Beth Definability in Infinitary LanguagesAuthor(s): John GregorySource: The Journal of Symbolic Logic, Vol. 39, No. 1 (Mar., 1974), pp. 22-26Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2272338 .

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THE JOURNAL OF SYMBOLIC LoGIc Volume 39, Number 1, March 1974

BETH DEFINABILITY IN INFINITARY LANGUAGES

JOHN GREGORY

Some negative results will be proved concerning the following for certain infinitary languages ?, and .2'2-

DEFINITION. Beth(Y1, y2) iff, for every'sentence 1(R) of Y1 and n-place relation symbols R and S such that S does not occur in +(R), if

k +(R) A O(S)-> Vxl * * x(R(xl, - - - , x.) S (x, *Xn))I

then there is an Yf2 formula e(xl, - *, x,,) such that

k +(R) -> *x1... , , xn) -* (x1, , x.))

and 8 is built up using only those constant and relation symbols of 0 other than R. That is, Beth(Y1, Y2) iff for every implicit Y, definition +(R) of relations, there

is a corresponding explicit 2 definition 0. Beth(,'Y , , O'O) was proved by Beth. Malitz proved that not Beth(G'lYl, S<1=Y ) (hence not Beth(Y.. , Y.9o'- )), but

Beth(YS,, Y ). In ?1, it is shown that Beth(Y.,2,,O,) is false. In ?2, we strengthen this by showing that, for every cardinal K, not Beth(Y.,, Y'). In fact, not Beth (Ylc+,w, .9UK) follows from property A(K) defined in ?2, and A(K) is known for regular K > w (unpublished result of Morley).

More information on infinitary Beth and Craig theorems is given in [2] and [3]. We assume that the reader is acquainted with the languages YcA which allow con- junctions over <K formulas and quantifiers over -<A variables. Thus, we assume that the reader is acquainted with the back and forth argument for showing that two structures are =::, (Y,,-elementarily equivalent). Our notation is fairly standard.

The following are a few of the well-known facts used later. If 21 23 and F is an assignment from fewer than K new constants to elements of 21, then there is an assignment G. to elements of 23, such that (21, F)-= ? (23, G). For each structure 21 there is a Ye,,, sentence 0 such that, for all 23,23 Z iff 23 =_ ., 21.

An element b of 21 is said to be Y.,, definable iff there is an Y.,, formula @(x) such that b is the unique element for which 21 k 0(b). By the above, b is undefinable (if and) only if there is a c # b such that (21, b)-,,c (21, c). Indeed, suppose b is undefinable. There is an YS,< sentence +(d), d a new constant, such that, for all 23, 2 k +(d) iff 23- (21, b). el F +(b). Since +(x) cannot be a definition of b, there is c # b such that 21 F +(c). (21, c) F +(d). Thus, (2, c)-= (21, b).

Received May 8, 1972.

D 1974, Association for Symbolic Logic

22

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BETH DEFINABILITY IN INFINITARY LANGUAGES 23

?1. THEOREM 1. Beth(Y.,, Y.J is false. That this Beth definability for Y.,,, fails follows from the fact that there is a

structure 21 such that 21 has exactly one automorphism (necessarily the identity) and 21 has an L' ,, undefinable element, where L' is a language interpreted by 21. As an example, we will use a dense linear ordering - = <D, <> with exactly one automorphism and with cardinality 2c0. Such an ordering exists (see [4]). Note that all the elements of 21 are L' ,, undefinable, except for its end points, if any.

Let L be the language with the equality symbol, distinct binary relation symbols R1 and R2, and the set D of constants (we can assume 21 chosen so that its domain D is a set of constants). Let R. S be distinct 2-place relation symbols not of L. Let +(R) be the conjunction of the following, wherein F1(x) is 3y(R1(x, y) v R1(y, x)) and F2(x) is 3y(R2(x, y) v R2(y, x)).

1. Vx(F1(x) ?-V{x = d I d e D}). 2. VX-,(F1(X) A F2(X)) A VX(F1(X) V F2(X)). 3. R1(d1, d2) if d1 $ d2.

5. Vxy(R(x, y) -> F1(X) A F2(y)). 6. Vx(Fl(x) 3 !yR(x, y)). 7. Vx(F2(x) 3 !yR(y, x)). 8. Vxyuv(R(x, y) A R(u, v) -* (Rl(x, u) +-+ R2(y, v))).

