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# Advanced Topics in Foundations MATH 873

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This 20 page Class Notes was uploaded by Zechariah Hilpert on Thursday September 17, 2015. The Class Notes belongs to MATH 873 at University of Wisconsin - Madison taught by Arnold Miller in Fall. Since its upload, it has received 32 views. For similar materials see /class/205274/math-873-university-of-wisconsin-madison in Mathematics (M) at University of Wisconsin - Madison.

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Date Created: 09/17/15

In nite Ramsey Theory Math 873 Fall 1996 AMiller 1 Ramsey s Theorem Let u 0 1 2 and for any set A and n S to let 1A1 B AilBln where lBl is the cardinality of the set B So for example 411 is the set of all in nite subsets of to Theorem 11 Pidgeon Hole Principall Suppose f u a h Then there em39sts H E wlw such that f H is constant Theorem 12 Ramsey s Theorem for any mk lt u and f talk a m there em39sts H E 411 such that f l is constant proof The set H is said to be homogeneous for the function f We begin with the standard proof for h 2 Construct a0 lt a1 lt lt owl and Xn E wlw as follows Let an minXn1 and nd Xn E Xn1 an w so that for every a 3 an and uo E Xn fa7u fa7v39 We can construct such a set by iteratively applying the pidgeon hole principal as follows Note given any a E u and Y E wlw there is a Z 6 Ylw and 239 lt m such that for every 2 E Z we have faz 239 Now we just iterate this Xn1 20 221 2Z2 QZWXL taking care of all a 3 an so m an 1 Finally consider the set K an n E to It is tail homogeneous 1e for any uow distinct elements of K if u lt o and u lt w then fuo 1One of my colleagues told me about the pidgeon head principal If you get into an elevator and there are more buttons pressed than there are people in the elevator then there must be a pidgeon headi fuw Thus we can de ne g K a m by gu fuv for any 1 gt u in K By the pidgeon hole principle there exists H E Klw such that g l H is constant and therefore f l H2 is constant Instead of giving the standard proof for k gt 2 for novelty we give a proof2 using model theory Let f wlkfl a m be any function Consider the model Qt wv lt7 fv new By applying the compactness theorem we can nd a model B which is a proper elementary extension of 2L This means it contains a hyper nite integer H7 ie7 an element of the model B satisfying n lt H for every n E w We construct a sequence 10 lt 11 lt lt an in u with the following property For any ul lt uz lt uk 3 1W1 we have that fu1 who fault7 uk This can be done7 because if we de ne z ulmuk lt m by fBu17quot397uk7Hiu1muk then B i x gt an fun utz u1ltmltuk an71 namely z H so by elementarity Qt l Hm gt an M fu1 7mm z ulmuk u1ltmltuk an71 so now choose an to be any such z E w Now our set K an n E w is tail homogeneous7 ie7 given any ul ltu2lt ltuk andvwgtuk iane have fltu17 lm fltu17 lm since both are equal to fBu1 uk As before7 we de ne 9 i liq 1 m m by WhyWM fu177uk7H on which 9 is constant and then and apply induction to g to get H E K f l Hlki l is constant I 2This proof was found by my fellow graduate student Charlie Gray circa 1975 and is based on the idea of Simpson s model theoretic proof of the Erdos Rado Theoremi Corollary 13 th39te Ramsey Theorem For arty m7k7h lt u there em39sts N lt w such that for every f le a m there em39sts H 6 NW such that f i ts cortstartt proof Suppose there is no such N and let fN N for each N De ne k a m be a counterexample h1 9 i w a m by 9017 1277ak7b fbalazak where a1 lt a2 lt lt ak lt b By Ramsey7s Theorem7 there exists H E wlw such that g i Hlk1 is constant But if a1 lt a2 lt lt ah lt b are any h 1 elements of H then a1 7ah is a homogeneous set for fb a contradiction There is another proof which works by invoking the compactness theorem I Ramsey7s theorem is not a corollary of its nite version This follows from the fact that there is a recursive partition with no recursive homogeneous set7 a result due to Specker See7 for example7 Simpson 11and Jockusch 2 GalvinPrikry Theorem Let U C wlw be arbitrary but xed In what follows lower case st letters will refer to nite subsets of w and upper case letters XY to in nite subsets of w Given 5 E wltw and Y E wlw de ne o stdY iff s