Embedding lattices in the Kleene degrees

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1 F U N D A M E N T A MATHEMATICAE 62 (999) Embeddng lattces n the Kleene degrees by Hsato M u r a k (Nagoya) Abstract. Under ZFC+CH, we prove that some lattces whose cardnaltes do not exceed ℵ can be embedded n some local structures of Kleene degrees. 0. We denote by 2 E the exstental nteger quantfer and by χ A the characterstc functon of A,.e. x A χ A (x) =, and x A χ A (x) = 0. Kleene reducblty s defned as follows: for A, B ω ω, A K B ff there s a ω ω such that χ A s recursve n a, χ B, and 2 E. We ntroduce the followng notatons. K denotes the upper semlattce of all Kleene degrees wth the order nduced by K. For X, Y ω ω, we set X Y = { 0 x x X} { x x Y }. Then deg(x Y ) s the supremum of deg(x) and deg(y ). The superump of X s the set X SJ = { e x ω ω {e}((x) 0, (x), χ X, 2 E) }. Here, e x s the real such that ( e x)(0) = e and ( e x)(n + ) = x(n) for n ω. More generally, for m ω, e 0,..., e m x s the real such that ( e 0,..., e m x)(n) = e n for n m and ( e 0,..., e m x)(n + m + ) = x(n) for n ω. Further, (x) 0 = λn.x(2n) and (x) = λn.x(2n + ). We dentfy (x) 0, (x) wth x. An X-admssble set s closed under λx.ω X;x ff t s X SJ -admssble. The followng condtons () and (2) are equvalent to A K B ([8]). () There s y ω ω such that A s unformly -defnable over all (B; y)- admssble sets;.e. there are Σ (Ḃ) formulas ϕ 0 and ϕ such that for any (B; y)-admssble set M and for all x ω ω M, x A M = ϕ 0 (x, y) M = ϕ (x, y). (2) There are y ω ω and Σ (Ḃ) formulas ϕ 0 and ϕ such that for all x ω ω, x A L ω B;x,y [B; x, y] = ϕ 0 (x, y) L ω B;x,y[B; x, y] = ϕ (x, y). 99 Mathematcs Subect Classfcaton: 03D30, 03D65. [47]

2 48 H. Murak Here, we are thnkng of the language of set theory wth an addtonal unary predcate symbol Ḃ. A set M s sad to be (B; y)-admssble ff the structure M,, B M s admssble and y M. Next, L α [B; y] denotes the αth stage of the herarchy constructble from {y} relatve to a unary predcate B, and ω B;y denotes the least (B; y)-admssble ordnal. For K, K ω ω, we set K[K, K ] = {deg(x) K K X K K }. In 3, we wll prove that under ZFC+CH, for some K ω ω, lattces whose felds ω ω and whch are Kleene recursve n K SJ can be embedded n K[K, K SJ ]. Wthout CH, t s unknown whether our Theorem can be proved or not.. Smlarly to [3] and [6], we use lattce tables (lattce representatons n [6]), on whch lattces are represented by dual lattces of equvalence relatons. For every lattce L wth cardnalty 2 ℵ 0, we denote the feld of L also by L and regard L ω ω. We denote by 0 the dentcally 0 functon from ω to ω. Defnton. Let L be a lattce wth relatons L, L, and L. For a, b L ( ω ω) and l L, we defne a l b by a(l) = b(l). Θ L ( ω ω) s called an upper semlattce table of L ff Θ satsfes: (R.0) If there s the least element 0 L of L, then for all a Θ, a(0 L ) = 0. (R.) (Orderng) For all a, b Θ and, L, f L and a b, then a b. (R.2) (Non-orderng) For all, L, f L, then there are a, b Θ such that a b and a b. (R.3) (Jon) For all a, b Θ and,, k L, f L = k, a b, and a b, then a k b. In addton, f Θ satsfes (R.4) below, then Θ s called a lattce table of L: (R.4) (Meet) For all a, b Θ and,, k L, f L = k and a k b, then there are c 0, c, c 2 Θ such that a c 0 c c 2 b. For every lattce L wth relatons L, L, L, and L ω ω, we say that (L, L, L, L ) s Kleene recursve n X ω ω ff L {, L } {,, k L = k} {,, k L = k} K X. In ths paper, we need sutable restrctons n (R.2) and (R.4). Proposton.. Let L be a lattce wth relatons L, L, L, and L ω ω. Let X ω ω. If (L, L, L, L ) s Kleene recursve n X, then there are a lattce table Θ of L and F ω ω L ω ω such that Θ = {F [x] x ω ω}, F K X, and F satsfes: (R.2*) For all, L, f L, then there are a, b ω ω L ω, [, ] such that F [a] F [b] and F [a] F [b].

3 Embeddng lattces n the Kleene degrees 49 (R.4*) For all a, b ω ω and,, k L, f L = k and F [a] k F [b], then there are c 0, c, c 2 ω ω L ω a,b,,,k[a, b,,, k] such that F [a] F [c0] F [c] F [c2] F [b]. (R.5) For all a ω ω, Rng(F [a] ) L ω a [a]. Here, for x ω ω, we set F [x] = { l, y x, l, y F } and regard F [x] : L ω ω. P r o o f. We fx X and L as n the proposton. We assume that there s the least element 0 L of L. We wll construct Θ and F wth the requred propertes. For x ω ω and m ω, we defne the functon f 0,m x : L ω ω as follows: If x L or m 2, then { 0 f l = f 0,m x 0L, (l) = 0, m x otherwse. If x L and m = 2, then 0 f l = 0 L, f 0,2 x (l) = 0, x f 0 L l L x, 0, 2 x otherwse. For x ω ω and n, m ω, we defne the functon f n+,m x : L ω ω nductvely as follows: If x = a, b,,, k, a b, max{a(0), b(0)} = n,,, k L, L = k, L, L, f a (k) = f b (k), and m 2, then { f n+,0 x f (l) = a (l) f l L, n +, 0 x otherwse, f n+,0 x (l) f l L, f n+, x (l) = n +, x f l L and l L, n +, 2 x otherwse, f b (l) f l L, f n+,2 x (l) = n +, x f l L and l L, n +, 3 x otherwse. In the other case, f n+,m x (l) = { 0 f l = 0L, n +, m + x otherwse. We set Θ = {f x x ω ω} and F = { x, l, y ω ω L ω ω f x (l) = y}. Then F [x] = f x for x ω ω. (To defne f x for all x ω ω, we make Θ contan some excess elements.) We prove that Θ and F have the requred propertes. By defnton, Θ = {F [x] x ω ω}, F K X, and F satsfes (R.5). For n ω, we set Θ n = {f x x ω ω x(0) n}.

