Embedding lattices in the Kleene degrees

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

can be embedded in some local structures of Kleene degrees.

0. We denote by

E the existential integer quantifier and by χ

the characteristic function of A, i.e. x ∈ A ⇔ χ

(x) = 1, and x 6∈ A ⇔ χ

(x) = 0. Kleene reducibility is defined as follows: for A, B ⊆

ω, A ≤

B iff there is a ∈

ω such that χ

is recursive in a, χ

, and

E.

We introduce the following notations. K denotes the upper semilattice of all Kleene degrees with the order induced by ≤

. For X, Y ⊆

ω, 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

= {hei ∗ x ∈

ω | {e}((x)

, (x)

, χ

,

E)↓}. 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

, . . . , e

i ∗ x is the real such that (he

, . . . , e

i ∗ x)(n) = e

for n ≤ m and (he

, . . . , e

i ∗ x)(n + m + 1) = x(n) for n ∈ ω. Further, (x)

= λn.x(2n) and (x)

= λn.x(2n + 1). We identify h(x)

, (x)

i with x.

An X-admissible set is closed under λx.ω

iff it is X

-admissible.

The following conditions (1) and (2) are equivalent to A ≤

B ([8]).

(1) There is y ∈

ω such that A is uniformly ∆

-definable over all (B; y)- admissible sets; i.e. there are Σ

( ˙ B) formulas ϕ

and ϕ

such that for any (B; y)-admissible set M and for all x ∈

ω ∩ M ,

x ∈ A ⇔ M |= ϕ

(x, y) ⇔ M |= ¬ϕ

(x, y).

(2) There are y ∈

ω and Σ

( ˙ B) formulas ϕ

and ϕ

such that for all x ∈

ω,

x ∈ A ⇔ L

1

[B; x, y] |= ϕ

(x, y) ⇔ L

1

[B; x, y] |= ¬ϕ

(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

[B; y] denotes the αth stage of the hierarchy constructible from {y} relative to a unary predicate B, and ω

denotes the least (B; y)-admissible ordinal.

For K, K

⊆

ω, we set K[K, K

] = {deg(X) | K ≤

X ≤

K

}. In §3, we will prove that under ZFC+CH, for some K ⊆

ω, lattices whose fields

⊆

ω and which are Kleene recursive in K

can be embedded in K[K, K

].

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

, we denote the field of L also by L and regard L ⊆

ω. We denote by 0 the identically 0 function from ω to ω.

Definition. Let L be a lattice with relations ≤

, ∨

, and ∧

. For a, b ∈

L

(

ω) and l ∈ L, we define a ≡

b by a(l) = b(l). Θ ⊆

(

ω) is called an upper semilattice table of L iff Θ satisfies:

(R.0) If there is the least element 0

of L, then for all a ∈ Θ, a(0

) = 0.

(R.1) (Ordering) For all a, b ∈ Θ and i, j ∈ L, if i ≤

j and a ≡

b, then a ≡

b.

(R.2) (Non-ordering) For all i, j ∈ L, if i 6≤

j, then there are a, b ∈ Θ such that a ≡

b and a 6≡

b.

(R.3) (Join) For all a, b ∈ Θ and i, j, k ∈ L, if i ∨

j = k, a ≡

b, and a ≡

b, then a ≡

b.

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 ∧

j = k and a ≡

b, then there are c

, c

, c

∈ Θ such that a ≡

c

≡

c

≡

c

≡

b.

For every lattice L with relations ≤

, ∨

, ∧

, and L ⊆

ω, we say that (L, ≤

, ∨

, ∧

) is Kleene recursive in X ⊆

ω iff L ⊕ {hi, ji | i ≤

j}

⊕ {hi, j, ki | i ∨

j = k} ⊕ {hi, j, ki | i ∧

j = k} ≤

X.

In this paper, we need suitable restrictions in (R.2) and (R.4).

Proposition 1.1. Let L be a lattice with relations ≤

, ∨

, ∧

, and L ⊆

ω

ω. Let X ⊆

ω. If (L, ≤

, ∨

, ∧

) is Kleene recursive in X, then there are a lattice table Θ of L and F ⊆

ω × L ×

ω such that Θ = {F

| x ∈

ω}, F ≤

X, and F satisfies:

(R.2*) For all i, j ∈ L, if i 6≤

j, then there are a, b ∈

ω ∩ L

1

[i, j] such

that F

≡

F

and F

6≡

F

.

