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1 Addendum 2 1 0 2 A Cantor-Bendixson-like process which detects ∆∆∆0 2 n a SAMUELALEXANDER J 3 2 Abstract: ] O ForeachsubsetofBairespace,wedefine,inawaysimilartoacommonproofofthe L Cantor-BendixsonTheorem,asequenceofdecreasingsubsetsSαofN<N,indexed byordinals. We usethistoobtaintwonewcharacterizationsoftheboldface ∆∆∆0 . 2 h Borel pointclass. ADDENDUM: In January 2012 we learned that the notion of t a guessabilityappearedinanequivalentform,andevenwiththesamename,inthe m doctoraldissertationofWilliamWadge[4]. Asforthemainresultofthispaper, [ Wadge proved one direction and gave a proof for the other direction which he 2 attributed to Hausdorff. The proofs in this paper present an alternate means to v thoseresults. 0 7 2000MathematicsSubjectClassification03E15(primary);28A05(secondary) 4 2 Keywords: descriptivesettheory,Cantor-Bendixson,Borelhierarchy,guessability . 6 0 1 Please read the addendum in the above abstract for an important note on this paper’s 1 unoriginality. : v i The usual Cantor-Bendixson derivative “detects” countability, in the sense that the X perfect kernel of S ⊆ NN (the result of applying the Cantor-Bendixson derivative r a repeatedly until a fixed point is reached) is empty if and only if S is countable ([3], page 34). In this paper, I will show a process which “detects” ∆∆∆0: a process which 2 depends on S ⊆ NN andwhichreaches afixedpoint orkernel, akernel whichwillbe emptyifandonlyif S is ∆∆∆0. 2 Definition 1 Suppose S ⊆ NN. If X ⊆ N<N, let [X] denotethesetofinfinite sequenceswhoseinitialsegmentsareallinX. • Define Sα ⊆ N<N foreveryordinal α byinductionasfollows: S0 = N<N, S = ∩ S foranylimitordinalλ.Andfinally,foranyordinalβ define λ β<λ β Sβ+1 = {x ∈ Sβ : ∃x′,x′′ ∈ [Sβ] suchthatx⊆ x′,x ⊆ x′′,x′ ∈ S,andx′′ 6∈ S} • Letα(S) betheminimalordinalα suchthatSα = Sα+1. 2 SamuelAlexander • LetS∞ = Sα(S) (thekerneloftheaboveprocess). Throughout thepaper, S willdenote asubset of NN. If f : N → N and n ∈ N,Iwill use f ↾ n to denote (f(0),...,f(n−1)); f ↾ 0 will denote the empty sequence. My goalistoprovethatthefollowingareequivalent: • S is ∆∆∆0. 2 • S∞ =∅. • S =T\[T∞] forsome T. The reader might wonder why I define S to lie in N<N, rather than in NN as one α might expect by extrapolating from the classical Cantor-Bendixson derivative. Why notdefineanewderivative S∗ = {f ∈ S : f isalimitofpointsof S\{f} andalsoof Sc}, and then follow Cantor-Bendixson more directly? If we do this, we end up getting a kernel which does not detect ∆∆∆0. For example, let S = {f : ∀nf(n) 6= 0}, a ∆∆∆0 2 2 subset of NN. It’s easy tosee S∗ = S, whereas inorder for ourprocess todetect ∆∆∆0, 2 wewouldlike forittoreduce S to ∅. Thereader cancheck that S = N<N, S isthe 0 1 setoffinitesequences notcontaining 0,and S = ∅. 