Separating principles below Ramsey s Theorem for Pairs

Transcription

1 Sparating principls blow Ramsy s Thorm for Pairs Manul Lrman, Rd Solomon, Hnry Towsnr Fbruary 4, Introduction In rcnt yars, thr has bn a substantial amount of work in rvrs mathmatics concrning natural mathmatical principls that ar provabl from RT 2 2, Ramsy s Thorm for Pairs. Ths principls tnd to fall outsid of th big fiv systms of rvrs mathmatics and a complicatd pictur of subsystms blow RT 2 2 has mrgd. In this papr, w answr two opn qustions concrning ths subsystms, spcifically that ADS is not quivalnt to CAC and that EM is not quivalnt to RT 2 2. W bgin with a rviw of th dfinitions and known rsults for th rlvant systms blow RT 2 2, but will assum a gnral familiarity with rvrs mathmatics. W rfr th radr to Simpson [6] for background on rvrs mathmatics and to Hirschfldt and Shor [4] for background on th gnral pictur of subsystms blow RT 2 2. Unlss othrwis spcifid, w always work in th bas thory RCA 0. W will hav ordrings on a varity of structurs, but w typically rsrv th symbols < and for thr contxts: th usual ordr on N, xtnsions of forcing conditions and comparing sts. If F is a finit st and G is a (finit or infinit) st, w writ F < G to dnot max(f ) < min(g). Without loss of gnrality, w assum that th infinit algbraic structurs dfind blow hav domain N. Dfinition 1.1. A 2-coloring of [N] 2 (or simply a coloring), whr [N] 2 dnots th st of all two lmnt substs of N, is a function c : [N] 2 {0, 1}. A st H N is homognous for c if c is constant on [H] 2. (RT 2 2) Ramsy s Thorm for Pairs: Evry 2-coloring of [N] 2 has an infinit homognous st. W rfr to an infinit homognous st for a coloring c as a solution to c. W typically writ c(x, y), as opposd to c({x, y}), with implicit assumption that x < y. Dfinition 1.2. Lt c b a 2-coloring of [N] 2. Dfin A (c) (rspctivly B (c)) to b th st of numbrs which ar colord 0 (rspctivly 1) with all but finitly many othr numbrs. Th coloring c is stabl if A (c) B (c) = N. A (c) = {n x y > x (c(n, y) = 0)} B (c) = {n x y > x (c(n, y) = 1)} 1

2 (SRT 2 2) Stabl Ramsy s Thorm for Pairs: Evry stabl 2-coloring of [N] 2 has an infinit homognous st. Chong, Slaman and Yang [3] hav rcntly shown that SRT 2 2 is strictly wakr than RT 2 2. Dfinition 1.3. Lt M = (N, M ) b a post. For x, y M, w say that x and y ar comparabl if ithr x M y or y M x, and w say x and y ar incomparabl (and writ x M y) if x M y and y M x. S N is a chain in M if for all x, y S, x and y ar comparabl. S is an antichain in M if for all x y S, x and y ar incomparabl. (CAC) Chain-AntiChain: Evry infinit post M contains ithr an infinit chain or an infinit antichain. A solution to an infinit post M is an infinit st S such that S is ithr a chain or an antichain. It is straightforward to show that RT 2 2 CAC by transforming instancs of CAC into instancs of RT 2 2. Givn a partial ordr M = (N, M ), dfin th coloring c M by stting c M (x, y) = 0 if x and y ar comparabl and stting c M (x, y) = 1 othrwis. If H is an infinit homognous st for c M with color 0, thn H is an infinit chain in M. If H is an infinit homognous st with color 1, thn H is an infinit antichain in M. Hirschfldt and Shor [4] showd that on cannot giv a similar transformation of instancs of RT 2 2 into instancs of CAC by showing that CAC SRT 2 2. Dfinition 1.4. Lt M = (N, M ) b an infinit partial ordr. Dfin A (M) = {n x y > x (n M y)} B (M) = {n x y > x (n M y)} C (M) = {n x y > x (y M n)} M is stabl if ithr A (M) B (M) = N or C (M) B (M) = N. (SCAC) Stabl Chain-Antichain: Evry infinit stabl post M contains ithr an infinit chain or an infinit anti chain. Whn w work with SCAC latr, w will construct an infinit post M such that A (M) B (M) = N. Thus, our notations for A (M) and B (M) ar chosn to paralll th corrsponding notations for SRT 2 2. Although SRT 2 2 SCAC by th transformation givn abov, Hirschfldt and Shor [4] showd that SCAC CAC. Dfinition 1.5. Lt L = (N, < L ) b an infinit linar ordr. A function f : N L is an infinit ascnding squnc in L if for all n < m, f(n) < L f(m) and is an infinit dscnding squnc in L if for all n < m, f(n) > L f(m). (ADS) Ascnding or Dscnding Squnc: Evry infinit linar ordr L has an infinit ascnding squnc or an infinit dscnding squnc. Dfinition 1.6. An infinit linar ordr L is stabl if L has ordr typ ω + ω. That is, for vry x, thr is a y such that ithr z > y (x < L z) or z > y (z < L x). 2