4 50 H. Murak Lemma.2. () Θ 0 s an upper semlattce table of L. (2) F satsfes (R.2*). P r o o f. () We check that Θ 0 satsfes (R.0) (R.3). (R.0) By defnton, for all f x Θ 0, f x (0 L ) = 0. (R.) Suppose f 0,m x, f 0,m x Θ 0 and, L satsfy L and f 0,m x () = f 0,m x (). If f 0,m x = f 0,m x or = 0 L, then clearly f 0,m x () = f 0,m x (). Suppose f 0,m x f 0,m x and 0 L. Clearly 0 L. By defnton and f 0,m x () = f 0,m x (), we have {m, m } = {, 2}, x = x L, and L x (moreover, f 0,m x () = f 0,m x () = 0, x). Hence, L x and so f 0,m x () = 0, x = f 0,m x () by defnton. (R.2) Let, L and L. We choose f 0, and f 0,2 n Θ 0. Snce L, we have f 0, () = 0, 0, 2 = f 0,2 (). If = 0 L, then f 0, () = 0 = f 0,2 (), and f 0 L, then f 0, () = 0, = f 0,2 (). (R.3) Suppose f 0,m x, f 0,m x Θ 0 and,, k L satsfy L = k, f 0,m x () = f 0,m x (), and f 0,m x () = f 0,m x (). We may suppose f 0,m x f 0,m x and k 0 L. By defnton, we have {m, m } = {, 2}, x = x L, and, L x. Hence, k L x and so f 0,m x (k) = 0, x = f 0,m x (k) by defnton. (2) Snce 0,, 0, 2 L ω, [, ], (2) s clear from the proof of (R.2) n (). Lemma.3. For all n ω, f Θ n s an upper semlattce table of L, then Θ n+ s an upper semlattce table of L. P r o o f. By defnton, Θ n+ satsfes (R.0). Snce Θ n Θ n+, Θ n+ satsfes (R.2). It s routne to check that Θ n+ satsfes (R.) and (R.3). Below, we check (R.) n a few cases, and leave the check of (R.) n the other cases and of (R.3) to the reader. Suppose f m 0,m x, f m 0,m x Θ n+ and l, l L satsfy l L l and f m 0,m x (l ) = f m 0,m x (l ). We may assume f m 0,m x f m 0,m x and l 0 L. Snce Θ n s an upper semlattce table of L, we may also assume that f m 0,m x Θ n or f m 0,m x Θ n. We notce that f f m 0,m x or f m 0,m x s defned by In the other case n the constructon of Θ n+, then f m 0,m x (l ) = f m 0,m x (l ) does not occur. Case : f m 0,m x Θ n and there are a, b ω ω and,, k L such that m 0 = n +, m =, x = a, b,,, k, a b, max{a(0), b(0)} = n, L = k, L, L, and f a (k) = f b (k). Snce f m 0,m x Θ n, t follows that f m 0,m x (l )(0) n and so f n+, x (l )(0) n. Then, by defnton, l L, l L, and f n+, x (l ) = f n+,0 x (l ) = f a (l ). Hence f a (l ) = f m 0,m x (l ). Snce f a Θ n

5 Embeddng lattces n the Kleene degrees 5 and Θ n satsfes (R.), f a (l) = f m 0,m x (l). Clearly, l L L, hence f n+, x (l) = f n+,0 x (l) = f a (l) = f m 0,m x (l). Case 2: There are a, b, a, b ω ω and,, k,,, k L such that m 0 = m 0 = n +, m =, m = 2, x = a, b,,, k, x = a, b,,, k, a b, a b, max{a(0), b(0)} = max{a (0), b (0)} = n, L = k, L = k, L, L, L, L, f a (k) = f b (k), and f a (k ) = f b (k ). By defnton, we have two subcases. Subcase 2.: l L L L and f n+, x (l ) = f n+,0 x (l ) = f a (l ) = f b (l ) = f n+,2 x (l ). Then, smlarly to Case, we obtan f n+, x (l) = f a (l) = f b (l) = f n+,2 x (l). Subcase 2.2: l L, l L, x = x, and f n+, x (l ) = n+, x = f n+,2 x (l ). Then =, =, k = k, a = a, and b = b clearly. If l L, then f n+, x (l) = n +, x = f n+,2 x (l). Suppose l L. Snce l L L, f n+, x (l) = f a (l) and f n+,2 x (l) = f b (l). Snce L = k, f a (k) = f b (k), and Θ n satsfes (R.), we have f a (l) = f b (l). Hence, f n+, x (l) = f n+,2 x (l). By Lemmas.2 and.3, Θ s an upper semlattce table of L. Lemma.4. F satsfes (R.4*). Hence, Θ s a lattce table of L. P r o o f. Suppose a, b ω ω and,, k L satsfy L = k and f a (k) = f b (k). In the case of L or L, we set c 0 = c = c 2 = b or c 0 = c = c 2 = a, and then c 0, c, c 2 have the requred propertes. Suppose L, L, and a b. We set n = max{a(0), b(0)} and c m = n +, m a, b,,, k for m 2. Then c 0, c, c 2 L ω a,b,,,k[a, b,,, k]. By defnton, f a f c 0 f c and f c 2 f b. Snce L, we have f c f c 2. Ths completes the proof of Proposton.. 2. We start ths secton wth Lemma 2. (ZFC+CH). There s S ℵ such that ω ω L ℵ [S]. P r o o f. We take a becton f : ℵ ω ω and set S = {ξ ℵ γ ξ m, n ω( ξ = ω γ + 2 m 3 n f(γ)(m) = n)}. Notce that for all ξ < ℵ, there are unque γ ξ and unque k ω such that ξ = ω γ + k. Let x ω ω be arbtrary. We choose γ ℵ such that f(γ) = x; then x(m) = n ω γ + 2 m 3 n S for all m, n ω. Hence, x L ℵ [S]. We fx S ℵ such that ω ω L ℵ [S]. We defne the functon rk : ω ω ℵ by rk(x) = mn{α ℵ x L α+ [S]} for x ω ω. We set K 0 = {x WO o.t.(x) S} and