2 Definition2 SaythatS ⊆ NN isguessableifthereisafunctionG : N<N → N such thatforeveryf : N → N, 1,iff ∈ S; lim G(f ↾ n) = n→∞ (cid:26) 0,otherwise. Ifso,wesayG isaguesserforS. Theorem3 AsubsetofNN isguessableifandonlyifitis∆∆∆0. 2 This theorem is proved on page 11 of Alexander [1]. It is also a special case of the maintheorem ofAlexander[2]. Proposition4 SupposeS is∆∆∆0.ThenS = ∅. 2 ∞ Proof Contrapositively,suppose S 6= ∅. Iwillshow S isnon-guessable,hencenon- ∞ ∆∆∆0 byTheorem3. Assumenot,andlet G :N<N → Nbeaguesserfor S. Iwillbuilda 2 sequenceonwhoseinitialsegments Gdiverges,contrarytoDefinition2. Thereissome σ0 ∈ S∞. Now inductively suppose I’ve defined finite sequences σ0 ⊂6= ··· ⊂6= σk in S∞ such that for 0 < i ≤ k, G(σi) ≡ imod2. Since σk ∈ S∞ = Sα(S) = Sα(S)+1, ACantor-Bendixson-likeprocesswhichdetects ∆∆∆0 3 2 this means there are σ′,σ′′ ∈ [S ], extending σ , with σ′ ∈ S, σ′′ 6∈ S. Choose ∞ k σ ∈ {σ′,σ′′} with σ ∈ S iff k is even. Then lim G(σ ↾ n) ≡ k+1mod2. Let n→∞ σk+1 ⊂ σ properly extend σk such that G(σk+1) ≡ k+1mod2. Note σk+1 ∈ S∞ since σ ∈ [S ]. ∞ By induction, I’ve defined σ0 ⊂6= σ1 ⊂6= ··· such that for i > 0, G(σi) ≡ imod2. Thiscontradicts Definition2since lim G((∪σ)↾ n) oughttoconverge. n→∞ i i Fortheconverse weneedmoremachinery. Definition 5 Ifσ ∈ N<N,σ 6∈ S ,thenletβ(σ) denotetheleastordinalsuchthat ∞ σ 6∈ S . β(σ) Notethatwhenever σ 6∈ S , β(σ) isasuccessor ordinal. ∞ Lemma 6 Suppose σ ⊆ τ arefinitesequences. Ifτ ∈ S thenσ ∈ S . Andif ∞ ∞ σ 6∈ S ,thenβ(τ) ≤ β(σ). ∞ Proof It’s enough to show for any ordinal β if τ ∈ S then σ ∈ S . This is by β β inductionon β,thelimitcaseand β = 0 casebeingtrivial. Assume β issuccessor. If τ ∈ S ,thismeans τ ∈ S andthereare τ′,τ′′ ∈ [S ] extending τ with τ′ ∈ S, β β−1 β−1 τ′′ 6∈ S. Since τ′ and τ′′ extend τ, and τ extends σ, τ′ and τ′′ extend σ, and since σ ∈ S byinduction, thisshows σ ∈ S . β−1 β Lemma 7 Suppose f : N → N, f 6∈ [S ]. Thenthereissomei suchthatforall ∞ j≥ i,f ↾ j 6∈ S andβ(f ↾ j) = β(f ↾ i).Furthermore,f ∈ [S ]. ∞ β(f↾i)−1 Proof The first part of the lemma follows from Lemma 6 and the well-foundedness of ORD. For the second part we must show f ↾ k ∈ S for every k. If k ≤ i, β(f↾i)−1 then f ↾ k ∈ S by Lemma 6. If k ≥ i, then β(f ↾ k) = β(f ↾ i) and so β(f↾i)−1 f ↾ k ∈ S sinceitisin S bydefinitionof β. β(f↾i)−1 β(f↾k)−1 Proposition8 IfS = ∅ thenS is∆∆∆0. ∞ 2 Proof Assume S = ∅. I will define a function G : N<N → N which guesses S, ∞ whichissufficientbyTheorem3. Let σ ∈ N<N, we have σ 6∈ S (since S = ∅) and so σ ∈ S \S . Since ∞ ∞ β(σ)−1 β(σ) σ 6∈ S ,thismeansthatforeverytwoextensions σ′,σ′′ of σ in [S ],either σ′ β(σ) β(σ)−1 4 SamuelAlexander and σ′′ are both in S, or both are outside S. It might