Guessing, Mind-changing, and the Second Ambiguous Class

0.12 MB·

by  Samuel Alexander

SubmittedonSeptember3,2013totheNotreDameJournalofFormalLogic
Volume??,Number??,
4
1
0 Guessing, Mind-changing, and the Second Ambiguous
2
Class

n
a
J

9
Samuel Alexander

]
O
L Abstract Inhisdissertation, Wadge defined anotionof guessability on sub-
. setsoftheBairespaceandgavetwocharacterizationsofguessablesets.Asetis
h
t guessableiffitisinthesecondambiguousclass(DDD 02),iffitiseventuallyannihi-
a latedbyacertainremainder. Wesimplifythisremainderandgiveanewproof
m
ofthelatterequivalence. Wethenintroduceanotionofguessingwithanordinal
[ limitonhowoftenonecanchangeone’smind. Weshowthatforeveryordinal
a ,aguessablesetisannihilatedbya applicationsofthesimplifiedremainderif
1
andonlyifitisguessablewithfewerthana mindchanges. Weuseguessability
v
withfewerthana mindchangestogiveasemi-characterizationoftheHausdorff
4
9 difference hierarchy, and indicate how Wadge’s notion of guessability can be
8 generalized to higher-order guessability, providing characterizations of DDD 0a for
1 allsuccessorordinalsa >1.
.
1
0 1 Introduction
4
1 Let NN be the set of sequences s:N→N and let N<N be the set ∪nNn of finite
: sequences. Ifs∈N<N, wewillwrite[s] for{f ∈NN : f extendss}. We equipNN
v
with a second-countabletopologyby declaring[s] to be a basic openset whenever
i
X s∈N<N.
r Throughoutthe paper, S will denote a subset of NN. We say that S∈DDD 0 if S is
a 2
simultaneouslyacountableintersectionofopensetsandacountableunionofclosed
setsintheabovetopology. Inclassicterminology,S∈DDD 02 justincase SisbothGd
andFs .
ThefollowingnotionwasdiscoveredbyWadge[9] (pp.141–142)andindepen-
dentlybythisauthor[1]. 1
2010MathematicsSubjectClassification:Primary03E15
Keywords:guessability,differencehierarchy
1
2 S.Alexander
Definition1.1 WesaySisguessableifthereisafunctionG:N<N→{0,1}such
thatforevery f ∈NN,
1,if f ∈S,
limG(f ↾n)=c (f)=
n→¥ S (cid:26) 0,if f 6∈S.
Ifso,wesayGguessesS,orthatGisanS-guesser.
The intution behind the above notion is captured eloquently by Wadge (p. 142,
notationchanged):
Guessingsetsallowustoformanopinionastowhetheranelement f ofNN is
inSorSc,givenonlyafiniteinitialsegment f ↾nof f.
Game theoretically, one envisions an asymmetric game where II (the guesser) has
perfectinformation,I(thesequencechooser)haszeroinformation,andII’swinning
setconsistsofallsequences(a ,b ,a ,b ,...)suchthatb →1if(a ,a ,...)∈Sand
0 0 1 1 i 0 1
b →0otherwise.
i
Thefollowingresultwasprovedin[9](pp.144–145)byinfinitegame-theoretical
methods. The present author found a second proof [1] using mathematical logical
methods.
Theorem1.2 (Wadge)SisguessableifandonlyifS∈DDD 0.
2
Wadgedefined(pp.113–114)thefollowingremainderoperation.
Definition 1.3 For A,B⊆NN, define Rm0(A,B)=NN. For m >0 an ordinal,
define
Rmm (A,B)= Rmn (A,B)∩A∩Rmn (A,B)∩B .
n\<m (cid:16) (cid:17)
(Here•denotestopologicalclosure.)WriteRmm (S)forRmm (S,Sc).
By countability considerations, there is some (in fact countable) ordinal m , de-
pendingonS,such thatRmm (S)=Rmm ′(S)forall m ′ ≥m ; WadgewritesRmW (S)
forRmm (S)forsucham . Hethenprovesthefollowingtheorem:
Theorem1.4 (Wadge,attributedtoHausdorff)S∈DDD 02ifandonlyifRmW (S)=0/.
