Optimal Results on Recognizability for Infinite Time Register Machines

OPTIMAL RESULTS ON RECOGNIZABILITY FOR INFINITE TIME REGISTER MACHINES

MERLIN CARL

Abstract. Exploring further the properties ofITRM-recognizable reals started in [1], we provide a detailed analysis of recognizable reals and their distribution in G ¨odels constructible universeL. In particular, we show that new unrecognizable reals are generated at every index≥^CK . We give a machine- independent characterization of recognizability by proving that a realris recognizable if and only if it is Σ1-definable overL

CK,r

and thatr∈L

CK,r

for every recognizable realrand show that either every or norwithr∈LCK,r

generated over an index stageLis recognizable. Finally, the techniques developed along the way allow us to prove that the halting number forITRMs is recognizable and that the set of ITRM-computable reals is notITRM-decidable.

§1. Introduction. Infinite Time Register Machines (ITRMs) were introduced by Peter Koepke and Russell Miller in [9] and provide a transfinite analogue of classi- cal register machines in much the same way as the Infinite Time Turing Machines (ITTMs) introduced in [5] generalize classical Turing machines. We give a brief sketch of their behaviour here; detailed definitions can be found in [9] and [6].

These papers also contain most of the results we use in this paper.

Classical register machines, as e.g., described in [3], have as their ‘hardware’

finitely many registers, each of which can store a single natural number. A program for a register machine consists of finitely many numbered lines, each of which con- tains a single command, where the available commands are incrementing a register content by 1, setting a register content to 0, copying the content from one register to another, the oracle call (which replaces the contenti of some registerR with theith bit of an oraclex ⊆) and a conditional jump which proceeds to a fixed program line when two registers have the same content and otherwise proceeds with the next program line. When the machine arrives at a line index which is not part of the program, the machine halts.

To generalize these machines to the transfinite, we keep the ‘hardware’ - finitely many registers, each of which stores a single natural number - as well as the ‘soft- ware’, i.e., the notion of program. What is changed is the ‘working time’: While the

2010Mathematics Subject Classification. 03D65, 03D60, 03Exx, 03E45.

Key words and phrases. Infinite Time Computability, Generalized Computability, Infinite Time Register Machines, Recognizable Reals.

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https://dx.doi.org/10.1017/jsl.2015.8

working time of a halting classical register machine just a natural number,ITRMs run along arbitrary ordinals. To make sense of this, we need to explain what state the machine is supposed to assume at a limit ordinal. At successor ordinals, we define the computation in the same way as for classical register machines. Ifis a limit ordinal, we set the contentRiof thei-th registerRi at timeto lim inf<Ri

if and only if this limit< , and otherwise to 0. Likewise, the active program line Z to be carried out in theth step is lim inf<Z, (where the limit is always finite as the set of lines is finite).

Definition1.1. x ⊆ isITRM-computable in the oracle y ⊆ if and only if there exists an ITRM-programP such that, fori ∈ , P with oracley stops for every natural numberj in its first register at the start of the computation and returns 1 if and only ifj∈xand otherwise returns 0. A realITRM-computable in the empty oracle is simply calledITRM-computable.

The notion of the recognizability of a real was introduced by Hamkins and Lewis in [5] for Infinite Time Turing machines. The adaption toITRMs is straightforward:

Definition1.2. Let r ⊆ . Then r is recognizable if and only if there is an ITRM-program P such that P^x stops with output 1 if and only if x = r and otherwise stops with output 0.

In the last section of [6], we showed that there are recognizable reals that are not computable. This is analogous to the lost melody theorem for infinite time Turing machines (ITTMs) demonstrated in [5]. The example given there was a

<_L-minimal real coding an∈-minimalL_α modellingZF⁻ (see below). Here, we will give a much more natural example by showing that the real coding the halting problem for ITRMs is recognizable. After that, in [1], we obtained some more results on ITRM-recognizable reals and their relation to G ¨odel’s constructible universe L. In particular, we showed that all recognizable reals are constructible, but that the recognizable reals do not form an initial segment of the constructible reals in the canonical well-ordering <L of L. In fact, there are quite large gaps:

If is the supremum of the ordinals indexing levels of theL-hierarchy at which new recognizable reals appear andα < , then there is a<L-interval of length

> α in the constructible reals without a recognizable element, while there are recognizable reals that are<L-greater than all elements of that interval. Here, our goal is to give a more precise picture ofITRM-recognizability and their distribution among the constructible reals. For example, it is easy to see that all computable reals are recognizable, so that, by the theorem cited above, all reals in LCK

are

recognizable. Is there anyα > ^CKsuch that the reals inL_αare still all recognizable?

