MATHEMATICAE 162 (1999)
Embedding lattices in the Kleene degrees
by
Hisato M u r a k i (Nagoya)
Abstract. Under ZFC+CH, we prove that some lattices whose cardinalities do not
exceed ℵ
1can be embedded in some local structures of Kleene degrees.
0. We denote by
2E the existential integer quantifier and by χ
Athe characteristic function of A, i.e. x ∈ A ⇔ χ
A(x) = 1, and x 6∈ A ⇔ χ
A(x) = 0. Kleene reducibility is defined as follows: for A, B ⊆
ωω, A ≤
KB iff there is a ∈
ωω such that χ
Ais recursive in a, χ
B, and
2E.
We introduce the following notations. K denotes the upper semilattice of all Kleene degrees with the order induced by ≤
K. For X, Y ⊆
ωω, we set X ⊕ Y = {h0i ∗ x | x ∈ X} ∪ {h1i ∗ x | x ∈ Y }. Then deg(X ⊕ Y ) is the supremum of deg(X) and deg(Y ). The superjump of X is the set X
SJ= {hei ∗ x ∈
ωω | {e}((x)
0, (x)
1, χ
X,
2E)↓}. Here, hei ∗ x is the real such that (hei ∗ x)(0) = e and (hei ∗ x)(n + 1) = x(n) for n ∈ ω. More generally, for m ∈ ω, he
0, . . . , e
mi ∗ x is the real such that (he
0, . . . , e
mi ∗ x)(n) = e
nfor n ≤ m and (he
0, . . . , e
mi ∗ x)(n + m + 1) = x(n) for n ∈ ω. Further, (x)
0= λn.x(2n) and (x)
1= λn.x(2n + 1). We identify h(x)
0, (x)
1i with x.
An X-admissible set is closed under λx.ω
X;x1iff it is X
SJ-admissible.
The following conditions (1) and (2) are equivalent to A ≤
KB ([8]).
(1) There is y ∈
ωω such that A is uniformly ∆
1-definable over all (B; y)- admissible sets; i.e. there are Σ
1( ˙ B) formulas ϕ
0and ϕ
1such that for any (B; y)-admissible set M and for all x ∈
ωω ∩ M ,
x ∈ A ⇔ M |= ϕ
0(x, y) ⇔ M |= ¬ϕ
1(x, y).
(2) There are y ∈
ωω and Σ
1( ˙ B) formulas ϕ
0and ϕ
1such that for all x ∈
ωω,
x ∈ A ⇔ L
ωB;x,y1
[B; x, y] |= ϕ
0(x, y) ⇔ L
ωB;x,y1
[B; x, y] |= ¬ϕ
1(x, y).
1991 Mathematics Subject Classification: 03D30, 03D65.
[47]
Here, we are thinking of the language of set theory with an additional unary predicate symbol ˙ B. A set M is said to be (B; y)-admissible iff the structure hM, ∈, B ∩ M i is admissible and y ∈ M . Next, L
α[B; y] denotes the αth stage of the hierarchy constructible from {y} relative to a unary predicate B, and ω
1B;ydenotes the least (B; y)-admissible ordinal.
For K, K
0⊆
ωω, we set K[K, K
0] = {deg(X) | K ≤
KX ≤
KK
0}. In §3, we will prove that under ZFC+CH, for some K ⊆
ωω, lattices whose fields
⊆
ωω and which are Kleene recursive in K
SJcan be embedded in K[K, K
SJ].
Without CH, it is unknown whether our Theorem can be proved or not.
1. Similarly to [3] and [6], we use lattice tables (lattice representations in [6]), on which lattices are represented by dual lattices of equivalence relations. For every lattice L with cardinality ≤ 2
ℵ0, we denote the field of L also by L and regard L ⊆
ωω. We denote by 0 the identically 0 function from ω to ω.
Definition. Let L be a lattice with relations ≤
L, ∨
L, and ∧
L. For a, b ∈
L
(
ωω) and l ∈ L, we define a ≡
lb by a(l) = b(l). Θ ⊆
L(
ωω) is called an upper semilattice table of L iff Θ satisfies:
(R.0) If there is the least element 0
Lof L, then for all a ∈ Θ, a(0
L) = 0.
(R.1) (Ordering) For all a, b ∈ Θ and i, j ∈ L, if i ≤
Lj and a ≡
jb, then a ≡
ib.
(R.2) (Non-ordering) For all i, j ∈ L, if i 6≤
Lj, then there are a, b ∈ Θ such that a ≡
jb and a 6≡
ib.
(R.3) (Join) For all a, b ∈ Θ and i, j, k ∈ L, if i ∨
Lj = k, a ≡
ib, and a ≡
jb, then a ≡
kb.
In addition, if Θ satisfies (R.4) below, then Θ is called a lattice table of L:
(R.4) (Meet) For all a, b ∈ Θ and i, j, k ∈ L, if i ∧
Lj = k and a ≡
kb, then there are c
0, c
1, c
2∈ Θ such that a ≡
ic
0≡
jc
1≡
ic
2≡
jb.
For every lattice L with relations ≤
L, ∨
L, ∧
L, and L ⊆
ωω, we say that (L, ≤
L, ∨
L, ∧
L) is Kleene recursive in X ⊆
ωω iff L ⊕ {hi, ji | i ≤
Lj}
⊕ {hi, j, ki | i ∨
Lj = k} ⊕ {hi, j, ki | i ∧
Lj = k} ≤
KX.