6 52 H. Murak K = { m, n x ω ω w WO(rk(x) = o.t.(w) w WO(w < L[S] w o.t.(w ) rk(x)) w(m) = n)}. Here, WO denotes the set of all x ω ω whch code a well-orderng relaton on ω, and o.t.(w) denotes the order type of w. If e.g. n-determnacy (2 n ω) s assumed, then by the localzaton of the theorem of Solovay [7], for any n set K ω ω, K[K, K SJ ] = {deg(k), deg(k SJ )}. Under ZFC+CH (even f some determnacy axom s assumed), f K 0 K K ω ω, then K[K, K SJ ] {deg(k), deg(k SJ )} ([5]; n fact we can prove that K[K, K SJ ] contans many elements). To prove the Theorem n 3, we use K n addton to K 0. We note that under ZFC+CH, {d K deg(k 0 K ) K d} s dense, whch can be proved smlarly to [2] and [4]. Lemma 2.2 (ZFC+CH). Let K 0 K K K ω ω and T = S K. () For all x ω ω, L ω K;x[K; x] s S-admssble, and so T -admssble. (2) If M s K-admssble, then for all x ω ω M, rk(x) M. (3) For all x ω ω, x L ω T ;x [T ], hence L ω T ;x [T ; x] = L ω T ;x[t ]. (4) If M s T -admssble and On M = α, then ω ω M = {x ω ω rk(x) < α}. P r o o f. () It s suffcent to prove that S s over L ω K;x[K; x]. For all ξ ω K;x, snce there s an necton from ξ to ω n L ω K;x[K; x], there s w WO L ω K;x[K; x] whch codes a well-orderng of order type ξ. Hence, for all ξ ω K;x, ξ S L ω K;x[K; x] = w K 0 (o.t.(w) = ξ) L ω K;x[K; x] = w WO(o.t.(w) = ξ w K 0 ). Therefore, S s Σ and Π over L ω K;x[K; x]. (2) Let w be the L[S] -least element of WO such that o.t.(w) = rk(x). By defnton, for all m, n ω, w(m) = n m, n x K. Snce M s K -admssble, w M and hence rk(x) = o.t.(w) M. (3) Snce x L ω T ;x by (2). Snce L ω T ;x [T ; x] and L ω T ;x x L rk(x)+ [S], hence x L ω T ;x [T ; x] s K-admssble, rk(x) < ω T ;x [T ]. By defnton, [T ] s S-admssble, L rk(x)+ [S] L ω T ;x [T ]. (4) Suppose x ω ω and rk(x) < α. Snce M s S-admssble, L rk(x)+ [S] M, hence x M. Conversely, f x ω ω M, then snce M s K-admssble, rk(x) < α by (2).

7 Embeddng lattces n the Kleene degrees Let S, rk, K 0, and K be as n 2. Theorem (ZFC+CH). Let K 0 K K K ω ω. For any lattce L, f L ω ω and (L, L, L, L ) s Kleene recursve n K SJ, then L can be embedded n K[K, K SJ ]. Ths secton s entrely devoted to provng the Theorem. We use AC and CH wthout notce n the proof. We fx K ω ω such that K 0 K K K, and a lattce L such that L ω ω and (L, L, L, L ) s Kleene recursve n K SJ. We set T = S K. Then every T -admssble set s S-admssble and K-admssble, and ω ω L ℵ [T ]. We fx a lattce table Θ of L and F ω ω L ω ω whch are obtaned by Proposton.. For smplcty, we assume that (L, L, L, L ) s Kleene recursve n K SJ wth no addtonal real parameter and F K K SJ wth no addtonal real parameter. For x ω ω, we denote F [x] by f x as n the proof of Proposton.. We may assume that f 0 s dentcally 0 on L and 0 s the L[T ] -least real. For every total or partal functon p from ω ω to ω ω, we defne the proectons of p by P l = { x, f p(x) (l) x Dom(p)} for l L. We wll construct a total functon g : ω ω ω ω such that l L deg(k G l ) K[K, K SJ ] s a lattce embeddng. Recall that G l denotes the proecton of g on the coordnate l. By recurson, we defne a strctly ncreasng sequence τ α α ℵ of countable ordnals whch satsfes: (T.) τ α+ s the least T -admssble ordnal such that ω ω (L τα+ [T ] (T.2) L τα [T ]) s not empty. If α s a lmt ordnal, then τ α = β α τ β. The followng s proved by routne work. Lemma 3.. () The graph of τ α α ℵ s unformly Σ (T )-defnable over all T -admssble sets. (2) For any T -admssble set M, f α ℵ M and τ β β α M, then τ β β α M. Lemma 3.2. For all α ℵ and x ω ω (L τα+ [T ] L τα [T ]), we have L τα+ [T ] = L ω K;x[K; x]. P r o o f. By Lemma 2.2, x L ω T ;x[t ], hence t follows by the defnton of τ α+ that τ α+ ω T ;x. Snce L ω K;x[K; x] s T -admssble by Lemma 2.2, L τα+ [T ] L ω T ;x[t ] L ω K;x[K; x]. Conversely, snce L τα+ [T ] s (K; x)- admssble, we have L ω K;x[K; x] L τα+ [T ].

