . We say a structure M = ( M , A ) is amenable if u E M + A n u E M . Lemma 4. A structure M = ( M ,A ) is rud closed i f f the set M is rud closed and M is amenable.

Proof. By Lemma 3. Lemma 5. L e t A C V. Zff is rud in A , then f r M n is uniformly X o ( ( M ,A n M ) ) for all transitive, rud closed M = ( M , A nM ).

Proof. By Lemmas 2 and 3. The next result will be of considerable use to us later on. Lemma 6. Every rud function is a composition of some of the following rud functions

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Proof. Let Cdenote the class of all functions obtainable from F,, ..., Fg by composition. We must show thatfrud + f € C. For each €-formula cp(xl,..., xn), set

By induction on cp, we show that for all cp, t, E C. (The required result will then be proved using this fact.)

(a)

cp(x)-xiExj,

1
Write F,(y) for F3(x,y).Then t,(u) = ui-' X FL-'(F4(E f l u 2 , u"-j)) so t, E

(b)

c.

Let cpI(x),..., cp,(x) be such that tvl, ..., t

PP

E C.

Let p(x) be any propositional combination of cpl, ..., q p . Since x t, E

-

c.

(c)

y , x U y (= U {xJ}), x f3 y (= x

Let &,x)

be such that

tz E

-

-

(x - y ) )

€

C, we clearly have

C. Let ~ ( x ) 3y?(y,x) or VyF(y,x).

Clearly, t=~,,~(u) = dom (t,(u)) and tvYlp(u)= u n - dom(un - tF(u)). So in either case, t, E C. (d)

v(X)

xi = Xj .

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132

e.

By (a), (b), t, E But look, k=(u,E) q [ x ] iff (VY E U u ) ~ ~ ~ u " ( " u ) , E ) ~ t Y 7 x l l . Hence t,(u) = u n n tvy,(u U (Uu)), so t, E C?, by (c). Let O(y,x)= y E x j -y

(e)

Exi.

cp(x)zxj€xi,

I
-

Let J / ( y , z , x ) = y € z h y = x j ~ z = x i . B y ( a ) , ( b ) , ( d ) , E t J(?But I dx) ~ y w ( Y , z , x ) so , by (c), tp E e.

e.

Hence, for any €-formula cp, tV E I f f : V n+ V , definef* : V + V by f * ( u ) = f n u n .Using our above result, we prove by induction on the rud definition off, that f rud -+f * E C . This easily implies the required result.

(b)

f ( ~ ) =xj

~

xi.

f * ( u ) = f " u " = { x - y ~ x , y~ u }Letcp(z,x,y)-z . E X - y . LetF(u) = t,(u U ( U u ) ) n ( U u X u 2 ) = { ( z , x , y ) ( x , y E u ~ z E x - y } . T h e n f * ( u >= ~ ~ ( ~ ( u >E , u since ~ ) t, E

e,

e.

Let G ( u ) = U;=,gf(u), H(u)=h*(Uf=lgT(u)) = h*(G(u)),and K ( u ) = U" U C(u) U H(u). By hypothesis, G, H, K E C . By Lemma 1, let cp(y,x) be an €-formula equivalent to the formula

e.

Clearly, f*(u)=F8(([t,(K(u))] n [ H ( u ) X u n ] ) , u n ) E

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133

LetG(u)= { ( z , y , ~ ) ( ( 3 u E y ) [ z E g ( u , ~ )x] E u }A. s a b o v e G E C . But f * ( u ) = F ~ ( G ( ~u”+’) ) , E e. Hence f rud +f* E C, for all f . Finally, let f be rud. We show that f E C?. Setf((z)) = f ( a ) , T ( y )= @ i nall other cases. ThusTis rud. So by the above, E C?. Let P ( x ) = {(x)}. Thus P E But look, f ( x )= {{f ( x ) } }= U U ( ~ { ( X ) } }= U U F ~ ( ~ ( P ( X ) ) , P ( E X))

7*

e.

e.

uu

As an immediate corollary of Lemmas 3 and 6 we have: Lemma 7 . Let A C V and define F9 by F9(x,y ) = A n x. Every function rud in A may be expressed as a composition of some of the (rud in A ) functions Fo, ..., F9.

We shall make immediate use of Lemma 7 in investigating the logical comz far suitable M.We assume, once and for all, plexity of the predicates kMn that we have a futed arithmetization of our language. Lemma 8. bgois uniformly Ey for transitive, rud closed M = ( M A ,> .

Proof. Let d: be the language consisting of: (i) variables wi,i E w . (ii) function symbols (binary) f o , ...,f9. We shall assume we have a futed arithmetization of %. We also assume that the reader understands what is meant by a “term” of L. Henceforth, let M = ( M ,A ) be arbitrary, transitive, and rud closed. We first define precisely how d: is to be interpreted in M. Let Q be the set of functionsp mapping a finite subset of {wili E a} into M . We may clearly assume Q is rud. Let C be the (rud) function which to each term r of d: assigns the set of all component terms of r , including variables. Let Vbl, be the rud predicate defining the set {wiI i E w } . Let P be the

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predicate

ThusP is rud in A . We may now define the interpretation of a term

y

=

P[ p ]

-

“T

T

of d: at a “point” p E Q by

is an L-term” A p E c A gg[P(C(T), g , p ) A g(7) = y ]

Hence the function if 7 is an &term and p E Q otherwise is (uniformly) ?2? (for transitive, rud closed M). Since M is rud closed we can use the above result to define kE0 as an M-predicate. Let cp E Frnl‘o. By Lemma 2, pM is rud in A . Hence the function r defined by I , if ( ~ ~ 1 x 1

Iyx) =

i

0, otherwise

is rud in A . So, by Lemma 7, we may assume r = T ~ where , T is a term of L, under the above interpretation (i.e. with Fiinterpretingfi for each i). In fact, we may clearly pick a recursive function u mapping FmlEo into the terms of d: so that whenever cp E Fml’o, pM[x] [ U ~ ) ] ~ [= X 1. ] But by our above result, this implies that kzo is (uniformly) ?2 (for transitive, rud closed M).

-

As an immediate consequence of this result, we have

Lemma 9. Let n 2 1. Then M=(M.A).

&‘is uniformly ?2:

for transitive,’ rud closed

We conclude this section with a few miscellaneous results of use later. The first two are technical, and will often be used without mention.

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Lemma 10. Let M = ( M ,A ) be rud closed. If R C M is E,(M), there is a Xo(M) relation P such that R ( x ) 3x1Vx23x3 ... Q,x,P(x,xI, ...,x,).

Proof. SupposeR(x)

-

FM 3 0 ~ V 0 2 3 0 3... Q , o , ( P ( u , o..., ~ , v,)[x],

cp is a Co-formula. Using the rud functions (-,

-

..., -),

(-)?,

...,

where we

can easily obtain, via Lemma 1, a Co-formula J/ such that R ( x ) c--f bM 3ulVu2 ... Q,u,$(u,ul, ...,u,)[x]. ThenR(x) 3x,Vx, ... Qnxn [bM $ [ x , x ~ ..., , x,]], as required.

A ) be rud closed. If R C M is E,(M), there is a single Lemma 11. Let M = ( M , element p E M such that R is E ; ( (p}).

Proof. If R is Ey({pl, ..., p , } ) , thenR is also CF(((pl, ..., p , ) } ) . Let M = ( M A , ), n 2 0. Write X < x , M iff X C M and for every En formula cp and every x € X , ~ ( x , nX) A CP [XI

iff b~ CP [XI .

Clearly, i f X , M are transitive and X CM, we always have X < x o M. And for n>O,wehaveX
Pnxf0.