It turns out that this is not the case: Whenever > ^CK is such that L₊₁−L contains a real number at all, it also contains a nonrecognizable real. Furthermore, we give a machine-independent, purely set-theoretical characterization ofITRM- recognizability and lower estimates on the length of gaps in the recognizable reals and some information on ordinals starting them. The techniques used in these results will then be applied to give a characterization of ITRM-decidable sets of reals and show that the set ofITRM-computable reals is not decidable.

Most of our notation is standard.ZF⁻isZF set theory without the power set axiom in the version described in [4]. For anITRM-programP,P^x(i)↓=jmeans that the programP with oraclexwith initial inputi in its first register stops with outputjin register 1. We will writeP^x↓forP^x(0)↓. When we write thatP^x‘stops in less than αmany steps’, we mean thatP^x(i) stops in less thanαmany steps for everyi ∈ .Onis the class of ordinals. Unless stated otherwise, small Greek letters refer to ordinals. For, ∈ On, we denote by ^CK the smallest admissible ordinal greater than^CKwhen =+ 1 andsup{^CK | < }whenis a limit ordinal. When we consider admissible ordinals relative to a realx, we write^CK,x. ForX ⊆Lα, Σ^L₁^α{X}denotes the Σ1-Skolem hull ofX inLαand Σ^Lα{X}denotes the elementary hull ofX in Lα. When H is a Σ1-substructure of some Lα, then :H ∼=L and :H →coll Ldenote the Mostowski collapse of H toL with isomorphism. Throughout the paper,p:×→denotes the usual bijection between×and. At various places, we will code certain∈-structures by real numbers. This works as follows: Let (M,∈) be countably infinite and transitive, and letf : → M be a bijection. Then the real coding (M,∈) according to f is{p(i, j) | [f(i) ∈ f(j)]}. We say thatx codes (M,∈) if and only if there is a functionfsuch thatxcodes (M,∈) according tof. If (M,∈)∈L,cc(M) denotes the<L-smallest real coding (M,∈).

§2. ITRMs. We give here some basic results onITRMs and admissible set theory which are relevant for our further development.

Theorem2.1. LetPndenote the set ofITRM-programs using at mostnregisters, and let(P_i,n | i ∈ )enumeratePnin some natural way. Then the bounded halting problemH_n^x := {i ∈ | P_i,n^x ↓} is computable uniformly in the oracle x by an ITRM-program.

Furthermore, ifP ∈ Pn andP^x ↓, thenP_x halts in less than_n+1^CK,x many steps.

Consequently, ifPis a haltingITRM-program, thenP^xstops in less than^CK,xmany steps.

Proof. The corresponding results from [9] easily relativize.

Definition 2.2. X ⊆ P() isITRM-decidable if and only if there exists an ITRM-programPsuch thatP^x↓= 1 if and only ifx∈X andP^x ↓= 0, otherwise.

Corollary2.3. LetAbe anITRM-decidable set of reals and letPbe anITRM- program. ThenM1 := {x | P^x ↓},M2 :={x | ∀i ∈P^x(i) ↓}, andM3 :={x | P^xcomputes an element ofA}are decidable.

Proof. The decidability ofM₁andM₂is immediate from Theorem 2.1. ForM₃, letQbe a program for decidingA. To decideM₂, proceed as follows: Given some realxin the oracle, first check, whetherx∈M2. If not, thenx /∈A. Otherwise,P^x computes some realy, and we can useQto decide whethery ∈A.

Lemma2.4. Let(φ_i |i ∈)be a natural enumeration of the∈-formulas. There is anITRM-programP such that, for allx ⊆,i ∈ ,v = (v₁, . . . , v_n)a finite sequence of natural numbers of the appropriate length coded by a natural number

¯

v, P^x(i, v) ↓= 1 if and only ifx codes some Lα such thatφi(x1, . . . , xn)holds in Lα, wherex1, . . . , xnare the elements coded byv1, . . . , vn, respectively, and otherwise

P^x(i, v)↓= 0. The same holds with a recursive setSof formulas instead of one single formulaφ, wherei is then a code for a Turing program enumeratingS.

Proof.This is Corollary 13 of [1].

Definition2.5. An ordinalαis called Σ1-fixed if and only if there exists a Σ1- statementφsuch thatαis minimal with the property thatLα |=φ. Let denote the supremum of the Σ1-fixed ordinals.

Definition2.6. An ordinalis called an index if and only ifL+1−Lcontains a subset of.

Theorem2.7.Denote byRECOGthe set of recognizable reals.ThenRECOG⊆L. Furthermore, for each < , there existsα < such that[α, α+]contains unbound- edly many indices, but

(L_α+−L_α)∩RECOG=∅.

Proof.See [1].

Theorem2.8. Letx, y ⊆. ThenxisITRM-computable in the oracley if and only ifx∈LCK,y

[y].

Proof.This is a straightforward relativization of the main result of [8].

Lemma2.9. LetA =∅be anITRM-decidable set of reals such thata ∈ LCK,a

for alla ∈A. Then the<_L-minimal element ofAis recognizable.