InSection2, we introducea simplerremainder(S,a )7→Sa anduseitto givea
newproofofTheorem1.4.
InSection3,weintroducethenotionofSbeingguessablewhilechangingone’s
mindfewerthana manytimes(a ∈Ord)andshowthatthisisequivalenttoSa =0/.
InSection4, weshowthatfora >0,Sisguessablewhilechangingone’smind
fewerthana +1manytimesifandonlyifatleastoneofSorSc isinthea thlevel
ofthedifferencehierarchy.
In Section 5, we generalize guessability, introducing the notion of m th-order
guessability (1≤ m <w ). We show that S is m th-order guessable if and only if
1
S∈DDD 0m +1.
2 GuessableSetsandRemainders
InthissectionwegiveanewproofofTheorem1.4. We finditeasiertoworkwith
thefollowingremainder2whichiscloselyrelatedtotheremainderdefinedbyWadge.
ForX ⊆N<N,wewillwrite[X]todenotethesetofinfinitesequencesallofwhose
finiteinitialsegmentslieinX.
GuessingandMind-changing 3
Definition2.1 LetS⊆NN. WedefineSa ⊆N<N (a ∈Ord)bytransfiniterecur-
sionasfollows. We defineS0=N<N, andSl =∩b <l Sb foreverylimitordinall .
Finally,foreveryordinalb ,wedefine
Sb +1={x∈Sb : ∃x′,x′′∈[Sb ]suchthatx⊆x′,x⊆x′′,x′∈S,x′′6∈S}.
Wewritea (S)fortheminimalordinala suchthatSa =Sa +1,andwewriteS¥ for
Sa (S).
ClearlySa ⊆Sb wheneverb <a . ThisremaindernotionisrelatedtoWadge’sas
follows.
Lemma2.2 Foreachordinala ,Rma (S)=[Sa ].
Proof Since Sa ⊆ Sb whenever b < a , for all a , we have Sa = ∩b <a Sb +1
(with the convention that ∩0/ = N<N). We will show by induction on a that
Rma (S)=[Sa ]=[∩b <a Sb +1].
Suppose f ∈[∩b <a Sb +1]. Let b < a . Let U be an open set around f, we
can assume U is basic open, so U =[f ], f a finite initial segment of f. Since
0 0
f ∈[∩b <a Sb +1], f0 ∈Sb +1. Thus there are x′,x′′ ∈ [Sb ] extending f0 (hence in
U), x′ ∈S, x′′ 6∈S. In otherwords, x′ ∈[∩g<b Sg+1]∩S and x′′ ∈[∩g<b Sg+1]∩Sc.
By induction, x′ ∈ Rmb (S)∩S and x′′ ∈ Rmb (S)∩Sc. By arbitrariness of U,
f ∈Rmb (S)∩S∩Rmb (S)∩Sc. Byarbitrarinessofb , f ∈Rma (S).
Thereverseinclusionissimilar.
NotethatLemma2.2doesnotsaythatRma (S)=0/ ifandonlyif Sa =0/. Itis(at
leastapriori)possiblethatSa 6=0/ while[Sa ]=0/. Lemma2.2doeshoweverimply
that RmW (S)=0/ if and only if S¥ =0/, since it is easy to see that if [Sa ]=0/ then
Sa +1=0/. ThusinordertoproveTheorem1.4itsufficestoshowthatSisguessable
ifandonlyifS¥ =0/. The⇒directionrequiresnoadditionalmachinery.
Proposition2.3 IfSisguessablethenS¥ =0/.
Proof Let G : N<N → {0,1} be an S-guesser. Assume (for contradiction)
S¥ 6= 0/ and let s 0 ∈ S¥ . We will build a sequence on whose initial segments
G diverges, contrary to Definition 1.1. Inductively suppose we have finite se-
quences s 0 ⊂6= ··· ⊂6= s k in S¥ such that ∀0 < i ≤ k, G(s i) ≡ i mod2. Since
s k∈S¥ =Sa (S)=Sa (S)+1,thereares ′,s ′′∈[S¥ ],extendings k,withs ′∈S,s ′′6∈S.
Chooses ∈{s ′,s ′′}withs ∈Siffkiseven.Thenlimn→¥ G(s ↾n)≡k+1mod2.