3 (SADS) Stabl Ascnding or Dscnding Squnc: Evry infinit stabl linar ordr has an infinit ascnding squnc or an infinit dscnding squnc. A solution to an infinit linar ordr L is a function which is ithr an infinit ascnding squnc or an infinit dscnding squnc. As abov, on can show CAC ADS by transforming instancs of ADS into instancs of CAC. Givn an infinit linar ordr (N, < L ), dfin an infinit partial ordr M = (N, M ) by x M y if and only if x L y and x y. Lt S = {s 0 < s 1 < } b a solution to M and dfin f(n) = s n. If S is a chain in M, thn f is an ascnding chain in L. If S is an antichain in M, thn f is a dscnding chain in L. Hirschfldt and Shor [4] provd that SADS ADS, but lft opn th qustion of whthr ADS implis CAC or SADS implis SCAC. Our first rsult answrs both of ths qustions in th ngativ by sparating ADS from SCAC in an ω-modl. Thorm 1.7. Thr is a Turing idal I P(ω) such that th ω-modl (ω, I) satisfis ADS but not SCAC. Thrfor, ADS dos not imply SCAC. This thorm will b provd in Sction 2. Our scond rsult concrns infinit tournamnts and th Erdös-Mosr Thorm. Dfinition 1.8. A tournamnt T on a domain D N is an irrflxiv binary rlation on D such that for all x y D, xactly on of T (x, y) or T (y, x) holds. T is transitiv if for all x, y, z D, if T (x, y) and T (y, z) hold, thn T (x, z) holds. In kping with our trminology abov, an infinit tournamnt rfrs to a tournamnt T with domain N. An infinit transitiv subtournamnt of T (or a solution to T ) is an infinit st S N such that T rstrictd to domain S is transitiv. Th Erdös-Mosr Principl stats that such solutions always xist. (EM) Erdös-Mosr Principl: Evry infinit tournamnt contains an infinit transitiv subtournamnt. EM follows from RT 2 2 by transforming instancs of EM into instancs of RT 2 2. Lt T b an infinit tournamnt and dfin th coloring c T for x < y by c T (x, y) = 0 if T (x, y) holds and c T (x, y) = 1 if T (y, x) holds. Suppos H is an infinit homognous st for th color 0. Thn, H is transitiv in T bcaus for all x y H, T (x, y) holds if and only if x < y. Similarly, if H is homognous for th color 1, thn H is transitiv in T bcaus for all x y H, T (x, y) holds if and only if x > y. Sinc computabl instancs of RT 2 2 hav Π 0 2 solutions and hav low 2 solutions, it follows from this translation that computabl instancs of EM also hav Π 0 2 solutions and hav low 2 solutions. In Sction 3, w prsnt a proof du to Kach, Lrman, Solomon and Wbr that ths bounds ar bst possibl. Thorm 1.9 (Kach, Lrman, Solomon and Wbr). Thr is a computabl instanc of EM with no 0 2 solution, and hnc no Σ 0 2 solution or low solution. Similar tchniqus wr usd by Dzhafarov, Kach, Lrman and Solomon to diagonaliz against th xistnc of hyprimmun-fr solutions. 3

4 Thorm 1.10 (Dzhafarov, Kach, Lrman and Solomon). Thr is a computabl instanc of EM with no hyprimmun-fr solution. Formalizing Thorm 1.10 in rvrs mathmatics, which can b don in RCA 0 + BΣ 0 2, givs a lowr bound on th strngth of EM. Hirschfldt, Shor and Slaman [5] provd that th following vrsion of th Omitting Typs Thorm, dnotd OPT, is quivalnt to th statmnt that for vry X, thr is a function not dominatd by any X-rcursiv function (i.. thr is a dgr which is hyprimmun rlativ to X). (OPT) Omitting Partial Typs: Lt T b a complt thory and S b a st of partial typs of T. Thr is a modl of T that omits all th nonprincipal typs in S. Hnc, EM implis OPT ovr RCA 0 + BΣ 0 2. It rmains an opn qustion whthr EM implis BΣ 0 2. Bovykin and Wirmann [1] showd that on can transform an instanc c of RT 2 2 into an instanc T c of EM, but that xtracting th solution to c from th solution to T c rquirs an application of ADS. To s why ADS might b usful, notic that if S is a transitiv subtournamnt of an infinit tournamnt T, thn T dfins a linar ordr on S. Thorm 1.11 (Bovykin and Wirmann [1]). EM + ADS implis RT 2 2. Proof. Fix a coloring c : [N] 2 {0, 1}. Dfin an infinit tournamnt T c as follows. T c (x, y) holds if ithr x < y and c(x, y) = 1 or y < x and c(y, x) = 0. Lt S b an infinit transitiv subtournamnt of T c and lt S b th linar ordr on S inducd by T c. By ADS, lt f b an infinit ascnding squnc or an infinit dscnding squnc in (S, S ). By thinning out f, w can assum that f(0) < f(1) < f(2) < and hnc th rang of f xists in RCA 0. Suppos that f is an ascnding squnc in S. Fix n < m. Sinc f(n) < S f(m), th rlation T c (f(n), f(m)) holds. Bcaus f(n) < f(m) and T c (f(n), f(m)) holds, it follows that c(f(n), f(m)) = 1. Thrfor, th rang of f is homognous for c with color 1. Suppos that f is a dscnding squnc in S. Fix n < m. Sinc f(m) < S f(n), th rlation T c (f(m), f(n)) holds. Bcaus f(n) < f(m), it follows that c(f(n), f(m)) = 0. Thrfor, th rang of f is homognous for c with color 0. Corollary CAC dos not prov EM (and hnc ADS dos not prov EM ithr). Proof. Suppos for a contradiction that CAC implis EM. Sinc CAC also provs ADS, it follows from Thorm 1.11 that CAC provs RT 2 2. Howvr, by Hirschfldt and Shor [4], CAC dos not prov RT 2 2. Corollary EM implis RT 2 2 if and only if EM implis ADS. Proof. This follows immdiatly from Thorm 1.11 and th fact that RT 2 2 implis ADS. An infinit tournamnt T is stabl if for all x, thr is a y such that ithr T (x, z) holds for all z > y or T (z, x) holds for all z > y. (SEM) Stabl Erdös-Mosr Principl: Evry infinit stabl tournamnt contains an infinit transitiv subtournamnt. 4