Recall that if ( X ,E) satisfies the axiom of extensionality, there is a unique isomorphism n : (X,E) 1 (W,E), where W is a unique transitive set. Furthermore, if Z C X is transitive, then n 1 Z = id Z . In fact, n is defined by € induction thus: n ( x ) = { n ( y )I y E x n X } for each x EX. The next result is of considerable importance.

r

Lemma 12. Let M be transitive and rud closed. Let X < x : , M. Then ( X ,A nX ) satisfies the axiom of extensionality and is rud closed. Let n : ( X , A n X ) S ( W , B )where , Wistransitive.Letf:Mn+MberudinA. Then for all z E X , n( f(z)) = f ( n ( z ) ) .

Proof. Since M is transitive, M satisfies the axiom of extensionality. Hence as X < q M, so does ( X , A n X ) . Similarly, by Lemma 5, ( X , A n X ) is rud closed. Hence, in particular, z G X + f ( z ) E X for f : M n + M rud in A . By induction on the (rud in A ) definition off, ~(f(z)) =f(n(%)) for each I EX.

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$3.Admissible sets Let M = ( M , A ) be non-empty and transitive. We say M is admissible iff M is rud closed and satisfies the X o-ReplacementAxiom: for all Co formulas cp and all u E M , l=M [ V ~ 3 y c+ p V U ~ V ( Eu) V X ( 3 y Eu) cp] [ a ] . In case A = !J in the above, we call M an admissible set. More generally, M is X,-admissible iff M is rud closed and satisfies the (analogous) C,-Replucement Axiom. Likewise a Enadmissible set. We prove below that M is admissible iff M is C , -admissible. All our results extend trivially from admissibility t o C,-admissibility, with “En” everywhere replacing XI,etc. Roughly speaking, an admissible set (or structure) behaves like the universe as far as El concepts are concerned. We give a few elementary results which set the tone for the rest of this exposition.

Convention: For the whole of this paper, we shall adopt the following abuse of notation. Suppose M is a structure, cp(v) is a formula, and x EM. We shall write FM~ ( xrather ) than FM~ ( V ) [ X ] .Clearly, this is purely a notational convenience. Firstly, we give the promised “stronger” form of the admissibility definition.

,

Lemma 13 ( C -Replacement). Let M be admissible, and let cp be a X -formula,

a E M . Then

Proof. Let $ be a E,,-formula such that

bMcp(x,y, a ) t--, 3 2 $ ( x , y , z , U ) .Then

Convention: The essentially superfluous role played by a in the above theorem

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leads us to extend our previous convention slightly by allowing formulas to contain members of M as parameters. Again, this is clearly an avoidable convenience.

Lemma 14. L e t M be admissible. I f R ( x , y )is Xc,(M), so is (Vy €z)R(x,y).

-

Proof. Let cp be a Eo-formula with parameters from (“w.p.f.”) M such that W , y ) kM ~ w P ( ~ , YThen )-

So by Z0-Replacement,

which is z 1 ( M ) .

Lemma 15 (A l-Comprehension). Let M be admissible, P E A1(M). Then E M + P nU E M .

U

Proof. Let cp, $ be Eo-formulas w.p.f. M such that

So by Xl-Replacement there is u E M such that

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K . J . DEVLIN

But M is rud closed (SO satisfies what might be called the Eo-Comprehension Axiom), and therefore we conclude t h a t P n u = { z E u I k M ( 3 y E u ) $ ( y , z ) } E M . The next result has nothing specificalIy to do with admissibility, but is of considerable value. Let f : C M -+ M mean that f : X .+ M for some X C M. Lemma 16. Let M bearbitray, f : C M - + M b eX,(M).Ifdom(f)is then in fact fand dom (f)are A (M).

n,(M),

Proof.

It was necessary to state the above result explicitly because we shall frequently have to deal with functions which, though definable, are not total functions. A particular case of the above theorem would of course occur when dom(f) EM, (whence dom(f) is Z,(M)). As usual, we shall use the notatiolif(x) =g(x) for partial functions, with its usual meaning(i.e. f(s)is defined iffg(x) is defined, in which case f(x> = 'dx)). Lemma 17. Let M be admissible, f : C M u C dom (f),then f " u E M .

+

M be X I(M). If u E M and

Proof. Since M is rud closed andf"u = ran(fr u ) , it suffices to prove that fl'uEM.Now,asuEM,fI'uisA1(M)byLemma 16.Letcp(x,y)bea X -formula w.p.f. M such that f(x) = y +-+ kMcp(x,y). Then kMVx 3 y [(x E u A cp(x,y)) V ( x e u ) ] , so byX I-Replacement there is u E M such that kM(Vx E u ) ( 3 y E u ) cp(x,y). Hence f r u C u X u. So, by A -Cornprehension, f 1II = (f u ) n (u X u ) E M . Theorem 18 (Recursion Theorem). Let M be admissible. Let h : Mn+' -+A4 bea L1(M)fiinctionsuch thatforallx E M , {(z,$)l z E h ( y , x ) }is weN-

FINE STRUCTURE OF THE CONSTRUCTIBLE HIERARCHY

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founded. Let G = M n + 2 + M be El (M).Then there is a unique X (M)function F such that (i) ( y , x ) Edom(F) f3 {(z,x)lz E h ( y , x ) } C dom(F) (4 F ( y , x ) - G ( y , x , ( F ( z , x ) I z E h ( y , x ) ) ) . Proof. Let @ be the predicate

@(f,x)

“ f i s a function"^ (Vy E dom(f))(Vz E h ( y , x ) ) ( z E dom(f))

~ - - f

-

By Lemma 16, h , G are A1(M), so @ is Al(M). Let cp be a E1-formula w.p.f.M such that @ ( f , x ) p(f.x). Define a XI(M)predicate F by (using notation which will later be justified)

We verify (i) for this F. Suppose first that ( y , x )E dom ( F ) . Then, by definition, 3f[@(f,x)A y E dom (f)] . By definition of @, for such anf we must have(Vz Eh(y,x))(zEdom(f)). HencezEh(y,x)+(z,x)Edom(F). Now suppose that z E h ( y , x ) -+ ( z , x ) E dom (F). Note that a s M is transitive, h ( y , x ) C M . By our supposition,

so by Xl-Replacement,

Pick such a u . As @ is Al(M), by Al-Comprehension we see that w = u n {fl @(’,x)} EM. Hence u w EM.It is easily seen that @(UW,X).Noting that h ( y , x ) C dom(Uw), note that U w /‘h(y,x) EM. S e t f = U w 1h ( y , x ) U {(G(y,x, U w f h ( y , x ) ) , y ) } Clearly, . @ ( f , x ) so , ( y , x )E dom(F). Hence (i) holds for this F. We now show that F is a function and is unique. By (i), dom (F) is already uniquely determined, so for both of these it suffices to prove the following:

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To this end, suppose not. Then P = { y Iy E dom(f) n dom (f’) A f(y) # f ’ ( y ) }# 0. Let y o be an h-minimal element o f P . Since y o E P , f ( y o ) # f ‘ ( y o ) . But @ ( f , ~ )@ , ( f ’,x), so clearlyf(yO) =f’(yO) by the h-minimality o f y o EP. This contradiction suffices (and thus justifies our notation somewhat). Finally, it is trivial to note that (ii) must hold, virtually by definition. In view of the many set theoretic concepts defined by a recursion of the above type, it is clear that admissible sets play an important role in set theory. Say M is strongly admissible iff M is non-empty, transitive, rud closed, and satisfies the Strong CO-ReplacementAxiom: for all C o formulas cp w.p.f. M , kM Vu3u(VxEu)[3ycp(x,y)’(3y Eu)cp(x,y)].(Clearly, such a n M will also satisfy the “Strong El-Replacement Axiom”.) Strongly admissible structures M are (for reasons to be indicated later) also called rion-projectibleadmissible structures. The difference between admissibility and strong admissibility is closely connected with the difference between En predicates and An predicates, which is in turn closely connected with the difference between a function being partial and total. We shall have more to say on this matter later.