8 54 H. Murak Remember that for any K-admssble set N, N s closed under λx.ω K;x ff N s K SJ -admssble, and moreover N s closed under λx.ω K;x ff x ω ω N α On N(L α [K; x] s (K; x)-admssble) N. Hence the quantfers n the statement N s K SJ -admssble are bounded by N. Moreover, note that F s unformly over all K SJ -admssble sets, snce F K K SJ. Lemma 3.3. Let p be a partal functon from ω ω to ω ω, M be a T - admssble set, p M and l L M. If for all x Dom(p), there s σ On M such that L σ [T ] s K SJ -admssble and p(x), l L σ [T ], then P l M. P r o o f. By Σ -collecton, there exsts γ On M such that for all x Dom(p) there s σ < γ such that L σ [T ] s K SJ -admssble and p(x), l L σ [T ] (moreover f p(x) (l) L σ [T ] by (R.5)). Then for all x, y ω ω we have x, y P l M = x Dom(p) y L γ [T ] Hence, P l M by -separaton. σ < γ z L σ [T ](L σ [T ] s K SJ -admssble l, y L σ [T ] z = p(x) ( z, l, y F ) L σ[t ] ). We construct g α (α ℵ ) of the parts of g as follows: Stage 0. We set g 0 =. Stage α lmt. We set g α = β α gβ. Stage α +. Case : There s t ω ω L τα [T ] whch satsfes (G.) or (G.2) below: (G.) There are e ω, v ω ω,, L, and σ τ α such that t = 0, e v,,, L, L σ [T ] s K SJ -admssble, t L σ [T ], and x ω ω L τα [T ](χ G α (x) = {e}(x, v, χ K G α, 2 E)). (G.2) There are e 0, e ω, v 0, v ω ω,,, k L, and σ τ α such that t =, e 0, e v 0, v,,, k, L = k, L σ [T ] s K SJ - admssble, t L σ [T ], x ω ω L τα [T ]({e 0 }(x, v 0, χ K G α, 2 E) = {e }(x, v, χ K G α, 2 E)), and there s a partal functon p L τα+ [T ] from ω ω to ω ω such that g α p, Rng(p g α ) L σ [T ], and x ω ω L τα+ [T ]({e 0 }(x, v 0, χ K P 0, 2 E) = {e }(x, v, χ K P 0, 2 E)). Here, P l 0 = P l { y, 0 y ω ω Dom(p)} for l L. We choose the L[T ] -least t ω ω L τα [T ] whch satsfes (G.) or (G.2) and dstngush two subcases. Subcase.: t satsfes (G.). We choose the L[T ] -least z ω ω (L τα+ [T ] L τα [T ]) and the L[T ] -least a, b ω ω ω ω such that f a () = f b () and f a () f b () by (R.2). Notce that f σ s as n (G.), then

9 Embeddng lattces n the Kleene degrees 55 a, b, f a () L σ [T ] by (R.2*) and (R.5). We set z = z, f a () and defne partal functons p a, p b by { p a (x) (p b g (x) resp.) = α (x) f x Dom(g α ), a (b resp.) f x = z. Then P a = P b, z P a, and z P b. If {e}(z, v, χ K P a 0, 2 E) = 0, then we defne { p g α+ a (x) f x Dom(p a ), (x) = 0 f x ω ω L τα+ [T ] Dom(p a ), and f {e}(z, v, χ K P a 0, 2 E) = 0, then we defne { g α+ p (x) = b (x) f x Dom(p b ), 0 f x ω ω L τα+ [T ] Dom(p b ). Subcase.2: t satsfes (G.2). We choose the L[T ] -least partal functon p L τα+ [T ] as n (G.2) and defne { p(x) f x Dom(p), g α+ (x) = 0 f x ω ω L τα+ [T ] Dom(p). Case 2: Otherwse. We defne { g g α+ α (x) f x Dom(g α ), (x) = 0 f x ω ω L τα+ [T ] Dom(g α ). In the constructon at Stage α + above, notce that for l L, G α+ l = Pl a 0 L τ α+ [T ] or = Pl b 0 L τ α+ [T ] (Subcase.), or = P l 0 L τα+ [T ] (Subcase.2), or = G α l 0 L τα+ [T ] (Case 2) respectvely. We defne g = α ℵ g α. Then, for all α ℵ, g ω ω L τα [T ] = g α and g α : ω ω L τα [T ] ω ω L τα [T ]. Moreover g α+ : ω ω L τα+ [T ] ω ω L τα [T ] by defnton. If there s no σ τ α such that L σ [T ] s K SJ -admssble, then Rng(g α+ ) = {0}. As for proectons, for all α ℵ and l L L τα [T ], we have G l L τα [T ] = G α l. Lemma 3.4. Let ϱ ℵ and L ϱ [T ] be K SJ -admssble. () For all α < ℵ, f ϱ τ α, then there s σ τ α such that L σ [T ] s K SJ -admssble and Rng(g α+ g α ) L σ [T ]. (2) For all x ω ω, there s σ max{rk(x), ϱ} such that L σ [T ] s K SJ - admssble and g(x) L σ [T ]. P r o o f. () We dstngush three cases at Stage α +. Case : g α+ s constructed n Subcase. at Stage α +. We choose σ as n (G.). By defnton, there s c ω ω L σ [T ] (c = a or = b n Subcase