Q Y and maxs lt minY s7 o 5Y X E w stdX Q YU s7 0 Y accepts 5 iff 5Y Q U 0 Y rejects 5 iff HEX E Ylw X accepts 5 Proposition 21 If Y accepts rejects s artd X E Yw thert X accepts rejects 5 Proposition 22 Given arty Y artd 5 there em39sts X E Ylw such that either X accepts 5 or X rejects s Proposition 23 Giueri ariy Y arid 51 52 5 there epists X E Ylw such that for each i 17 7it either X accepts 51 or X rejects 5i proof lterate7 ie7 construct YY02Y12Y22YnX so that K either accepts 51 or rejects 51 Proposition 24 Giueri ariy Y E wlw there epists Z 6 Ylw such that for every 5 E Zlt either Z rejects s or Z accepts 5 proof Construct a0 lt a1 lt lt an rninYn with YY02Y12Y22m so that for every n and s Q at 7an we have that Yn1 either rejects or accepts 5 Let Z an n lt w lf 5 E Zlt 7 then let an rnaxs By construction Yn1 accepts or rejects 5 But 5 Z 57Yn1 since am 6 Yn1 for all m 2 n 1 It follows that if Yn1 accepts 57 then Z accepts 5 and if Yn1 rejects 57 then Z rejects 5 Call such a Z as in Proposition 24 decisive Proposition 25 Suppose Z is decisiue arid Z rejects the empty set Theri there epists Y E Z such that Y rejects s for every 5 E Ylt In order to prove this proposition we have the following clairn Claim Suppose Z is decisive and Z rejects t for every t Q 5 Then for all but nitely many it E Z7 for every t Q s U Z rejects t proof Suppose not Then there are in nitely many it E Z such that for some 5 C s U we have that Z accepts 5 Since Z rejects all subsets of s it must be that 5 tn U for some tn Q 5 Since 5 is a nite set there must be Y E Z and t Q s such that for every n E Y we have that Z accepts tU and hence t U nZ Q U Without loss we may assume maxs lt minY But in UtUnY n 6 Y and t U nY Q t U ii7 Z Q L1 This means that Y accepts t But Z rejects t Contradiction I To prove the proposition construct 0070177an71 E Z inductively so that Z rejects every t Q 104117 7cowl We can get started because Z rejects the empty set Let Y 104117 I De ne o M Q wlw is Ramsey iff for every X E wlw there exists Y E Xlw such that either Ylw Q U or Ylw u Q 0 U Q wlw is Completely Ramsey iff for every X E wlw and s E wltw there exists Y E Xlw such that either 5Y Q U or 5Y u 0 Since Ylw MY it is clear that Completely Ramsey implies Ramsey The usual topology on wlw is the topology it inherits by being considered as follows wlw Q Pw E 2w A basic open set in the usual or product topology has the form7 57w The Ellentuck Topology has as its basic open sets those of the form 5 X where s E wlw and X E 44 To be a basis as opposed to subbasis it is neccessary that the intersection of any two basic open sets is a union of basic open sets Proposition 26 Suppose 57X and t7Y are two basic open sets Then eitheiquot they are disjoint 07quot 5X m m sUtX Y proof If neither stdt or tdes then they are disjoint since then there are no Z with stdZ and tdeZ So suppose stdt Now if it is not true that t s Q X then again they are disjoint since t Q Z for every Z 6 tY Finally if X Y is nite then they are disjoint If none of these things happen then s U t t t s Q X and both sides can described as the set of in nite Z such that tdeZ and Z t Q X Y I Lemma 27 flI is open in the Ellehtueh Topology thehlj is Ramsey proof Given X E wlw apply the Proposition 24 and nd Z 6 Xlw such Z is decisive Now if Z accepts the empty set then le 0Z Q U and we are done If Z rejects the empty set then apply Proposition 25 and obtain Y E Z such that Y rejects all of its nite subsets To nish the proof it is enough to prove the following Claim Ylw ll Q proof If not there is some Z 6 Ylw u In the Ellentuck topology the sets of the form Z 71 Z form a neighborhood basis for Z Since U is open it must be that for some n Z6 2mg go But this means that Z accepts Z n contradicting the fact that Y rejects Z n I Lemma 28 flI is open in the Ellehtueh Topology then M is Completely Ramsey proof We can either say just do the whole proof over again but start with 5 Y instead of Y or we can use the following argument Fix 5 and Y with maxs lt minY Let h wlw a 5Y be de ned as follows Let Y yn n lt to be written in increasing order For each X E wlw let hX s U yn n E X It is easy to check that h is a 6 homeomorphism in the Ellentuck topology7 in fact7 it takes basic open sets to basic