Lets k+1⊂s properlyextends k suchthatG(s k+1)≡k+1mod2. Notes k+1∈S¥
sinces ∈[S¥ ].
By induction, there are s ⊂ s ⊂ ··· such that for i>0, G(s )≡imod2.
0 6= 1 6= i
ThiscontradictsDefinition1.1sincelimn→¥ G((∪is i)↾n)oughttoconverge.
The⇐directionrequiresalittlemachinery.
Definition 2.4 If s ∈ N<N, s 6∈ S¥ , let b (s ) be the least ordinal such that
s 6∈Sb (s ).
Notethatwhenevers 6∈S¥ ,b (s )isasuccessorordinal.
Lemma2.5 Supposes ⊆t are finitesequences. Ift ∈S¥ thens ∈S¥ . And if
s 6∈S¥ ,thenb (t )≤b (s ).
4 S.Alexander
Proof It is enough to show that ∀b ∈Ord, if t ∈Sb then s ∈Sb . This is by
induction on b , the limit and zero cases being trivial. Assume b is successor. If
t ∈Sb , this meanst ∈Sb −1 and there are t ′,t ′′ ∈[Sb −1] extendingt with t ′ ∈S,
t ′′ 6∈S. Since t ′ and t ′′ extendt , and t extendss , t ′ and t ′′ extends ; and since
s ∈Sb −1(byinduction),thisshowss ∈Sb .
Lemma2.6 Suppose f :N→N, f 6∈[S¥ ]. Thereissomeisuchthatforall j≥i,
f ↾ j6∈S¥ andb (f ↾ j)=b (f ↾i). Furthermore, f ∈[Sb (f↾i)−1].
Proof The first part follows from Lemma 2.5 and the well-foundedness of
Ord. For the second part we must show f ↾k ∈Sb (f↾i)−1 for every k. If k ≤i,
then f ↾k ∈Sb (f↾i)−1 by Lemma 2.5. If k ≥i, then b (f ↾k)= b (f ↾ i) and so
f ↾k∈Sb (f↾i)−1sinceitisinSb (f↾k)−1bydefinitionofb .
Definition 2.7 If S¥ = 0/ then we define GS : N<N → {0,1} as follows. Let
s ∈N<N. Since S¥ =0/, s 6∈S¥ , so s ∈Sb (s )−1\Sb (s ). Since s 6∈Sb (s ), this
meansforeverytwoextensionsx′,x′′ofs in[Sb (s )−1],eitherx′,x′′∈Sorx′,x′′∈Sc.
Soeitherallextensionsofs in[Sb (s )−1]areinS,orallsuchextensionsareinSc.
(i) If there are no extensions of s in [Sb (s )−1], and length(s ) > 0, then let
G (s )=G (s −)wheres − isobtainedfroms byremovingthelastterm.
S S
(ii) Iftherearenoextensionsofs in[Sb (s )−1],andlength(s )=0,letGS(s )=0.
(iii) If there are extensions of s in [Sb (s )−1] and they are all in S, define
G (s )=1.
S
(iv) If there are extensions of s in [Sb (s )−1] and they are all in Sc, define
G (s )=0.
S
Proposition2.8 IfS¥ =0/ thenGS guessesS.
Proof Assume S¥ =0/. Let f ∈S. I will show GS(f ↾n)→1 as n→¥ . Since
f 6∈[S¥ ],letibeasinLemma2.6. IclaimGS(f ↾ j)=1whenever j≥i. Fix j≥i.
Wehaveb (f ↾ j)=b (f ↾i)bychoiceofi,and f ∈[Sb (f↾i)−1]=[Sb (f↾j)−1]. Since
f ↾ jhasoneextension(namely f itself)inboth[Sb (f↾j)−1]andS,GS(f ↾ j)=1.
Identicalreasoningshowsthatif f 6∈Sthenlimn→¥ GS(f ↾n)=0.
Theorem2.9 S∈DDD 02 ifandonlyifS¥ =0/. Thatis,Theorem1.4istrue.
Proof BycombiningPropositions2.3and2.8andTheorem1.2.
3 Guessingwithoutchangingone’sMindtoooften
In this section our goal is to tease out additional information about DDD 0 from the
2
operationdefinedinDefinition2.1.