5 Corollary SEM + SADS implis SRT 2 2. Proof. Lt c b a stabl coloring and dfin T c as in Thorm W show that T c is a stabl tournamnt. Fix x. Lt y > x and i {0, 1} b such that c(x, z) = i for all z > y. Suppos that i = 0. For vry z > y, w hav x < z and c(x, z) = 0, and hnc T c (z, x) holds. On th othr hand, suppos i = 1. For all z > y, w hav x < z and c(x, z) = 1, w hav T c (x, z) holds. Thrfor, T c is stabl. By SEM, thr is an infinit transitiv subtournamnt S of T c. Th corollary follows onc w show that th linar ordr inducd by T c on S is stabl. Fix x S. Sinc T c is stabl, thr is a y > x such that ithr T c (x, z) holds for all z > y (and hnc x < S z for all z > y with z S) or T c (z, x) holds for all z > y (and hnc z < S x for all z > y with z S). Thrfor, (S, S ) is a stabl linar ordr and SADS suffics to xtract an infinit ascnding or dscnding chain in S. Our scond rsult, to b provd in Sction 4, is that EM dos not imply SRT 2 2, and hnc th inclusion of ADS in Thorm 1.11 cannot b rmovd. Thorm Thr is a Turing idal I P(ω) such that th ω-modl (ω, I) satisfis EM but not SRT 2 2. Thrfor, EM dos not imply SRT 2 2. Corollary EM dos not imply SADS (and hnc nithr EM nor SEM implis ithr ADS or SADS). Proof. Suppos for a contradiction that EM implis SADS. Sinc EM implis SEM, and SEM + SADS implis SRT 2 2, w hav EM implis SRT 2 2, contradiction Thorm ADS dos not imply SCAC 2.1 Outlin In this sction, w prov Thorm 1.7 by constructing a Turing idal I P(ω) such that (ω, I) ADS and I contains a stabl partial ordr M = (N, M ) but dos not contain a solution to M. Th construction procds in two stps; w us a ground forcing to build M followd by an itratd forcing to add solutions to infinit linar ordrs without adding a solution to M. Rcall that for an infinit post M, A (M) is th st of lmnts which ar blow almost vry lmnt and B (M) is th st of lmnts which ar incomparabl with almost vry lmnt. In th ground forcing, w spcify A (M) and B (M) as w construct M so that A (M) B (M) = N and hnc M is stabl. W satisfy two typs of rquirmnts. First, to nsur that M cannot comput a solution to itslf it suffics to nsur that if Φ M is infinit, thn Φ M (a) = Φ M (b) = 1 for som a A (M) and b B (M). Sinc w ar dfining A (M) and B (M) as w construct M, ths ar asy to satisfy. Scond, w satisfy ground lvl rquirmnts which guarant that rquirmnts for th first lvl of th itration forcing ar appropriatly dns (in a sns dfind blow). 5

6 For th first lvl of th itration forcing, w bgin with M, A (M) and B (M) alrady dfind. W fix an indx such that Φ M is an infinit linar ordr and attmpt to add a solution f for Φ M to I so that M f dos not comput a solution to M. As abov, th stratgy is to show that if Φ M f is infinit, thn thr ar lmnts a A (M) and b B (M) such that Φ M f (a) = Φ M f (b) = 1. Howvr, sinc A (M) and B (M) ar alrady dfind, implmnting this stratgy rquirs using th fact that th ground forcing nsurd that rquirmnts for th itratd forcing ar appropriatly dns. This dnsity will man that as f is dfind, if thr ar lots of options to forc larg numbrs into Φ M f, thn thr must b numbrs from A (M) and B (M) in Φ M f. In addition to handling ths diagonalization stratgis, w nd to guarant that th rquirmnts for th nxt lvl of th itration forcing ar appropriatly dns. In th construction blow, w xplain th itration forcing first (assuming M, A (M) and B (M) hav alrady bn constructd) bcaus it allows us to introduc th dnsity notions that hav to b forcd at th ground lvl. Aftr xplaining th itration forcing, w prsnt th ground forcing to construct M, A (M) and B (M). Bfor starting th construction, w rstrict th collction of infinit linar ordrs for which w nd to add solutions to I. Dfinition 2.1. A linar ordring (N, ) is stabl-ish if thr is a non-mpty initial sgmnt V which has no maximum undr, and such that N \ V is non-mpty and has no minimum undr. Not that thr is no rquirmnt that th st V b computabl from. Lmma 2.2. If (N, ) is not stabl-ish thn thr is a solution to (N, ) computabl from. Proof. Assum (N, ) is not stabl-ish. Not that if V is a non-mpty initial sgmnt with no maximum lmnt, thn V can comput an infinit ascnding squnc. Lt a 1 V b arbitrary. Givn a n, thr must b infinitly many lmnts x V such that a n x, so simply sarch (ffctivly in V ) for such an lmnt and st a n+1 = x. If thr is a non-mpty initial sgmnt V with no maximum, obsrv that sinc is not stabl-ish, ithr N \ V =, in which cas V is computabl, or N \ V has a minimal lmnt b, in which cas V = {x x b}. In ithr cas, V is computabl from, and so thr is an infinit ascnding squnc computabl from. So suppos thr is no such V. Thn vry non-mpty initial sgmnt has a maximum lmnt. Lt V b th st of lmnts with finitly many prdcssors. V is ithr mpty or finit, sinc if V wr infinit, it would not hav a maximal lmnt. Thus N \ V is computabl from, and can hav no minimal lmnt. (Any minimal lmnt would hav only th finitly many lmnts of V as prdcssors, and would thrfor blong to V.) Thrfor, by an argumnt similar to th on abov, N \ V contains an infinit dscnding squnc computabl from. W nd this subsction by fixing som notation and convntions. If σ and δ ar finit strings, thn σ δ dnots th concatnation of σ and δ. W writ σ τ to dnot that σ is an initial sgmnt of τ (i.. τ = σ δ for som string δ). If is a linar ordr on N, σ is a 6