open sets Let V h 1l Then V is open and hence by Lemma 27 it is Ramsey7 and so there exists H E wlw such that either le Q V or le V 0 Let Z yn n E Then either 5Z Q U or pZl ll 0 Lemma 29 The Completely Ramsey sets form arr algebra le afamz39ly of sets closed under takmg compliments and taking countable um39ohs proof It is easy to see that the compliment of a Completely Ramsey set is Com pletely Ramsey It also easy to see that the union of two Completely Ramsey sets is Completely Ramsey So it suf ces to prove that the countable union of an increasing union of Completely Ramsey sets is Completely Ramsey Let U Lyman be an increasing union of Completely Ramsey sets We begin by showing that U is Ramsey The acceptance rejection terminology is with respect to a xed background set V so we write modulo V77 to indicate which one For any X E wlw there exists Y E Xlw such that either 0 Y accepts the empty set modulo l1 and so Ylw Q U or 0 Y rejects all of its nite subsets modulo M In the rst case7 we are done7 so we must analize the second case If Y rejects s modulo M then since an is smaller it must also reject s modulo Mn But an is Completely Ramsey7 so there must be Z E Y such that 5Z ln 0 Lemma 210 Suppose for every m39te s E Y and Z 6 Y there epists W E Z such that 5Z ln 0 Then there epists Z 6 Ylw such that 2 UTKWUW Q proof Construct a0 lt a1 lt lt alkl lt an minYn with YY02Y12 Apply the hypothesis to obtain Yn1 E Ynlw with the property that an lt minYn1 and so that for every 5 Q 107017 7an 57Yn1l an Now let Z an n E to We claim that Z UnwlIn Q If not7 there exists W E Z Mn for some 71 Since the Mn are increasing7 we may by choosing n larger7 Mn is an increasing sequence assume that an E W But then letting s W an 1 W E 57 g l57Yn1l contradicting the fact that 57Yn1l an I This proves that the countable union of Completely Ramsey sets is Ram sey The proof that it is Completely Ramsey is the same but done by rela tivizing the entire argument to a xed 5 Y Corollary 211 Galvin P7quotth The Borel subsets of wlw are Ramsey le for any Borel set E Q wlw there em39sts an H E wlw such that either le Q B or le B Q In fact the Borel subsets of wlw m the Ellehtuek topology are Completely Ramsey Ramsey7s Theorem is also a corollary of the Galvin Prikry Theorem Given f 44 a 2 de ne B Q wlw by B X E wlw f1n 0 where x17 an is the rst 71 elements X Then B is Borel in fact clopen and a homogeneous set for it7 is homogeneous for f 3 Rosenthal s Theorem A sequence of subsets of a set X ltAn n E wgt converges i for any x E X we have that z E An for all but nitely many 71 or x An for all but nitely 8 many n This is the same as saying that the characteristic functions are pointwise convergent A sequence of sets7 ltAn n E wgt is independent iff given any two disjoint nite subsets of u s and t the set H An N An n69 net is nonempty N An is the compliment of An in the set X Theorem 31 Rosenthal Given ltAn n E wgt sequence of subsets of a set X there eists a C E w such that either ltAn n E Cgt is convergent 0r ltAn n E Cgt is independent proof This proof was found by Farahat see also Lindenstrauss and Tzafriri 5 page 100 De ne Q Q wlw by Y E Q iff Y yo lt yl lt and for every n E w A0 A1 A2 A3 m A2n1 A2n7 That is7 we take the compliment of every other one Then Q is a closed set Therefore7 by the Galvin Prikry Theorem7 there exists an in nite H Q in such that either 1 HF Q Q or 2 Hw Q In the second case take 0 H Then it must be that ltAn n E Hgt is convergent7 otherwise we could nd z E X and an in nite subsequence of H7 say K ho lt k1 lt 7 with z 6 Am and z AWL for each n7 but then K 6 Q7 contradiction In the rst case7 le Q Q7 let H hn n lt w and take 0 h2n1n lt w Then ltAn n E Cgt is independent This is because7 given any disjoint s and t in 0 we can nd K 6 H by lling in on the even ones as needed so that sQh2nnltw andtQh2n1nltw But since K E Q this implies A mm 197 n69 net 4 Ellentuck s Theorem Silver 10 generalized the Galvin Prikry Theorem Thm 211 by showing the 2 sets are Ramsey in fact Completely Ramsey His proof was meta mathematical Ellentuck 1 gave a more general and simplier proof of Silver7s Theorem In order to state Ellentuck7s result we begin by reviewing the notion of Property of Baire or Baire Property Let X be any