Proof.Letabe the<_L-minimal element of such anA. By Theorem 2.8, there is anITRM-programPsuch thatP^acomputes a code for someL-levelL_αcontaining a. LetQ_Abe anITRM-program decidingA. Nowacan be recognized as follows:

Given somex ⊆in the oracle, first check whetherP^x(i)↓ for alli ∈ , using a halting problem solver forP which exists by Theorem 2.1. If not, then x = a.

Otherwise, test whetherP^x computes a codec for an L-level containing x. This can be done using Lemma 2.3 and Lemma 2.4. If not, thenx=a. Otherwise, test whether QA(x) ↓= 1. If not, thenx /∈ A, so x = a. Otherwise, usec to search through all reals belowxin <L for a realz <L x such thatQA(z) ↓= 1. If such

a real is found, thenx=a. Otherwise,x=a.

We will need the following result of Jensen and Karp:

Theorem 2.10. Let α ∈ On be a limit of admissible ordinals, and let φ be a Σ1-statement. Thenφis absolute betweenLαandVα.

Proof.See [7].

§3. Unrecognizable Reals Everywhere. The goal of this section is to show the result announced above: If > ^CK is such thatL₊₁−Lcontains a real number, then it already contains a nonrecognizable real number. To show this, we will use Cohen-forcing over models ofKP. A general reference for the forcing technique is [10]. For forcing over set theories weaker thanZFC, we refer the reader to [12], [2], and [11].

Theorem3.1. LetP ∈L be a notion of forcing, where is indecomposable, let > be admissible and letGbeP-generic overL. ThenL[G]in the sense of relative constructibility coincides withL[G]in the sense of generic extensions.

Proof.This follows from the proof of Proposition 9.5 of [11].

Theorem 3.2. Let x ∈ RECOG. Then x ∈ L^CK,x . Consequently, if x is recognizable, but notITRM-computable, we have^CK,x> ^CK.

Proof. LetPbe a program that recognizesx.

Then L^CK,x[x] |= ∃yP^y ↓= 1. Now φ ≡ ∃yP^y ↓= 1 is a (set theoretical) Σ1- statement, stating that there are a real y and a set c such that c codes the P- computation in the oracley and ends with 1. By Jensen-Karp (see Theorem 2.10), this is absolute between V_α andL_α wheneverαis a limit of admissible ordinals.

Now^CK,x= sup{^CK,x_i |i ∈}is a limit of admissible ordinals, soφis absolute betweenV_αandLCK,x

. Also, asφis Σ₁, it is upwards absolute. HenceLCK,x [x]|= φ =⇒ V^CK,x

|=φ =⇒ L^CK,x

=⇒ L^CK,x

[x], soφis absolute betweenL^CK,x

[x]

and L^CK,x . As φ holds in L^CK,x [x], it follows that φ holds in L^CK,x . So L^CK,x contains a realrsuch thatP^r ↓= 1. By absoluteness of computations,P^r↓= 1 also holds inV. SoP^r ↓= 1. AsPrecognizesx, it follows thatx=r. Hencex∈L^CK,x . Now letxbe recognizable, but not computable. Asxis not computable, we have x /∈L^CK . By the first part of the claim,x∈L^CK,x . Hence^CK,x> ^CK .

This immediately leads to the following dichotomy:

Corollary3.3. Ifx ⊆ is such that^CK,x = ^CK , then eitherx isITRM- computable orxis notITRM-recognizable.

Proof. IfxisITRM-computable, thenxis clearlyITRM-recognizable (see [1]).

Ifxis notITRM-computable, thenx /∈L^CK , hencex /∈L^CK,x if^CK,x=^CK.

By Theorem 3.2 then,xis not recognizable.

For the next lemma, we need the following result of Mathias:

Theorem3.4. IfMis admissible,P∈MandGis an(M,P)-generic filter meeting each dense open subclass ofMthat is the union of aΣ₁(M)and aΠ₁(M)class, then M^P[G]is admissible.

Proof. This is Theorem 10.11 of [11].

Lemma3.5. Letαbe admissible,(P,≤) ∈Lαbe a notion of forcing andG be a filter onPsuch thatP∩D =∅for every dense subsetDofPsuch thatD ∈Lα+1. ThenLα[G]is admissible.

Proof. This follows from Theorem 3.4, since unions of Σ1(Lα) and Π1(Lα)- definable subsets ofPare clearly elements ofLα+1. Corollary3.6. Let ≥ ^CK , let (P,≤P)be the notion of forcing for adding a Cohen real (i.e.,Pconsists of the finite partial functions fromto2andx≤Pyif and only ify⊆x) and letGbe a filter on(P,≤)which intersects every denseD⊆P such thatD∈L. ThenL^CK

i [G]is admissible for everyi ∈. Proof. This is immediate from Lemma 3.5 as L ⊇ LCK

⊇ LCK

i +1 for all

i ∈.