In this paper, we need suitable restrictions in (R.2) and (R.4).
Proposition 1.1. Let L be a lattice with relations ≤
L, ∨
L, ∧
L, and L ⊆
ω
ω. Let X ⊆
ωω. If (L, ≤
L, ∨
L, ∧
L) is Kleene recursive in X, then there are a lattice table Θ of L and F ⊆
ωω × L ×
ωω such that Θ = {F
[x]| x ∈
ωω}, F ≤
KX, and F satisfies:
(R.2*) For all i, j ∈ L, if i 6≤
Lj, then there are a, b ∈
ωω ∩ L
ωi,j1
[i, j] such
that F
[a]≡
jF
[b]and F
[a]6≡
iF
[b].
(R.4*) For all a, b ∈
ωω and i, j, k ∈ L, if i ∧
Lj = k and F
[a]≡
kF
[b], then there are c
0, c
1, c
2∈
ωω ∩ L
ωa,b,i,j,k1
[a, b, i, j, k] such that F
[a]≡
iF
[c0]≡
jF
[c1]≡
iF
[c2]≡
jF
[b].
(R.5) For all a ∈
ωω, Rng(F
[a]) ⊆ L
ωa1[a].
Here, for x ∈
ωω, we set F
[x]= {hl, yi | hx, l, yi ∈ F } and regard F
[x]: L →
ωω.
P r o o f. We fix X and L as in the proposition. We assume that there is the least element 0
Lof L. We will construct Θ and F with the required properties.
For x ∈
ωω and m ∈ ω, we define the function f
h0,mi∗x: L →
ωω as follows: If x 6∈ L or m 6= 2, then
f
h0,mi∗x(l) =
0 if l = 0
L, h0, mi ∗ x otherwise.
If x ∈ L and m = 2, then
f
h0,2i∗x(l) =
0 if l = 0
L, h0, 1i ∗ x if 0
L6= l ≤
Lx, h0, 2i ∗ x otherwise.
For x ∈
ωω and n, m ∈ ω, we define the function f
hn+1,mi∗x: L →
ωω inductively as follows: If x = ha, b, i, j, ki, a 6= b, max{a(0), b(0)} = n, i, j, k ∈ L, i ∧
Lj = k, i 6≤
Lj, j 6≤
Li, f
a(k) = f
b(k), and m ≤ 2, then
f
hn+1,0i∗x(l) =
f
a(l) if l ≤
Li, hn + 1, 0i ∗ x otherwise,
f
hn+1,1i∗x(l) =
f
hn+1,0i∗x(l) if l ≤
Lj,
hn + 1, 1i ∗ x if l ≤
Li and l 6≤
Lj, hn + 1, 2i ∗ x otherwise,
f
hn+1,2i∗x(l) =
f
b(l) if l ≤
Lj,
hn + 1, 1i ∗ x if l ≤
Li and l 6≤
Lj, hn + 1, 3i ∗ x otherwise.
In the other case,
f
hn+1,mi∗x(l) =
0 if l = 0
L,
hn + 1, m + 1i ∗ x otherwise.
We set Θ = {f
x| x ∈
ωω} and F = {hx, l, yi ∈
ωω × L ×
ωω | f
x(l) = y}.
Then F
[x]= f
xfor x ∈
ωω. (To define f
xfor all x ∈
ωω, we make Θ contain some excess elements.)
We prove that Θ and F have the required properties. By definition, Θ = {F
[x]| x ∈
ωω}, F ≤
KX, and F satisfies (R.5).
For n ∈ ω, we set Θ
n= {f
x| x ∈
ωω ∧ x(0) ≤ n}.
Lemma 1.2. (1) Θ
0is an upper semilattice table of L.
(2) F satisfies (R.2*).
P r o o f. (1) We check that Θ
0satisfies (R.0)–(R.3).
(R.0) By definition, for all f
x∈ Θ
0, f
x(0
L) = 0.
(R.1) Suppose f
h0,mi∗x, f
h0,m0i∗x0∈ Θ
0and i, j ∈ L satisfy i ≤
Lj and f
h0,mi∗x(j) = f
h0,m0i∗x0(j). If f
h0,mi∗x= f
h0,m0i∗x0or i = 0
L, then clearly f
h0,mi∗x(i) = f
h0,m0i∗x0(i). Suppose f
h0,mi∗x6= f
h0,m0i∗x0and i 6= 0
L. Clearly j 6= 0
L. By definition and f
h0,mi∗x(j) = f
h0,m0i∗x0(j), we have {m, m
0} = {1, 2}, x = x
0∈ L, and j ≤
Lx (moreover, f
h0,mi∗x(j) = f
h0,m0i∗x0(j) = h0, 1i ∗ x). Hence, i ≤
Lx and so f
h0,mi∗x(i) = h0, 1i ∗ x = f
h0,m0i∗x0(i) by definition.
(R.2) Let i, j ∈ L and i 6≤
Lj. We choose f
h0,1i∗jand f
h0,2i∗jin Θ
0. Since i 6≤
Lj, we have f
h0,1i∗j(i) = h0, 1i ∗ j 6= h0, 2i ∗ j = f
h0,2i∗j(i). If j = 0
L, then f
h0,1i∗j(j) = 0 = f
h0,2i∗j(j), and if j 6= 0
L, then f
h0,1i∗j(j) = h0, 1i ∗ j = f
h0,2i∗j(j).