(R.4*) For all a, b ∈

ω and i, j, k ∈ L, if i ∧

j = k and F

≡

F

, then there are c

, c

, c

∈

ω ∩ L

1

[a, b, i, j, k] such that F

≡

F

≡

F

≡

F

≡

F

.

(R.5) For all a ∈

ω, Rng(F

) ⊆ L

[a].

Here, for x ∈

ω, we set F

= {hl, yi | hx, l, yi ∈ F } and regard F

: L →

ω.

P r o o f. We fix X and L as in the proposition. We assume that there is the least element 0

of L. We will construct Θ and F with the required properties.

For x ∈

ω and m ∈ ω, we define the function f

: L →

ω as follows: If x 6∈ L or m 6= 2, then

f

(l) =

 0 if l = 0

, h0, mi ∗ x otherwise.

If x ∈ L and m = 2, then

f

(l) =

 



0 if l = 0

, h0, 1i ∗ x if 0

6= l ≤

x, h0, 2i ∗ x otherwise.

For x ∈

ω and n, m ∈ ω, we define the function f

: L →

ω inductively as follows: If x = ha, b, i, j, ki, a 6= b, max{a(0), b(0)} = n, i, j, k ∈ L, i ∧

j = k, i 6≤

j, j 6≤

i, f

(k) = f

(k), and m ≤ 2, then

f

(l) =

 f

(l) if l ≤

i, hn + 1, 0i ∗ x otherwise,

f

(l) =

 



f

(l) if l ≤

j,

hn + 1, 1i ∗ x if l ≤

i and l 6≤

j, hn + 1, 2i ∗ x otherwise,

f

(l) =

 



f

(l) if l ≤

j,

hn + 1, 1i ∗ x if l ≤

i and l 6≤

j, hn + 1, 3i ∗ x otherwise.

In the other case,

f

(l) =

 0 if l = 0

,

hn + 1, m + 1i ∗ x otherwise.

We set Θ = {f

| x ∈

ω} and F = {hx, l, yi ∈

ω × L ×

ω | f

(l) = y}.

Then F

= f

for x ∈

ω. (To define f

for all x ∈

ω, we make Θ contain some excess elements.)

We prove that Θ and F have the required properties. By definition, Θ = {F

| x ∈

ω}, F ≤

X, and F satisfies (R.5).

For n ∈ ω, we set Θ

= {f

| x ∈

ω ∧ x(0) ≤ n}.

Lemma 1.2. (1) Θ

is an upper semilattice table of L.

(2) F satisfies (R.2*).

P r o o f. (1) We check that Θ

satisfies (R.0)–(R.3).

(R.0) By definition, for all f

∈ Θ

, f

(0

) = 0.

(R.1) Suppose f

, f

∈ Θ

and i, j ∈ L satisfy i ≤

j and f

(j) = f

(j). If f

= f

or i = 0

, then clearly f

(i) = f

(i). Suppose f

6= f

and i 6= 0

. Clearly j 6= 0

. By definition and f

(j) = f

(j), we have {m, m

} = {1, 2}, x = x

∈ L, and j ≤

x (moreover, f

(j) = f

(j) = h0, 1i ∗ x). Hence, i ≤

x and so f

(i) = h0, 1i ∗ x = f

(i) by definition.

(R.2) Let i, j ∈ L and i 6≤

j. We choose f

and f

in Θ

. Since i 6≤

j, we have f

(i) = h0, 1i ∗ j 6= h0, 2i ∗ j = f

(i). If j = 0

, then f

(j) = 0 = f

(j), and if j 6= 0

, then f

(j) = h0, 1i ∗ j = f

(j).

(R.3) Suppose f

, f

∈ Θ

and i, j, k ∈ L satisfy i ∨

j = k, f

(i) = f

(i), and f

(j) = f

(j). We may suppose f

6= f

and k 6= 0

. By definition, we have {m, m

} = {1, 2}, x = x

∈ L, and i, j ≤

x. Hence, k ≤

x and so f

(k) = h0, 1i ∗ x = f

(k) by definition.

(2) Since h0, 1i ∗ j, h0, 2i ∗ j ∈ L

1

[i, j], (2) is clear from the proof of (R.2) in (1).