7 finit squnc which is ascnding in and τ is a finit squnc which is dscnding in, thn σ τ mans that σ( σ 1) τ( τ 1) (i.. th last lmnt in σ is strictly blow th last lmnt in τ in th ordr). For any computation in which part of th oracl is a finit string, for xampl Φ X σ k, w follow th standard convntion that if Φ X σ k (y) convrgs, thn both y and th us of th computation ar boundd by σ. 2.2 Itration Forcing Assum that w hav alrady usd th ground forcing to construct our stabl post (M, M ) along with A (M) and B (M). Th gnral contxt for on stp of th itration forcing will b a fixd st X and an indx mting th following conditions: M T X; X dos not comput a solution to M; Φ X is th charactristic function for a stabl-ish linar ordr X on N; and ach rquirmnt K X,A (M),B (M) is uniformly dns (dfind blow). Th ground forcing will crat this contxt for X = M. Our goal is to find a gnric solution G for X (ithr an infinit ascnding or dscnding squnc) such that X G dos not comput a solution to M and such that for ach stabl-ish linar ordr X G, th rquirmnts K X G,A (M),B (M) ar uniformly dns. W add G to th Turing idal and not that for any indx such that X G is a stabl-ish linar ordr, w hav cratd th contxt for th itration forcing to continu with X G. Bfor giving th spcifics of our forcing notion, w dscrib th basic intuition for constructing a solution G for X whil diagonalizing against computing a solution to M from X G. W work with pairs (σ, τ) whr σ is a finit ascnding squnc in X, τ is a finit dscnding squnc in X and σ X τ. W viw this pair as a simultanous attmpt to build an infinit ascnding solution and an infinit dscnding solution to X. Th goal is to construct an infinit nstd squnc of such pairs (σ k, τ k ) such that w succd ithr with G = σ = σ k or with G = τ = τ k. Suppos w hav constructd a pair (σ k, τ k ). A typical diagonalization rquirmnt is spcifid by a pair of indics m and n. To mt this rquirmnt, w nd to ithr find an ascnding squnc σ k+1 xtnding σ k such that σ k+1 X τ k and thr xists a pair of lmnts a A (M), b B (M) such that Φ X σ k+1 m (a) = Φ X σ k+1 m (b) = 1; or find a dscnding squnc τ k+1 xtnding τ k such that σ k X τ k+1 and thr xists a pair of lmnts a A (M), b B (M) such that Φ X τ k+1 n (a) = Φ X τ k+1 n (b) = 1. That is, w xtnd our approximation to an ascnding solution to X in a mannr that diagonalizs or w xtnd our approximation to a dscnding solution to X in a mannr that diagonalizs. If w can always win on th ascnding sid, thn G = σ k is an infinit 7

8 ascnding solution to X such that X G cannot comput a solution to M. Othrwis, thr is an indx m for which w cannot win on th ascnding sid. In this cas, w must win on th dscnding sid for vry indx n (whn it is paird with m) and hnc G = τ k is an appropriat infinit dscnding solution to X. In gnral, thr is no rason to think w can mt ths rquirmnts without som additional information about X. It is th fact that ach rquirmnt K X,A (M),B (M) is uniformly dns which allows us to mt ths rquirmnts. W first focus on formalizing ths diagonalization rquirmnts in a gnral contxt and thn w show why this gnral contxt also forcs th rquirmnts K X G,A (M),B (M) to b uniformly dns at th nxt lvl. W bgin by dfining th following sts, ach computabl from X. A X = {σ σ is a finit ascnding squnc in X } D X = {τ τ is a finit dscnding squnc in X } P X = {(σ, τ) σ A X τ D X σ X τ} P X is our st of forcing conditions. For p P X, w lt σ p and τ p dnot th first and scond componnts of p. For p, q P X, w say q p if σ p σ q and τ p τ q. To dfin th gnric G, w construct a squnc p 0 p 1 p 2 of conditions p n = (σ n, τ n ) P X. At th (n + 1)st stp, w dfin p n+1 p n to mt th highst priority rquirmnt K X,A (M),B (M) which is not yt satisfid. Mting this rquirmnt will mak progrss ithr towards making σ = n σ n our dsird infinit ascnding solution to X or towards making τ = n τ n our dsird infinit dscnding solution to X. In th nd, w show that on of G = σ or G = τ satisfis all th rquirmnts. Bfor dfining th rquirmnts, thr is on obvious worry w nd to addrss. During this procss, w nd to avoid taking a stp which liminats ithr sid from bing xtndibl to a solution of X. Bcaus X is stabl-ish, w fix a st V for X as in Dfinition 2.1. W dfin V X = {(σ, τ) P X σ V τ N \ V }. For (σ, τ) V X, σ is an initial sgmnt of an incrasing solution to X and τ is an initial sgmnt of a dcrasing solution to X. Thrfor, as long as w choos our gnric squnc to li within V X, w will nvr limit ithr sid from bing xtndibl to a solution to X. Howvr, working strictly in V X has th disadvantag that V X is not computabl from X. W rconcil th advantags of working in P X (which is computabl from X) with working in by using split pairs. V X Dfinition 2.3. A split pair blow p = (σ p, τ p ) is a pair of conditions q 0 = (σ p σ, τ p ) and q 1 = (σ p, τ p τ ) such that σ X τ. Lmma 2.4. If p V X and q 0, q 1 is a split pair blow p thn ithr q 0 V X or q 1 V X. Proof. Lt q 0 = (σ p σ, τ p ) and q 1 = (σ p, τ p τ ). Suppos q 0 V. Sinc σ p σ X τ p, it must b that σ ovrflows from V into N \ V. Thrfor, sinc σ X τ, q 1 V. W will us Lmma 2.4 as follows. Each rquirmnt K X,A (M),B (M) will hav th proprty that whn w nd to mt K X,A (M),B (M) blow an lmnt p n in our gnric squnc, thr 8

9 will b a split pair q 0, q 1 (from P X ) blow p n in K X,A (M),B (M). Thrfor, if p n V X by induction, thn w can mt K X,A (M),B (M) within V X by choosing p n+1 to b whichvr of q 0 and q 1 is in V X. Thus, by starting with th mpty squnc p 0 (which is in V X ), w can assum that our gnric squnc is chosn in V X. W hav two typs of rquirmnts: half rquirmnts and full rquirmnts. For uniformity of prsntation, it is asist to dal with a gnral dfinition for th full rquirmnts, although in th nd, th only full rquirmnts w nd to mt ar thos mad up of a pair of half rquirmnts. Dfinition 2.5. W dfin th following typs of rquirmnts and half-rquirmnts. A rquirmnt is a downward closd st K X,A (M),B (M) P X K X,A (M),B (M) = {p P X for som rlation K X (x, y, z) computabl in X. such that a A (M) b B (M) (K X (p, a, b))} which is closd undr xtn- An A-sid half rquirmnt is a st R X,A (M),B (M) A X sions such that R X,A (M),B (M) = {σ A X for som rlation R X (x, y, z) computabl in X. A D-sid half rquirmnt is a st S X,A (M),B (M) D X such that S X,A (M),B (M) = {τ D X for som rlation S X (x, y, z) computabl in X. a A (M) b B (M) (R X (σ, a, b))} which is closd undr xtnsions a A (M) b B (M) (S X (τ, a, b))} If R X,A (M),B (M) is an A-sid half rquirmnt and S X,A (M),B (M) is a D-sid half rquirmnt, thn J X,A (M),B (M) R,S is th rquirmnt J X,A (M),B (M) R,S = {p P X σ p R X,A (M),B (M) τ p S X,A (M),B (M) }. W say R X,A (M),B (M) is a half rquirmnt to man that it is ithr an A-sid or a D-sid half rquirmnt. Each rquirmnt and half rquirmnt is c.. in X A (M) B (M) and th dpndnc on A (M) and B (M) is positiv. Exampl 2.6. Fix a pair of indics m and n. Th formal vrsion of our basic diagonalization stratgy is givn by th following half rquirmnts: A X,A (M),B (M) m = {σ A X a A (M) b B (M) (Φ X σ m n = {τ D X a A (M) b B (M) (Φ X τ n D X,A (M),B (M) Ths half rquirmnts combin to form th rquirmnt J X,A (M),B (M) A m,d n = { p P X σ p A X,A (M),B (M) m Notic that if σ A X,A (M),B (M) m τ D X,A (M),B (M) n and τ G, thn Φ X G and σ G, thn Φ X G m n is not a solution to M. 9 (a) = Φ X σ m (b) = 1)}, (a) = Φ X τ n (b) = 1)}. } τ p D X,A (M),B (M) n. is not a solution to M. Similarly, if