$4. The Jensen hierarchy Let X be a set. The rudimentary closure of X is the smallest set Y 3 X such that Y is rud closed. Lemma 19. If U is transitive, so is its rud closure.

Proof. Let W be the rud closure of U . Since rud functions are closed under composition, we clearly have W = {f(x)lx E U A f is rud}. An easy induction on the rud definition of any rudfshows that x E U + TC (f(x))C W . Hence W is transitive. (TC denotes the transitive closure function.) For U transitive, let rud(li) = the rud closure of U U { U ] . Of crucial importance is:

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Lemma 20. L e t U be transitive. Then 9(u>n rud(U) = Zu(U). Proof. Clearly, P(U) n Z;,(UU { U } ) = X u ( U ) , so it suffices to show that

3 ( u ) n z , ( ~ u{ ~ ) ) = P ( ~ ) n n t d ( ~ ) . ~ e t ~ ~ P ({u}). ~)nz,(uu Then, exactly as in the proof of Lemma 2 , X E rud ( U ) (by induction on the Z o definitionofX).NowletX€P(U)nrud(U). T h e n X i s a X (rud(U)) subset of 0. By Lemma 1, we may in fact assume thatXisE:'gUd( )(UU { U } ) .

8

transitive, s o x i s

Also very relevant is:

Lemma 21. There is a rud function S such that whenever U is transitive, S(U) is transitive, U U { U } C S(U)and U,,,Sn(U)= rud(U). Proof. Set S(U) = ( U U { U } ) U (Ui8,,F;(UU Lemma 6.

{U})2).The result follows by

Lemma 22. There is a rud function Wo such that whenever r is a well-ordering of u, Wo (r, u ) is an end-extension of r which well-orders S(u). Proof. Define iu, j f , j ; by:iu(x)=t h e l e a s t i s 8 such t h a t ( 3 x 1 , x 2 E u ) [ F i ( x 1 , x 2 ) = x ] jy(x) = the r-least x1 E u such that (3x2 E u)[Fiu~,)(xl,x2) = x] jT(x) = the r-least x2 E u such that Fju(&f(x), x2) = x. Clearly, iu, j r , j ; are rud functions of u , x . Define Wo(r,u)= { ( x , y ) [ x , y E u~ x r y {}( x~, y ) l x ~ uA Y @ U } u{(x,y)lx@u

A ~ @ U [Ai u ( x ) < i u ( y )

viu(x)=

The Jensen hierarchy, (J,I a € OR), is defined as follows: J, =

b

J,+l = rud(J,)

J, =

u,,

J,,

if lim (A).

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Lemma 23. (i) Each J, is transitive. (ii) a Ip + J, C Jp (iii) rank (J,) = OR n J, = WLY. Proof. (i) By Lemma 19. (ii) Immediate. (iii) By induction: rank(J,+l)= rank(rud(J,))= rank(J,) (by an earlier remark, this last step is easily verified).

+w

To facilitate our handling of the hierarchy, we “stratify” the J,’s by defining an auxiliary hierarchy
Clearly, the J,’s are just the limit points of this sequence. In fact:

Lemma 24. (i) Each S, is transitive (ii) a LO S, C Sp (iii) J, = u v < w , S, = sw,. +

Proof. (i) By Lemma 21. (ii) Immediate. (iii) By induction: Jol+l= rud(J,) = “nEw

Lemma 25. (S,I

SW,+H

= +,sw

u,,

Sn(J,) =

u,,,

Sn(S,,)=

= Sw(a+lY

u < m a ) is uniformly

X k for all a.

Proof. Set c p ( f ) ~ ‘ ~ i s a f u n c t i o n ” dAo m ( f ) € O R

Af(O)=OA

(VuEdom(f))[(succ(u)+f(u)=S(f(u-l)))

b ( v ) + f ( v ) = u,,.f(.>lI

A

.

Clearly, is uniformly X k .And by definition, y = S, +-+ 3f(@(f)A y =f(v)). Thus it suffices to show that for any a, u < wa, the existential quantifier here

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can be restricted to J,. In other words, we must show that whenever T < oa, then (S,I v < T) E J,. This is proved by induction on a. For a = 0 it is trivial. For limit a the induction step is immediate. So assume a = p + 1 and that T < wp -+ (S,I v < 7)E Jp. Then, by our above remarks, it is clear that (S,I v < 00) is Zip. So by Lemma 20, (S,I v < u p ) E J,. Thus for all n < o, (S,I v < o p + n ) = (S,I v < U P ) U {(Sm(Jp),UP + m)l m < n } E J,, as J, is rud closed.

Lemma 26. (J,I v < a ) is uniformly E k for all‘a. Proof. By an easy induction, (ovlv < a ) is uniformly E p for all a. Since J, = S,,, The result follows by Lemma 25. Lemma 27. There are well-orderings<, of the S, suck that: C <%; (i) v1 < v2 -+ (ii) is uniformly E+ for all a. Proof. We use Lemma 22. Set <,, = (4, and by induction:

(i) and (ii) are immediate and (iii) is proved like Lemma 25.

Lemma 28. There are well-orderings <,J of the J, suck that : (i) a1 < “2
k;

Proof. Set <, =
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Lemmas 12 and 26 enable us to prove the following extremely powerful result (due in its original form to Godel, the present version being Jensen’s):

Theorem 29 (Condensation Lemma). Let X < c l J,. Then for some 0 I a, X z Jp. Proof. L e t X < c l J,. Then by Lemma 12, let n : X % W , where W is transitive. We prove by induction on a that W = Jp for /3 = n”(Xn a). Assume, therefore, that whenever v < a and X u < z1J,, the unique isomorphism n u o f X V onto a transitive set W’ yields W’ = J,uft(xvnv). Note that, as (J,I v < a) is EP, v E X n a ++ J, EX.

Claim 1 : For all v E X n a,?(J,) = Jn(,]. To see this, note first that for v E X n a,X f7 J,
For v < a,define rudx(Jv) = the rud closure of X

fl(J,,

U {J,}).

Claim 2 : X = UvEXn, rudx(J,). To establish this claim, note that as X < q J,, X is rud closed, so 3 is obvious. For the converse, let x E X . Then x E J, = U,<,rud(J,), so for some rud functionf, F J a ( 3 v ) ( 3 p EJ,)(x =f(p,J,)). ButX
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Note. It is easily seen that we may regard the following as part of the statement of Theorem 29: If Y C X is transitive, then r 1 Y = id / Y . And for v E X n a,n(v) 2 v, and for all x E X, r(x) 2J, X. By an argument well known to all set theorists, it is easily shown that J = UaEOR J, is a model of ZFC. (In fact, setting
$5. On the fine structure of the Jensen hierarchy As mentioned in the introduction, a theory similar to the one following can be developed for the usual Lhierarchy, if desired. Central in our discussion will be the concept of a “uniformising function” for a relation, which is a sort of “choice function” for a given relation. Specifically, a function r is said to uniformise a relation R iff dom (r) = dom ( R ) and for a l l x , 3yR(y,x)-R(r(x),x). Let M = ( M , A ) , n 2 1. We say M is E:,-unifomzisabZe iff every Zn(M) relation on M is uniformised by a Xn(M) function. A few moments reflection will reveal that En-uniformisability is a very strong condition to demand of an arbitrary structure M,since in the more obvious cases, the definition of a uniformising function for a given relation would appear to increase the logical complexity by one or more quantifier switches. However, it will turn out that for all a,all n 2 1, J, i s Z,-uniformisable. For n = 1, this will be easy to prove, but for n > 1 , the corresponding argument will only work when J, is E:,-l-admissible, so a more indirect approach will be necessary. We shall outline the approach required after we dispose of some of the more easy results. First, E l-uniformisability. The E1(J,) well-ordering of each J, gives

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us this with little effort. in fact, we have a much stronger result, of importance in applications of E l-uniformisability. Let F be a class of structures M = ( M A , ) , n 2 1. Say F is uniformly Enuniformisable if, whenever p is a En-formula w.p.f. n {MI M € F } such that (PM is a relation on M for each M E F , there is a En-formula $ (w.p.f. n {MIM E F } ) such that for each M E F, $M is a function uniformising $1. Theorem 30. (J,, A ) is El-uniformisable.In fact, the class of all (J,, A ) is uniformly X -uniformisable. Proof. Let p be a X -formula w.p.f. J, such that [cpbx)]‘ , JaA)is a E relation on J,. By contraction of quantifiers, we can, in a uniform way, find a X o formula J/ (w.p.f. J,) such that k ( J , , A ) p b , ~ )c-,3 z $ ( z , y , x ) . Define g by: g(x)= the
Set r(x)= (g(x))l. Then r is (unifcrmly) E l((J,, A ) ) and clearly uniformises [ d Y ,x>l(J,A).