Lemma 1.3. For all n ∈ ω, if Θ

is an upper semilattice table of L, then Θ

is an upper semilattice table of L.

P r o o f. By definition, Θ

satisfies (R.0). Since Θ

⊆ Θ

, Θ

satisfies (R.2). It is routine to check that Θ

satisfies (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

, f

∈ Θ

and l, l

∈ L satisfy l ≤

l

and f

(l

) = f

(l

). We may assume f

6= f

and l 6= 0

. Since Θ

is an upper semilattice table of L, we may also assume that f

6∈ Θ

or f

6∈ Θ

. We notice that if f

or f

is defined by “In the other case” in the construction of Θ

, then f

(l

) = f

(l

) does not occur.

Case 1: f

∈ Θ

and there are a, b ∈

ω and i, j, k ∈ L such that m

= n + 1, m

= 1, x = ha, b, i, j, ki, a 6= b, max{a(0), b(0)} = n, i ∧

j = k, i 6≤

j, j 6≤

i, and f

(k) = f

(k).

Since f

∈ Θ

, it follows that f

(l

)(0) ≤ n and so f

(l

)(0) ≤ n. Then, by definition, l

≤

j, l

≤

i, and f

(l

)

= f

(l

) = f

(l

). Hence f

(l

) = f

(l

). Since f

∈ Θ

and Θ

satisfies (R.1), f

(l) = f

(l). Clearly, l ≤

i ∧

j, hence f

(l) = f

(l) = f

(l) = f

(l).

Case 2: There are a, b, a

, b

∈

ω and i, j, k, i

, j

, k

∈ L such that m

= m

= n + 1, m

= 1, m

= 2, x = ha, b, i, j, ki, x

= ha

, b

, i

, j

, k

i, a 6= b, a

6= b

, max{a(0), b(0)} = max{a

(0), b

(0)} = n, i ∧

j = k, i

∧

j

= k

, i 6≤

j, j 6≤

i, i

6≤

j

, j

6≤

i

, f

(k) = f

(k), and f

(k

) = f

(k

).

By definition, we have two subcases.

Subcase 2.1: l

≤

i ∧

j ∧

j

and f

(l

) = f

(l

) = f

(l

) = f

(l

) = f

(l

). Then, similarly to Case 1, we obtain f

(l) = f

(l) = f

(l) = f

(l).

Subcase 2.2: l

≤

i, l

6≤

j, x = x

, and f

(l

) = hn+1, 1i∗x = f

(l

). Then i = i

, j = j

, k = k

, a = a

, and b = b

clearly. If l 6≤

j, then f

(l) = hn + 1, 1i ∗ x = f

(l). Suppose l ≤

j.

Since l ≤

i ∧

j, f

(l) = f

(l) and f

(l) = f

(l). Since i ∧

j = k, f

(k) = f

(k), and Θ

satisfies (R.1), we have f

(l) = f

(l).

Hence, f

(l) = f

(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 ∈

ω and i, j, k ∈ L satisfy i ∧

j = k and f

(k) = f

(k). In the case of i ≤

j or j ≤

i, we set c

= c

= c

= b or c

= c

= c

= a, and then c

, c

, c

have the required properties. Suppose i 6≤

j, j 6≤

i, and a 6= b. We set n = max{a(0), b(0)} and c

= hn + 1, mi ∗ ha, b, i, j, ki for m ≤ 2. Then c

, c

, c

∈ L

1

[a, b, i, j, k]. By definition, f

≡

f

≡

f

and f

≡

f

. Since i 6≤

j, we have f

≡

f

.

This completes the proof of Proposition 1.1.

2. We start this section with

Lemma 2.1 (ZFC+CH). There is S ⊆ ℵ

such that

ω ⊆ L

[S].

P r o o f. We take a bijection f : ℵ

→

ω and set

S = {ξ ∈ ℵ

| ∃γ ≤ ξ∃m, n ∈ ω( ξ = ω · γ + 2

· 3

∧ f (γ)(m) = n)}.

Notice that for all ξ < ℵ

, there are unique γ ≤ ξ and unique k ∈ ω such that ξ = ω · γ + k. Let x ∈

ω be arbitrary. We choose γ ∈ ℵ

such that f (γ) = x; then x(m) = n ⇔ ω · γ + 2

· 3

∈ S for all m, n ∈ ω. Hence, x ∈ L

[S].