10 56 H. Murak.) such that Rng(g α+ g α ) = {c, 0}. Snce 0 L σ [T ], Rng(g α+ g α ) L σ [T ]. Case 2: g α+ s constructed n Subcase.2 at Stage α+. We choose the L[T ] -least partal functon p and σ as n (G.2). By (G.2), Rng(p g α ) L σ [T ], hence Rng(g α+ g α ) L σ [T ]. Case 3: g α+ s constructed n Case 2 at Stage α +. By defnton, Rng(g α+ g α ) = {0} L ϱ [T ]. (2) We choose α < ℵ such that x L τα+ [T ] L τα [T ]. By Lemma 2.2, τ α rk(x). If ϱ τ α, then by () there s σ rk(x) such that L σ [T ] s K SJ - admssble and g(x) = g α+ (x) L σ [T ]. If τ α < ϱ, then snce Rng(g α+ ) L τα [T ], we have g(x) L ϱ [T ]. Snce L ℵ [T ] s K SJ -admssble and ω ω L ℵ [T ], for all x ω ω there exsts ϱ < ℵ such that L ϱ [T ] s K SJ -admssble and x L ϱ [T ] (usng the Löwenhem Skolem Theorem). For x ω ω, we set ϱ(x) = mn{σ < ℵ L σ [T ] s K SJ -admssble and x L σ [T ]}. Lemma 3.5. Let α ℵ and l L. () For any T -admssble set M, f τ α M, then g α M. (2) For any T -admssble set M, f τ α, ϱ(l) M, then G α l M. (3) If ϱ(l) < τ α+, then L τα+ [T ] s G l -admssble. P r o o f. () We prove α ℵ M : T -admssble set (τ α M g β β α M) by nducton. If α = 0, then ths s clear. Let 0 < α ℵ. We assume that for all β α and every T -admssble set M we have (τ β M g γ γ β M). Let M be a T -admssble set and τ α M. Let α = β + for some β. By assumpton, g β L τα [T ]. In the constructon at Stage β +, p a, p b n Subcase. and p n Subcase.2 are elements of L τα [T ]. Snce L τα [T ] M, by defnton g β+ M. Hence g β β α M. Let α be a lmt ordnal. For every lmt ordnal β α, snce g γ γ β L τβ+ [T ], the constructon at Stage β can be expressed over L τβ+ [T ]. And for every β + α, snce the condtons of every case at Stage β + can be expressed over L τβ+ [T ] (notce that f t =......,,,... and ϱ(t) τ β, then G β, Gβ L τ β+ [T ] by Lemmas 3.4 and 3.3, hence we can express (G.) (G.2); otherwse, we proceed to Case 2 mmedately), the constructon at Stage β + can be expressed over L τβ+2 [T ]. Thus, g β β α s - defnable over M wth parameter τ β β α, hence g β β α M. (By

11 Embeddng lattces n the Kleene degrees 57 Lemma 3., τ β β α M.) Therefore, by defnton, g α M, and so g β β α M. (2) By (), g α M. For all x Dom(g α ), snce rk(x) M, there s σ On M such that L σ [T ] s K SJ -admssble and g α (x), l L σ [T ] by Lemma 3.4. Hence, G α l M by Lemma 3.3. (3) By (2), G α l L τα+ [T ]. In the constructon at Stage α +, p a, p b n Subcase. and p n Subcase.2 are elements of L τα+ [T ], hence smlarly to (2), Pl a, P l b, P l L τα+ [T ] by Lemma 3.3. Snce G α+ l = Pl a 0 L τα+ [T ] or = Pl b 0 L τ α+ [T ] or = P l 0 L τα+ [T ] or = G α l 0 L τ α+ [T ], we see that L τα+ [T ] s G α+ l -admssble and so G l -admssble. Lemma 3.6. For all l L, G l K K SJ, hence deg(k G l ) K[K, K SJ ]. P r o o f. For α ℵ, smlarly to Lemma 3.5, the constructon of g α (.e. constructons tll Stage α) and the condtons of every case at Stage α + can be expressed over L τα+ [T ]. Hence, there are formulas ψ and ψ 2 such that: L τα+ [T ] = ψ (p, α) There s t ω ω L τα [T ] whch satsfes (G.) or (G.2) at Stage α + and let t be the L[T ] -least such real, f t = 0, e v,, satsfes (G.) and z, a, b, p a, p b are as n Subcase. L τα+ [T ] = ψ 2 (p, α) then {e}( z, f a (), v, χ K P a, 2 E) = 0 p = p a or {e}( z, f a (), v, χ K P a, 2 E) = 0 p = p b, and f t =, e 0, e v 0, v,,, k satsfes (G.2), then p s the L[T ] -least partal functon as n (G.2). There s no t ω ω L τα [T ] whch satsfes (G.) or (G.2) at Stage α + and p = g α. Here, ψ and ψ 2 correspond to Case and Case 2 respectvely. We choose r WO such that o.t.(r) = ϱ(l). We prove G l K K SJ va r usng (2) of 0. Let x, y ω ω be arbtrary and M = L ω K SJ ;x,y,r[k SJ ; x, y, r]. Notce that f x L τα+ [T ] L τα [T ], then by Lemma 3.2 and K SJ -admssblty of M, we have L τα+ [T ] = L ω K;x[K; x] M. By Lemma 3.4, there s σ max{rk(x), ϱ(l)} such that L σ [T ] s K SJ -admssble and g(x), l L σ [T ]; moreover, f g(x) (l) L σ [T ]. Hence,

12 58 H. Murak x, y G l M = α ω K;x p L ω K;x[K; x] (L ω K;x[K; x] = L τα+ [T ] x L τα [T ] L τα+ [T ] = ψ (p, α) ψ 2 (p, α) ( σ max{rk(x), ϱ(l)}(x Dom(p) p(x), l L σ [T ] L σ [T ] s K SJ -admssble (y = f p(x) (l)) L σ[t ] ) (x Dom(p) y = 0))). Notce that the quantfers n the statement ω K;x = τ α+ are bounded by L ω K;x[K; x], snce ω K;x = τ α+ ff τ ω K;x (τ α < τ τ satsfes (T.)) L ω K;x [K;x]. Hence x, y Gl s over M. Therefore, G l K K SJ. Lemma 3.7. () G 0L K. (2) For all, L, f L, then K G K K G. (3) For all,, k L, f L = k, then (K G ) (K G ) K K G k. P r o o f. () By defnton, G 0L = { x, f g(x) (0 L ) x ω ω} = { x, 0 x ω ω} K. (2) We choose r WO such that o.t.(r) = ϱ(, ). To prove K G K K G, t s suffcent to prove that for all x, y ω ω, x, y G M = σ max{rk(x), ϱ(, )} a, z L σ [T ] (L σ [T ] s K SJ -admssble, L σ [T ] x, z G (f a () = z f a () = y) L σ[t ] ), where M = L K G ω ;,,x,y,r[k G ;,, x, y, r]. Suppose x, y G. By Lemma 2.2, rk(x) M. By Lemma 3.4, there s σ max{rk(x), ϱ(, )} such that L σ [T ] s K SJ -admssble and g(x),, L σ [T ]. By (R.5), we have f g(x) (), f g(x) () L σ [T ]. Thus, f we set a = g(x) and z = f a (), then snce y = f a () and F K K SJ, the rght-hand sde holds. Conversely, suppose that x, y ω ω satsfy the rght-hand sde. Let a, z be as n the rght-hand sde. By x, z G, f g(x) () = z = f a (). Then, by (R.), f g(x) () = f a (). Hence, y = f g(x) (), and so x, y G. (3) By (2), K G G K K G k. We choose r WO such that o.t.(r) = ϱ(,, k). To prove K G k K K G G, t s suffcent to prove that for all x, y ω ω, x, y G k M = σ max{rk(x), ϱ(,, k)} a, z, z L σ [T ] (L σ [T ] s K SJ -admssble,, k L σ [T ] x, z G x, z G (f a () = z f a () = z f a (k) = y ) L σ[t ] ), where M = L K G ω G ;,,k,x,y,r[k G G ;,, k, x, y, r].