(R.3) Suppose f
h0,mi∗x, f
h0,m0i∗x0∈ Θ
0and i, j, k ∈ L satisfy i ∨
Lj = k, f
h0,mi∗x(i) = f
h0,m0i∗x0(i), and f
h0,mi∗x(j) = f
h0,m0i∗x0(j). We may suppose f
h0,mi∗x6= f
h0,m0i∗x0and k 6= 0
L. By definition, we have {m, m
0} = {1, 2}, x = x
0∈ L, and i, j ≤
Lx. Hence, k ≤
Lx and so f
h0,mi∗x(k) = h0, 1i ∗ x = f
h0,m0i∗x0(k) by definition.
(2) Since h0, 1i ∗ j, h0, 2i ∗ j ∈ L
ωi,j1
[i, j], (2) is clear from the proof of (R.2) in (1).
Lemma 1.3. For all n ∈ ω, if Θ
nis an upper semilattice table of L, then Θ
n+1is an upper semilattice table of L.
P r o o f. By definition, Θ
n+1satisfies (R.0). Since Θ
n⊆ Θ
n+1, Θ
n+1satisfies (R.2). It is routine to check that Θ
n+1satisfies (R.1) and (R.3).
Below, we check (R.1) in a few cases, and leave the check of (R.1) in the other cases and of (R.3) to the reader.
Suppose f
hm0,m1i∗x, f
hm00,m01i∗x0∈ Θ
n+1and l, l
0∈ L satisfy l ≤
Ll
0and f
hm0,m1i∗x(l
0) = f
hm00,m01i∗x0(l
0). We may assume f
hm0,m1i∗x6= f
hm00,m01i∗x0and l 6= 0
L. Since Θ
nis an upper semilattice table of L, we may also assume that f
hm0,m1i∗x6∈ Θ
nor f
hm00,m01i∗x06∈ Θ
n. We notice that if f
hm0,m1i∗xor f
hm00,m01i∗x0is defined by “In the other case” in the construction of Θ
n+1, then f
hm0,m1i∗x(l
0) = f
hm00,m01i∗x0(l
0) does not occur.
Case 1: f
hm00,m01i∗x0∈ Θ
nand there are a, b ∈
ωω and i, j, k ∈ L such that m
0= n + 1, m
1= 1, x = ha, b, i, j, ki, a 6= b, max{a(0), b(0)} = n, i ∧
Lj = k, i 6≤
Lj, j 6≤
Li, and f
a(k) = f
b(k).
Since f
hm00,m01i∗x0∈ Θ
n, it follows that f
hm00,m01i∗x0(l
0)(0) ≤ n and so f
hn+1,1i∗x(l
0)(0) ≤ n. Then, by definition, l
0≤
Lj, l
0≤
Li, and f
hn+1,1i∗x(l
0)
= f
hn+1,0i∗x(l
0) = f
a(l
0). Hence f
a(l
0) = f
hm00,m01i∗x0(l
0). Since f
a∈ Θ
nand Θ
nsatisfies (R.1), f
a(l) = f
hm00,m01i∗x0(l). Clearly, l ≤
Li ∧
Lj, hence f
hn+1,1i∗x(l) = f
hn+1,0i∗x(l) = f
a(l) = f
hm00,m01i∗x0(l).
Case 2: There are a, b, a
0, b
0∈
ωω and i, j, k, i
0, j
0, k
0∈ L such that m
0= m
00= n + 1, m
1= 1, m
01= 2, x = ha, b, i, j, ki, x
0= ha
0, b
0, i
0, j
0, k
0i, a 6= b, a
06= b
0, max{a(0), b(0)} = max{a
0(0), b
0(0)} = n, i ∧
Lj = k, i
0∧
Lj
0= k
0, i 6≤
Lj, j 6≤
Li, i
06≤
Lj
0, j
06≤
Li
0, f
a(k) = f
b(k), and f
a0(k
0) = f
b0(k
0).
By definition, we have two subcases.
Subcase 2.1: l
0≤
Li ∧
Lj ∧
Lj
0and f
hn+1,1i∗x(l
0) = f
hn+1,0i∗x(l
0) = f
a(l
0) = f
b0(l
0) = f
hn+1,2i∗x0(l
0). Then, similarly to Case 1, we obtain f
hn+1,1i∗x(l) = f
a(l) = f
b0(l) = f
hn+1,2i∗x0(l).
Subcase 2.2: l
0≤
Li, l
06≤
Lj, x = x
0, and f
hn+1,1i∗x(l
0) = hn+1, 1i∗x = f
hn+1,2i∗x0(l
0). Then i = i
0, j = j
0, k = k
0, a = a
0, and b = b
0clearly. If l 6≤
Lj, then f
hn+1,1i∗x(l) = hn + 1, 1i ∗ x = f
hn+1,2i∗x0(l). Suppose l ≤
Lj.
Since l ≤
Li ∧
Lj, f
hn+1,1i∗x(l) = f
a(l) and f
hn+1,2i∗x0(l) = f
b(l). Since i ∧
Lj = k, f
a(k) = f
b(k), and Θ
nsatisfies (R.1), we have f
a(l) = f
b(l).
Hence, f
hn+1,1i∗x(l) = f
hn+1,2i∗x0(l).
By Lemmas 1.2 and 1.3, Θ is an upper semilattice table of L.