5(R) A O(S) logically implies Vxlx2(R(xl, x2) +-+ S(x1, x2)). Indeed, consider any model of O(R) A O(S). We can assume each constant d e D to be interpreted by itself (because of conjuncts 4). Then R1 is necessarily interpreted by the ordering < of 21 (conjuncts 1, 3, 4). R and S are each interpreted as isomorphisms of 21 onto the field of the interpretation of R2. Call these isomorphisms R, S. respectively. Then S-'R is an automorphism of c21, so S-'R must be the identity. Thus, S is R and Vxlx2(R(xl, x2) + S(x1, x2)) is satisfied.

There is no L, formula 8(x1, x2) *such that F '(R)- Vx1x2(R(x1, x2) 8(x1, x2)). For suppose there were such a 6. Find an isomorphism R of <D, < > onto some KD2, ? 2> such that D2 is disjoint from D. Let R, ', ' 2 interpret R, R1, R2, respectively. Let each d e D interpret itself. Then b(R) is satisfied by the interpretation 23 just determined. By the choice of 21, there is a d e D such that d is not definable relative to 21 by an L' ,, formula. Thus, <D, <, d> is Y, equivalent to some KD, <, c> such that c # d. It follows that <D2, < 2, R(d)> is ?'U,, equiva- lent to <D2, < 2, R(c)> since R is an isomorphism. From this and the fact that <D, <, e, d>ec-D is Y.oco-elementarily equivalent to itself, it follows, since cardinal sums preserve Y,,-elementary equivalence (see [2]), that

<D u D2, <, e, d, < , R(d)>eeD -.oa <D u D2, <, e, d, ? 2, R(C)>eeD

I.e., (23 I L, d, R(d))- , (3 I L, d, R(c)). By choice of 0, 0 defines R relative to 23 L. Thus, 23 L k 0(d, R(d)). By the last , equivalence, 23 L k 0(d, R(c)). Since R(c) 0 R(d), 6 does not define R. Thus there is no such 6.

By the above, Beth definability fails for Y... There is a modification of the above such that R is unary and O(R) is a conjunction of Yc,1(0 sentences. Note that q(R) is a complete sentence, since it has exactly one model up to isomorphism.

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24 JOHN GREGORY

?2. THEOREM 2. (1) Beth(Y.?C., Y,,f)failsfor every cardinal K. (2) If uncountable K is regular, then Beth(yK . (,,, S.2) fails.

We will prove part (2). Part (1) follows easily from (2). Compare (2) with the known facts that, for every infinite cardinal K, Beth(2'K + Y g(2K)+ X+) [3], and that if cof(K) = c, Beth(9K + 1,S (2K)+ K) holds (Chang [1] generalized the result of Lopez-Escobar for K = w). In [2], Friedman proved that, if cof(K) > c, then not Beth(.F+,e, .,YK+ c+). (The author does not know whether Beth(.F+ ,,, ASK) when CO < COf(K) < K, but this would follow from A(K) below.) By "a structure of car- dinality K, we mean its domain has cardinal K and it has no more than K constants and relation symbols. A rigid structure is one which has exactly one automorphism.

Consider the following for infinite cardinal K.

A(K): There exist two structures which are of cardinality K, are .O'K-elementarily equivalent, but are not isomorphic.

B(K): There exist two rigid structures which are of cardinality K, are Y0K- elementarily equivalent, but are not isomorphic.

C(K): Not Beth(2,, + i,, S0K).

We will prove that A(K) implies B(K), and B(K) implies C(K). By the result of Morley, A(K) is true for all uncountable regular K. Thus, for uncountable regular K,

C(K) is true; i.e., part (2) of the theorem is true. Note that, if there exist two structures as in A(K), then there are 2K such structures.

Indeed, suppose 210 = <AO, RO,>, and 21, = <Al, R,,>, are two structures as in A(K). For each Fe 21', define 2F as follows. (21F is sometimes called a strong cardinal sum.) For each a < K, there is a unary relation U~, = {a} x AF(,). For corresponding relations Roy and R,, of to and 21,, respectively, there is the relation

RY = {<<a, a,>, - -, <a, a?>> I <a,, - - , an> E R.(,), y}- Let domain A = U jUa I a < K}. 21F is the structure <A, U4, R,>,, ,. Then for dis- tinct elements F and G of 2K, 2JF is of cardinality K, 2OF - -K , and 2[F is not iso- morphic to 2G

THEOREM 3. A(K) implies B(K).