13 Embeddng lattces n the Kleene degrees 59 Suppose x, y G k. Smlarly to (2), we set a = g(x), z = f a (), z = f a () and choose σ max{rk(x), ϱ(,, k)} such that L σ [T ] s K SJ - admssble and g(x),,, k L σ [T ]. Then the rght-hand sde holds. Conversely, suppose that x, y ω ω satsfy the rght-hand sde. Let a, z, z be as n the rght-hand sde. Smlarly to (2), we have f g(x) (k) = f a (k) = y by (R.3), and so x, y G k. Lemma 3.8. Let α ℵ and t ω ω L τα [T ] be the L[T ] -least real whch satsfes (G.) or (G.2) at Stage α +. () If t = 0, e v,, satsfes (G.), then there s x ω ω L τα+ [T ] such that χ G α+ (x) = {e}(x, v, χk G α+, 2 E) and so χ G (x) = {e}(x, v, χk G, 2 E). (2) If t =, e 0, e v 0, v,,, k satsfes (G.2), then there s x ω ω L τα+ [T ] such that {e 0 }(x, v 0, χ K G α+, 2 E) = {e }(x, v, χ K G α+, 2 E) and so {e 0 }(x, v 0, χ K G, 2 E) = {e }(x, v, χ K G, 2 E). P r o o f. Both n () and n (2) (.e. n (G.) and n (G.2)), snce ϱ(t) τ α, L τα+ [T ] s G -admssble and G -admssble by Lemma 3.5. () We choose the L[T ] -least z ω ω (L τα+ [T ] L τα [T ]) and the L[T ] -least a, b ω ω ω ω such that f a () = f b () f a () f b (). We set z = z, f a (). Then z L τα+ [T ]. Let p a and p b be as n Subcase. at Stage α +. Case : {e}(z, v, χ K P a 0, 2 E) = 0. Then, for l {, }, G l L τα+ [T ] = = Pl a 0 L τα+ [T ] by defnton. Snce L τα+ [T ] s (G ; v, z )-admssble, {e}(z, v, χ K G, 2 E) = {e}(z, v, χ K G α+, 2 E) = 0. By defnton, z G α+ G. Hence, G α+ l {e}(z, v, χ K G α+ and {e}(z, v, χ K G, 2 E) = χg (z )., 2 E) = = χg α+(z ) Case 2: {e}(z, v, χ K P a 0, 2 E) = 0. Smlarly to Case, {e}(z, v, χ K G, 2 E) = {e}(z, v, χ K G α+, 2 E) = 0. Snce g(z) = g α+ (z) = b and f b () f a (), we have z G α+ and z G. Hence, {e}(z, v, χ K G α+ and {e}(z, v, χ K G, 2 E) = χg (z )., 2 E) = 0 = χg α+(z )

14 60 H. Murak (2) We choose the L[T ] -least partal functon p L τα+ [T ] from ω ω to ω ω as n (G.2). Then, for l {, }, G l L τα+ [T ] = G α+ l = P l 0 L τα+ [T ]. Hence, by (G.2), there s x ω ω L τα+ [T ] such that {e 0 }(x, v 0, χ K G α+, 2 E) = {e }(x, v, χ K G α+, 2 E) and hence {e 0 }(x, v 0, χ K G, 2 E) = {e }(x, v, χ K G, 2 E). Lemma 3.9. For all t ω ω, {α ℵ t satsfes (G.) or (G.2) at Stage α + } s countable. Hence t< L[T ] s {α ℵ t satsfes (G.) or (G.2) at Stage α + } s countable and so bounded for all s ω ω (snce {t ω ω t < L[T ] s} s countable). P r o o f. We set X t = {α ℵ t satsfes (G.) or (G.2) at Stage α + } for t ω ω. We prove that for all t ω ω, X t s countable by nducton on t. Let t ω ω and assume that for all u ω ω, f u < L[T ] t then X u s countable. Suppose that, on the contrary, X t s uncountable. By the nductve assumpton u< L[T ] t X u s countable, hence we can take β X t u< L[T ] t X u. Then t s the < L[T ] -least real whch satsfes (G.) or (G.2) at Stage β +. Snce X t s uncountable, there s α X t such that β + α. Case : t satsfes (G.) at Stage β+. There are e ω, v ω ω, and, L such that t = 0, e v,,. By Lemma 3.8, there s x ω ω L τβ+ [T ] ( L τα [T ]) such that χ G β+(x) = {e}(x, v, χk G β+, 2 E). Then, smlarly to the proof of Lemma 3.8, snce G α l L τ β+ [T ] = G β+ l for l {, } and L τβ+ [T ] s G -admssble, we have χ G α (x) = {e}(x, v, χk G α, 2 E). Hence, t does not satsfy (G.) at Stage α +. Moreover, snce t(0) = 0, t does not satsfy (G.2) at Stage α +. Ths contradcts α X t. Case 2: t satsfes (G.2) at Stage β +. There are e 0, e ω, v 0, v ω ω, and,, k L such that t =, e 0, e v 0, v,,, k. Smlarly to Case, there s x ω ω L τβ+ [T ] such that {e 0 }(x, v 0, χ K G α, 2 E) = {e }(x, v, χ K G α, 2 E). Hence, t does not satsfy (G.2) at Stage α+. Moreover, snce t(0) =, t does not satsfy (G.) at Stage α +. Ths contradcts α X t. Lemma 3.0. For all, L, f L, then K G K K G. P r o o f. Assume L and G K K G. We choose e ω and v ω ω such that for all x ω ω, χ G (x) = {e}(x, v, χ K G, 2 E). We set t = 0, e v,,. By Lemma 3.9, we can choose α ℵ such that for all u < L[T ] t, u does not satsfy (G.) or (G.2) (takng u n place of t) at Stage α +. Choosng α suffcently large, we may assume that there s α < α such that α = α + and ϱ(t) τ α. Then, by Lemma 3.5, L τα [T ]