Definition3.1 ForeachfunctionGwithdomainN<N,ifG(f ↾(n+1))6=G(f ↾n)
(f ∈NN, n∈N), wesayGchangesitsmindon f ↾(n+1). Nowleta ∈Ord. We
say S is guessable with <a mind changesif there is an S-guesser G along with a
functionH:N<N→a suchthatthefollowinghold,where f ∈NNandn∈N.
(i) H(f ↾(n+1))≤H(f ↾n).
(ii) IfGchangesitsmindon f ↾(n+1),thenH(f ↾(n+1))<H(f ↾n).
ThisnotionbearssomeresemblancetothenotionofasetZ⊆Nbeing f-c.e.in
[4],org-c.a.in[7].
GuessingandMind-changing 5
Theorem3.2 Fora ∈Ord, Sisguessablewith<a mindchangesifandonlyif
Sa =0/.
Proof
(⇒) Assume S is guessable with <a mind changes. Let G,H be as in Definition
3.1. We claimthatforallb ∈Ord,ifs ∈Sb thenH(s )≥b . Thiswillprove(⇒)
becauseitimpliesthatifSa 6=0/ thenthereissomes withH(s )≥a ,absurdsince
codomain(H)=a .
Weattacktheclaimbyinductiononb . Thezeroandlimitcasesaretrivial. As-
sumeb =g +1.Supposes ∈Sg+1.Therearex′,x′′∈[Sg]extendings ,x′∈S,x′′6∈S.
Pickx∈{x′,x′′}sothatc (x)6=G(s )andpicks +∈N<Nwiths ⊆s +⊆xsuchthat
S
G(s +)=c S(x)(somesuchs + existssinceGguessesS). Sincex∈[Sg], s +∈Sg.
Byinduction,H(s +)≥g . ThefactG(s +)6=G(s )impliesH(s +)<H(s ),forcing
H(s )≥g +1.
(⇐) Assume Sa =0/. For all s ∈N<N, define H(s )=b (s )−1 (by definition of
b (s ), since Sa =0/, clearly H(s )∈a ). I claim GS,H witness that S is guessable
with<a mindchanges.
By Proposition 2.8, G guesses S. Let f ∈ NN, n ∈ N. By Lemma 2.5,
S
H(f ↾ (n+1)) ≤ H(f ↾ n). Now suppose G changes its mind on f ↾ (n+1),
S
we must show H(f ↾(n+1))<H(f ↾n). Assume, for sake of contradiction,that
H(f ↾(n+1))=H(f ↾n). Assume G (f ↾n)=0, the other case is similar. By
S
definitionofGS, (∗)foreveryinfiniteextension f′ of f ↾n, if f′ ∈[Sb (f↾n)−1] then
f′ ∈Sc. SinceG changesitsmindon f ↾(n+1), G (f ↾(n+1))=1. Thus(∗∗)
S S
for every infinite extension f′′ of f ↾(n+1), if f′′ ∈[Sb (f↾(n+1))−1] then f′′ ∈S.
And f ↾(n+1)doesactuallyhavesomesuchinfiniteextension f′′,becauseifithad
none,thatwouldmakeG (f ↾(n+1))=G (f ↾n)bycase1ofthedefinitionofG
S S S
(Definition2.7). Beinganextensionof f ↾(n+1), f′′ alsoextends f ↾n;andbythe
assumptionthatH(f ↾(n+1))=H(f ↾n), f′′∈[Sb (f↾n)−1]. By(∗), f′′∈Sc,andby
(∗∗), f′′∈S. Absurd.
It is not hard to show S is a Boolean combination of open sets if and only if S is
guessablewith<w mindchanges,soTheorem3.2andLemma2.2giveanewproof
ofaspecialcaseofthemaintheorem(p.1348)of[3](seealso[2]).
4 MindChangingandtheDifferenceHierarchy
Werecallthefollowingdefinitionfrom[5](p.175,statedingreatergenerality—we
specializeittotheBairespace). Inthisdefinition,SSS 0(NN)isthesetofopensubsets
1
ofNN,andtheparityofanordinalh istheequivalenceclassmodulo2ofn,where
h =l +n,l alimitordinal(orl =0),n∈N.