We fix S ⊆ ℵ

such that

ω ⊆ L

[S]. We define the function rk :

ω

ω → ℵ

by rk(x) = min{α ∈ ℵ

| x ∈ L

[S]} for x ∈

ω. We set

K

= {x ∈ WO | o.t.(x) ∈ S} and

K

= {hm, ni ∗ x ∈

ω | ∃w ∈ WO(rk(x) = o.t.(w) ∧ ∀w

∈ WO(w

<

w

⇒ o.t.(w

) 6= rk(x)) ∧ w(m) = n)}.

Here, WO denotes the set of all x ∈

ω which code a well-ordering relation on ω, and o.t.(w) denotes the order type of w.

If e.g. ∆

-determinacy (2 ≤ n ∈ ω) is assumed, then by the localiza- tion of the theorem of Solovay [7], for any ∆

set K ⊆

ω, K[K, K

] = {deg(K), deg(K

)}. Under ZFC+CH (even if some determinacy axiom is assumed), if K

≤

K ⊆

ω, then K[K, K

] 6= {deg(K), deg(K

)} ([5];

in fact we can prove that K[K, K

] contains many elements). To prove the Theorem in §3, we use K

in addition to K

. We note that under ZFC+CH, {d ∈ K | deg(K

⊕ K

) ≤

d} is dense, which can be proved similarly to [2]

and [4].

Lemma 2.2 (ZFC+CH). Let K

⊕ K

≤

K ⊆

ω and T = S ∪ K.

(1) For all x ∈

ω, L

1

[K; x] is S-admissible, and so T -admissible.

(2) If M is K-admissible, then for all x ∈

ω ∩ M , rk(x) ∈ M . (3) For all x ∈

ω, x ∈ L

1

[T ], hence L

1

[T ; x] = L

[T ].

(4) If M is T -admissible and On ∩M = α, then

ω ∩ M = {x ∈

ω | rk(x) < α}.

P r o o f. (1) It is sufficient to prove that S is ∆

over L

1

[K; x]. For all ξ ∈ ω

, since there is an injection from ξ to ω in L

1

[K; x], there is w ∈ WO ∩L

1

[K; x] which codes a well-ordering of order type ξ. Hence, for all ξ ∈ ω

,

ξ ∈ S ⇔ L

1

[K; x] |= “∃w ∈ K

(o.t.(w) = ξ)”

⇔ L

1

[K; x] |= “∀w ∈ WO(o.t.(w) = ξ ⇒ w ∈ K

)”.

Therefore, S is Σ

and Π

over L

[K; x].

(2) Let w be the ≤

-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

. Since M is K

-admissible, w ∈ M and hence rk(x) = o.t.(w) ∈ M .

(3) Since x ∈ L

1

[T ; x] and L

1

[T ; x] is K-admissible, rk(x) < ω

by (2). Since L

1

[T ] is S-admissible, L

[S] ⊆ L

1

[T ]. By definition, x ∈ L

[S], hence x ∈ L

1

[T ].

(4) Suppose x ∈

ω and rk(x) < α. Since M is S-admissible, L

[S]

⊆ M , hence x ∈ M . Conversely, if x ∈

ω∩M , then since M is K-admissible,

rk(x) < α by (2).

3. Let S, rk, K

, and K

be as in §2.

Theorem (ZFC+CH). Let K

⊕ K

≤

K ⊆

ω. For any lattice L, if L ⊆

ω and (L, ≤

, ∨

, ∧

) is Kleene recursive in K

, then L can be embedded in K[K, K

].

This section is entirely devoted to proving the Theorem. We use AC and CH without notice in the proof.

We fix K ⊆

ω such that K

⊕ K

≤

K, and a lattice L such that L ⊆

ω

ω and (L, ≤

, ∨

, ∧

) is Kleene recursive in K

. We set T = S ∪ K. Then every T -admissible set is S-admissible and K-admissible, and

ω ⊆ L

[T ].

We fix a lattice table Θ of L and F ⊆

ω × L ×

ω which are obtained by Proposition 1.1. For simplicity, we assume that (L, ≤

, ∨

, ∧

) is Kleene recursive in K

with no additional real parameter and F ≤

K

with no additional real parameter. For x ∈

ω, we denote F

by f

as in the proof of Proposition 1.1. We may assume that f

is identically 0 on L and 0 is the ≤

-least real.