Lemma 1.4. F satisfies (R.4*). Hence, Θ is a lattice table of L.
P r o o f. Suppose a, b ∈
ωω and i, j, k ∈ L satisfy i ∧
Lj = k and f
a(k) = f
b(k). In the case of i ≤
Lj or j ≤
Li, we set c
0= c
1= c
2= b or c
0= c
1= c
2= a, and then c
0, c
1, c
2have the required properties. Suppose i 6≤
Lj, j 6≤
Li, and a 6= b. We set n = max{a(0), b(0)} and c
m= hn + 1, mi ∗ ha, b, i, j, ki for m ≤ 2. Then c
0, c
1, c
2∈ L
ωa,b,i,j,k1
[a, b, i, j, k]. By definition, f
a≡
if
c0≡
jf
c1and f
c2≡
jf
b. Since i 6≤
Lj, we have f
c1≡
if
c2.
This completes the proof of Proposition 1.1.
2. We start this section with
Lemma 2.1 (ZFC+CH). There is S ⊆ ℵ
1such that
ωω ⊆ L
ℵ1[S].
P r o o f. We take a bijection f : ℵ
1→
ωω and set
S = {ξ ∈ ℵ
1| ∃γ ≤ ξ∃m, n ∈ ω( ξ = ω · γ + 2
m· 3
n∧ f (γ)(m) = n)}.
Notice that for all ξ < ℵ
1, there are unique γ ≤ ξ and unique k ∈ ω such that ξ = ω · γ + k. Let x ∈
ωω be arbitrary. We choose γ ∈ ℵ
1such that f (γ) = x; then x(m) = n ⇔ ω · γ + 2
m· 3
n∈ S for all m, n ∈ ω. Hence, x ∈ L
ℵ1[S].
We fix S ⊆ ℵ
1such that
ωω ⊆ L
ℵ1[S]. We define the function rk :
ω
ω → ℵ
1by rk(x) = min{α ∈ ℵ
1| x ∈ L
α+1[S]} for x ∈
ωω. We set
K
0= {x ∈ WO | o.t.(x) ∈ S} and
K
1= {hm, ni ∗ x ∈
ωω | ∃w ∈ WO(rk(x) = o.t.(w) ∧ ∀w
0∈ WO(w
0<
L[S]w
⇒ o.t.(w
0) 6= rk(x)) ∧ w(m) = n)}.
Here, WO denotes the set of all x ∈
ωω which code a well-ordering relation on ω, and o.t.(w) denotes the order type of w.
If e.g. ∆
1n-determinacy (2 ≤ n ∈ ω) is assumed, then by the localiza- tion of the theorem of Solovay [7], for any ∆
1nset K ⊆
ωω, K[K, K
SJ] = {deg(K), deg(K
SJ)}. Under ZFC+CH (even if some determinacy axiom is assumed), if K
0≤
KK ⊆
ωω, then K[K, K
SJ] 6= {deg(K), deg(K
SJ)} ([5];
in fact we can prove that K[K, K
SJ] contains many elements). To prove the Theorem in §3, we use K
1in addition to K
0. We note that under ZFC+CH, {d ∈ K | deg(K
0⊕ K
1) ≤
Kd} is dense, which can be proved similarly to [2]
and [4].
Lemma 2.2 (ZFC+CH). Let K
0⊕ K
1≤
KK ⊆
ωω and T = S ∪ K.
(1) For all x ∈
ωω, L
ωK;x1
[K; x] is S-admissible, and so T -admissible.
(2) If M is K-admissible, then for all x ∈
ωω ∩ M , rk(x) ∈ M . (3) For all x ∈
ωω, x ∈ L
ωT ;x1
[T ], hence L
ωT ;x1
[T ; x] = L
ωT ;x 1[T ].
(4) If M is T -admissible and On ∩M = α, then
ωω ∩ M = {x ∈
ωω | rk(x) < α}.
P r o o f. (1) It is sufficient to prove that S is ∆
1over L
ωK;x1
[K; x]. For all ξ ∈ ω
1K;x, since there is an injection from ξ to ω in L
ωK;x1
[K; x], there is w ∈ WO ∩L
ωK;x1
[K; x] which codes a well-ordering of order type ξ. Hence, for all ξ ∈ ω
K;x1,
ξ ∈ S ⇔ L
ωK;x1
[K; x] |= “∃w ∈ K
0(o.t.(w) = ξ)”
⇔ L
ωK;x1
[K; x] |= “∀w ∈ WO(o.t.(w) = ξ ⇒ w ∈ K
0)”.
Therefore, S is Σ
1and Π
1over L
ωK;x 1[K; x].
(2) Let w be the ≤
L[S]-least element of WO such that o.t.(w) = rk(x).
By definition, for all m, n ∈ ω, w(m) = n ⇔ hm, ni ∗ x ∈ K
1. Since M is K
1-admissible, w ∈ M and hence rk(x) = o.t.(w) ∈ M .
(3) Since x ∈ L
ωT ;x1
[T ; x] and L
ωT ;x1
[T ; x] is K-admissible, rk(x) < ω
1T ;xby (2). Since L
ωT ;x1
[T ] is S-admissible, L
rk(x)+1[S] ⊆ L
ωT ;x1
[T ]. By definition, x ∈ L
rk(x)+1[S], hence x ∈ L
ωT ;x1
[T ].