Definition 4.1 Let (Ah )h <q be an increasing sequence of subsets of NN with
q ≥1. DefinethesetDq ((Ah )h <q )⊆NNby
x∈Dq ((Ah )h <q ) ⇔ x∈ Ah & the least h <q with x∈Ah has parity
h[<q
oppositetothatofq .
Let
Dq (SSS 01)(NN)={Dq ((Ah )h <q ) : Ah ∈SSS 01(NN),h <q }.
6 S.Alexander
ThishierarchyoffersaconstructivecharacterizationofDDD 0: itturnsoutthat
2
DDD 02=∪1≤q <w 1Dq (SSS 01)(NN)
(seeTheorem22.27of[5],p.176,attributedtoHausdorffandKuratowski).
Forbrevity,wewillwriteDa forDa (SSS 01)(NN).
Theorem4.2 (Semi-characterizationofthedifferencehierarchy)Leta >0. The
followingareequivalent.
(i) Sisguessablewith<a +1mindchanges.
(ii) S∈Da orSc∈Da .
WewillproveTheorem4.2byasequenceofsmallerresults.
Definition4.3 Fora ,b ∈Ord,writea ≡b toindicatethata andb havethesame
parity(thatis,2|n−m,wherea =l +nandb =k +m,n,m∈N,l alimitordinal
or0,k alimitordinalor0).
Proposition4.4 Let a >0. If S∈Da , say S=Da ((Ah )h <a )(Ah ⊆NN open),
thenSisguessablewith<a +1mindchanges.
Proof Define G : N<N → {0,1} and H : N<N → a +1 as follows. Suppose
s ∈N<N. Ifthereisnoh <a suchthat[s ]⊆Ah ,letG(s )=0andletH(s )=a .
Ifthereisanh <a (wemaytakeh minimal)suchthat[s ]⊆Ah ,thenlet
0, ifh ≡a ;
G(s )= H(s )=h .
(cid:26) 1, ifh 6≡a ,
Let f :N→N.
Claim1 limn→¥ G(f ↾n)=c S(f).
If f 6∈∪h <a Ah , then f 6∈Da ((Ah )h <a )=S, and G(f ↾n) will always be 0, so
limn→¥ G(f ↾n)=0=c S(f). Assume f ∈∪h <a Ah ,andleth <a beminimumsuch
that f ∈Ah . SinceAh isopen,thereissomen0 solargethat∀n≥n0,[f ↾n]⊆Ah .
Foralln≥n0,byminimalityofh ,[f ↾n]6⊆Ah ′ foranyh ′<h ,soG(f ↾n)=0if
andonlyifh ≡a . Thefollowingareequivalent.
f ∈Siff f ∈Da ((Ah )h <a )
iffh 6≡a
iffG(f ↾n)6=0
iffG(f ↾n)=1.
Thisshowslimn→¥ G(f ↾n)=c S(f).
Claim2 ∀n∈N,H(f ↾(n+1))≤H(f ↾n).
IfH(f ↾n)=a , thereis nothingtoprove. If H(f ↾n)<a , thenH(f ↾n)=h
where h is minimal such that [f ↾n]⊆Ah . Since [f ↾(n+1)]⊆[f ↾n], we have
[f ↾(n+1)]⊆Ah ,implyingH(f ↾(n+1))≤h .
Claim3 ∀n∈N,ifG(f ↾(n+1))6=G(f ↾n),thenH(f ↾(n+1))<H(f ↾n).
Assume (for sake of contradiction) H(f ↾ (n+1)) ≥ H(f ↾ n). By Claim
2, H(f ↾ (n+1)) = H(f ↾ n). By definition of H this implies that ∀h < a ,
[f ↾(n+1)]⊆Ah ifandonlyif[f ↾n]⊆Ah . ThisimpliesG(f ↾(n+1))=G(f ↾n),
contradiction.
ByClaims1–3,GandHwitnessthatSisguessablewith<a +1mindchanges.
GuessingandMind-changing 7
Corollary4.5 Leta >0. IfS∈Da orSc∈Da thenSisguessablewith<a +1
mindchanges.
Proof IfS∈Da thisisimmediatebyProposition4.4. IfSc∈Da thenProposition
4.4saysSc isguessablewith<a +1mindchanges,andthisclearlyimpliesthatS
istoo.