For every total or partial function p from

ω to

ω, we define the pro- jections of p by

P

= {hx, f

(l)i | x ∈ Dom(p)} for l ∈ L.

We will construct a total function g :

ω →

ω such that l ∈ L 7→

deg(K ⊕ G

) ∈ K[K, K

] is a lattice embedding. Recall that G

denotes the projection of g on the coordinate l.

By recursion, we define a strictly increasing sequence hτ

| α ∈ ℵ

i of countable ordinals which satisfies:

(T.1) τ

is the least T -admissible ordinal such that

ω ∩ (L

[T ] − L

[T ]) is not empty.

(T.2) If α is a limit ordinal, then τ

= S

β∈α

τ

. The following is proved by routine work.

Lemma 3.1. (1) The graph of hτ

| α ∈ ℵ

i is uniformly Σ

(T )-definable over all T -admissible sets.

(2) For any T -admissible set M , if α ∈ ℵ

∩ M and hτ

| β ∈ αi ⊆ M , then hτ

| β ∈ αi ∈ M .

Lemma 3.2. For all α ∈ ℵ

and x ∈

ω ∩ (L

[T ] − L

[T ]), we have L

[T ] = L

1

[K; x].

P r o o f. By Lemma 2.2, x ∈ L

1

[T ], hence it follows by the definition of τ

that τ

≤ ω

. Since L

1

[K; x] is T -admissible by Lemma 2.2, L

[T ] ⊆ L

1

[T ] ⊆ L

1

[K; x]. Conversely, since L

[T ] is (K; x)- admissible, we have L

1

[K; x] ⊆ L

[T ].

Remember that for any K-admissible set N , N is closed under λx.ω

iff N is K

-admissible, and moreover N is closed under λx.ω

iff ∀x ∈

ω

ω ∩ N ∃α ∈ On ∩N (L

[K; x] is (K; x)-admissible)

. Hence the quantifiers in the statement “N is K

-admissible” are bounded by N . Moreover, note that F is uniformly ∆

over all K

-admissible sets, since F ≤

K

.

Lemma 3.3. Let p be a partial function from

ω to

ω, 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

[T ] is K

-admissible and p(x), l ∈ L

[T ], then P

∈ M .

P r o o f. By Σ

-collection, there exists γ ∈ On ∩M such that for all x ∈ Dom(p) there is σ < γ such that L

[T ] is K

-admissible and p(x), l ∈ L

[T ] (moreover f

(l) ∈ L

[T ] by (R.5)). Then for all x, y ∈

ω we have

hx, yi ∈ P

⇔ M |= “x ∈ Dom(p) ∧ y ∈ L

[T ]

∧ ∃σ < γ∃z ∈ L

[T ](L

[T ] is K

-admissible

∧ l, y ∈ L

[T ] ∧ z = p(x) ∧ (hz, l, yi ∈ F )

)”.

Hence, P

∈ M by ∆

-separation.

We construct g

(α ∈ ℵ

) of the parts of g as follows:

Stage 0. We set g

= ∅.

Stage α limit. We set g

= S

β∈α

g

. Stage α + 1.

Case 1: There is t ∈

ω ∩ L

[T ] which satisfies (G.1) or (G.2) below:

(G.1) There are e ∈ ω, v ∈

ω, i, j ∈ L, and σ ≤ τ

such that t = h0, ei ∗ hv, i, ji, i 6≤

j, L

[T ] is K

-admissible, t ∈ L

[T ], and ∀x ∈

ω ∩ L

[T ](χ

(x) ∼ = {e}(x, v, χ

,

E)).

(G.2) There are e

, e

∈ ω, v

, v

∈

ω, i, j, k ∈ L, and σ ≤ τ

such that t = h1, e

, e

i ∗ hv

, v

, i, j, ki, i ∧

j = k, L

[T ] is K

- admissible, t ∈ L

[T ], ∀x ∈

ω ∩ L

[T ]({e

}(x, v

, χ

,

E)

∼ = {e

}(x, v

, χ

,

E)), and there is a partial function p ∈ L

[T ] from

ω to

ω such that g

⊆ p, Rng(p − g

)

⊆ L

[T ], and ∃x ∈

ω ∩ L

[T ]({e

}(x, v

, χ

,

E) 6∼

= {e

}(x, v

, χ

,

E)). Here, P

∗ 0 = P

∪ {hy, 0i | y ∈

ω

ω − Dom(p)} for l ∈ L.