(4) Suppose x ∈
ωω and rk(x) < α. Since M is S-admissible, L
rk(x)+1[S]
⊆ M , hence x ∈ M . Conversely, if x ∈
ωω∩M , then since M is K-admissible,
rk(x) < α by (2).
3. Let S, rk, K
0, and K
1be as in §2.
Theorem (ZFC+CH). Let K
0⊕ K
1≤
KK ⊆
ωω. For any lattice L, if L ⊆
ωω and (L, ≤
L, ∨
L, ∧
L) is Kleene recursive in K
SJ, then L can be embedded in K[K, K
SJ].
This section is entirely devoted to proving the Theorem. We use AC and CH without notice in the proof.
We fix K ⊆
ωω such that K
0⊕ K
1≤
KK, and a lattice L such that L ⊆
ω
ω and (L, ≤
L, ∨
L, ∧
L) is Kleene recursive in K
SJ. We set T = S ∪ K. Then every T -admissible set is S-admissible and K-admissible, and
ωω ⊆ L
ℵ1[T ].
We fix a lattice table Θ of L and F ⊆
ωω × L ×
ωω which are obtained by Proposition 1.1. For simplicity, we assume that (L, ≤
L, ∨
L, ∧
L) is Kleene recursive in K
SJwith no additional real parameter and F ≤
KK
SJwith no additional real parameter. For x ∈
ωω, we denote F
[x]by f
xas in the proof of Proposition 1.1. We may assume that f
0is identically 0 on L and 0 is the ≤
L[T ]-least real.
For every total or partial function p from
ωω to
ωω, we define the pro- jections of p by
P
l= {hx, f
p(x)(l)i | x ∈ Dom(p)} for l ∈ L.
We will construct a total function g :
ωω →
ωω such that l ∈ L 7→
deg(K ⊕ G
l) ∈ K[K, K
SJ] is a lattice embedding. Recall that G
ldenotes the projection of g on the coordinate l.
By recursion, we define a strictly increasing sequence hτ
α| α ∈ ℵ
1i of countable ordinals which satisfies:
(T.1) τ
α+1is the least T -admissible ordinal such that
ωω ∩ (L
τα+1[T ] − L
τα[T ]) is not empty.
(T.2) If α is a limit ordinal, then τ
α= S
β∈α
τ
β. The following is proved by routine work.
Lemma 3.1. (1) The graph of hτ
α| α ∈ ℵ
1i is uniformly Σ
1(T )-definable over all T -admissible sets.
(2) For any T -admissible set M , if α ∈ ℵ
1∩ M and hτ
β| β ∈ αi ⊆ M , then hτ
β| β ∈ αi ∈ M .
Lemma 3.2. For all α ∈ ℵ
1and x ∈
ωω ∩ (L
τα+1[T ] − L
τα[T ]), we have L
τα+1[T ] = L
ωK;x1
[K; x].
P r o o f. By Lemma 2.2, x ∈ L
ωT ;x1
[T ], hence it follows by the definition of τ
α+1that τ
α+1≤ ω
1T ;x. Since L
ωK;x1
[K; x] is T -admissible by Lemma 2.2, L
τα+1[T ] ⊆ L
ωT ;x1
[T ] ⊆ L
ωK;x1
[K; x]. Conversely, since L
τα+1[T ] is (K; x)- admissible, we have L
ωK;x1
[K; x] ⊆ L
τα+1[T ].
Remember that for any K-admissible set N , N is closed under λx.ω
1K;xiff N is K
SJ-admissible, and moreover N is closed under λx.ω
1K;xiff ∀x ∈
ω
ω ∩ N ∃α ∈ On ∩N (L
α[K; x] is (K; x)-admissible)
N. Hence the quantifiers in the statement “N is K
SJ-admissible” are bounded by N . Moreover, note that F is uniformly ∆
1over all K
SJ-admissible sets, since F ≤
KK
SJ.
Lemma 3.3. Let p be a partial function from
ωω to
ωω, M be a T - admissible set, p ∈ M and l ∈ L ∩ M . If for all x ∈ Dom(p), there is σ ∈ On ∩M such that L
σ[T ] is K
SJ-admissible and p(x), l ∈ L
σ[T ], then P
l∈ M .
P r o o f. By Σ
1-collection, there exists γ ∈ On ∩M such that for all x ∈ Dom(p) there is σ < γ such that L
σ[T ] is K
SJ-admissible and p(x), l ∈ L
σ[T ] (moreover f
p(x)(l) ∈ L
σ[T ] by (R.5)). Then for all x, y ∈
ωω we have
hx, yi ∈ P
l⇔ M |= “x ∈ Dom(p) ∧ y ∈ L
γ[T ]
∧ ∃σ < γ∃z ∈ L
σ[T ](L
σ[T ] is K
SJ-admissible
∧ l, y ∈ L
σ[T ] ∧ z = p(x) ∧ (hz, l, yi ∈ F )
Lσ[T ])”.
Hence, P
l∈ M by ∆
1-separation.
We construct g
α(α ∈ ℵ
1) of the parts of g as follows:
Stage 0. We set g
0= ∅.
Stage α limit. We set g
α= S
β∈α
g
β. Stage α + 1.
Case 1: There is t ∈
ωω ∩ L
τα[T ] which satisfies (G.1) or (G.2) below:
(G.1) There are e ∈ ω, v ∈
ωω, i, j ∈ L, and σ ≤ τ
αsuch that t = h0, ei ∗ hv, i, ji, i 6≤
Lj, L
σ[T ] is K
SJ-admissible, t ∈ L
σ[T ], and ∀x ∈
ωω ∩ L
τα[T ](χ
Gαi(x) ∼ = {e}(x, v, χ
K⊕Gαj,
2E)).