Lemma4.6 SupposeSisguessablewith<a mindchanges.LetG:N<N→{0,1},
H :N<N→a be a pair offunctionswitnessing asmuch (Definition3.1). There is
an H′ : N<N → a such that G,H′ also witness that S is guessable with < a
mind changes, with H′(0/)=H(0/), and with the additionalproperty that for every
f :N→Nandeveryn∈N,
H(f ↾(n+1))≡H(f ↾n)ifandonlyifG(f ↾(n+1))=G(f ↾n).
Proof DefineH′(s )byinductiononthelengthofs asfollows.LetH′(0/)=H(0/).
If s 6= 0/, write s = s ⌢ n for some n ∈ N (⌢ denotes concatenation). If
0
G(s ) = G(s ), let H′(s ) = H′(s ). Otherwise, let H′(s ) be either H(s ) or
0 0
H(s )+1,whicheverhasparityoppositetoH′(s ).
0
By construction H′ has the desired parity properties. A simple inductive argu-
mentshowsthat(∗)∀s ∈N<N,H(s )≤H′(s )<a . Iclaimthatforall f :N→N
and n ∈ N, H′(f ↾ (n+1)) ≤ H′(f ↾ n), and if G(f ↾ (n+1)) 6= G(f ↾ n) then
H′(f ↾(n+1))<H′(f ↾n).
IfG(f ↾(n+1))=G(f ↾n),thenbydefinitionH′(f ↾(n+1))=H′(f ↾n)andthe
claimistrivial.NowassumeG(f ↾(n+1))6=G(f↾n).IfH′(f ↾(n+1))=H(f↾(n+1))
thenH′(f ↾(n+1))<H(f ↾n)≤H′(f ↾n)andwearedone.Assume
H′(f ↾(n+1))6=H(f ↾(n+1)),
whichforcesthat(∗∗)H′(f ↾(n+1))=H(f ↾(n+1))+1. Toseethat
H′(f ↾(n+1))<H′(f ↾n),
assumenot(∗∗∗).ByDefinition3.1,H(f ↾(n+1))<H(f ↾n),so
H(f ↾n)≥H(f ↾(n+1))+1 (Basicarithmetic)
=H′(f ↾(n+1)) (By(∗∗))
≥H′(f ↾n) (By(∗∗∗))
≥H(f ↾n). (By(∗))
Equalityholdsthroughout,andH′(f ↾(n+1))=H′(f ↾n).Contradiction:wechose
H′(f ↾(n+1))withparityoppositetoH′(f ↾n).
Definition 4.7 For all G,H as in Definition 3.1, f ∈ NN, write G(f) for
limn→¥ G(f ↾n) (so G(f) = c S(f)) and write H(f) for limn→¥ H(f ↾ n). Write
G ≡ H to indicate that ∀f ∈ NN, G(f) ≡ H(f); write G 6≡ H to indicate that
∀f ∈NN,G(f)6≡H(f)(wepronounceG6≡H as“GisanticongruenttoH”).
Lemma4.8 SupposeG:N<N→{0,1}andH:N<N→a witnessthatSisguess-
ablewith<a mindchanges.ThereisanH′:N<N→a suchthatG,H′witnessthat
Sisguessablewith<a mindchanges,andsuchthatthefollowinghold.
IfG(0/)≡a thenH′6≡G. IfG(0/)6≡a thenH′≡G.
8 S.Alexander
Proof Iclaimthatwithoutlossofgenerality,wemayassumethefollowing(∗):
IfG(0/)≡a thenH(0/)6≡G(0/). IfG(0/)6≡a thenH(0/)≡G(0/).
To seethis, supposenot: eitherG(0/)≡a andH(0/)≡G(0/), orelse G(0/)6≡a and
H(0/)6≡G(0/).Ineithercase,H(0/)≡a .IfH(0/)≡a thenH(0/)+16=a ,andso,since
H(0/)<a ,H(0/)+1<a ,meaningwemayadd1toH(0/)toenforcetheassumption.
Having assumed (∗), we may use Lemma 4.6 to construct H′ :N<N →a such
thatG,H′witnessthatSisguessablewith<a mindchanges,H′(0/)=H(0/),andH′
changesparitypreciselywhenGchangesparity.Thelatterfacts,combinedwith(∗),
provethelemma.