The following will be used to show that, for each index ≥ ^CK, L+1−L

contains an unrecognizable real.

Theorem3.7. Let≥^CKbe an index. Then there exists

x ∈ L+1 −L with _i^CK,x = _i^CK for all i ∈ . In particular, this implies that ^CK,x=^CK.

Proof.LetPbe the notion of forcing for adding a Cohen real (see above). LetG be anL-generic filter onP (i.e., G intersects every dense subset ofP which lies insideL). By Corollary 3.6,L^CK

i [G] is then admissible for alli ∈. Now letx :=

G ∈P(). We show thatLCK

i [G] =LCK

i [x]: Asx = G ∈ LCK

i [G], we haveLCK

i [x]⊆LCK

i [G] andG ∈LCK

i [x] (sinceG is definable from x), hence alsoL^CK

i [G]⊆L^CK

i [x]. More generally, ifαis additively indecomposable (which certainly holds forαadmissible) and if we havex∈Lα[y] andy ∈Lα[x], thenLα[x] =Lα[y]. To see this, letz ∈ L[x] ( < α) andx ∈L[y] ( < α).

Thenz∈L[x]∈L++1[y]⊆Lα[y], henceLα[x]⊆Lα[y].Lα[y]⊆Lα[x] now follows by symmetry.

Now it follows thatL^CK

1 [x] |= KP, i.e., ^CK₁ isx-admissible, so that₁^CK ≥ ₁^CK,x ≥₁^CK. Consequently, we get^CK,x₁ =₁^CK. Now assume inductively that _i^CK =_i^CK,xfor somei ∈ . It then follows thatL^CK

i+1[x]|=KP, hence_i+1^CK is x-admissible and thus_i+1^CK > _i^CK,x = _i^CK. But then^CK_i+1 ≥ _i+1^CK,x ≥^CK_i+1, so _i+1^CK,x=^CK_i+1. This now gives us_i^CK,x=_i^CKfor alli ∈, so that^CK,x=^CK.

Next, we demonstrate thatG - and hencex =

G - is definable overL and hence elements of L+1. This can be seen as follows: As is an index, there is f : → L surjective such that f ∈ L+1 and hence definable overL. Now define g : (^<2, ) →^< 2 thus: Let g(s , i) be the lexically minimal element of f(i), of whichs is a subsequence iff(i)⊆^< 2 is a dense subset ofP otherwise let g(s, i) =s. Now, define recursively:h(0) = ∅,h(i + 1) =g(h(i), i+ 1). This recursion can be carried out definably overLas follows: Set (fori ∈)h(i) =x if and only if

(i = 0∧x = ∅)∨(i ≥ 1∧ ∃(s₀, . . . , s_i−1)[∀j ∈i((j = 0∧s₀ = ∅)∨(s_j = g(s_j−1 , j)))∧x =g(s_i−1 , i)]). This is definable overL, asg is definable overL and all finite sequences of elements ofP are contained inL, and hence certainly inL^CK. It follows thatx∈L+1. That_i^CK,x =^CK_i holds for alli ∈was seen above.

Finally, we show thatx /∈L: Roughly, this follows immediately from the fact that G is definable fromxand thatG is generic overLas in the case of Cohen-forcing forZFCmodels. More precisely, letz∈P()∩L. Also, let < be minimal such thatz∈L+1−L. ThenDz :={b ∈^<2| ∃i ∈b(i)=z(i)}is dense inPand definable overL, hence an element ofL. Consequently, everyDzhas nonempty intersection with everyL-generic filterG, so that

G =z. As this holds for all z ∈L, we getx=

G /∈L.

We can now show that new unrecognizable reals appear wherever possible, i.e., are generated at every index stage:

Theorem 3.8. Let ≥ ^CK be an index. Then L+1 − L contains an unrecognizable real.

Proof.By Theorem 3.7,L₊₁−L contains a realxsuch that^CK,x =^Ck.

By Corollary 3.3,xis not recognizable.

§4. The halting number is recognizable. We obtain a very natural lost melody by showing that the halting number forITRMs is in fact recognizable. Fix a natural

well-ordering (Pi |i ∈) of theITRM-programs in order typeby e.g., sorting the programs lexicographically. The halting numberhforITRMs is then defined as h:={i ∈|Pi ↓}. This realhis natural insofar its definition is purely internal to ITRMs (e.g., not in any way related toL) and it is also arguably the first noncom- putable real coming to mind.

We start by showing that, givenh, there is a universalITRM:

Lemma 4.1. There is an ITRM-programP such that, for every (i, j) ∈ ², we haveP^h(p(i, j))↓=k+ 1ifP_i(j)↓=kandP^h(p(i, j))↓= 0ifP_i(j)↑. That is,P can, giveni, compute the function computed byP_i.