(G.2) There are e
0, e
1∈ ω, v
0, v
1∈
ωω, i, j, k ∈ L, and σ ≤ τ
αsuch that t = h1, e
0, e
1i ∗ hv
0, v
1, i, j, ki, i ∧
Lj = k, L
σ[T ] is K
SJ- admissible, t ∈ L
σ[T ], ∀x ∈
ωω ∩ L
τα[T ]({e
0}(x, v
0, χ
K⊕Gαi,
2E)
∼ = {e
1}(x, v
1, χ
K⊕Gαj,
2E)), and there is a partial function p ∈ L
τα+1[T ] from
ωω to
ωω such that g
α⊆ p, Rng(p − g
α)
⊆ L
σ[T ], and ∃x ∈
ωω ∩ L
τα+1[T ]({e
0}(x, v
0, χ
K⊕Pi∗0,
2E) 6∼
= {e
1}(x, v
1, χ
K⊕Pj∗0,
2E)). Here, P
l∗ 0 = P
l∪ {hy, 0i | y ∈
ω
ω − Dom(p)} for l ∈ L.
We choose the ≤
L[T ]-least t ∈
ωω ∩ L
τα[T ] which satisfies (G.1) or (G.2) and distinguish two subcases.
Subcase 1.1: t satisfies (G.1). We choose the ≤
L[T ]-least z ∈
ωω ∩
(L
τα+1[T ] − L
τα[T ]) and the ≤
L[T ]-least ha, bi ∈
ωω ×
ωω such that f
a(j) =
f
b(j) and f
a(i) 6= f
b(i) by (R.2). Notice that if σ is as in (G.1), then
a, b, f
a(i) ∈ L
σ[T ] by (R.2*) and (R.5). We set z
0= hz, f
a(i)i and define partial functions p
a, p
bby
p
a(x) (p
b(x) resp.) =
g
α(x) if x ∈ Dom(g
α), a (b resp.) if x = z.
Then P
ja= P
jb, z
0∈ P
ia, and z
06∈ P
ib. If {e}(z
0, v, χ
K⊕Pja∗0,
2E) ∼ = 0, then we define
g
α+1(x) =
p
a(x) if x ∈ Dom(p
a),
0 if x ∈
ωω ∩ L
τα+1[T ] − Dom(p
a), and if {e}(z
0, v, χ
K⊕Pja∗0,
2E) 6∼ = 0, then we define
g
α+1(x) =
p
b(x) if x ∈ Dom(p
b),
0 if x ∈
ωω ∩ L
τα+1[T ] − Dom(p
b).
Subcase 1.2: t satisfies (G.2). We choose the ≤
L[T ]-least partial function p ∈ L
τα+1[T ] as in (G.2) and define
g
α+1(x) =
p(x) if x ∈ Dom(p),
0 if x ∈
ωω ∩ L
τα+1[T ] − Dom(p).
Case 2: Otherwise. We define g
α+1(x) =
g
α(x) if x ∈ Dom(g
α),
0 if x ∈
ωω ∩ L
τα+1[T ] − Dom(g
α).
In the construction at Stage α + 1 above, notice that for l ∈ L, G
α+1l= P
la∗ 0 ∩ L
τα+1[T ] or = P
lb∗ 0 ∩ L
τα+1[T ] (Subcase 1.1), or = P
l∗ 0 ∩ L
τα+1[T ] (Subcase 1.2), or = G
αl∗ 0 ∩ L
τα+1[T ] (Case 2) respectively.
We define g = S
α∈ℵ1
g
α. Then, for all α ∈ ℵ
1, gd
ωω ∩ L
τα[T ] = g
αand g
α:
ωω∩L
τα[T ] →
ωω∩L
τα[T ]. Moreover g
α+1:
ωω∩L
τα+1[T ] →
ωω∩L
τα[T ] by definition. If there is no σ ≤ τ
αsuch that L
σ[T ] is K
SJ-admissible, then Rng(g
α+1) = {0}. As for projections, for all α ∈ ℵ
1and l ∈ L ∩ L
τα[T ], we have G
l∩ L
τα[T ] = G
αl.
Lemma 3.4. Let % ∈ ℵ
1and L
%[T ] be K
SJ-admissible.
(1) For all α < ℵ
1, if % ≤ τ
α, then there is σ ≤ τ
αsuch that L
σ[T ] is K
SJ-admissible and Rng(g
α+1− g
α) ⊆ L
σ[T ].
(2) For all x ∈
ωω, there is σ ≤ max{rk(x), %} such that L
σ[T ] is K
SJ- admissible and g(x) ∈ L
σ[T ].
P r o o f. (1) We distinguish three cases at Stage α + 1.
Case 1: g
α+1is constructed in Subcase 1.1 at Stage α + 1. We choose σ
as in (G.1). By definition, there is c ∈
ωω ∩ L
σ[T ] (c = a or = b in Subcase
1.1) such that Rng(g
α+1− g
α) = {c, 0}. Since 0 ∈ L
σ[T ], Rng(g
α+1− g
α) ⊆ L
σ[T ].