10 W nxt dscrib whn an A-sid half rquirmnt R X,A (M),B (M) is satisfid by an infinit ascnding squnc Λ in X. (With th obvious changs, this dscription applis to a D-sid half rquirmnt S X,A (M),B (M) and an infinit dscnding squnc Λ.) R X,A (M),B (M) is spcifid by an indx i such that R X,A (M),B (M) = {σ A X a A (M) b B (M) (Φ X i (σ, a, b) = 1)} whr Φ X i is total. For any (typically finit) sts A and B (givn by canonical indics), w lt R X,A,B = {σ A X a A b B (Φ X i (σ, a, b) = 1)}. Unlik R X,A (M),B (M), th st R X,A,B is not ncssarily closd undr xtnsions. Howvr, for any finit sts A and B, w hav R X,A,B T X. W writ R X to indicat th opration mapping A, B to R X,A,B. Dfinition 2.7. R X is ssntial in Λ if for vry n and vry x, thr is a finit st A > x such that for vry y, thr is a finit st B > y and an m > n so that Λ m R X,A,B. W say th infinit ascnding squnc Λ satisfis R X,A (M),B (M) if ithr R X is not ssntial in Λ, or thr is an n such that Λ n R X,A (M),B (M). Exampl 2.8. Considr th A-sid diagonalization half rquirmnt A X,A (M),B (M) m = {σ A X a A (M) b B (M) (Φ X σ m (a) = Φ X σ m (b) = 1)} and an infinit ascnding squnc Λ in X. infinit. Thrfor, A X,A (M),B (M) m xists a A (M) and b B (M) such that Φ X Λ m solution to X such that Φm X Λ is not a solution to M. A X m is ssntial in Λ if and only if Φ X Λ m is is finit or thr is satisfid by Λ if and only if ithr Φ X Λ m (a) = Φ X Λ m (b) = 1. In ithr cas, Λ is a This xampl dos not xplain why w nd th quantifir altrnations in Dfinition 2.7. This quantifir altrnation will b rflctd in a similar dfinition for full rquirmnts and th rason for it will bcom clar in th ground forcing. W nd similar notions in th contxt of our (full) rquirmnts. Each rquirmnt K X,A (M),B (M) is spcifid by an indx i such that K X,A (M),B (M) = {p P X a A (M) b B (M) (Φ X i (p, a, b) = 1)} whr Φ X i is total. For any (typically finit) sts A and B, w lt K X,A,B = {p P X a A b B (Φ X i (p, a, b) = 1)}. As abov, th st K X,A,B nd not b downward closd in P X, but is computabl from X whn A and B ar finit. 10

11 Dfinition 2.9. K X is ssntial blow p P X if for vry x, thr is a finit st A > x such that for vry y, thr is a finit st B > y and a split pair q 0, q 1 blow p such that q 0, q 1 K X,A,B. K X,A (M),B (M) is uniformly dns if whnvr K X is ssntial blow p, thr is a split pair q 0, q 1 blow p blonging to K X,A (M),B (M). Exampl Lt J X,A (M),B (M) A m,d n b th rquirmnt from Exampl 2.6 and fix a condition p = (σ p, τ p ). Lt q 0 = (σp σ, τ p ) and q 1 = (σ p, τp τ) b a split pair blow p. For finit sts A and B, q 0 J X,A,B A m,d n if a A b B ( Φ X σ p σ m (a) = Φ X σ p σ m (b) = 1 Φ X τp n (a) = Φn X τp (b) = 1 ). For A > τ p, th scond disjunct cannot occur by our us convntion, and hnc q 0 J X,A,B A m,d n a A b B ( Φ X σ p σ m (a) = Φ X σ p σ m (b) = 1 ). Similarly, if B > σ p, thn q 1 J X,A,B A m,d n a A b B ( Φ X τ p τ n (a) = Φ X τ p τ n (b) = 1 ). Thus th dfinition of JA X m,d n bing ssntial blow p formalizs a notion of having lots of options to forc larg numbrs into a potntial solution to M. Informally, th dfinition of J X,A (M),B (M) A m,d n bing uniformly dns says that whnvr thr ar lots of options to forc larg numbrs into a potntial solution to M, thn thr is an xtnsion which forcs numbrs from both A (M) and B (M) into th potntial solution. Dfinition W say an infinit squnc p 0 > p 1 > of conditions satisfis K X,A (M),B (M) if ithr thr ar cofinitly many p i such that K X is not ssntial blow p i, or thr is som p n K X,A (M),B (M). W hav now mad all th inductiv hypothss on X prcis and can giv th formal construction of our gnric squnc of conditions. Lt K X,A (M),B (M) n, for n ω, b a list of all rquirmnts. (As w will s blow, it suffics for this list to consist of all rquirmnts formd from pairs of half rquirmnts.) Lmma Thr is a squnc of conditions p 0 > p 1 > from V X K X,A (M),B (M) n. which satisfis vry Proof. Lt p 0 = (σ 0, τ 0 ) whr both σ 0 and τ 0 ar th mpty squnc and not that p 0 V X. Givn p n, lt m b th last indx such that Km X is ssntial blow p n and for all i n, p i K X,A (M),B (M) m. By assumption K X,A (M),B (M) m is uniformly dns, so w may apply Lmma 2.4 to obtain p n+1 p n such that p n+1 K X,A (M),B (M) m and p n+1 V X. 11