Remark. We call the above construction the canonical X l-uniformisation procedure. Observe that if R ( y , x ) is a E l((J,, A )) predicate, then the canonical Xl-unifomisation o f R is a function whose El((J,,A)) definition involves only those parameters which occur in the definition of R. Let us take a little time off to examine the above construction more closely. SupposeR(y,x) is a given X I relation, sayR(y,x) * 3 z P ( z , y , x ) , where P is X o . To obtain the Z; uniformisation of R , we first obtain a Z; uniformisation of the T o relation {(w,x)( P((W)~, ( W ) ~ , X ) } and , then simply pick out the requisite component of the result as our required function. And since < is a X well-order of J, the result is also E 1 . However, returning now to the notation of Theorem 30, we see that, if we try to extend this procedure to the case n > 1, we cannot conclude that the function g is En, the problem being the last conjunct in the explicit definition of g. Let \k(w,x)

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denote the predicate [l$ ( ( W ) ~ , ( W ) ~ , X ) ] ‘ ~ “ ~F’o. r n = 1, there was no problem, since if \k(w,x)is so is (Vw E t)\k(w,x).However, for n > 1, \k(w,x) is I:,-l, and we can only conclude that (Vw E t)\k(w,x)is I:,-, if (J,,A) is Z,-l-admissible. Otherwise it is merely II, of course, and so the resulting uniformisation of the original X, relation turns out to be En+,. So, in order to establish the general I:,-uniformisation lemma, it is not altogether unreasonable to try and “reduce” all En relations on an arbitrary J, to C, relations on some I:,-l-admissible Jp, for which we have a I:,-uniformisation procedure. In practice, it will turn out that this hint is slightly off target, but in its general tone it is worth bearing in mind. Closely connected with Z,-uniformisability is the notion of a “2, Skolem function”. Let M =( M , A )be transitive and rud closed. By a 2 , Skolem function for M we mean a &(M) function h with dom (h)C w X M , such that for some p EM, h is Z:({p}), and wheneverP E Z r ( { x , p } )for some x EM, then 3 y P ( y )-+ (3E w)P(h(i,x)).(With h, p as above, we say that p is a good parameter for h.) Note that C, Skolem functions need not be (and in general are not) total! As far as existence of I:, Skolem functions is concerned, we can get away with slightly less than might first appear. In fact:

zo,

Lemma 31. Let M = ( M ,A ) be transitive and rud closed. Let h be a Z r ( { p } ) function with dom ( h )C w X M. Suppose that whenever P E I:,,”({x}) for some x E M, then 3yP(y)-+(3i E w )P(h (i, x)). Then M has a X Skolem finction.

,

Proof. S e t i ( i , x ) = h ( i , (x,p)).It is easily seen that h“ is a C, Skolem function for M. Note that in the above, if h is actually I:,: then h” = h. This is used in establishing the following result:

Lemma 32. If (J,, A ) is amenable, then it has a E Skolem function. In fact, for (Ja, A ) which is uniformly I::Jff$” there is a C Skolem function for all amenable (J,, A ) .

Proof. Let ( p i [i < w ) be a recursive enumeration of Fml’l. Let (J,,A) be amenable. By Lemma 9, i=(I; is (uniformly) E:J,A’. Let h = h a J be the

JLJA)

canonical I: l-uniformisation of the I::Jae) relation {(y,i,x)l k $ ~ e ) p i ~ . x ] } .

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(By Lemma 30 and the ensuing remark, h.is thus uniformly E i J a P ) for amenable (Ja, A ) .) By the remark following Lemma 3 1, it is clear that h is a El Skolem function for (Ja, A ) . We refer to h, as the canonical E Skolem finction for (amenable) (J,, A ) . By a similar argument, we have: Lemma 33. If (J,, A ) is amenable and X,-uniformisable, it has a X, Skolem function. The following lemmas indicate our reason for using the word "Skolem" here. Lemma 34. Let M be transitive and rud closed, and let h be a I;, Skolem function for M. Then whenever x E M , x E h " ( o X {x})< c 'M. n

Proof. Set X = h " ( o X {x}). Clearly, x EX. Let P E X F ( X ) , P f @. We must show t h a t P n X # $ . L e t p be agoodparameter f o r h , a n d p i c k y I , ...,y, E X with P E E;({yl, ..., y,}). By definition of X,there are j l , ..., j, E w such that y1 = h(jl,x), ...,y, = h(j,,x). Since h is Z F ( { p } ) , it follows that PEXF({p,x}). Hence, P # @-+ 3 y P ( y ) - + ( 3 i E o ) P ( h ( i , x ) ) + ( 3 y E X ) P ( y ) . Lemma 35. Let M be a transitive and rud closed, and let h be a E n Skolem function for M. I f X C M is closed under ordered pairs, then X C h"(U XX)
Proof. Set Y = h"(w X X ) . By Lemma 34, X C Y . Let P E E,"( Y ) , P # @. We must show that P n Y # 0. Let p be a good parameter for h , and pick y l , ...,y m E Y w i t h P E X F ( { y l , ...,ym}).Pickjl ,...,j,Ew a n d x l , ...,x , E X such t h a t y l = h ( j l , x l ) ,..., y, =h(j,,x,).Letx=(xl ,..., x,).Byassumption, x EX. But clearly, ash is ZF({p}), P i s then Ey({p,x}), so P f @ + 3yP(y) + (3iE o)P(h(i,x)) + ( 3 y E Y ) P ( y ) . Corollary 36. Let M,h be as above. Let X C M and suppose h"(w X X ) is closed under ordered pairs. Then X C h"(w X X ) i M. n

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Proof. Let Y = h”(w X X ) . Clearly, Y = h”(w X Y ) ,so the result follows by the lemma. Lemma 37 (Godel). There is a bijection @J : OR2 *OR such that Iwa is uniformly C for all a.

@(a,P)2 a,P for all a,P, and @-’

Proof. Define a well-order <* of OR2 by

Let @ : <* OR. By induction on a,@-’ 1 wa is C p (uniformly).

Lemma 38. There is a I: l(J,) map of w a onto (oa)*for all a. Proof. Let Q = (a1@(O, a) = a}.Then Q is closed and unbounded in OR. In fact, Q = (a1@ : a2 ++a}, so wa E Q + wa = a.We prove the lemma by induction on a.Assume it is true for all v < a.

Case 1 : wa E Q. Then W’ 1 a suffices.

+ 1. I f P = 0, then w a = w E Q , so we are done by Case 1. Hence we may assume 0 2 1. Then clearly, there is a X1(J,) map j : wa u p . By hypothesis, there is a E l(Jp) map of wp onto so there is certainly a one-one into wb. Theng E rud (Jp) = J,, so for I: l(Jp) map g of v,y E wa, define Case2: a = P

++

Then f is E1(Ja) and f maps ( ~ aone-one ) ~ into 06. Now r a n 0 = ran(g) E J,, so if we define h by (for v E wa)

i

f-’(v),

h(v) =

if v E ran (f)

(0,O) , otherwise

we see that h is C1(J,) and h : wa-

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Case 3 : lim(a)

A

o a 4 Q. In this case let ( u , T )

= @-'(ma).