We choose the ≤

-least t ∈

ω ∩ L

[T ] which satisfies (G.1) or (G.2) and distinguish two subcases.

Subcase 1.1: t satisfies (G.1). We choose the ≤

-least z ∈

ω ∩

(L

[T ] − L

[T ]) and the ≤

-least ha, bi ∈

ω ×

ω such that f

(j) =

f

(j) and f

(i) 6= f

(i) by (R.2). Notice that if σ is as in (G.1), then

a, b, f

(i) ∈ L

[T ] by (R.2*) and (R.5). We set z

= hz, f

(i)i and define partial functions p

, p

by

p

(x) (p

(x) resp.) =

 g

(x) if x ∈ Dom(g

), a (b resp.) if x = z.

Then P

= P

, z

∈ P

, and z

6∈ P

. If {e}(z

, v, χ

,

E) ∼ = 0, then we define

g

(x) =

 p

(x) if x ∈ Dom(p

),

0 if x ∈

ω ∩ L

[T ] − Dom(p

), and if {e}(z

, v, χ

,

E) 6∼ = 0, then we define

g

(x) =

 p

(x) if x ∈ Dom(p

),

0 if x ∈

ω ∩ L

[T ] − Dom(p

).

Subcase 1.2: t satisfies (G.2). We choose the ≤

-least partial function p ∈ L

[T ] as in (G.2) and define

g

(x) =

 p(x) if x ∈ Dom(p),

0 if x ∈

ω ∩ L

[T ] − Dom(p).

Case 2: Otherwise. We define g

(x) =

 g

(x) if x ∈ Dom(g

),

0 if x ∈

ω ∩ L

[T ] − Dom(g

).

In the construction at Stage α + 1 above, notice that for l ∈ L, G

= P

∗ 0 ∩ L

[T ] or = P

∗ 0 ∩ L

[T ] (Subcase 1.1), or = P

∗ 0 ∩ L

[T ] (Subcase 1.2), or = G

∗ 0 ∩ L

[T ] (Case 2) respectively.

We define g = S

α∈ℵ1

g

. Then, for all α ∈ ℵ

, gd

ω ∩ L

[T ] = g

and g

:

ω∩L

[T ] →

ω∩L

[T ]. Moreover g

:

ω∩L

[T ] →

ω∩L

[T ] by definition. If there is no σ ≤ τ

such that L

[T ] is K

-admissible, then Rng(g

) = {0}. As for projections, for all α ∈ ℵ

and l ∈ L ∩ L

[T ], we have G

∩ L

[T ] = G

.

Lemma 3.4. Let % ∈ ℵ

and L

[T ] be K

-admissible.

(1) For all α < ℵ

, if % ≤ τ

, then there is σ ≤ τ

such that L

[T ] is K

-admissible and Rng(g

− g

) ⊆ L

[T ].

(2) For all x ∈

ω, there is σ ≤ max{rk(x), %} such that L

[T ] is K

- admissible and g(x) ∈ L

[T ].

P r o o f. (1) We distinguish three cases at Stage α + 1.

Case 1: g

is constructed in Subcase 1.1 at Stage α + 1. We choose σ

as in (G.1). By definition, there is c ∈

ω ∩ L

[T ] (c = a or = b in Subcase

1.1) such that Rng(g

− g

) = {c, 0}. Since 0 ∈ L

[T ], Rng(g

− g

) ⊆ L

[T ].

Case 2: g

is constructed in Subcase 1.2 at Stage α+1. We choose the

≤

-least partial function p and σ as in (G.2). By (G.2), Rng(p − g

) ⊆ L

[T ], hence Rng(g

− g

) ⊆ L

[T ].

Case 3: g

is constructed in Case 2 at Stage α + 1. By definition, Rng(g

− g

) = {0} ⊆ L

[T ].

(2) We choose α < ℵ

such that x ∈ L

[T ] − L

[T ]. By Lemma 2.2, τ

≤ rk(x). If % ≤ τ

, then by (1) there is σ ≤ rk(x) such that L

[T ] is K

- admissible and g(x) = g

(x) ∈ L

[T ]. If τ

< %, then since Rng(g

) ⊆ L

[T ], we have g(x) ∈ L

[T ].