PROOF. Assume A(K). Then there are 2K structures as in A(K), hence there are K

such structures, 210, P < K. We can assume 2t = <K, R8. >.. Define Z = <D, R, Ra>a where D, R, R. are as follows. D is the set of all nonempty finite sequences of ele- ments of K. (That is, D is the set of all functions f: n-> K such that 0 # n E co.) R(f, g) iff, for m + 1 = dom(g), f = g r m. Ra(f - - - ,fjn) iff, for some f, for m + 1 = dom(f), and for P = f(m), R(ff) for each i and

R8a(f1(m + 1),.. fn(m + 1))-

Put Df = {g I R(f, g)} and 93f = <Df, Ra | Df >a. Then 23 21(f r when m + 1 =

dom(f). Z is of cardinality K.

Every automorphism G of i; is the identity. Indeed, we show by induction on dom(f) that G(f) = f.

Case 1. dom(f) = 1. Then 1 -, 3xR(x,f), hence 2 k -,3xR(x, G(f)). Thus, dom(G(f)) = 1. Zf W 2to) and ZG(f) 2 tG(f)(O)- G r Df is an isomorphism of B5f onto ZGUf) since G(Df) = DG(f). Thus, 21f(o) - G(f)(O) and so f(O) = G(f)(O)_ Therefore, G(f) = f.

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BETH DEFINABILITY IN INFINITARY LANGUAGES 25

Case 2. dom(f) = m + 1 > 1. Then R(f r m,f). Since G is an automorphism, R(G(f i m), G(f)). By the induction hypothesis, G(f r m) = f r m. Thus,

R(f r m, G(f)), i.e., G(f) r m = f r m and dom(G(f)) = m + 1. As in Case 1,

2f(m) -

Xf G(f) G 'G(f)(m) SO f(m) = G(f)(m). Thus, G(f) = f. Define X similarly to Z by changing RB<e to R + 1, in the above definition of Rn.

As above, XE is of cardinality K and has no proper automorphism. X f 2 since, for

f = {<0K O>},I f - 40 g for all g E D. Forf e D and assignments F, define Ff to be F r F- Df . X = Z. Indeed, this clearly follows from the following, which will be proved

by induction on formulas q. LEMMA. For all 5P-K sentences 0 with at most set s -< K of new constants,

(Z3, F) f iff (X(, G) k 0 whenever (A) F, G are assignments of s to Z3, X, respectively,

the ranges of F and G are closed under restriction, and if F(ai) = fA and G (ai) =g

for i = 1, 2, then f1 = .f2 if g1 = g2, R(fi, f2) iff R(g1, g2), and (Zfl, Ff,1)OD

(Xg1' Gg9). PROOF. Assume the induction hypothesis that the Lemma holds for all sub-

stitution instances of -<K constants into subformulas of 0, to show the Lemma

holds for 0. Consider any F and G satisfying clause (A). Atomic case. If 0 is of the form a, = a2 or R(al, a2), then clause (A) gives

(Z3, F) f 0 iff (X, G) k 0. Suppose 0 is Rz(al,** , an), fi = F(ai), gi = G(ai). We

show that Z3 k Rz(fi, - - - , f) implies X k Rr(gl1,.- - , gn) (the converse is proven

similarly). Assume Z3 k Raz(fi, - - ,fn). Then, for some f E rng(F), fA, * * *, fin E Df and 93, k R,,(fi, - - -, fn). But iff = F(b), then, by clause (A), (93,, Ff) _K (9g, Gg)

where g = G(b). So (g k R,(g1, * *, gn) and therefore X k R(gl, . .., n)-

Propositional case. If 0 is a conjunction or negation, then the desired result

follows easily from the induction hypothesis. Quantificational case. Suppose 0 is 3v/(s, v) where v is a set of -<c variables.