T h u s v , T < w a . S e t c = { z l z < * ( v , ~ ) } ( € J , ) . T h u s @ } c m a p s c one-one onto wa, and is I:l(J,). Pick y < a with v, T < oy.Then @-I 1 w a is a C '(J,) map of o a one-one into oy.And by assumption, there is a map g E J, mapping (or12 one-one into wy. SO, settingf((L, 6 )) = g((g(@-'(L)), g(W1(0)))), I , 0 < wa, we see that f i s a X l(J,) map of ( 0 ~ one-one ) ~ into d = g " ( g " ~ )But ~ . d E J,, so we can define a X1(J,) map h on wa by

I

f - ' ( ~ ) , if

h(0)=

(0, O),

e ~d

otherwise.

onto

Clearly, h = o a --+( ~ a ) ~ . The lemma is proved. Using this lemma, we may now establish the following important result: Theorem 39. There is a E l(J,) map of o a onto J, for all a.

be Ep({p}), wherep E J, is the
( v l , v 2 ) = f2(7). Then { ( y 1 , y 2 )is} a E k ( { ~ , p } predicate, ) so by definition of

h , ( y 1 , y 2 ) E X , asclaimed.

So by Corollary 36, X
8 5 a. Since w a c X , we clearly have 8 = a here.

Claim 2. For all i E w , x E X , n(h(i,x))= h(i, n(x)).

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To see this, observe first that as h is X p , there is a rud function H such that

x

y = h ( i , x ) *( 3 t E J , ) [ H ( t , i , x , y ) = 11. Now let i € w , x EX. Since X < J,, 1 y = h ( i , x )E X (if defined). Thus, by the above, since x,y E X
n ( y ) ) =l].Sincen”X= J,, thiscan be rewritten as(3tEJa)[H(t,i,n(x),n(y)) = 11. Thus n ( y )= h(i,n(x)),as claimed. Now, f C ( ~ a so ) ~as ,n 1 o a = id 1 oa,n’y=f.And by isomorphism, n‘yis E p ( { n ( p ) } ) .So as n ( p ) IJp , the choice of p shows that n ( p ) = p . So, by Claim 2, if i E a,v E wa, n(h(i, (v, p))) =h ( i , (u, p ) ) , which is to say ntX=idrX.ThusX=J,. Now define i: ( w ~ r+ )~ J, by setting

I

y , if ( 3 t E S,) [H(t,i , (v, p ) ,y )

i(i,U,7) =

8,

11

otherwise.

Thenzis El(J,), andclearlyi”(oa)3 =h“(oX(oaX{p}))=X=J,. Therefore, f 3 is as required by the theorem. Observe that in Lemmas 38,39, the maps constructed generally have parameters in their definitions. Note also that, being total, these maps are in fact A,(J,). Recalling the results of $ 3 , we now investigate those ordinals cy for which J, is an admissible set. Let us call an ordinal a admissible iff a = o p and Jp is an admissible set. Theorem 40. wa is admissible iff there is no E l(J,) map of any y < wa cofinally into we. (Note that such a map, having domain y E ,J would in fact be Al(J,).) Proof. (+). Let y < wa and suppose f : y + wa is El(J,). Then ( V i E y)(3( E oa)(f(.$)= (). If J, is admissible, then by X l-Replacement, ( 3 ~ € w a ) ( ~ i E ~ ) ( 3 ( € ~ ) ( fsofisnotcofinalinoa. (~)=(), (+). Assume w a is not admissible. If cy = fl + 1, then the Z1(J,) map ((wflt n , n > l n E o } maps o cofinally into wa, so we are done. Assume then that lim (a).Since J, is not admissible, there must be a E I(J,) relation R and a ~1 E J, such that (Vx E u) (3y)R(x,y) but for all z E J,,

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l ( V x E u ) ( 3 y E z ) R ( x , y ) .Take y < a with u E J,. By Theorem 39, let f be a C1(J,) map of wy onto J,. Thus f E J,, and u C f " o y . Define g : o y -+ o a

the least T such that ( 3 y E S,)R( f ( v ) , y ) , if f(v) E u

0, if f ( v ) $ u . Theng is a C1(J,) map of o y cofinally into wa. Recalling our discussion at the end of 8 3 , let us call an ordinal a strongly admissible (or non-projectibleadmissible) iff a = 00 and Jp is strongly admissible. Imitating the proof of Theorem 40, we have: Theorem 41. wa is strongly admissible iff there is no Z ,(Jay> map of a bounded subset of wa cofinally into war. The above two results illustrate our earlier remark concerning the difference between a function being partial and being total, and the corresponding difference between a predicate being En and being A n . The next two results, which strengthen the last two, and are also due to Kripke and Platek, also highlight this distinction. Theorem 42. The following are equivalent: (i) wci is admissible. (ii) (J,,A)isamenable forallA E A1(J,). (iii) There is no (J,) function mapping a y < o a onto J, (Of course, any such function would in fact be A I(J,).)

-

~

Proof. (i) + (ii). By Lemma 15 ( A , -Comprehension). onto (ii) -+ (iii). Assume (ii) h l(iii). Let y < wa, and let f : y J, be C1(J,).

-

Thenfis A,(J,),sod= { v l v $ f(v))is A,(J,).Thus b y ( i i ) , d = d n y E J , . So, d = f ( v ) for some v < y, so v E f ( v ) ++ v E d v 4f ( v ) , a contradiction. (iii) (i). Assume (iii) A l ( i ) . If a = 0 + 1, we can easily construct a Zl(J,) map of wp onto wa, so Theorem 39 yields the required contradiction. Assume lim (a). By Theorem 40, there must be a T < wa! and a Z1(Ja) m a p f o f T cofinally into ma. Let f be Ek((p}). Pick y < a with T,p E J,. Let h = h, be the canonical El Skolem function for J,. Set X = h"(w X J,). As J, is closed -+

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under ordered pairs, Lemma 35 tells us that X < z1J,. Let n : X Jp. Thus IT r J, = id F J,. By an argument as in Theorem 39, n X = id X,so X = Jp. Now, f isXF({p}) a n d p € X - ( q J,, soXisclosedunderf. B u t T C X a n d so f ’’7C X , which means, since f ”7 is cofinal in wa and X = Jp is transitive, that wa C J p . Thus@= a, a n d X = J,. Define a Zl(J,) map i: w X T X J , + J, as follows. Let H be a X k relation such t h a t y = k ( i , x ) -(3t€J,)H(t,i,x,y). Set y , if ( 3 t E S f , v ) ) H ( t , i , x , y ) K(i,v,x) = 0, otherwise.

i

Then h is total on w X T X J,, and i ” ( w X 7 X {x}) = kf’(w X {x})for any x , as f “ T is cofinal in we. Hence h”(oX T X J,) = X = J,. By Theorem 39 there isg E J,, g : o y w X 7 X J,. Then K e g is a Z1(J,) map of wy onto J,, contrary to (iii). Theorem 43. The following are equivalent: (i) wa is strongly admissible. (ii) (J,, A ) is amenable for all A E C,(J,). (iii) Tkere is no X l(J,) function mapping a bounded subset of wa onto J,.