12 It rmains to show that for ithr G = σ = σ n or G = τ = τ n, G satisfis th ncssary inductiv conditions: X G dos not comput a solution to M and all rquirmnts K X G,A (M),B (M) ar uniformly dns. W do this in two stps. First w xplain th connction btwn satisfying half rquirmnts and satisfying full rquirmnts. Scond, w show that th satisfaction of th appropriat half rquirmnts forcs ths conditions for X G. Lmma Lt R X,A (M),B (M) and S X,A (M),B (M) b half-rquirmnts and p 0 > p 1 > b an infinit squnc of conditions with p n = (σ n, τ n ). Lt σ = i σ i, τ = i τ i. If R X is ssntial in σ and S X is ssntial in τ, thn J X R,S is ssntial blow vry p n. Proof. Fix p n. To show JR,S X is ssntial blow vry p n, fix x. Lt A 0 > x witnss that R X is ssntial in σ and lt A 1 > x witnss that S X is ssntial in τ. A 0 A 1 will b our witnss that JR,S X is ssntial blow p n. Fix y. Lt B 0 > y witnss that R X is ssntial in σ and lt B 1 > y witnss that S X is ssntial in τ. B 0 B 1 will b our witnss that JR,S X is ssntial blow p n. Fix m 0 > n such that σ m0 R X,A 0,B 0 and fix m 1 > n such that τ m1 S X,A 1,B 1. Bcaus th dpndnc on A 0, A 1, B 0 and B 1 in ths sts is positiv, it follows that σ m0 R X,A 0 A 1,B 0 B 1 and τ m1 S X,A 0 A 1,B 0 B 1. Thus th conditions (σ m0, τ n ) and (σ n, τ m1 ) ar in J X,A 0 A 1,B 0 B 1 R,S and form a split pair blow p n. Putting ths pics togthr, w obtain th following: Lmma Suppos that for ach pair of half-rquirmnts R X,A (M),B (M) and S X,A (M),B (M), th rquirmnt J X,A (M),B (M) R,S is uniformly dns. Thn thr is an infinit squnc (σ 0, τ 0 ) > (σ 1, τ 1 ) > of conditions such that, stting σ = i σ i and τ = i τ i, ithr σ satisfis vry A-sid half-rquirmnt or τ satisfis vry D-sid half-rquirmnt. Proof. Lt p 0 > p 1 > b chosn as in Lmma Sinc ach J X,A (M),B (M) R,S is uniformly dns, this squnc satisfis vry rquirmnt J X,A (M),B (M) R,S. If σ satisfis vry halfrquirmnt, w ar don. So suppos thr is som R X,A (M),B (M) not satisfid by σ, and not that R X must b ssntial in σ. W show that τ satisfis vry S X,A (M),B (M). Fix S X,A (M),B (M) and assum that S X is ssntial in τ (othrwis this half rquirmnt is trivially satisfid). By Lmma 2.13, JR,S X is ssntial for vry (σ n, τ n ), and sinc th squnc of conditions satisfis J X,A (M),B (M) R,S, thr must b som condition (σ n, τ n ) J X,A (M),B (M) R,S. W cannot hav σ n R X,A (M),B (M), sinc thn σ would satisfy R X,A (M),B (M), so τ n S X,A (M),B (M). W st G = σ if σ satisfis all th A-sid half rquirmnts and w st G = τ othrwis. By Lmma 2.14, G satisfis vry half rquirmnt (on th appropriat sid). It rmains to show that X G dos not comput a solution to M and that ach rquirmnt K X G,A (M),B (M) is uniformly dns. W work undr th hypothsis that G = σ and hnc rstrict our attntion to A-sid half rquirmnts. Th sam argumnts, with th obvious changs, giv th corrsponding rsults if G = τ working with D-sid half rquirmnts. Lmma If G satisfis vry A X,A (M),B (M) m comput a solution to M. half rquirmnt, thn X G dos not 12

13 Proof. Fix an indx m. If Φm X G is finit, thn w ar don. So, suppos Φ X G m is infinit. W claim that A X m is ssntial in G. To prov this claim, fix n and x. Lt a 0 > x b such that Φ X G m (a 0 ) = 1 and st A = {a 0 }. Fix y, lt b 0 > y b such that Φ X G m (b 0 ) = 1 and st B = {b 0 }. St n > n b gratr than th us of ithr of ths computations. By dfinition, G n A X,A,B m and hnc A X m is ssntial in G. Sinc G satisfis A X,A (M),B (M) m, thr must b an n such that G n A X,A (M),B (M) m. Thrfor, for som a A (M) and b B (M), w hav Φ X G m (a) = Φ X G m (b) = 1, complting th proof. Finally, w show that for vry indx such that X G is a stabl-ish linar ordr, ach rquirmnt K X G,A (M),B (M) P X G is uniformly dns. Rcall that K X G,A (M),B (M) is spcifid by an indx i such that K X G,A (M),B (M) = {p P X G a A (M) b B (M) (Φ X G i (p, a, b) = 1)} whr Φ X G i is total. As w construct G, w do not know which indics will rsult in X G bing a stabl-ish linar ordr and, for ach such indx, which indics i will corrspond to rquirmnts K X G,A (M),B (M) P X G. Thrfor, w dfin th following A-sid half rquirmnts for vry pair of indics and i. (Of cours, w also dfin th corrsponding D-sid half rquirmnts and all proofs that follow work qually wll on th D-sid.) Dfinition Fix σ A X and an indx. For a pair of finit strings q = (σ q, τ q ), w say q P X σ if for all i < j < σ q, σ q (i) X σ σ q (j), for all i < j < τ q, τ q (j) X σ τ q (i) and σ p ( σ p 1) X σ τ p ( τ p 1). W say σ forcs q P X G if ithr thr ar i < j < σ q such that σ q (j) X σ σ q (i) or thr ar i < j < τ q such that τ q (i) X σ τ q (j) or τ p ( τ p 1) X σ σ p ( σ p 1). Not that this dfinition dos not match th usual mthod for forcing th ngation of a statmnt. By th us convntion, P X σ is finit and w can X-computably quantify ovr this finit st. Furthrmor, w can X-computably dtrmin whthr σ forcs q P X G. Dfinition For ach pair of indics and i and ach q = (σ q, τ q ), w dfin th A- sid half rquirmnt T X,A (M),B (M),i,q to b th st of all σ A X such that ithr σ forcs q P X G or thr xist strings σ and τ such that q 0 = (σq σ, τ q ) and q 1 = (σ q, τq τ ) satisfy q 0, q 1 P X σ and a 0, a 1 A (M) b 0, b 1 B (M) (Φ X σ i (q 0, a 0, b 0 ) = Φ X σ i (q 1, a 1, b 1 ) = 1) (i.. σ forcs th xistnc of a split pair blow q which lis in K X,A (M),B (M) ). Lt G b th gnric constructd by our itratd forcing as in Lmma 2.14 and assum G = σ. Thus, G satisfis vry A-sid half rquirmnt T X,A (M),B (M),i,q. Fix an indx such that X G is a stabl-ish linar ordr and fix an indx i spcifying a rquirmnt K X G,A (M),B (M) = {q P X G a A (M) b B (M) (Φ X G i (q, a, b) = 1)} Th following lmma (and its D-sid countrpart) complt our vrification of th proprtis of th itration forcing. 13