topological space De ne o N Q X is nowhere dense iff its closure has no interior Or equivalently for any nonempty basic open set U there exists a nonempty basic open set V Q Usuch that V N o M Q X is meager iff M is the countable union of nowhere dense subsets of X Meager is also refer to as rst category77 in X o G Q X is comeager iff X G is meager o B Q X has the Property of Baire iff there exists an open set U and a meager set M such that B UAM where UAM U M U M U is the symmetric di ference lt equivalent to say BAU M Note that a set is nowhere dense iff its closure is nowhere dense Thus any meager set can be covered by a meager Ff7 ie7 a countable union of closed nowhere dense sets Also any subset of a nowhere dense set is nowhere dense7 and hence the meager subsets of X form a U ideal One terminology for sets with the Baire Property77 is to refer to them as sets which are almost open Another equivalent de nition is the following a set E Q X has the Baire property iff there exists a comeager set G and an open set U such that B G U G Thus when we restrict B to a comeager subset7 it is open 10 Theorem 41 Batre The sets with Property of Batre m X form a o algebra 239e if B has the Property of Batre thert so does X B artd if En n E wgt each have the Batre property thert so does UMWBn proof If U is any open set let V be the interior of XU and note that XUUV is nowhere dense This is true because for any nonempty open W either W meets U and so W U Q W is a nonempty open set missing X U U V or W is disjoint from U and hence contained in V7 the interior of the closed set X U7 and so W already misses X U U Suppose B G U G where U is open and G is comeager Then XB GV G where V is the interior of X U and G G U U V is comeager lf Bn Gn Un Gn where each Gn is comeager and Un open7 then UerBn nQuGrl UerUrL nltan39 Theorem 42 Ellerttueh A set E Q wlw ts Completely Ramsey t it has the Batre Property in the Ellerttueh topology proof This will follow easily from the following lemma Lemma 43 Arty meager set in the Ellerttueh topology is rtowhere dehse Before proving the Lemma let us deduce from it7 Theorem 42 Suppose that B Q wlw is Completely Ramsey Let U UsX 57X Q B ie7 U is the interior of B in the Ellentuck topology To see7 that B has the property of Baire7 it is enough to show that 8V1 is nowhere dense Since U is open it is Completely Ramsey7 hence 8V1 is Completely Ramsey Given any 57A there exists B E A such that either 57B Q 8V1 or 57B BZ Q 11 But the rst cannot happen by the de nition of U so the second 5B 8V1 0 must happen And since 5A is an arbitrary basic open set it follows that 8V1 is nowhere dense Conversely suppose B has the property of Baire in the Ellentuck topology Then there exists an open set U and meager set M such that BALI Q M By Lemma 43 There is a closed nowhere dense C with M Q C Now since open sets are Completely Ramsey for any 5X there exists Y E X such that 5Y Q U or 5Y u 0 Since closed sets are Completely Ramsey there exists Z 6 Ylw such that either 5Z Q C or 5Z C Q But the rst doesn7t happen because C is nowhere dense It follows that 5Z Q B or 5 Z 78 Q according to what was true for U And so 8 is Completely Ramsey Proof of Lemma 43 This proof is somewhat analogous to the proof that the countable union of Completely Ramsey sets is Completely Ramsey Suppose that 9 for n E w are nowhere dense in the Ellentuck topology Since their closures are also nowhere dense and the nite union of nowhere dense set is nowhere dense we may assume without loss that they are closed and increasing Since they are closed each 9 is completely Ramsey Now given any A E w we can construct a sequence 10 lt 11 lt lt an1 lt minAn as follows Let A0 A Given a0lta1 lt ltanminAn nd An Q An with the property that for every 5 Q 10 an 5An1 9 0 Now consider B an n lt w We claim that Br m wag If not there exists 0 and n with an E C and C E Q But if sC a0an then C E 5An1 contradicting the fact that 5An1 9 0 By relativiZing this argument to any 5 A it follows that Unltw 9 is nowhere dense and the lemma is proved I 5 Souslin Operation The Souslin operation is the following Given AS s 6 Low de ne SltAS s E wltwgt U nltwAfW few This is the Souslin operation also called operation A Given a family of sets A de ne SA SKAS