15 Embeddng lattces n the Kleene degrees 6 s G -admssble, and so by the choce of e, v, for all x ω ω L τα [T ], we have χ G α (x) = {e}(x, v, χ K G α, 2 E). Hence, t satsfes (G.) at Stage α +. Moreover, t s the L[T ] -least real whch satsfes (G.) or (G.2) at Stage α +. Therefore, by Lemma 3.8, there s x ω ω L τα+ [T ] such that χ G (x) = {e}(x, v, χk G, 2 E). Ths s a contradcton. Lemma 3.. Let,, k L, L = k, α ℵ, e 0, e ω, and v 0, v ω ω. Assume that there are partal functons p, p L τα+ [T ] from ω ω to ω ω, σ τ α, and x ω ω such that g α p, p, Dom(p) = Dom(p ), P k = P k, L σ [T ] s K SJ -admssble,,, k L σ [T ], Rng(p g α ), Rng(p g α ) L σ [T ], and {e 0 }(x, v 0, χ K P 0, 2 E) = {e }(x, v, χ K P 0, 2 E). Then there s a partal functon p L τα+ [T ] from ω ω to ω ω such that g α p, Rng(p g α ) L σ [T ], and {e 0 }(x, v 0, χ K P 0, 2 E) = {e }(x, v, χ K P 0, 2 E). P r o o f. We set D = Dom(p) Dom(g α ). Snce P k = P k, for all y D, f p(y) (k) = f p (y) (k). By (R.4*), for all y D there are c y 0, cy, cy 2 ω ω L σ [T ] such that f p(y) f cy 0 f cy f cy 2 f p (y). Snce p, p, D, L σ [T ] L τα+ [T ] and F K K SJ, there exsts c y 0, cy, cy 2 y D L τ α+ [T ] such that for all y D, c y 0, cy, cy 2 ω ω L σ [T ] and f p(y) f cy 0 f cy f cy 2 f p (y) by -separaton. We defne p n : Dom(p) ω ω (n 3) by { g p n α (y) f y Dom(g α ), (y) = c y n f y D. Then p n L τα+ [T ] and Rng(p n g α ) L σ [T ] for n 3. By defnton, P = P 0, P 0 = P, P = P 2, and P 2 = P. If we assume that for all n 3, {e 0 }(x, v 0, χ K P n 0, 2 E) = {e }(x, v, χ K P n 0, 2 E), then we obtan {e 0 }(x, v 0, χ K P 0, 2 E) = {e }(x, v, χ K P 0, 2 E), a contradcton. So there s n 3 such that {e 0 }(x, v 0, χ K P n, 2 E) = {e }(x, v, χ K P n, 2 E). Lemma 3.2. For all,, k L, f L = k, then deg(k G k ) s the K -nfmum of deg(k G ) and deg(k G ). P r o o f. It s suffcent to prove that for all X ω ω, f X K K G and X K K G, then X K K G k. We fx X ω ω such that X K K G and X K K G, and choose e 0, e ω and v 0, v ω ω such that for all x ω ω, χ X (x) = {e 0 }(x, v 0, χ K G, 2 E) = {e }(x, v, χ K G, 2 E). We set t =, e 0, e v 0, v,,, k. By Lemma 3.9, we choose γ ℵ such that sup( u< L[T ] t {α ℵ u satsfes (G.) or (G.2) at Stage α + }) < γ and ϱ(t) τ γ. Clam. For all α ℵ, f γ α, then there s no partal functon p L τα+ [T ] from ω ω to ω ω as n (G.2) at Stage α +.

16 62 H. Murak P r o o f. Assume γ α ℵ and there s a partal functon p L τα+ [T ] from ω ω to ω ω as n (G.2) at Stage α +. Then t satsfes (G.2) at Stage α + by the choce of e 0, e, v 0, v. Snce γ α, t s the L[T ] -least real whch satsfes (G.) or (G.2) at Stage α +. Thus, by Lemma 3.8, there s x ω ω such that {e 0 }(x, v 0, χ K G, 2 E) = {e }(x, v, χ K G, 2 E). Ths s a contradcton and completes the proof of Clam. Clam 2. For all α ℵ wth γ α and for all partal functons p, p L τα+ [T ] from ω ω to ω ω, f g α p, p, Dom(p) = Dom(p ), P k = P k, and there s σ τ α such that L σ [T ] s K SJ -admssble, t L σ [T ], and Rng(p g α ), Rng(p g α ) L σ [T ], then for all x ω ω L τα+ [T ], {e 0 }(x, v 0, χ K P 0, 2 E) = {e 0 }(x, v 0, χ K P 0, 2 E). P r o o f. Assume γ α < ℵ and Clam 2 does not hold for some partal functons p, p. Then there s x ω ω L τα+ [T ] such that {e 0 }(x, v 0, χ K P 0, 2 E) = {e0 }(x, v 0, χ K P 0, 2 E). Snce Clam mples that p s not as n (G.2) at Stage α +, Hence {e 0 }(x, v 0, χ K P 0, 2 E) = {e }(x, v, χ K P 0, 2 E). {e 0 }(x, v 0, χ K P 0, 2 E) = {e }(x, v, χ K P 0, 2 E). Thus, by Lemma 3., there s a partal functon p L τα+ [T ] as n (G.2) at Stage α +. Ths contradcts Clam and completes the proof of Clam 2. Clam 3. For all α ℵ wth γ α, set H α = G α { x, y ω ω x L τα [T ] a ω ω(y = f a () a s the L[T ] -least real such that x, f a (k) G k )}. Then: () H α s unformly -defnable over all T, G k -admssble sets of whch τ α s an element. (2) For all x ω ω L τα+ [T ], {e 0 }(x, v 0, χ K G, 2 E) = {e 0 }(x, v 0, χ K Hα, 2 E). P r o o f. () It s suffcent to prove that H α G α s unformly - defnable over all T, G k -admssble sets of whch τ α s an element. By Lemma 3.4, for all x ω ω L τα [T ] (notce ϱ(t) τ γ τ α rk(x)), there s σ rk(x) such that L σ [T ] s K SJ -admssble and g(x),, k L σ [T ], and moreover f a s the L[T ] -least real such that x, f a (k) G k, then snce a L[T ] g(x), we have a L σ [T ] and so f a (k), f a () L σ [T ]. Hence, for any T, G k -admssble set M wth τ α M and for all x, y ω ω M,