Proof. Pworks as follows: Giveni andj, first usehto check whetherPi(j)↓.

IfPi(j) ↑,P returns 0. Otherwise, we carry out the following procedure for each k ∈: Compute (which can be done with a standard register machine, in fact) an indexl = l(i, j, k) such thatP_l ↓if and only ifPi(j)↓=k.Pl will use a halting problem solver forP_i(which can be easily obtained fromPi), i.e., a sub-programQ such thatQ(j)↓= 1 if and only ifP_i(j)↓andQ(j)↓= 0, otherwise. If it turns out thatQ(j) = 0, thenP_l enters an infinite loop. Otherwise, we wait untilP_i(j) has stopped and check whether the outcome isk. If it is, we stop, otherwise we enter an infinite loop. (Note thatP is not required to do all this; it is only required thatP can compute a code for a program that does this, which is in fact easy).

Usingl andh, we can easily check whetherPi(j)↓=k. If so, we returnk+ 1.

Otherwise, we continue withk+ 1.

AsPi(j)↓is already clear at this point, this has to lead to the value ofPi(j) after

finitely many iterations.

The next step is that, usingh, a code forLCK

i can be computed uniformly ini. Corollary 4.2. There is an ITRM-program Q such that, for every i ∈ , Q^h(i) computes a code for L^CK

i . (I.e.: Q^h(n) halts for every n ∈ and {j ∈ |Q^h(p(i, j)↓= 1}will be a code forLCK

i .) Proof. First note that codes forL^CK

i are uniformly recognizable ini, i.e., there is a programR such that, for every i ∈ , x ⊆ , R^x(i) ↓= 1 if and only if x codes L^CK

i and otherwise R^x(i) ↓= 0. This can be obtained using the well- foundedness checker combined with the first-order checker described in [6] for V =L+KP+‘There are exactlyi−1 admissible ordinals’.

Usingh, we can now run through, first testing whetherPk(j) will halt for each j ∈and then, usingPfrom the last lemma, whetherPkwill compute a code for LCK

i . (We can evaluatePk(j) for everyjusingPfrom the last lemma and then use Rto recognize whether the computed number is a code.)

AsL^CK

i hasITRM-computable codes, the minimal indexlsuch thatPlcomputes a code forL^CK

i will eventually be found in this way.

After that, we can, again usingP from the last lemma, evaluatel to compute the

desired code.

These bits can now be put together to form a code forL^CK. This code will be a bit different from the codes considered so far, as we allow one element of the coded structure to be represented by arbitrary many elements of.

Definition 4.3. Let (X,∈) be a transitive ∈-structure. Furthermore, let f:→X be surjective. Then {p(i, j) | f(i) ∈ f(j)} is called an odd code for (X,∈).

Odd codes can be evaluated in the same way that the codes we used so far could.

The possibility of elements appearing repeatedly hinders none of those methods.

It is helpful, however, to note that the equality is computable:

Proposition4.4. There is anITRM-programT¯such that, for every odd codexfor a well-founded, transitive∈-structure(X,∈)(with associated functionf : → X) and alli, j ∈,T¯^x(p(i, j))↓= 1if and only iff(i) =f(j)andT¯^x ↓= 0, otherwise.

Furthermore, there is anITRM-programTsuch that, for every two odd codesxand yfor well-founded, transitive∈-structures(X,∈)and(Y,∈)(with associated functions f1andf2),T^x⊕y(p(i, j))↓= 1if and only iff1(i) =f2(j)andT^x⊕y(p(i, j))↓= 0, otherwise.

Finally, there is anITRM-programT_∈such that, for every two odd codesxandy for well-founded, transitive∈-structures(X,∈)and(Y,∈)(with associated functions f1andf2),T_∈^x⊕y(p(i, j))↓= 1if and only iff1(i)∈f2(j)andT_∈^x⊕y(p(i, j))↓= 0, otherwise.

Proof.This is an easy application of the techniques developed in [6]. We give a brief impression how this works: To decide, given x, y ⊆ coding transitive

∈-structures andi, j ∈ , whetheri andj represent the same element, we use a stack. Initially, the stack containsp(i, j). We then need to decide whetherf1(i)⊆ f2(j) andf2(j)⊆f1(i). To see the former, we usexto successively consider the natural numbers coding elements of f1(i). For each such natural number n, we search, usingy, through all natural numbers coding elements of f2(j). For each such elementm, we putp(n, m) on our stack and decide whetherf1(n) =f2(m).

When such anmis found, we replace the content of the stack register withp(i, j) (this ensures that the lim inf s will be as desired). When allmhave been tried and no m withf₂(m) = f1(n) has been found, thenf1(i) = f2(j). Otherwise, we continue with the nextn. When anmwithf₂(m) =f1(n) has been found for every suchn, we havef₁(i) =f2(j). That the algorithm must terminate follows from the

well-foundedness of (X,∈) and (Y,∈).