s E wltwgt AS s 6 Low Q A Typically A is the family of all closed subsets of a topological space X and then SA is known as the family of Souslin sets Theorem 51 Let A be an arbitrary family of sets Then 1 A Q SA 2 8A is closed under countable unions ie Xn E SA for each n implies Unltw Xn E 5A 3 SA is closed under countable intersections and 4 55A 5A proof The rst item is obvious since if AS A for all s then A SltAS s E wlt gt For countable unions note that if Xk Uf6W nltwA W then U Xi U nltwaln kltw feww where BS Ajitg SW for each s E Lon For countable intersections a coding argument is required Let ltn mgt E u be a bijective map between LUZ and to satisfying i ltj implies ltkigt lt ltkjgt Given Xk Uf6W nltwA m then X U mMWBf kltw few 13 where B5 A where t and k are determined by s as follows Let s 507 sn then 71 km and t t0tm where tz skz for each 239 S m This encoding is based on the idea Vk Eu 3f 6 w 902k i af 6w Vk e w6fkk where M fltk7ngt The fact that SSA 5A is left to the reader7 see 8 for example I Another way to de ne the Souslin sets is in terms of the projection oper ation For B Q U gtlt X let pronB x E X Eu 6 U um E B For B a family of subsets of U gtlt X let PTOjXB pTOjXB 3 B E B For X a topological space let clX be the family of closed subsets of X Theorem 52 For my topological space X pr0jx01ww x X sltc1ltXgtgt proof If B Uf6W nltw0 n where each 05 is closed in X7 then let CfxVnltw xEC n wwgtltX Then G is closed and B pronC Suppose B pronC where C Q o gtlt X is closed For each s E wltw de ne AS closurex E X 3f 2 5 x 6 We claim that B U HMWAM few 14 If x E B pronC then for some f E u we have that fa E C and therefore z E Afr for each n E u Contrarywise if z E SAS s 6 Low then for some g E u for each m lt w x closurep X3ng m fx EC But this implies that gz E 0 since 0 is closed and therefore gx 0 would imply that there exists an m and open U Q X with 9796 6 l9 lml X U and g l m gtlt U disjoint from C contradiction I The Borel subsets of X BorelX is the smallest o algebra containing the open subsets of X Theorem 53 Suppose X is a topological space and every closed subset of X is a countable intersection of open sets More generally assume every open set is in SclX Then the following classes are all the same 1 SclX 2 SBorelX 3 projxltc1ltw x X 4 pronltBore1ltw x X proof Obviously 1 implies 2 3 implies 4 and we already know 1 and 3 are equivalent The fact that 2 implies 4 is proved similarly to Theorem 52 So we only need to see that 4 implies 3 Let Y u gtlt X Then BorelY Q SclY This is true because SclY is closed under count able union and countable intersections and contains the closed sets So it is enough to see that every open subset of Y is in SclY But because to has countable base every open subset of Y is a countable union of open rectangles ie set of the form U gtlt V where U Q to is open and V Q X is open It follows that pronBorelw gtlt X Q pronprojwnxXclw gtlt u gtlt X 15 pronclw gtlt w gtlt X pronclww gtlt I This class of sets also includes pronBorelZ gtlt X for all suf ciently nice Z For more on projective sets see Miller Marczewski proved a general theorem which gives a suf cient condition under which a o eld is closed under the Souslin operation A o eld B of subsets of a set X is a family of sets closed under complimentation and countable unions A o ideal I in B is a subfamily of B which is closed under countable unions and taking subsets7 ie7 if Z 6 I and W Q Z then W E I Theorem 54 Marezewshi See Kuratowshi Suppose the o eld B on the set X and a o idealI in B satisfy the following minimal covering prop erty For every Y Q X there epists B E B such that Y Q B and for every068 ifYQCQB thenBC I Then 88 B ie B is closed under the Souslin operation proof Suppose AS 6 B and A U mWA n few For each s E wltw de ne A9 U mWA n ngeww Note 1 A Altgt where is the empty sequence7 2 A9 Q A97 3 A9 UWAM Apply the minimal cover assumption to all As to get BS 2 A5 with BS 6 B a minimal cover Using 2 we may as well assume BS Q As By using 3 we may replace BSA with BSA B5 so without loss we may assume BSA Q BS for each s and 71 Now note that since As UnltwASAn g UnltwBsm C B9 we have that by the minimal cover property that B9 UnltwBsWLgt E To nish the proof since A Altgt Altgt Q B0 and BO 6 