Case 2: g
α+1is constructed in Subcase 1.2 at Stage α+1. We choose the
≤
L[T ]-least partial function p and σ as in (G.2). By (G.2), Rng(p − g
α) ⊆ L
σ[T ], hence Rng(g
α+1− g
α) ⊆ L
σ[T ].
Case 3: g
α+1is constructed in Case 2 at Stage α + 1. By definition, Rng(g
α+1− g
α) = {0} ⊆ L
%[T ].
(2) We choose α < ℵ
1such that x ∈ L
τα+1[T ] − L
τα[T ]. By Lemma 2.2, τ
α≤ rk(x). If % ≤ τ
α, then by (1) there is σ ≤ rk(x) such that L
σ[T ] is K
SJ- admissible and g(x) = g
α+1(x) ∈ L
σ[T ]. If τ
α< %, then since Rng(g
α+1) ⊆ L
τα[T ], we have g(x) ∈ L
%[T ].
Since L
ℵ1[T ] is K
SJ-admissible and
ωω ⊆ L
ℵ1[T ], for all x ∈
ωω there exists % < ℵ
1such that L
%[T ] is K
SJ-admissible and x ∈ L
%[T ] (using the L¨owenheim–Skolem Theorem). For x ∈
ωω, we set %(x) = min{σ < ℵ
1| L
σ[T ] is K
SJ-admissible and x ∈ L
σ[T ]}.
Lemma 3.5. Let α ∈ ℵ
1and l ∈ L.
(1) For any T -admissible set M , if τ
α∈ M , then g
α∈ M . (2) For any T -admissible set M , if τ
α, %(l) ∈ M , then G
αl∈ M . (3) If %(l) < τ
α+1, then L
τα+1[T ] is G
l-admissible.
P r o o f. (1) We prove
∀α ∈ ℵ
1∀M : T -admissible set (τ
α∈ M ⇒ hg
β| β ≤ αi ∈ M ) by induction.
If α = 0, then this is clear.
Let 0 < α ∈ ℵ
1. We assume that for all β ∈ α and every T -admissible set M we have (τ
β∈ M ⇒ hg
γ| γ ≤ βi ∈ M ). Let M be a T -admissible set and τ
α∈ M .
Let α = β + 1 for some β. By assumption, g
β∈ L
τα[T ]. In the construc- tion at Stage β +1, p
a, p
bin Subcase 1.1 and p in Subcase 1.2 are elements of L
τα[T ]. Since L
τα[T ] ∈ M , by definition g
β+1∈ M . Hence hg
β| β ≤ αi ∈ M . Let α be a limit ordinal. For every limit ordinal β ∈ α, since hg
γ| γ ≤ βi
∈ L
τβ+1[T ], the construction at Stage β can be expressed over L
τβ+1[T ]. And
for every β + 1 ∈ α, since the conditions of every case at Stage β + 1 can be
expressed over L
τβ+1[T ] (notice that if t = h. . .i∗h. . . , i, j, . . .i and %(t) ≤ τ
β,
then G
βi, G
βj∈ L
τβ+1[T ] by Lemmas 3.4 and 3.3, hence we can express (G.1)
(G.2); otherwise, we proceed to Case 2 immediately), the construction at
Stage β + 1 can be expressed over L
τβ+2[T ]. Thus, hg
β| β ∈ αi is ∆
1-
definable over M with parameter hτ
β| β ≤ αi, hence hg
β| β ∈ αi ∈ M . (By
Lemma 3.1, hτ
β| β ≤ αi ∈ M .) Therefore, by definition, g
α∈ M , and so hg
β| β ≤ αi ∈ M .
(2) By (1), g
α∈ M . For all x ∈ Dom(g
α), since rk(x) ∈ M , there is σ ∈ On ∩M such that L
σ[T ] is K
SJ-admissible and g
α(x), l ∈ L
σ[T ] by Lemma 3.4. Hence, G
αl∈ M by Lemma 3.3.
(3) By (2), G
αl∈ L
τα+1[T ]. In the construction at Stage α + 1, p
a, p
bin Subcase 1.1 and p in Subcase 1.2 are elements of L
τα+1[T ], hence similarly to (2), P
la, P
lb, P
l∈ L
τα+1[T ] by Lemma 3.3. Since G
α+1l= P
la∗ 0 ∩ L
τα+1[T ] or = P
lb∗ 0 ∩ L
τα+1[T ] or = P
l∗ 0 ∩ L
τα+1[T ] or = G
αl∗ 0 ∩ L
τα+1[T ], we see that L
τα+1[T ] is G
α+1l-admissible and so G
l-admissible.
Lemma 3.6. For all l ∈ L, G
l≤
KK
SJ, hence deg(K ⊕G
l) ∈ K[K, K
SJ].
P r o o f. For α ∈ ℵ
1, similarly to Lemma 3.5, the construction of g
α(i.e.
constructions till Stage α) and the conditions of every case at Stage α + 1 can be expressed over L
τα+1[T ]. Hence, there are formulas ψ
1and ψ
2such that:
L
τα+1[T ] |= ψ
1(p, α)
⇔ There is t ∈
ωω ∩ L
τα[T ] which satisfies (G.1) or (G.2) at Stage α + 1 and let t be the ≤
L[T ]-least such real,
if t = h0, ei ∗ hv, i, ji satisfies (G.1) and z, a, b, p
a, p
bare as in Subcase 1.1
then {e}(hz, f
a(i)i, v, χ
K⊕Pja,
2E) ∼ = 0 ∧ p = p
aor {e}(hz, f
a(i)i, v, χ
K⊕Pja,
2E) 6∼ = 0 ∧ p = p
b, and if t = h1, e
0, e
1i ∗ hv
0, v
1, i, j, ki satisfies (G.2),
then p is the ≤
L[T ]-least partial function as in (G.2).