Lemma4.5. There is anITRM-programS such thatS^h computes an odd codec forL^CK.

Proof.The idea is to reservebits for coding LCK

i ; in one portion (the i-th portion), we useQ^h to compute a codec_i forLCK

i with corresponding surjection f_i. Then we useTfrom the last proposition to relate the portions.

More precisely, we construct c as follows: First, for all i, j, k ∈ , we let p(p(j, i), p(k, i)) ∈ x if and only if p(j, k) ∈ c_i. It remains to decide the bits of the form p(p(j, i1), p(k, i2)) with i1 = i2. This corresponds to the question whetherfi1(j)∈fi2(k), which can be answered usingT_∈from Proposition 4.4.

Theorem4.6. Let h := {i ∈ | Pi ↓}be the set of indices of halting ITRM- programs. Thenh∈RECOG.

Proof.We first claim that the set of odd codes for L^CK is decidable. To test whether a real x is an odd code of L^CK, check first whether x codes a well- founded relation and then whether the structure coded byxsatisfies the statement

V =L∧‘There is no largest admissible ordinal and there is no limit of admissible ordinals’∧‘Everyxis contained in some admissibleLα’. The techniques from the proof of Corollary 4.2 are easily adapted to this task, using Proposition 4.4.

Now letxbe the real in the oracle, and letSbe as in Lemma 4.5. As the set of odd codes forL^CK is decidable, we can, by the third statement of Lemma 2.3, decide the set{y ⊆|P^ycomputes an odd code forL^CK }. Consequently, we can check whetherS^x computes an odd codecforL^CK. If not, return 0. Checking whether certain programs halt amounts to checking whether certain first-order statements (expressing thatPihalts, i.e., there is a set coding a halting computation according toP) hold inL^CK , which can be done usingcby Lemma 2.4. We can then compare the results of the computation with the oracle numberx. If the agree, thenx=h,

otherwisex=h. This identifiesh.

As a consequence, the halting numberhis a minimal lost melody with respect to ITRM-reducibility:

Corollary4.7. Letx ∈RECOG−COMP. Then there is anITRM-program Psuch that∀i ∈P^x(i)↓=h(i).

Proof. Asx ∈ RECOG−COMP, it follows from Theorem 3.2 that^CK,x >

^CK. Consequently, there isi ∈such that_i^CK,x > ^CK. Ashis definable over LCK

, we have h ∈ LCK

+1, so h ∈ LCK,x

i ⊆ LCK,x

i [x] ⊆ LCK,x

[x]. Hence h is

ITRM-computable fromx, as desired.

§5. Potential recognizability. We saw above (via Jensen-Karp, see Theorem 3.2) thatx ∈RECOG implies thatx ∈L^CK,x . Reals without this property are hence ruled out, we concentrate on those that have it.

Definition 5.1. x ⊆ is potentially recognizable if and only ifx ∈ L^CK,x . We denote the set of potentially recognizable reals byPRECOG.

Theorem5.2. Letbe an index. Then either all potentially recognizable elements ofL+1−Lare recognizable or none is.

Proof. Supposea ∈(L+1−L)∩RECOGandx∈(L+1−L)∩PRECOG. We want to show thatx ∈RECOG. Pick a programQthat recognizesa. Asx∈ PRECOG, there isi ∈such thatx∈L^CK,x

i . In particular, we haveL+1 ∈L^CK,x

i . Hencec=cc(L+1), the<L-minimal real code forL+1, is computable fromx. Let Pbe a program that computescfromx.

To identify whethery =x(withyin the oracle), we first use a halting problem solver (see Theorem 2.1) forPto check whetherP^y(i)↓for alli ∈. If not, then y=x. If yes, we check whetherP^ycomputes a coded for anL-level containingy.

If not, theny =x. If yes, we check, as in the proof of the Lost Melody Theorem in [6], whetherd is<_L-minimal with that property (this is possible asx ∈L₊₁).

If not, theny=x. If yes, we check whether the structure coded bydcontains a real rsuch thatQ^r ↓= 1. This can be done Lemma 2.3. If there is no suchr, theny =x.

If there is, we check whether the structure coded bydcontains anL-level that also containsr(this checks the minimality of). If so, theny=x. Otherwise, we know thatP^ycomputesc. But inc,xis coded by some fixed natural numbernwhich can be given to the program in advance. All that remains is hence to check whetheryis

the number coded byninc. If not, theny=x, otherwisey=x. So this procedure

recognizesx, hencex∈RECOG.

This allows us to give a characterization of recognizability in purely set-theoretical vocabulary, without reference to machines:

Theorem5.3. Letx∈PRECOG. Thenx∈RECOG if and only if there exists aΣ1-formulaφof set theory without parameters such thatxis the unique witness for φinL^CK,x.