17 Embeddng lattces n the Kleene degrees 63 x, y H α G α M = x L τα [T ] σ rk(x) a ω ω L σ [T ] (L σ [T ] s K SJ -admssble, k L σ [T ] z L σ [T ]((z = f a (k)) L σ[t ] x, z G k ) b, z ω ω L σ [T ]((b < L[T ] a z = f b (k)) L σ[t ] x, z G k ) (y = f a ()) L σ[t ] ),.e. the quantfers n the formula whch states x, y H α G α are bounded by L σ [T ] and rk(x). (2) By defnton, there s a partal functon p L τα+ [T ] such that g α p and { p(x) f x Dom(p), g α+ (x) = 0 f x ω ω L τα+ [T ] Dom(p). Then P 0 L τα+ [T ] = G α+ = G L τα+ [T ]. By Lemma 3.4, there s σ τ α such that L σ [T ] s K SJ -admssble, t L σ [T ], and Rng(p g α ) L σ [T ]. We defne p : Dom(p) ω ω by g α (x) f x Dom(g α ), p (x) = the L[T ] -least a ω ω such that x, f a (k) G k f x Dom(p) Dom(g α ). Then for all x Dom(p) we have f p (x) (k) = f g(x) (k) = f p(x) (k), and so P k = P k. Snce p (x) L[T ] g(x) for all x Dom(p), t follows that Rng(p g α ) L σ [T ]. Snce L τα+ [T ] s G k -admssble, smlarly to (), p s over L τα+ [T ] and so p L τα+ [T ] by -separaton. Moreover, P 0 L τ α+ [T ] = H α L τα+ [T ] by defnton (notce the assumpton that 0 s the L[T ] -least real). Thus, by Clam 2, for all x ω ω L τα+ [T ], {e 0 }(x, v 0, χ K P 0, 2 E) = {e 0 }(x, v 0, χ K P 0, 2 E) and hence {e 0 }(x, v 0, χ K G, 2 E) = {e 0 }(x, v 0, χ K Hα, 2 E). Ths completes the proof of Clam 3. Let x ω ω L τγ [T ] and n 2, and M = L K G ω k ;x[k G k ; x]. By Lemma 2.2, M s T -admssble, and f x L τα+ [T ] L τα [T ], then γ α and τ α rk(x) τ α+ M. Hence by Clam 3, χ X (x) = n α ℵ (x L τα+ [T ] L τα [T ] {e 0 }(x, v 0, χ K Hα, 2 E) = n) M = α rk(x)(τ α rk(x) τ rk(x)(τ α < τ τ satsfes (T.)) {e 0 }(x, v 0, χ K Hα, 2 E) = n).

18 64 H. Murak Therefore, X L τγ [T ] and ( ω ω X) L τγ [T ] are unformly Σ -defnable over all (K G k ; w)-admssble sets, where w s a real n WO such that o.t.(w) = τ γ. Snce L τγ [T ] s countable, X K K G k. Ths completes the proof of the Theorem. Remark. In the Theorem, we may replace (L, L, L, L ) s Kleene recursve n K SJ by (L, L, L, L ) s Kleene recursve n the fnte tmes superump of K. Concernng, for example, (K SJ ) SJ, for any K-admssble set N, N s closed under λx.ω K;x and λx.ω KSJ ;x ff N s (K SJ ) SJ -admssble, and the quantfers n the statement N s closed under λx.ω KSJ ;x are bounded by N as x ω ω N α On N(L α [K; x] s (K; x)-admssble y ω ω L α [K; x] β < α(l β [K; y] s (K; y)-admssble)) N. Replacng L σ [T ] s K SJ -admssble by L σ [T ] s (K SJ ) SJ -admssble n the proof of the Theorem, we can prove the followng: Theorem (ZFC+CH). Let K 0 K K K ω ω. For any lattce L, f L ω ω and (L, L, L, L ) s Kleene recursve n (K SJ ) SJ, then L can be embedded n K[K, K SJ ]. References [] K. J. Devln, Constructblty, Sprnger, 984. [2] K. Hrbáček, On the complexty of analytc sets, Z. Math. Logk Grundlag. Math. 24 (978), [3] M. Lerman, Degrees of Unsolvablty, Sprnger, 983. [4] H. Murak, Local densty of Kleene degrees, Math. Logc Quart. 43 (995), [5], Non-dstrbutve upper semlattce of Kleene degrees, J. Symbolc Logc 64 (999), [6] R. A. S h o r e and T. A. S l a m a n, The p-t degrees of the recursve sets: lattce embeddngs, extensons of embeddngs and the two-quantfer theory, Theoret. Comput. Sc. 97 (992), [7] R. Solovay, Determnacy and type 2 recurson (abstract), J. Symbolc Logc 36 (97), 374. [8] G. Wetkamp, Kleene recurson over the contnuum, Ph.D. Thess, Pennsylvana State Unv., 980. Department of Mathematcs Nagoya Unversty Chkusa-ku, Nagoya , Japan E-mal: murak@math.nagoya-u.ac.p Receved 2 May 998; n revsed form 8 May 999