A Cantor-Bendixson-like process which detects Delta_2^0

0.05 MB·English

by  Samuel Alexander

1 Addendum
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2 A Cantor-Bendixson-like process which detects ∆∆∆0
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Abstract:

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O ForeachsubsetofBairespace,wedefine,inawaysimilartoacommonproofofthe
L Cantor-BendixsonTheorem,asequenceofdecreasingsubsetsSαofN<N,indexed
byordinals. We usethistoobtaintwonewcharacterizationsoftheboldface ∆∆∆0
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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.
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2000MathematicsSubjectClassification03E15(primary);28A05(secondary)
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Keywords: descriptivesettheory,Cantor-Bendixson,Borelhierarchy,guessability
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1 Please read the addendum in the above abstract for an important note on this paper’s
1 unoriginality.
:
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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
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depends on S ⊆ NN andwhichreaches afixedpoint orkernel, akernel whichwillbe
emptyifandonlyif S is ∆∆∆0.
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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.
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• 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.
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• 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
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subset of NN. It’s easy tosee S∗ = S, whereas inorder for ourprocess todetect ∆∆∆0,
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wewouldlike forittoreduce S to ∅. Thereader cancheck that S = N<N, S isthe
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setoffinitesequences notcontaining 0,and S = ∅.
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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.
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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 = ∅.
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Proof Contrapositively,suppose S 6= ∅. Iwillshow S isnon-guessable,hencenon-
∞
∆∆∆0 byTheorem3. Assumenot,andlet G :N<N → Nbeaguesserfor S. Iwillbuilda
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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
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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
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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.
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Theorem9 S is∆∆∆0 ifandonlyifS = ∅.
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Proof Bycombining Propositions 4and8.
Wewillclosebygivingonemorecharacterization of ∆∆∆0.
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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
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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
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