Proof.Ifx∈RECOG, thenx∈PRECOGby Theorem 3.2 above. Now, ifPis a program that recognizesx, thenP^x ↓= 1 is Σ₁-expressable overLCK,x

. By upwards preservation of Σ₁-statements, ifLCK,x

|=P^y ↓= 1 for somey =x, thenP^y↓= 1 in the real world, which contradicts the assumption thatPrecognizesx.

On the other hand, ifx is definable as above, then let L be the first L-level containing x such that L |= φ(x). Then < ^CK,x, so c := cc(L) can be computed fromx, say by programQ. Given a realy andcin the oracle, we can check whethery∈LandL|=φ(y) hold.

Checking whether some oracle numbery is equal tox then works as follows:

Check (using Lemma 2.3 and the techniques from the proof of the Lost Melody theorem in [6]) whetherQ^ycomputes a minimal code for anL-level containingy, then check whetherφ(y) holds in thatL-level and then whether it fails in all earlier L-levels. If all of this holds, theny =x(since Σ1is preserved upwards), otherwise

y =x.

The same argument can be adapted to characterizeITRM-decidable sets of reals (a set A ⊆ P() is ITRM-decidable if and only if there is anITRM-program P such thatP^x ↓= 1 if and only ifx ∈ AandP^x ↓= 0, otherwise, for allx ∈ P()).ITRM-decidable sets of reals were considered and demonstrated to be Δ¹₂ in [9].

Theorem5.4. LetA⊆P(). ThenAisITRM-decidable if and only if there exist n ∈and a parameter-freeΣ1-formulaφ(y)with a single free variableysuch that A={x∈P()|LCK,x

n [x]|=φ(x).

Proof.Assume first that Ais ITRM-decidable. Let P be anITRM-program that decides A. The computation is contained in L^CK,x

n [x] ifn is larger than the number of registers used byP. Hence membership ofxinAcan be expressed as the existence of a computation byPwith inputxthat terminates with output 1, which is Σ1overL_n^CK,x[x].

On the other hand, ifx ∈ A is characterized as described in the assumptions by a Σ1-formula overL^CK,x_n [x], we canITRM-decideAby computing a code for L^CK,x_n [x] fromx, then identifying the natural number representingxin this code and finally using the code for evaluatingφinL^CK,x_n [x]. The only new step here is the computation of a (not necessarily <_L-minimal) code forLCK,x

n [x] uniformly in x with a program that always halts. Using a halting problem solver H as in Theorem 2.1 forITRMs withn+ 3 registers, this can be done as follows: Givenx in the oracle and taking (Pj |j ∈) to be a natural enumeration of theITRM- programs using at mostn+ 3 registers, we perform the following routine for every i ∈: First, check whetherP_i^xcomputes a real, i.e., whetherP_i^x(k) halts for every

k ∈. This can be done usingH. If this is not the case, restart the routine with i+ 1. Otherwise, check, with the usual techniques, whether the realrcomputed by P_i^xis a code forL_n^CK,x[x] (i.e., an admissible level of theL[x]-hierarchy containing exactlyn−1x-admissible ordinals). If not, restart the routine withi+ 1. Otherwise,

an algorithm computing the desired code is found.

Theorem 5.5. For allx ⊆ , x is recognizable if and only ifx ∈ L^CK,x and LCK,x

|= RECOG(x). In particular, if L_α |= ZF⁻ and x ∈ P()∩L_α, then x∈RECOGholds if and only ifLα|=RECOG(x).

Proof.Suppose first thatx∈RECOG, and letPbe a program that recognizesx.

Thenx∈L^CK,x

by Theorem 3.2. By [2], ifz∈Land⁺is the smallest admissible ordinal greater than, then₁^CK,z ≤⁺. Inductively, we get that^CK,z_i ≤⁺ⁱ, where ⁺ⁱis theith admissible ordinal above. Inductively, it follows that^CK,z ≤^CK,x for allz∈LCK,x

whenxis such thatx∈LCK,x

. This implies thatP^zstops after at most^CK,xmany steps for allz ∈LCK,x

and hence thatP^zcan be carried out inside LCK,x

for allz ∈LCK,x

. Hence, sinceP recognizesx, we haveLCK,x

|=P^z ↓= 0 for allz =xand furthermoreL^CK,x |=P^x ↓= 1. HenceL^CK,x |=RECOG(x).

On the other hand, assume that x ∈ L^CK,x and thatL^CK,x |= RECOG(x).

HenceP^x ↓= 1 andP^z ↓= 0 for all z <L x. Now let Qbe a program such that Q^x computes the<L-minimal code of the firstL-level containingx. Thenx can be recognized as follows: Given some realrin the oracle, first check, using Lemma 2.3, whetherP^r ↓= 1. If not, thenr =x. Otherwise check - applying Lemma 2.3 to Q- whether Q^r(i) ↓ for all i ∈ . If not, thenr = x. If yes, check whether Q^r codes a minimalL-level containingr. If not, thenr=x. If yes, check whether Q^r is<_L-minimal with this property, using the usual strategy. If not, thenr = x.