Proposition4.9 SupposeG:N<N→{0,1}andH:N<N→a +1witnessthatS
isguessablewith<a +1mindchanges.IfG(0/)=0thenS∈Da .
Proof ByLemma4.8wemaysafelyassumethefollowing:
IfG(0/)≡a +1thenH 6≡G. IfG(0/)6≡a +1thenH≡G.
Inotherwords,
(∗)IfG(0/)≡a thenH≡G. (∗∗)IfG(0/)6≡a thenH6≡G.
Foreachh <a ,let
Ah ={f ∈NN : H(f)≤h }. (H(f)asinDefinition4.7)
IclaimS=Da ((Ah )h <a ),whichwillprovethepropositionsinceeachAh isclearly
open.
Suppose f ∈S,Iwillshow f ∈Da ((Ah )h <a ). Since f ∈S,H(f)6=a ,because
if H(f) were =a , this would imply that G never changes its mind on f, forcing
limn→¥ G(f ↾n)=limn→¥ G(0/)=0,contradictingthefactthatGguessesS.
SinceH(f)6=a ,H(f)<a . Itfollowsthatforh =H(f)wehave f ∈Ah andh
isminimalwiththisproperty.
Case 1: G(0/) ≡ a . By (∗), H ≡ G. Since f ∈ S, limn→¥ G(f ↾ n) = 1, so
h = limn→¥ H(f ↾ n) ≡ 1. Since a ≡ G(0/) = 0, this shows h 6≡ a , putting
f ∈Da ((Ah )h <a ).
Case 2: G(0/) 6≡ a . By (∗∗), H 6≡ G. Since f ∈ S, limn→¥ G(f ↾ n) = 1,
so h = limn→¥ H(f ↾ n) ≡ 0. Since a 6≡ G(0/) = 0, this shows h 6≡ a , so
f ∈Da ((Ah )h <a ).
Conversely,suppose f ∈Da ((Ah )h <a ), Iwillshow f ∈S. Leth beminimalsuch
that f ∈Ah (bydefinitionofAh ,h =H(f)). BydefinitionofDa ((Ah )h <a ),h 6≡a .
Case1:G(0/)≡a .By(∗),H≡G.Sincelimn→¥ H(f ↾n)=H(f)=h 6≡a ≡G(0/)=0,
weseelimn→¥ H(f ↾n)=1.SinceH≡G,limn→¥ G(f ↾n)=1,forcing f ∈Ssince
GguessesS.
Case2: G(0/)6≡a . By(∗∗),H6≡G. Since
limH(f ↾n)=H(f)=h 6≡a 6≡G(0/)=0,
n→¥
we see limn→¥ H(f ↾n)=0. Since H 6≡G, limn→¥ G(f ↾n)=1, again showing
f ∈S.
Corollary 4.10 If S is guessable with < a +1 mind changes, then S ∈Da or
Sc∈Da .
GuessingandMind-changing 9
Proof LetG,HwitnessthatSisguessablewith<a +1mindchanges.IfG(0/)=0
thenS∈Da byProposition4.9. Ifnot,then(1−G),H witnessthatSc isguessable
with<a +1mindchanges,and(1−G)(0/)=0,soSc∈Da byProposition4.9.
CombiningCorollaries4.5and4.10provesTheorem4.2.
5 Higher-orderGuessability
Inthissectionweintroduceanotionthatgeneralizesguessabilitytoprovideachar-
acterizationforDDD 0m +1 (1≤m <w 1). WewillshowthatS∈DDD 0m +1 ifandonlyifSis
m th-orderguessable.Throughoutthissection,m denotesanordinalin[1,w ).
1
Definition 5.1 Let S = (S0,S1,...) be a countably infinite tuple of subsets
S ⊆NN.
i
(i) Forevery f ∈NN,writeS(f)forthesequence(c (f),c (f),...)∈{0,1}N.