Proof. (i) + (ii) + (iii). Similar to the above. (iii) + (i). Assume (iii) n l ( i ) and proceed much as before. So, we assume lim (a),f i s (by Theorem 41) a E l(J,) map of some a C r < wa cofinally into ma, f E Zp({p}), and T < oy, p E J,, y < a.As before, if k = k , and X = k ” ( o X J,), then X = J,. Now, since we do not need to bother about functions being total, we can easily contradict (iii). By Theorem 39, let g E J,, g : wyw X J,. Setf(v) =h(g(v)).ThenSis a Z1(Ja) map of a subset of w y onto J,. Note that an immediate corollary of Theorem 42 is: Theorem 44. If K is a cardinal, then K is an admissible ordinal. Using admissibility theory, we can give a quick proof that L = UaEORJ,. Theorem 45. I f

MYis admissible, then

J, = L,

.,

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Proof. If a = 1, then J , = L, = the hereditarily finite sets. Assume a > 1. Thus o E J,. Since J, is admissible, the recursion theorem tells us that rud(x) = Un<,Sn(x) is C1(J,). But if u is transitive, then C,(u) = P ( u ) n rud(u). war). So, by the recur-’ Hence - . -the map Ly * X,(Ly) = L,+l is E1(J,)(y< sion theorem again, we see that (L,l v < war) is l(Ju). Hence L, = L, C J,. For the converse inclusion, it suffices to show that L,, is admissible. (For then, by the recursion theorem, (S, I v < oar)is Cl(L,J, so J,=U,<,,S,CL,,.)LetR beXO(L,,),xEL,,,andassume (Vz E x ) 3yR(y,z). Since (L,I v < war) is El(Ja), we may define a X1(J,) predicate R’ by R ’ ( v , z ) z E x A (3y E L,)R(y,z). Since J, is admissible, t h e r e i s T < w a with(Vz E : x ) ( 3 v < ~ ) R ’ ( v , z ) .Hence(Vz€x)(3yEL7)R(y,z). So as L, E L,, L,, satisfies the X O-Replacementaxiom. Since lim (a), it follows easily that L, is admissible.

uv.,w,

-

Let a,n 2 0. The E,-projectum of a, p:, is the largest p < _ asuch that (J,,A) is amenable for allA E C,(J,) fl 3 ( J ) P’ Roughly speaking, our reason for introducing the X, projectum is this. We have seen that, for example, we can reasonably handle C,(J,) predicates when J, is C,-admissible. This is because Cn-admissibility is a sort of “hardness” condition on J, for C, predicates. For, if we take an arbitrary J,, it may be “soft” for Xn(J,) predicates; we may, for instance, find Zn(J,) subsets of members of J, which are not themselves members of J,, or even E,(J,) functions which project a subset of a member of J, onto all of J,. But if J, is C,-admissible, none of these situations can arise. Thus, we try to isolate that part of J, which is “hard” for E,(J,) predicates, a sort of “Enadmissible core” of J,. One natural way of formalising these ideas is provided by the Z,l projectum. Clearly, J is a reasonable interpretation of the notion P” of a “C,-hard core” of J,. We sh%l eventually give two characterisations of the x,-projectuni which make it appear even more reasonable - if not inevitable. One of these is that p z is the smallest p a for which there is a X,(J,) map of a subset of u p onto J,. Then, since we clearly have, for wa admissible, that o a is strongly admissible iff p: = a,we obtain some justification for our alternative name of “non-projectible admissible” for strong admissibility. I t is convenient, at this point, for us to define an obvious generalisation of the notion of the C -projecturn of an ordinal. Let (J,,A) be amenable. The C,-projectum of (J,,A), p z p , is the largest

<

,

FINE STRUCTURE OF THE CONSTRUCTIBLE HIERARCHY

155

5 a such that (J,,B)

is amenable for all B E C,((J,,A)). is always strongly admissible. Note that by Theorem 43, We shall make strong use of the X,-projectum in proving that every J, is Z,-unifomisable, all n 2 1. Since most of the following lemmas are directed towards this goal, it is worth indicating briefly our strategy. We already know that (J,,A) is Zl-uniformisable for all (J,,A). What we shall do is attempt to “reduce” Zn(J,) predicates to C1((Jpn,-4)) predicates for someA C J which is itself En(J,). To carry out this re%uction, we G need to have at our disposal a Cn(J,) map of a subset (at least) of J onto 4 J,. Thus, what we shall do is to simultaneously prove, by induction on n , a, the following two propositions: p

(P 1) J, is Z,+l-uniformisable (P 2) There is a Cn(J,) map of a subset of o p : onto J,. The proof of (P 1) goes roughly as follows. Let R be a Cn+l(Ja) predicate J, be C (J ). Now, f is a T,(J,) relation, so by on J,. Let f : C u p : assuming x,-uniformisability, f-’cg be ‘‘shrunk” to a X,(J,) map of J, into u p : . This reducesR to a X‘n+l(J,) predicate R’ on J p n . NOWfind a C,(J,) predicateA C J such that R’ is in fact XI((Jpn,&). UnifomiseR’ P: by a C1((Jpn,A)) function, and then reverse the proced&e to recover a Cn+l(J,) uiformising function for R. There is one doubtful point in the above outlbe. Can we in fact find a setA as required. That we can has to be proved as we proceed, so we shall in fact simultaneously prove three propositions, (P I), (P 2 ) , and a proposition (P 3 ) to be formulated precisely later.

-’

Lemma 46. Let n 2 1, and assume J, is X,-unijiormisable. Let y 5 (Y be the J,. Then there is a X,(J,) map of I least ordinal such that ;P(wy) (Xn(J,) a subset of o y onto J,. Proof. By Lemma 33, J, has a C n Skolem function, h. Let h be Z::((p)). We may assume p is the
,-

x.

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Clearly,X ( { n ( p ) } ) En Skolem function for J,, so by choice o f p , n ( p ) = p . But then h , h' are both defined by the same X n formula (with parameter p ) in J,, so h = h'. It follows immediately that a * h = i ,of course. So for i E a, x E J,, n - h ( i , x )=-h *n(i,x) =h(i,x). Thus n 1 X = id 1 X , a n d X = J,.

-

Lemma 46 plays a direct part in the proof of (P 1)-(P 3). The next lemma, however, is only used during the proof of the lemma which follows it, and may, on first sight, appear somewhat uninspiring.

Lemma47.Let(J,,A)beamenable,p=p~JI. I f B C J , isXl((Ja,A)), then E,((Jp,B)) C Z2((Ja,A)). Proof. Case 1. There is a Zl((J , A ) ) map of some y < wp cofinally into wa. or_ Let g be such a map, and let B be Eo((J,, A ) ) such that B(x) t--, 3zB(z,x) for eachx E J Define B ' by B'((v,x)) e ( 3 z E S g ( v ) ) B ( ~ , x )for , v €7, p; x E J p .ThusB IS A,(
e

-

P(x,fl(x), ...,f,(X),B'nf,+,(x), ..., B'nf,+,(x)). HenceR(x)3 ~ 1..., , 3ykIv1=B'nfm+l(x)A . . . A y k =B'nf,+,(x)AP(x,f,(x),... ...,f , ( x ) , y ...,y k ) ] .Now P is certainly L o(Ja), and f l , ...,fm+k are rud in

parameterp, so it suffices to show that the function b(u) = B' n u is E2((J,,A)). It is in fact n,((J,,A)), because: y = b(u) t--f Vx [x E y * x E u A B'(x)] , aud B' is Al((J,, A ) ) . Case 2. Otherwise. As before, we must show that Zo((Jp,B)) C Z2((J,,A)). Again as before, this reduces, by the amenability of (Jp,B), to proving that the function b(u) = B n u is Z2((Ja, A ) ) on J,. Now, we clearly have

FINE STRUCTURE OF THE CONSTRUCTIBLE HIERARCHY

y = b(u) * (Vx E y ) ( x E u A B(x)) A (vx E u ) (B(x)

+

157 X

Ey)

Now, the second conjunct here is Ill((J,,A)). We show that the first conjunct is Zl((J,,A)), which is sufficient. It reduces to showing that (Vx Ey)B(x) is Xl((J,,A)). But look, we know that Case 1 fails to hold, so this is proved just as in Lemma 14. The next lemma is the key step involved in proving, by induction, the as yet unformulated (P 3). Lemma 48. Let (J,,A) be amenable, p = pAA. Suppose there is a Z,((J,,A)) map of a subset of u p onto J,. Then there is a B C J,, B E Zl((J,,A)), suck ((J, ,A ) )for all n 2 1. that Z (, (J,, B)) = P (J, ) n Z Proof. L e t u C u p , a n d l e t f : u * J , be Z,((J,,A)).PickpEJ, such that f i s C:Ja’A’({p}).Let (qli < w )be a recursive enumeration of Fml’I. Set C B = ( ( i , x ) l i € ~ A x E J ,A k(Jk,A)pi[x,P]}.