Otherwise, useQ^r (and the halting problem solver forP) to check whether there is any realy <L xsuch thatP^y ↓= 1. If that is the case, thenr=x. If it isn’t, thenr

is<L-minimal withP^r ↓= 1 and hencer=x.

§6. More on gaps.

Definition6.1(See[1]). A strong substantial gap is an ordinal interval [α, ] with > αsuch that every∈[α, ] is an index and such thatL−Lαcontains no recognizable reals, whileL+1−L andLα−Ldo contain recognizable reals for every < α. A weak substantial gap is an ordinal interval [α, ] such thatαis an index, the set of indices in that interval is unbounded inand such thatL+1−Lα

contains no recognizable reals.

Note that it is shown in [1] that strong substantial gaps exist (it suffices to see that there is some index ∈Onwith (L₊₁−L)∩RECOG = ∅). We can now show that gaps in the recognizable reals must be infinite. The same reasoning in fact supports much stronger conclusions, as we shall presently see. A minor technical issue arises in the following arguments due to the fact that we consider levels of the L-hierarchy with only very weak closure properties, so that we cannot exclude the possibility that some real coding such a level (or its ordinal height) will actually be contained in it as an element, while our arguments need codes arising only at a later stage. We hence make the following definition:

Definition6.2. If < 1, we denote bycc(L) the<L-minimal real codingL

which is not an element ofL.

Theorem6.3. There are no strong substantial gaps of finite length. Furthermore, strong gaps always start with limit ordinals.

Proof.Assume for a contradiction that there is a strong substantial gap of length i, wherei ∈. Letα∈Onbe minimal such that [α, α+i] is a strong substantial gap of lengthi. It is easy to see thatcc(L_α+i) is recognizable by the usual arguments:

Given x, check whetherx codes an L-level at which a strong substantial gap of lengthi ends. This can be done by Lemma 2.4. The minimality ofxcan then be checked by the routines for evaluating truth predicates described in the last section of [6]. there. By the results on the computational strength ofITRMs, one readily obtains that from the<L-minimal codecofLαwhich is not an element ofLα, we can computecc(Lα+i), say by programP. Note that, by the definition of a substantial strong gap. we will havecc(Lα+i)∈Lα+i+1. But this allows us to recognizec: Given the oraclex, first check (applying Lemma 2.3 toP) whetherP^xcomputescc(Lα+i) - which is possible ascc(Lα+i) is recognizable. Now, incc(Lα+i),c is represented by some integerj. It hence only remains to see whetherxis the number represented byjincc(L_α+i), which is also easy to do.

This implies thatc is recognizable. But, by definition,c ∈ L_α+1−L_α. Hence (L_α+i−L_α)∩RECOG=∅, which contradicts the assumption thatαstarts a gap.

To see that, ifαstarts a strong substantial gap,αhas to be a limit ordinal, we proceed as follows: Assume for a contradiction thatαstarts a strong substantial gap and α = + 1. Since α starts the gap, Lα−L contains a recognizable realr. We argue thatcc(Lα) ∈Lα+1−Lαis recognizable, which contradicts the assumption that α starts a gap. A procedure for recognizing cc(Lα) works as follows: Given x, simply check whetherx is the <L-minimal code of a minimal L-level containingrwhich is not contained in that level. This is possible sinceris

recognizable.

Theorem6.4. Ifαstarts a weak substantial gap[α, ], then ≥^CK,cc(α). Proof.Assume thatαstarts a weak substantial gap [α, ] where < ^CK,cc(α), so that < _i^CK,cc(α) for some minimal i ∈ . By definition, α is an index, so that cc(α) ∈ Lα+1. By definition of cc, we may assume that cc(α) ∈/ Lα. We now want to argue thatcc(α)∈RECOG, which will be a contradiction to the assumption thatα starts a gap. Fromcc(α), one can computecc(L

^CK,cc(α)_i ) by Theorem 2.8. LetPbe anITRM-program computingcc(L

^CK,cc(α)_i ) in the oracle cc(α). Since i ∈ is a fixed natural number, we can use i together with P to determine, for an arbitrary oraclex, whetherP^x computes a<_L-minimal code for LCK,x

i . We can hence also compute the<_L-minimal code forL₊₁ in the oracle cc(α), using program P, say. By our assumption that ends the gap, we must haveRECOG∩(L₊₁−L)=∅; sayr ∈RECOG ∩(L₊₁−L), and letQbe a program for recognizingr. Now, givenxin the oracle, we can determine whether P^xcomputes the minimal code for anL-level containing a realzsuch thatQ^z ↓= 1.

(This can be achieved by searching through the coded structure; sinceris recognized by Q, the calculation Q^z will terminate for all realsz from the coded structure.)