Our goal is to show that ∆1
1is exactly the collection of Borel sets with recursive codes. That will follow from the following two results and Σ1
1-Bounding.
Theorem 7.6 If A ⊆ Y is a recursively coded Borel set and f : X → Y is computable, thenf−1(A) is a recursively coded Borel set.
Proposition 7.7 If α < ωck
1 , thenWFα is a recursively coded Borel set. Corollary 7.8 Suppose A⊆X. The following are equivalent:
i)A is∆1 1;
ii) Ais a recursively coded Borel set.
Proof We have already shown that every recursively coded Borel set is ∆1 1. SupposeAis ∆1
1. SinceA is Π11, there is a computablef :X →T r such that x∈Aif and only iff(x)∈WF. The set
f(A) ={y:∃x x∈A∧f(x) =y} is a Σ1
1-subset of WF. By Σ11-Bounding, there isα < ωck1 such thatf(A)⊆WFα.
By 7.7 WFαis recusively coded, and by 7.6A=f−1(WFα) is recursively coded.
For notational simplicity we will assume X = N, but all our arguments generalize easily.
LetBCrec={(e, x) :φx
e is a total function andφxe ∈BC}. Then
(e, x)∈BCrec ⇔φx
e is total ∧ ∀z(∀n φxe(n) =z(n))→z∈BC).
ThusBCrec is Π1 1.
Ife∈BCrec, thenBrec(e, x) is the Borel set coded byφx
e. A similar argument
shows that there areRrec ∈Σ1
1andSrec ∈Π11 such that if (e, x)∈BCrec then y∈Brec(e, x)⇔Rrec(e, y)⇔Srec(e, x)
We saye∈BCrec andx∈Brec(e) if (e,∅)∈BCrec andx∈Brec(e,∅). The proofs of both 7.6 and 7.7 will use the Recursion Theorem to do a transfinite induction.
We begin with the base case of the induction
Lemma 7.9 There is a recursive function F : N×SY → N such that if
f :X →Y is computable and e is a code for the program computing f , then
Brec(F(e, i)) =f−1(N
η).
Proof For notational simplicity we assume X = Y = N, this is no loss of generality. Let
W ={ν∈N<ω:∀m <|η| ∃s≤ |ν|φν
e(m)↓s=η(m)}.
Then W is recursive and f−1(N
η) =Sν∈WNν. Let ν0, ν1, . . . be a recursive
enumeration of N<ω. Let T = {∅} ∪ {hni : σn ∈ W} and let l(∅) = h0,1i,
l(hni) =h1, νi. Thenx=hT, liis a recusive code. Giveneandη we can easily computeF(e, η) =isuch thatφi=x.
Lemma 7.10 i) There is a total recursive function Hc : N→ N such that if
e∈BC, thenBrec(Hc(e)) =N \Brec(e).
ii) There is a total recursive functionHu:N→Nsuch that ifφe(n)∈BCrec
Proof i)φeis a code for a pairhT, li. Let
T0={∅} ∪ {0bη:η∈T}
and l0(∅) = h0,0i, l0(0bη) = l(η). It is easy to find Hc such that Hc(e) codes
hT0, l0iand that ife∈BCrec, thenHc(e) is a code for the complement.
ii) Supposeφe(n) code a pairhTn, lni. Let
T ={∅} ∪ {nbσ:σ∈Tn}
and letl(∅) =h0,1i andl(nbσ) = ln(σ). It is easy to find Hu such thatHu(e)
codeshT, li. If eachhTn, lniis a Borel code, thenhT, licodes their union.
Theorem 7.6 follows from the next lemma.
Lemma 7.11 Ifx =hT, liis a recursive Borel code, there is a recursive func- tionG:N×T →Nsuch that if f:N → N is a computed by programPe, then
G(e, σ)∈BCrec is a Borel code forf−1(B(hT
σ, lσi)for allσ∈T. Proof
We define a recursive functiong:N×N×T →Nas follows: i) Ifl(σ) =h1, ηi, theng(i, e, σ) =F(e, η);
ii) Ifl(σ) =h0,0i, theng(i, e, σ) =Hc(φi(e, σb0));
iii) Suppose l(σ) = h0,1i. Choose j such that φj(n) = φi(e, σbn). Then
g(i, e, σ) =Hu(j).
By the Recursion Theorem, there isbi such that φi(e, σ) = g(bi, e, σ) for all e, σ. LetG(e, σ) =φi(e, σ).
We prove by induction onT, thatG(e, σ) is a code forf−1(B(hT
σ, lσi). By
i) this is clear ifl(σ) =h1, η). We assume the claim is true for allτ⊃σ. Ifl(σ) =h0,0i, then
G(e, σ) =g(bi, e, σ) =Hc(φi(e, σ)) =Hc(G(e, σb0)).
By inducition,Hc(G(e, σbn) is a code for
f−1(B(hTσ, lσi) =X\f−1(B(hTσ 0, lσ 0i).
Ifl(σ) =h0,1i, thenG(e, σbn) is a Borel code forAn=f−1(B(hTσbn, lσ ni).
We choosej such thatφj(n) is a code forAn andG(e, σ) =Hu(j) is a code for S
An.
Theorem 7.7 follows from the next lemma.
Lemma 7.12 If T is a recursive well founded tree, then there is a recursive functionG:T →BCrec, such thatBrec(G(σ)) ={S∈T r:ρ(S)≤ρ(Tσ)}. Proof Forσ∈N<ω letf
Note thatρ(S)≤ρ(T) if and only if for alln∈Nthere ism∈Nsuch that ρ(Shni)≤ρ(Thmi). Thus {S∈T r:ρ(S)≤ρ(Tσ)}= \ n∈N [ m∈N fh−n1i({S∈T r:ρ(S)≤ρ(Tσ m)}).
Fixc such that Brec(c) =∅. We define a recursive functiong:N×T →N
as follows.
i) Ifσ6∈T, theng(i, σ) =c.
ii) Otherwiseg(i, σ) is a Borel code for
\ n
[ m
fh−n1i(Brec(φi(σbm))).
We can do this using the functions F, Hu andHc above. Of course for somei,
this may well be undefined.
By the Recursion Theorem there isbisuch thatφi(σ) =g(bi, σ) for allσ. An easy induction shows thatG=φi is the desired function.