B it is enough to show BO Altgt E I Bltgt V10 Bltgt U nltwAfln Q Bltgt U nltwaW few feww This follows from the assumption that BS Q AS so we are subtracking off a smaller set BO U nltwaln Q U B9 nltwBs n39 few sewltw Since the last set is in I we are done This inclusion is true because if z not in Useww BS nltwBsAn then whenever z 6 BS there is an n lt in such that z E BEA Hence if z 6 BO but not in USEUKABS nltwBsAn we can construct f E w such that z is in nltwaW I The two main examples for which Marczewski7s result holds are 1 B is the family of sets with the property of Baire in some topological space X and I is the ideal of meager sets and 2 I is a ccc ideal in B ie there does not an uncountable disjoint family of sets in B I The second property holds for example when 8 is the family of measurable sets and I is the U ideal of measure zero sets where M is some nite countably additive measure on X Theorem 55 IfI is a 000 ideal in B then they satisfy mihimal cover prop erty see hypothesis of Theorem 54 17 proof Given Y Q X arbitrary7 try to construct disjoint Ba as follows Given Ba 04 lt 6 consider if there exists B E B I such that 1 B A and 2 B Ba forallozlt If there is such a B let B be any such If there is no such B stop the construction Since I is ccc the construction must eventually stop at some countable stage 60 lt m Let B X U04lt oBDt39 Then B is a minimal cover of A7 because if A Q C Q B and C E B then if B 0 I it would be a candidate for B o which however never got de ned I The obvious generalization of the above proof is to h elds and H ideals and the K chain condition Theorem 56 UK is the family of sets with the property of Baire in some topological space X aridI the meager ideal theri they satisfy miriimal cover property as in Theorem 54 proof Claim Suppose U is family of open sets and for every U E L we have that Y U is meager Then UM Y is meager proof First assume the family U is pairwise disjoint Then for each U E U there would exists NE nowhere dense so that Y m U UMng But because the U E U are disjoint it is easy to check that each Nn UUelNg is nowhere dense Given an arbitrary family of open sets l1 let V be a maximal family of pairwise disjoint open sets which re nes l1 ie7 for every V E V there exists 18 U E U with V Q U So we know by the rst case that UV Y is meager But the maximality assumtion implies that UM UV is nowhere dense hence UM Y Q UM UV u UV Y is meager This proves the Claim Now to prove the theorem let Y Q X be arbitrary and put U U U open and U Y meager Let B X UU U UU Y Then B has the property of Baire since it is the union of a closed set and a meager set Also Y Q B If Y Q C Q B and C has the Baire property then so does B 0 So let BoltUM1gtuM2 where U is open and M1M2 are meager Since B C is disjoint from Y U Y Q M1 Hence U E M It follows that U B Q UM Y and therefore that U is meager and so B C is meager This result would follow trivially from the ccc case if X had a countable base but in the case we are interested in the Ellentuck topology the space is not second countable Corollary 57 Silver 10 Every 2 set is Ramsey References 1 EEllentuck A new proof that analytic sets are Ramsey J Symbolic Logic 391974163 165 2 FGalvin and KPrikry Borel sets and Ramsey7s theorem J Symbolic Logic 197338 193 198 l3l E E l H 2 UN CJockusch Ramsey7s theorem and recursion theory Journal of Sym bolic Logic 371972 268 280 KKuratowski Topology vol 1 Academic Press 1966 JLindenstrauss and L Tzafriri Classical Banach Spaces I Se quence Spaces Springer Verlag 1977 AMiller Projective subsets of separable metric space Annals of Pure and Appliced Logic 501990 53 69 FPRamsey On a Problem of Formal Logic Proceedings of the London Mathematical Society 301930 264 286 CARogers JEJayne K analytic sets in Analytic Sets edited by CARogers Academic Press 1980 1 181 HRosenthal A characterization of Banach spaces containing l1 Pro ceedings of the National Academy of Sciences USA 711974 2411 2413 JSilver Every analytic set is Ramsey Journal of Symbolic Logic 351970 60 64 SSimpson Reverse Mathematics in Recursion Theory Proceed ings of Symposia in Pure Mathematics edited by ANerode and RShore American Mathematical Society 421982 461 472 20

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