14 Lmma If G satisfis T X,A (M),B (M),i,q dns in P X G. for vry q, thn K X G,A (M),B (M) is uniformly Proof. Fix q P X G and assum that K X G is ssntial blow q. W claim that T X,i,q is ssntial in G. Bfor proving th claim, notic that this claim suffics to prov th lmma. Sinc G satisfis T X,A (M),B (M),i,q and T X,i,q is ssntial in G, thr is an n such that G n T X,A (M),B (M),i,q. By th dfinition of T X,A (M),B (M),i,q, sinc q P X G, thr must b a split pair q 0, q 1 P X G n blow q and a 0, a 1 A (M) and b 0, b 1 B (M) such that Φ X G n i (q 0, a 0, b 0 ) = Φ X G n i (q 1, a 1, b 1 ) = 1. Thus q 0, q 1 giv th dsird split pair blow q in K X G,A (M),B (M). It rmains to prov th claim that T X,i,q is ssntial in G. Fix n and x. Fix A > x witnssing that K X G is ssntial blow q. Fix y and lt B > y and th split pair q 0, q 1 blow q b such that q 0, q 1 K X G,A,B. Thus, a 0, a 1 A b 0, b 1 B (Φ X G i (q 0, a 0, b 0 ) = Φ X G i (q 1, a 1, b 1 ) = 1). Lt m > n b such that m is gratr than th uss of ths computations and such that q, q 0, q 1 P X G m. Thn w hav G m T X,A,B,i,q as rquird. 2.3 Ground Forcing In this sction, w dfin th ground forcing to build (M, A (M), B (M)) such that M dos not comput a solution to itslf (i.. it dos not comput an infinit subst of A (M) or B (M)) and ach rquirmnt K M,A (M),B (M) is uniformly dns. Our ground forcing conditions F ar tripls (F, A, B ) satisfying F is a finit partial ordr such that dom(f ) is an initial sgmnt of ω and for all x, y dom(f ), x F y implis x < y, and A B dom(f ), A is downwards closd undr F, B is upwards closd undr F and A B =. W say (F, A, B ) (F 0, A 0, B 0) if: F xtnds F 0 as a partial ordr (i.. dom(f 0 ) dom(f ) and for all x, y dom(f 0 ), x F0 y if and only if x F y), A 0 A, B 0 B, whnvr a A 0 and x dom(f ) \ dom(f 0 ), a F x, whnvr b B 0 and x dom(f ) \ dom(f 0 ), b F x (which implis x is incomparabl with b sinc b < x and hnc x F b). 14

15 In what follows, w will typically writ x M rathr than x dom(m). W dfin a gnric squnc of conditions (F 0, A 0, B 0) > (F 1, A 1, B 1) > and lt M = F n. W nd to satisfy th following proprtis: (C1) For all i, thr is an n such that i A n B n. (Togthr with th dfinitions of our conditions and xtnsions of conditions, this proprty guarants that A (M) = A n and B (M) = B n and that M is stabl.) (C2) For all, if Φ M is infinit, thn thr ar a A (M) and b B (M) such that Φ M (a) = Φ M (b) = 1. (C3) If M is a stabl-ish linar ordr and K M,A (M),B (M) P M is a rquirmnt (as dfind in th prvious sction), thn for all p P M, ithr K M is not ssntial blow p or thr is a split pair q 0, q 1 blow p in K M,A (M),B (M). Th nxt thr lmmas show that th appropriat st of conditions forcing ths proprtis ar dns. For (C1), w us th following lmma. Lmma Th st of (F, A, B ) such that i A B is dns in F. Proof. Fix (F, A, B ) and i ω. Without loss of gnrality, w assum i F. If i A, thn i F a for all a A by th downwards closur of A. Lt F 0 = F, A 0 = A and B0 = B {c F i F c}. Thn i B0 and (F 0, A 0, B0) xtnds (F, A, B ). For (C2), w us th following standard forcing dfinitions (with G dnoting th gnric variabl). W say F Φ G is finit if k (F 0, A 0, B0) (F, A, B ) x (Φ F 0 (x) = 1 x k). W say F Φ G A (G) Φ G B (G) if a A b B (Φ F (a) = 1 Φ F (b) = 1). Lmma For ach indx, th st of conditions which ithr forc Φ G Φ G A (G) Φ G B (G) is dns in F. is finit or forc Proof. Fix and (F, A, B ) and assum that (F, A, B ) has no xtnsion forcing Φ G is finit. Fix x > F and an xtnsion (F 0, A 0, B0) (F, A, B ) such that Φ F 0 (x) = 1. Without loss of gnrality, w can assum that A 0 = A and B0 = B, so x A 0 B0. By th dfinition of xtnsions, w know b F0 x for all b B0. Thrfor, th condition (F 1, A 1, B1) dfind by F 1 = F 0, A 1 = A 0 {c F 0 c F0 x} and B1 = B0 is an xtnsion of (F, A, B ) such that x A 1 and Φ F 1 (a) = 1. Sinc (F 1, A 1, B1) dos not forc Φ G is finit, w can rpat this ida. Fix y > F 0 and an xtnsion (F 2, A 2, B2) (F 1, A 1, B1) such that Φ F 2 (y) = 0. Again, without loss of gnrality, w can assum that A 2 = A 1 and B2 = B1, and hnc that y F2 a for any a A 2. Th condition (F 3, A 3, B3) dfind by F 3 = F 2, A 3 = A 2 and B3 = B2 {c F 2 y F2 c} is an xtnsion of (F, A, B ) forcing Φ G A (G) Φ G B (G). 15