Since L

[T ] is K

-admissible and

ω ⊆ L

[T ], for all x ∈

ω there exists % < ℵ

such that L

[T ] is K

-admissible and x ∈ L

[T ] (using the L¨owenheim–Skolem Theorem). For x ∈

ω, we set %(x) = min{σ < ℵ

| L

[T ] is K

-admissible and x ∈ L

[T ]}.

Lemma 3.5. Let α ∈ ℵ

and l ∈ L.

(1) For any T -admissible set M , if τ

∈ M , then g

∈ M . (2) For any T -admissible set M , if τ

, %(l) ∈ M , then G

∈ M . (3) If %(l) < τ

, then L

[T ] is G

-admissible.

P r o o f. (1) We prove

∀α ∈ ℵ

∀M : T -admissible set (τ

∈ M ⇒ hg

| β ≤ αi ∈ M ) by induction.

If α = 0, then this is clear.

Let 0 < α ∈ ℵ

. We assume that for all β ∈ α and every T -admissible set M we have (τ

∈ M ⇒ hg

| γ ≤ βi ∈ M ). Let M be a T -admissible set and τ

∈ M .

Let α = β + 1 for some β. By assumption, g

∈ L

[T ]. In the construc- tion at Stage β +1, p

, p

in Subcase 1.1 and p in Subcase 1.2 are elements of L

[T ]. Since L

[T ] ∈ M , by definition g

∈ M . Hence hg

| β ≤ αi ∈ M . Let α be a limit ordinal. For every limit ordinal β ∈ α, since hg

| γ ≤ βi

∈ L

[T ], the construction at Stage β can be expressed over L

[T ]. And

for every β + 1 ∈ α, since the conditions of every case at Stage β + 1 can be

expressed over L

[T ] (notice that if t = h. . .i∗h. . . , i, j, . . .i and %(t) ≤ τ

,

then G

, G

∈ L

[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

[T ]. Thus, hg

| β ∈ αi is ∆

-

definable over M with parameter hτ

| β ≤ αi, hence hg

| β ∈ αi ∈ M . (By

Lemma 3.1, hτ

| β ≤ αi ∈ M .) Therefore, by definition, g

∈ M , and so hg

| β ≤ αi ∈ M .

(2) By (1), g

∈ M . For all x ∈ Dom(g

), since rk(x) ∈ M , there is σ ∈ On ∩M such that L

[T ] is K

-admissible and g

(x), l ∈ L

[T ] by Lemma 3.4. Hence, G

∈ M by Lemma 3.3.

(3) By (2), G

∈ L

[T ]. In the construction at Stage α + 1, p

, p

in Subcase 1.1 and p in Subcase 1.2 are elements of L

[T ], hence similarly to (2), P

, P

, P

∈ L

[T ] by Lemma 3.3. Since G

= P

∗ 0 ∩ L

[T ] or = P

∗ 0 ∩ L

[T ] or = P

∗ 0 ∩ L

[T ] or = G

∗ 0 ∩ L

[T ], we see that L

[T ] is G

-admissible and so G

-admissible.

Lemma 3.6. For all l ∈ L, G

≤

K

, hence deg(K ⊕G

) ∈ K[K, K

].

P r o o f. For α ∈ ℵ

, similarly to Lemma 3.5, the construction of g

(i.e.

constructions till Stage α) and the conditions of every case at Stage α + 1 can be expressed over L

[T ]. Hence, there are formulas ψ

and ψ

such that:

L

[T ] |= ψ

(p, α)

⇔ There is t ∈

ω ∩ L

[T ] which satisfies (G.1) or (G.2) at Stage α + 1 and let t be the ≤

-least such real,

if t = h0, ei ∗ hv, i, ji satisfies (G.1) and z, a, b, p

, p

are as in Subcase 1.1

then {e}(hz, f

(i)i, v, χ

,

E) ∼ = 0 ∧ p = p

or {e}(hz, f

(i)i, v, χ

,

E) 6∼ = 0 ∧ p = p

, and if t = h1, e

, e

i ∗ hv

, v

, i, j, ki satisfies (G.2),

then p is the ≤

-least partial function as in (G.2).

L

[T ] |= ψ

(p, α)

⇔ There is no t ∈

ω ∩ L

[T ] which satisfies (G.1) or (G.2) at Stage α + 1 and p = g

.