L
τα+1[T ] |= ψ
2(p, α)
⇔ There is no t ∈
ωω ∩ L
τα[T ] which satisfies (G.1) or (G.2) at Stage α + 1 and p = g
α.
Here, ψ
1and ψ
2correspond to Case 1 and Case 2 respectively.
We choose r ∈ WO such that o.t.(r) = %(l). We prove G
l≤
KK
SJvia r using (2) of §0. Let x, y ∈
ωω be arbitrary and M = L
ω1KSJ;x,y,r
[K
SJ; x, y, r].
Notice that if x ∈ L
τα+1[T ]−L
τα[T ], then by Lemma 3.2 and K
SJ-admissibi- lity of M , we have L
τα+1[T ] = L
ωK;x1
[K; x] ∈ M . By Lemma 3.4, there is σ ≤ max{rk(x), %(l)} such that L
σ[T ] is K
SJ-admissible and g(x), l ∈ L
σ[T ];
moreover, f
g(x)(l) ∈ L
σ[T ]. Hence,
hx, yi ∈ G
l⇔ M |= “∃α ∈ ω
K;x1∃p ∈ L
ωK;x 1[K; x]
(L
ωK;x1
[K; x] = L
τα+1[T ] ∧ x 6∈ L
τα[T ]
∧ L
τα+1[T ] |= ψ
1(p, α) ∨ ψ
2(p, α)
∧ (∃σ ≤ max{rk(x), %(l)}(x ∈ Dom(p) ∧ p(x), l ∈ L
σ[T ]
∧ L
σ[T ] is K
SJ-admissible ∧ (y = f
p(x)(l))
Lσ[T ])
∨ (x 6∈ Dom(p) ∧ y = 0)))”.
Notice that the quantifiers in the statement “ω
K;x1= τ
α+1” are bounded by L
ωK;x1
[K; x], since ω
1K;x= τ
α+1iff ¬∃τ ∈ ω
K;x1(τ
α< τ ∧ τ satisfies (T.1))
LωK;x1 [K;x]. Hence “hx, yi ∈ G
l” is ∆
1over M . Therefore, G
l≤
KK
SJ.
Lemma 3.7. (1) G
0L≡
K∅.
(2) For all i, j ∈ L, if i ≤
Lj, then K ⊕ G
i≤
KK ⊕ G
j.
(3) For all i, j, k ∈ L, if i∨
Lj = k, then (K ⊕G
i)⊕(K ⊕G
j) ≡
KK ⊕G
k. P r o o f. (1) By definition, G
0L= {hx, f
g(x)(0
L)i | x ∈
ωω} = {hx, 0i | x ∈
ωω} ≡
K∅.
(2) We choose r ∈ WO such that o.t.(r) = %(i, j). To prove K ⊕ G
i≤
KK ⊕ G
j, it is sufficient to prove that for all x, y ∈
ωω,
hx, yi ∈ G
i⇔ M |= “∃σ ≤ max{rk(x), %(i, j)}∃a, z ∈ L
σ[T ] (L
σ[T ] is K
SJ-admissible ∧ i, j ∈ L
σ[T ]
∧ hx, zi ∈ G
j∧ (f
a(j) = z ∧ f
a(i) = y)
Lσ[T ])”, where M = L
ωK⊕Gj ;i,j,x,y,r 1
[K ⊕ G
j; i, j, x, y, r].
Suppose hx, yi ∈ G
i. By Lemma 2.2, rk(x) ∈ M . By Lemma 3.4, there is σ ≤ max{rk(x), %(i, j)} such that L
σ[T ] is K
SJ-admissible and g(x), i, j ∈ L
σ[T ]. By (R.5), we have f
g(x)(i), f
g(x)(j) ∈ L
σ[T ]. Thus, if we set a = g(x) and z = f
a(j), then since y = f
a(i) and F ≤
KK
SJ, the right-hand side holds. Conversely, suppose that x, y ∈
ωω satisfy the right-hand side. Let a, z be as in the right-hand side. By hx, zi ∈ G
j, f
g(x)(j) = z = f
a(j). Then, by (R.1), f
g(x)(i) = f
a(i). Hence, y = f
g(x)(i), and so hx, yi ∈ G
i.
(3) By (2), K ⊕ G
i⊕ G
j≤
KK ⊕ G
k. We choose r ∈ WO such that o.t.(r) = %(i, j, k). To prove K ⊕ G
k≤
KK ⊕ G
i⊕ G
j, it is sufficient to prove that for all x, y ∈
ωω,
hx, yi ∈ G
k⇔ M |= “∃σ ≤ max{rk(x), %(i, j, k)}∃a, z, z
0∈ L
σ[T ] (L
σ[T ] is K
SJ-admissible ∧ i, j, k ∈ L
σ[T ]
∧ hx, zi ∈ G
i∧ hx, z
0i ∈ G
j∧ (f
a(i) = z ∧ f
a(j) = z
0∧ f
a(k) = y )
Lσ[T ])”, where M = L
ωK⊕Gi⊕Gj ;i,j,k,x,y,r 1