S0 S1
(ii) WesaythatSisguessablebasedonS ifthereisafunction
G:{0,1}<N→{0,1}
(calledanS-guesserbasedonS)suchthat∀f ∈NN,
limG(S(f)↾n)=c (f).
n→¥ S
Gametheoretically,weenvisionagamewhereI (thesequencechooser)haszero
informationandII (theguesser)haspossiblybetter-than-perfectinformation: II is
allowedtoask (onceperturn)whetherI’ssequencelies invariousS. ForeachS,
i i
playerI’sact(byansweringthequestion)ofcommittingtoplayasequenceinS or
i
inScissimilartotheact(describedin[6],p.366)ofchoosingaI-imposedsubgame.
i
Example5.2 IfS enumeratesthesetsoftheform{f ∈NN : f(i)= j},i,j∈N
thenitisnothardtoshowthatSisguessable(inthesenseofDefinition1.1)ifand
onlyifSisguessablebasedonS.
Definition5.3 WesaySism th-orderguessableifthereissomeS =(S0,S1,...)
asinDefinition5.1suchthatthefollowinghold.
(i) SisguessablebasedonS.
(ii) ∀i,Si∈DDD 0mi+1 forsomem i<m .
Theorem5.4 Sism th-orderguessableifandonlyifS∈DDD 0m +1.
In order to prove Theorem 5.4 we will assume the following result, which is a
specialization and rephrasing of Exercise 22.17 of [5] (pp. 172–173, attributed to
Kuratowski).
Lemma5.5 Thefollowingareequivalent.
(i) S∈DDD 0m +1.
(ii) Thereisasequence(Ai)i∈N,eachAi∈DDD 0mi+1 forsomem i<m ,suchthat
S= A = A .
m m
[n m\≥n \n m[≥n
ProofofTheorem5.4
10 S.Alexander
(⇒) Let S = (S ,S ,...) and G witness that S is m th-order guessable (so each
0 1
Si∈DDD 0mi+1forsomem i<m ). Foralla∈{0,1}andX ⊆NN,define
X, ifa=1;
Xa=(cid:26) NN\X, ifa=0.
For notational convenience, we will write “G(~a) = 1” as an abbreviation for
“0 ≤ a ,...,a ≤ 1 and G(a ,...,a ) = 1,” provided m is clear from con-
0 m−1 0 m−1
text. Observethatforall f ∈NNandm∈N,G(S(f)↾m)=1ifandonlyif
m−1
f ∈ Saj.
j
G([~a)=1 j\=0
Now,given f :N→N, f ∈S ifandonlyifG(S(f)↾n)→1, whichistrueifand
onlyif∃n∀m≥n,G(S(f)↾m)=1. Thus
f ∈Siff∃n∀m≥n,G(S(f)↾m)=1
m−1
iff∃n∀m≥n, f ∈ Saj
j
G([~a)=1 j\=0
m−1
iff f ∈ Saj.
j
[n m\≥nG([~a)=1 j\=0
So
m−1
S= Saj.
j
[n m\≥nG([~a)=1 j\=0
Atthesametime,sinceG(S(f)↾m)→0whenever f 6∈S,wesee f ∈Sifandonly
if∀n∃m≥nsuchthatG(S(f)↾m)=1. Thusbysimilarreasoningtotheabove,
m−1
S= Saj.
j
\n m[≥nG([~a)=1 j\=0
Foreachm, m−1Saj isafiniteunionoffiniteintersectionsofsetsinDDD 0
G(~a)=1 j=0 j m ′+1
forvariousmS′<m ,thTus G(~a)=1 mj=−01Sajj itselfisinDDD 0mm+1forsomem m<m .Letting
Am= G(~a)=1 mj=−01Sajj,SLemmaT5.5saysS∈DDD 0m +1.
(m⇐<)AmSs,ssuumchetShTa∈tDDD 0m +1. ByLemma5.5,thereare(Ai)i∈N,eachAi∈DDD 0mi+1forsome
i
S= A = A . (∗)
m m
[n m\≥n \n m[≥n
IclaimthatSisguessablebasedonS =(A ,A ,...). DefineG:{0,1}<N→{0,1}
0 1
byG(a ,...,a )=a ,IwillshowthatGisanS-guesserbasedonS.
0 m m
Suppose f ∈S.By(∗),∃ns.t.∀m≥n, f ∈A andthusc (f)=1.Forallm≥n,
m Am
G(S(f)↾(m+1))=G(c (f),...,c (f))
A0 Am
=c (f)
Am
=1,
so limn→¥ G(S(f) ↾ n) = 1. A similar argument shows that if f 6∈ S then
limn→¥ G(S(f)↾n)=0.