be that there is no extension of σ lying in [S ]. In that case, arbitrarily define G(σ) = 0. But if there are such β(σ)−1 extensions, let G(σ) = 1 ifallofthose extensions arein S, andlet G(σ) = 0 ifallof thoseextensions areoutside S. I claim G guesses S. To see this, let f ∈ S. I will show G(f ↾ n) → 1 as n → ∞. Since f 6∈ [S ], let i be as in Lemma 7. I claim G(f ↾ j) = 1 whenever j ≥ i. Fix ∞ j ≥ i. We have β(f ↾ j) = β(f ↾ i) by choice of i, and f ∈ [S ] = [S ]. β(f↾i)−1 β(f↾j)−1 Bythepreviousparagraph,ifanyinfinitesequenceextends f ↾ j andliesin [S ], β(f↾j)−1 then either all such sequences are in S, or all are outside S. One such sequence is f, anditisinside S,andtherefore,allsuchsequencesareinside S,whereby G(f ↾ j)= 1 asdesired. Identical reasoning shows that if f 6∈ S then lim G(f ↾ n) = 0. So G guesses S, n→∞ S isguessable, andbyTheorem3, S is ∆∆∆0. 2 Theorem9 S is∆∆∆0 ifandonlyifS = ∅. 2 ∞ Proof Bycombining Propositions 4and8. Wewillclosebygivingonemorecharacterization of ∆∆∆0. 2 Theorem10 S is∆∆∆0 ifandonlyifS = T\[T ] forsomeT ⊆ NN. 2 ∞ Proof ByTheorem9,if S is ∆∆∆0 then S = S\[S ]. Fortheconverse,itsufficestolet 2 ∞ S bearbitrary andprove S\[S ] is ∆∆∆0. ∞ 2 By Theorem 3, it is enough to exhibit a guesser G : N<N → N for S\[S ]. Let ∞ σ ∈ N<N. If σ ∈S ,let G(σ) = 0. Otherwise,if σ hasatleastoneinfiniteextension ∞ in [S ], and all such extensions are also in S, then let G(σ) = 1. In any other β(σ)−1 case,let G(σ) = 0. Weclaim G guesses S\[S ]. ∞ Case1: f ∈S\[S ]. ByLemma7,findan i suchthatforall j ≥ i wehave f ↾ j 6∈ S ∞ ∞ and β(f ↾ j) = β(f ↾ i) and f ∈ [S ]. Thus for any j ≥ i, f ↾ j does have one β(f↾i)−1 extension in [S ],namely f itself, and f isin S. Allothersuchextensions must β(f↾j)−1 also be in S, or else we would have f ↾ j ∈ S , violating the definition of β. So β(f↾j) G(f ↾ j)= 1,showingthat G(f ↾ n) → 1 as n → ∞. Case2: f ∈ [S ]. Thenforevery n, f ↾ n ∈ S andthusbydefinition G(f ↾ n) = 0. ∞ ∞ Case 3: f 6∈ S and f 6∈ [S ]. As in Case 1, find i such that for all j ≥ i, f ↾ j 6∈ S ∞ ∞ and β(f ↾ j) = β(f ↾ i), and f ∈ [S ]. Forany j ≥ i, f ↾ j hasone extension in β(f↾j)−1 [S ],namely f,and f 6∈ S;sobydefinition G(f ↾ j) = 0. β(f↾j)−1 ACantor-Bendixson-likeprocesswhichdetects ∆∆∆0 5 2 References [1] SamuelAlexander(2011).OnGuessingWhethera Sequencehasa Certain Property. TheJournalofIntegerSequences14. [2] Samuel Alexander (preprint). Highly lopsided information and the Borel hierarchy. arXiv: 1106.0196 [3] AlexanderKechris(1995).Classicaldescriptivesettheory.Springer. [4] William Wadge (1983). Reducibility and Determinateness on the Baire Space. PhD thesis,U.C.Berkeley. 231W.18thAvenue,Columbus,OH43210 [email protected] http://www.math.osu.edu/ alexander ~ Received: aabb20YY Revised: ccdd20ZZ