Now, (J,,A) is amenable, and hence rud closed, so by Lemma 9, B E Cl((J,,A)). And of courseB C J,. Commencing with Lemma 47, an easy induction shows that for all n 2 1, c Z,+I((J,>A)). For the converse, letR(x) be a X,+l((J,,A)) relation on J,, n 2 1. Assume, for the sake of argument, that n is even. L e t P be a Cl((J,,A)) relation such that, f o r x E J,, R(x) +-+ 3y1Vy2 ... Vy,P(y,x). Definepby p(z,x) * [z,x E J, A P ( f ( z ) , x)] . By choice off, any x E J, is XiJe’A’({p,v}) for some v < u p , so by definition of B, pis rud in B and some parameter v < u p . In particular,pis Al((J,,B)). Again, D = dom(f) is rud in B and some parameter T < u p , so D is also A,((J,, B)). But for x E J,, R ( x ) +-+ (32, E D)(Vz2 E D ) ... (Vz, E D)F(z,x), which is thus

~,((J,,fw

q(J,m.

We are now ready to formulate (P 3) and prove our promised uniformisation theorem. Let a,II 2 0. A C , master code for J, is a setA C J ,, A E Z,(J,), such that whenever m 2 1, C,((JpB,A)) = 9(J ,) n Z,+,&). Qff

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Theorem 49. Let a, n 2 0. m e n : (P I ) J, is En+l-uniformisable. (P 2 ) There is a Xn(J,) map of a subset of u p : onto J,. (Unless n = 0 , when it is X l(J,).) (P 3 ) J, has a master code.

xn

Proof. We prove the theorem (for all n ) by induction on a. For (Y = 0, it is trivial. So assume (Y > 0 and that (P 1)-(P 3) hold (for all n ) for all 0 < a . We prove (P 1)-(P 3) at a by induction on n. Case I : n = 0. (P 1) is already proved (Theorem 30). (P 2 ) p : = a, so (P 2 ) is already proved (Theorem 39) (P 3) since p: = a , A = 6 is a Eo master code for J,.

Case 2 : n = m t 1, m

2 0. Let p = p:

for convenience.

We first prove that p is the least ordinal such that some Xn(Ja) function maps a subset of w p onto J,. To this end, let 6 be the least such ordinal. Suppose first that 6 < p . Then B = {t E w61 g 4 f(g)} is a En(J,) subset of J,, so by definition of p , (J,,B) is amenable. Thus, as 6 < p , B = B n w6 E J, CI J,. S O B = f(g) for some E w 6 , whence g Ef([) ++ [ E B * $f(E), which is absurd. Hence p 5 6. Suppose p <6. By definition of p , this means that for some En(J,) set B C J , , (J,,B)is not amenable. Since (Jl,B>must be amenable. 6 > 1. I f 6 = y + 1, then since there is a E l(J,) map of o y onto w6, there i s a Xn(J,) map of a subset of wy onto J,, contrary to the choice of 6. Hence lim (6). I t follows, since (J,,B) is not amenable, that there is T < 6 with B f3 J, @ J,. By induction hypothesis, J, is En-uniformisable. So as 7 < 6. Lemma 46 implies that Y(w7)n En(Ja) C J,. But there is 3 J ~ so, this implies Y(J,) n E ~ ( J , )c J,. In particular, 11 E J,, h : R 0 J, E J,. Hence for some 0 < a, B n J, is Jp-definable. Let p be the least such, and let r be least such that B n J, is Zr(JP). By definition, ( J * , B n J,) PP is amenable, so if T < p i , then B n J, = (B f3 J,) fl J, E J r C Jp, contrary PP to the choice of p. Hence T >_ p‘p. By induction hypothesis, there is a X,(Jp) m a p g from a subset of wpi onto JP. And since B n J, E JP+l and B n J, 4 J,, t 1 > 6 , or p >_ 6. Hence there is a E,.(Jp) map g’ from a subset of u p ; onto U S . Then f - g ’ is a Xn(J,) map of a subset of u p ; onto J,. But we have established that p i 5 T < 6, so this contradicts the choice of 6. Hence 6 =p. (P 2) follows immediately from the above result of course.

FINE STRUCTURE OF THE CONSTRUCTIBLE HIERARCHY

159

We turn now to (P 3). By induction hypothesis, let A be a Ern master code for J,. Set 77 = p r for convenience. By the above, let f be a ZJJ,) map of a subsetofup ontoJ,.Bychoice ofA, f ’ = f t(f-l”J,)isaC1((Jq,A))map of a subset of wp onto J,. By choice of A , it is clear that p = p t = P 1, , ~ . Finally, of course, (J,,A) is amenable. So, we may apply Lemma 48 to (J,,A) to obtain a Zl((Jv,A)) setB C J, such that C,((J,,B)) = 3(J,) n Z,+l((Jq,A)) for all r 2 1. By choice ofA, B E Zn(Ja) and Z,((J,,B)) = 3(J,) n X,+JJ,) for all 1-2 1. Hence B is a Enmaster code for J,. Finally we prove (P 1). Let B be, as above, a C, master code for J,. Let R ( y , x ) be a Zn+l(Ja) relation on J,. Define, with f as above, a relation R on J, by a r , x > - [ Y , X E J, A R ( f ( Y ) , f ( % ) ) l -ThenE is Z*+l(JJ and hence Zl((J,,B)). Let? be a Zl((J,,B)) function uniformising R. Sincefis Z,(J,), so is f But J, is C,-uniformisable, by induction hypothesis, so we can 1etf”be a C,(J,) function uniformising f-l. Set r = f a ? f It is clear that r is a Zn+’(Ja) function which uniformises R. The proof is complete. N

-’.

-’.

The above results give us two (intuitive) equivalent formulations of the

E n-projecturn:

Theorem 50. Let a , n 2 0. Let 6 be the least ordinal such that some Z,(J,) function maps a subset of w6 onto J,. Let y be the least ordinal such that 3 (wy) n C,(J,) Q J,. Then 6 = y = p z . Proof. That 6 = p i was actually proved during the proof of Theorem 49. Since we now know that J, is Z,-unifonnisable, Lemma 46 tells us that 6 5 y. Assume 6 < y. Now by definition, let u C w6, and let f : u J, be Zn(Ja). Let Z = (51 [ ($f(l)}.Then 2 C 0 6 and Z E Z,(J,), so by definition of y, 2 E J,. Thus Z = f([) for some 5, so [ E f([) * ( 4 f([), which is absurd. Hence 6 = y.

There is, of course, a concept which, for A, predicates, plays the role that the En projectum plays for X, predicates. And, as might be expected, there is a corresponding ‘‘total function” or A, equivalent of Theorem 50 for this concept . Let a,n 2 0. The A,-projectum of a (sometimes called the weak Enprojectum), Q:, is the largest Q a such that (J,,A) is amenable for all An(J,) setsA C J,.