Assume (93, F) k 3vb(s, v), to show (W, G) k 3vb(s, v). (The proof of the converse

is similar.) Then for some assignment E from a set r of fewer than K new constants

into 93, (93, F u E) k t(s, r). We can assume rng(F u E) is closed under restriction

and r disjoint from s. There exists H such that F u E and G u H satisfy clause (A). Indeed, we now

start to construct such an H. Put Dn = {fI dom(f) = n + 1}, En = E E-'Dn, = F F'Dn, and

Gn= G r G 'D,. Note that since F and G satisfy clause (A), we have dom(Fn) =

dom(Gn) for all n Ec w. By definition, rng(Fn) c Dn and rng(Gn) c Dn. Hence the

structures <Dn, Fn> and <Dn, Gn> have the same similarity type; they both interpret

formulas of .2O,, having only the equality predicate and individual constants from

dom(Fn). In fact it follows from clause (A) that <Dn, Fn>F=> <Dn, Gn> for all n.

By induction on n, we will construct functions Hn such that (1) Hn is an assign-

ment from dom(En) into Dn, (2) <Dn, n u En>F = <Dn, Gn u Hn>, and (3)

whenever m = n + 1 and f e rng(Fn u En), there corresponds a, g such that

f = (Fn u En)(a), g = (Gn u Hn)(a), and (Zf, (F u E), =,,C (X9, QG u (Hm)g) (by

(2), g is uniquely determined by f). It is straightforward to show that properties

(1), (2), (3) imply that H = UnHn is an assignment from r into (E such that

H - l Dn = Hn and F u E, G u H satisfy clause (A).

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26 JOHN GREGORY

We have <Do, F0> K oc <Do, Go>. So there exists Ho such that 'Do, Fo U E0>- Wi <Do, Go u Ho> and dom(Ho) = dom(Eo).

Suppose Hn has been defined. Consider any fe rng(F,, u EJ). If f e rng(F,,), choose a so thatf = F,,(a) and put g = G,,(a); then (O3, Ff) -= K (X9, Gg) by clause (A), so one can choose Lf such that (23, Ff u En)-=, (Xg, Gg U Lf) and dom(Lf) = dom(Ef). If f rng(F,,), choose a so that f = En(a) and put g = H,(a); then 93f (g, so one can choose Lf such that (O,, E,) -x (9g, Lf) (here Ff = 0 = Gg since f 0 rng(F,,)). Let H,, 1 = U{Lf I fe rng(F,, U E,,)}.

It is straightforward to show that H,, is defined and has properties (1), (2) above. Property (3): Assume m = n + 1 and f E rng(F,, u E,,). Then in the construction of Hm we chose a, g, Lf such that f = (Fn, u E,,)(a), g = (G7, U H,,)(a), and (93f, (F U E)f)= (9g, Gg u L,). Thus, it suffices to observe that (Hm)g =

Hm 1 H;'D9 = Lf. By induction, the H,,, n E w, are defined and have properties (1), (2), (3). So

H = U,,Hn is such that F u E and G u H satisfy clause (A). Thus, the induction hypothesis for b implies (S, F U E) k #(s, r) iff (X, G U H) k #(s, r). Therefore, (W, G U H) k a(s, r) and ((:, G) k 3vb(s, v). Q.E.D. Lemma and Theorem 3.

THEOREM 4. B(K) implies C(K). PROOF. Assume there are two structures 21 and 23 as in B(K). Then 21 must have

an y, undefinable element, for otherwise there would be an .2OK formula (e.g., the conjunction of

3x1... x,,(defai(xi) A A defa.(x,) A R(x,* - x,,)) whenever R(a1,* , an),

3X1 xn(defa1(xI) A A A defan(x,,) A _ R(x1, . .. , xn)) whenever not R(al, an),

VxV{defa(x) I element a of Zl}, where defd(x) is a formula defining d relative to 21) determining 21 up to isomorphism which would contradict 21 i; Z A. By the method of proof of ?1, C(K) follows.

REFERENCES

[1] C. C. CHANG, Two interpolation theorems, Symposia Mathematica, vol. V, Academic Press, London, New York, 1971, pp. 5-19.

[2] H. FRIEDMAN, The Beth and Craig theorems in infinitary languages (mimeographed). [3] J. MALITZ, Problems in the model theory of infinite languages, Doctoral Dissertation,

University of California, Berkeley, California, 1966. [4] T. OHKUMA, Sur quelques ensembles ordonnds lineairement, Fundamenta Mathematica,

vol. 43 (1965), pp. 326-337.

UNIVERSITY OF MANITOBA

WINNIPEG R3T 2N2, MANITOBA, CANADA

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