Here, ψ

and ψ

correspond to Case 1 and Case 2 respectively.

We choose r ∈ WO such that o.t.(r) = %(l). We prove G

≤

K

via r using (2) of §0. Let x, y ∈

ω be arbitrary and M = L

ω₁^KSJ;x,y,r

[K

; x, y, r].

Notice that if x ∈ L

[T ]−L

[T ], then by Lemma 3.2 and K

-admissibi- lity of M , we have L

[T ] = L

1

[K; x] ∈ M . By Lemma 3.4, there is σ ≤ max{rk(x), %(l)} such that L

[T ] is K

-admissible and g(x), l ∈ L

[T ];

moreover, f

(l) ∈ L

[T ]. Hence,

hx, yi ∈ G

⇔ M |= “∃α ∈ ω

∃p ∈ L

[K; x]

(L

1

[K; x] = L

[T ] ∧ x 6∈ L

[T ]

∧ L

[T ] |= ψ

(p, α) ∨ ψ

(p, α)

∧ (∃σ ≤ max{rk(x), %(l)}(x ∈ Dom(p) ∧ p(x), l ∈ L

[T ]

∧ L

[T ] is K

-admissible ∧ (y = f

(l))

)

∨ (x 6∈ Dom(p) ∧ y = 0)))”.

Notice that the quantifiers in the statement “ω

= τ

” are bounded by L

1

[K; x], since ω

= τ

iff ¬∃τ ∈ ω

(τ

< τ ∧ τ satisfies (T.1))

. Hence “hx, yi ∈ G

” is ∆

over M . Therefore, G

≤

K

.

Lemma 3.7. (1) G

≡

∅.

(2) For all i, j ∈ L, if i ≤

j, then K ⊕ G

≤

K ⊕ G

.

(3) For all i, j, k ∈ L, if i∨

j = k, then (K ⊕G

)⊕(K ⊕G

) ≡

K ⊕G

. P r o o f. (1) By definition, G

= {hx, f

(0

)i | x ∈

ω} = {hx, 0i | x ∈

ω} ≡

∅.

(2) We choose r ∈ WO such that o.t.(r) = %(i, j). To prove K ⊕ G

≤

K ⊕ G

, it is sufficient to prove that for all x, y ∈

ω,

hx, yi ∈ G

⇔ M |= “∃σ ≤ max{rk(x), %(i, j)}∃a, z ∈ L

[T ] (L

[T ] is K

-admissible ∧ i, j ∈ L

[T ]

∧ hx, zi ∈ G

∧ (f

(j) = z ∧ f

(i) = y)

)”, where M = L

ωK⊕Gj ;i,j,x,y,r 1

[K ⊕ G

; i, j, x, y, r].

Suppose hx, yi ∈ G

. By Lemma 2.2, rk(x) ∈ M . By Lemma 3.4, there is σ ≤ max{rk(x), %(i, j)} such that L

[T ] is K

-admissible and g(x), i, j ∈ L

[T ]. By (R.5), we have f

(i), f

(j) ∈ L

[T ]. Thus, if we set a = g(x) and z = f

(j), then since y = f

(i) and F ≤

K

, the right-hand side holds. Conversely, suppose that x, y ∈

ω satisfy the right-hand side. Let a, z be as in the right-hand side. By hx, zi ∈ G

, f

(j) = z = f

(j). Then, by (R.1), f

(i) = f

(i). Hence, y = f

(i), and so hx, yi ∈ G

.

(3) By (2), K ⊕ G

⊕ G

≤

K ⊕ G

. We choose r ∈ WO such that o.t.(r) = %(i, j, k). To prove K ⊕ G

≤

K ⊕ G

⊕ G

, it is sufficient to prove that for all x, y ∈

ω,

hx, yi ∈ G

⇔ M |= “∃σ ≤ max{rk(x), %(i, j, k)}∃a, z, z

∈ L

[T ] (L

[T ] is K

-admissible ∧ i, j, k ∈ L

[T ]

∧ hx, zi ∈ G

∧ hx, z

i ∈ G

∧ (f

(i) = z ∧ f

(j) = z

∧ f

(k) = y )

)”, where M = L

ωK⊕Gi⊕Gj ;i,j,k,x,y,r 1

[K ⊕ G

⊕ G

; i, j, k, x, y, r].