<

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K.J. DEVLIN

Thus the A,-projectum of a represents the "hard core" of J, with regards to A, predicates on J,. Clearly, 77; 2 p t . We do not, however, necessarily have equality here. For example, let Q be the first admissible ordinal > w . Then it is easily seen that 77; = a, whereas p; = w . Corresponding to Lemma 46, we have:

Lemma 51. Let n 2 1, and let y be the least ordinal such that 3(oy) n A n ( J a ) J,. Then there is a Xn(J,) (and hence A,(J,)) map o f w r onto J,. Proof. Let n = rn + 1, n 2 0. Since Xm(J,) C A,(J,) C Z,(J,), Theorem 50 implies that p: 6 y 5 p';y".Theorem 50 also implies that there is a X m ( J a ) map of a subset of u p : onto J,. So, we can clearly define a C,(J,) map of w p z itself onto J,. This reduces our problem to showing that there is a Z,(J,J map of o y onto u p : . As a first step, we have the: Claim. There is a X n ( J a ) map g from w y cofinaliy into u p ? . LetA be a E m master code for J,. By hypothesis, let b C wy, b E A,(J,), b @ J,. By choice o f A , b is A ~ ( ( J ~ ~ , ASuppose )). b is in fact defined by:

whereBo,B1 areXO((J m , A ) ) .Then pol

there can be But ( J p E , A ) is amenable, and hence rud closed, so as b 4 J PF' no r < u p : such that

Define g : o y + u p : by

Clearly, g is Z n ( J a ) and cofinal in u p : , proving the claim.

FINE STRUCTURE OF THE CONSTRUCTIBLE HIERARCHY

161

We now prove the lemma. Since p : Iy, there must be a Xn(J,) map f from a subset of o y onto wp; (Iwa), and of course such anfwill then be E1((Jpm,A)). Definey: ( w ~2 )onto --.up! as follows. L e t f b e given by f(v) = 7 , - 3yF0,,7,v),where F is E0((Jp,,A)). Set ff

-

f (v,7)=

8 , if

(3YESy(,))F(y,8,4

0, otherwise.

Then S i s X1((JpmlA)), and hence Xn(J,).

u p : , as g cofina? m u p ; .

AndSclearly maps ( ~ 7 onto ) ~

Since we have (by Lemma 38) a X1(Ja) map of wy onto lemma follows.

the

Corresponding to Theorem 50, we have: Theorem 52. Let a, n'> 0. Let 6 be the least ordinal such that some X,(J,) (and hence A,(J,)) function maps 0 6 onto J,. Let y be the least ordinal such that P ( o y ) n A,(J,) Q J,. Then 6 = y = Q:,

Proof. Suppose y < q:. Let B C oy,B E A,(J,),B 4J,. Then wy n B = B 4Jqn, contrary to (Jgz,B)being amenable. Suptose now that Q: < y. Then there isA C J,, A E A,(J,), such that (J,,A) is not amenable. In particular, y > 1. Suppose y = 5' + 1. There is then a Xl(J,) map of w( onto wy, so by Lemma 51 there is a Zn(Ja) map,f, of 05' onto J,. Then Z = { L E wE I L 4f(~)}is a A,(J,) subset of w [ . Clearly, Z 4 J,, so this contradicts the choice of y. Hence lim(y). Thus as (J,,A) is not amenable, there must be T < y withA n J, 4J,. But by choice o f y , T < y -+A n J, E J,, so for some 0 < a ,A n J, is J,-definable. Let 0 be the least such.ThenA n J , E P ( J , ) n A m ( J e ) f o r s o m e m E w , a n d A nJ,$Je. Thus by Lemma 5 1 there is a E,(J,) mapf of WT onto J,. (Actually the hypotheses of Lemma 5 1 require that we have a A,(Jo) subset of OT not in Je , whereas we have only exhibited a subset of J, with these properties. However, since there is available a X l(J,) map of W T onto J ,, this point causes no problem.) Since 8 < a , f E J,. But A n J, 4 J,, andA n J,E Je+l, so 19 2 y, and there is thus a map f ' E J, of OT onto wy. By Lemma 5 1, again, this gives us a Xn(Ja) map k of 07 onto J,. Then, clearly, K =

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{ 1 I i @ K(1)) is a A,(J,) subset of W T not lying in J,, contrary to T < y. Hence y = 77:. Now, by Lemma 5 I, we have 6 5 y.Suppose 6 < y. Let f : w6%J, beZ,(J,). LetZ={vlv$f(v)}.ThenZE3(w6)n An(J,) J,. But this contradicts the choice of y. Hence 6 = y. ~

Remark. Lemmas 46 and 5 1 can be regarded as much sharper versions of the following, much earlier theorem of Putman: L,. Then contains a Suppose 9 ( y ) n well-ordering of y of order type a.(For y >_ 0.) Putman actually proved this result for the case p = w , but his proof works in the general case. The methods described above have, of course, many uses, We give just one, very general, example, showing that (in certain circumstances) it is possible to carry out Lowenhein-Skolem arguments for non-regular ordinals a which can generally only be done when a is actually a regular cardinal. More precisely, the following theorem, is well-known: Theorem 53. L e t K be a regular cardinal. L e t y < K 2 u p , and suppose that Y C Jp, I Y I < K . Then there is X < Jp such that Y U y C X a n d K n X E K .

To prove this, one simply forms an a-chain Xo< X I < ...4X , 4 ... 4 Jp of elementary submodels of J p , takingXo as the Skolem hull of Y U y in Jp, and Xn+l as the Skolem hull of X , U sup ( K f' X , ) in Jp, and then X = U n < w X n is the required submodel of Jp. By construction, K n X is transitive, and hence an ordinal, and since K is regular, I XI < K , so K nX E K . It should be observed that K being regular is a necessary condition for the above procedure to work (in general). However, providing we can, in some way, ensure that for each n , sup ( K n X,) < K , then we can, of course, get by with just cf(K) > w . The theorem below shows that, in certain cases we can do just this, providing we relax our demands somewhat. Let n 2 1, a 5 wp. We say that a is E,-reguZar at p iff there is no X ,(Jp) map of a bounded subset of a cofinally into a. For example, by Theorem 43, wa is strongly admissible iff wa is X 1 regular at a.

FINE STRUCTURE OF THE CONSTRUCTIBLE HIERARCHY

163

Theorem 54. Let n 2 1, wp 2 a 2 1. Suppose a is X,-regular at p. Let Y C Jp, w 2 1 Y 1 < cf (a),and let y < a- Then there is an X
Proof. Since cf (a)> w , it clearly suffices to prove that, under the stated hypotheses, there is X i r , Jp such that Y U y C X and sup (a n X ) < a. Let h be a E n Skolemfunction for Jp (by Theorem 49 and Lemma 33). Since w 5 I Y 1 < cf(a), we may, without loss of generality, assume that Y is closed under ordered pairs. Furthermore, let @ be the function defined in Lemma 37. Since a is X:,-regular at 0, a is certainly strongly admissible. Hence, by Lemma 37, { .$ E a I @ ' I t 2 C .$}is unbounded in a. It follows that we may y 2 is also, without loss of generality, assume that arty2 C y. Recall that zip.

Let X = h t t ( oX (Y X 7)). Then we claim that X i s closed under ordered pairs. To see this,letxl,x2 EX, sayxl =h(il,(yl,vl)),x2=h(i2,(y2,v2)). Lety = (y1,y2)E Y a n d v = @((vl,v2))Ey. Clearly {(xI,x2)} is t: P ( {p ,( y, v ) } ) , where p is a good parameter for h. Thus for some i E w , (x1,x2)= h ( i , ( y , v ) )E X , as required. So, by Corollary 36, X < z H Jp. And of course, we clearly have Y U y C X . We show that sup (a nX) < a. F o r y E Y , iEw,definehi,y: C y + a b y h . (v)l.h(i,(y,v)).Thushi,, 'Jn is Xn(Jp), and so as a is En-regular at 0, sup (hi,y y) y ( i , y ) < a. Since I Y I < cf(a), it follows that supyEy y ( i , y ) N- y( i ) < a . Since cf(a) > w , we conclude finally that supjEwy(i) < a. But clearly, supiEw y(i) = sup(& n X ) , so we are done.

-

The above lemma may be used to prove that if V = L and K is a regular uncountable, non-weakly compact cardinal, then there is a Souslin K-tree. (Jensen.)