16 W turn to (C3). Fix an indx for a potntial stabl-ish linar ordr G. For p = (σ, τ), w say (F, A, B ) p P G if σ is a F ascnding squnc, τ is a F dscnding squnc and σ F τ. W say (F, A, B ) p P G if no xtnsion of (F, A, B ) forcs p P G. Obviously, th st of conditions which ithr forc p P G or forc p P G is dns. Along with th indx, fix an indx i for a potntial rquirmnt K G,A (G),B (G) P G. That is, w want to considr th potntial rquirmnt {q P G a A (G) b B (G) (Φ G i (q, a, b) = 1)}. Suppos (F, A, B ) p P G for p = (σ, τ). W say (F, A, B ) thr is a split pair q 0, q 1 blow p in K G,A (G),B (G) if thr ar σ and τ such that for q 0 = (σ σ, τ) and q 1 = (σ, τ τ ) w hav (F, A, B ) q 0, q 1 P G σ σ F τ τ and a 0, a 1 A b 0, b 1 B (Φ F i (q 0, a 0, b 0 ) = Φ F i (q 1, a 1, b 1 ) = 1). Finally, w say that (F, A, B ) K G is not ssntial blow p if for any stabl partial ordr ( M, A ( M), B ( M)) with dom( M) = ω xtnding (F, A, B ) such that x M y implis that x < y, M is a stabl-ish partial ordr and K M,A ( M),B ( M) is a rquirmnt, w hav that K M is not ssntial blow p. Lmma Fix a pair of indics and i and lt K G,A (G),B (G) b th potntial rquirmnt spcifid by ths indics. For any p, thr is a dns st of (F, A, B ) such that ithr: (F, A, B ) p P G, or (F, A, B ) K G is not ssntial blow p, or (F, A, B ) thr is a split pair blow p in K G,A (G),B (G). Proof. Fix (F, A, B ) and p = (σ, τ). If thr is any (F, A, B ) (F, A, B ) forcing that p P G thn w ar don. So assum not, and assum that (F, A, B ) p P G. Suppos thr is an xtnsion (F, A, B ) (F, A, B ), sts B 0 > A 0 > A B and a split pair q 0, q 1 blow p such that (F, A, B ) q 0, q 1 K F,A 0,B 0. Lt A b th downwards closur of A 0 in F and B th upwards closur of B 0 in F. W claim that A is disjoint from B B. Fix x A and a A 0 such that x F a. First, suppos for a contradiction that x B and hnc x F. If a F, thn x F a and hnc a B bcaus B is closd upwards in F. But, a A 0 and A 0 > B giving a contradiction. If a F, thn a F \F, so x F a sinc x B and (F, A, B ) (F, A, B ), again giving a contradiction. Thrfor, x B. Scond, suppos for a contradiction that x B. Thn 16

17 y F x for som y B 0 and hnc y F a. Thrfor, y a which contradicts B 0 > A 0. Thrfor, A is disjoint from B B. W also claim that A is disjoint from B B. Fix x A and not that x B sinc (F, A, B ) is a condition and hnc A B =. Suppos for a contradiction that x B. Thr is a y B 0 such that y F x and hnc y x, which contradicts B 0 > A. Thrfor, A is disjoint from B B. Takn togthr, our claims show that A A is disjoint from B B. Sinc A A is downwards closd and B B is upwards closd, (F, A A, B B) (F, A, B ) is a condition forcing th xistnc of a split pair blow p in K G,A (G),B (G). If thr is no such (F, A, B ) (F, A, B ), w claim (F, A, B ) alrady forcs that K G is not ssntial blow p: lt M b any compltion of F to a stabl partial ordring satisfying th appropriat conditions from abov, and suppos K M wr ssntial blow p. Thn in particular, thr would b som A 0 > max(a B ), som B 0 > max A 0, and a split pair q 0, q 1 ovr p such that q 0, q 1 K M,A 0,B 0. But thn thr would hav bn som finit rstriction F = M [0, m] witnssing this, contradicting our assumption. Having vrifid that any gnric for th ground forcing satisfis (C1), (C2) and (C3), w can giv th proof of Thorm 1.7. Proof. W itrativly build a Turing idal I containing a partial ordr M, containing a solution to vry infinit linar ordr in I, but not containing any solution to M. Lt M b a partial ordring gnric for th ground forcing. M is stabl by (C1), M dos not comput a solution to itslf by (C2) and for ach stabl-ish linar ordr M, ach rquirmnt K M,A (M),B (M) P M is uniformly dns by (C3). Thus, w hav stablishd th initial conditions for th itratd forcing with X = M. For a fixd indx such that M is a stabl-ish linar ordr, lt G b a gnric solution to M obtaind from th itration forcing. By Lmmas 2.14, 2.15 and 2.18, M G dos not comput a solution to M and for vry stabl-ish linar ordr M G, ach rquirmnt K M G,A (M),B (M) P M G is uniformly dns. Itrating this procss (and choosing stabl-ish partial ordrs systmatically to nsur that w vntually considr ach on) givs an idal I with th proprty that whnvr is a linar ordr in I, ithr is stabl-ish, and thrfor w addd a solution to I at som stag, or is not stabl-ish, and so a solution is computabl from, and thrfor blongs to I. W hav nsurd that M I but that no solution to M blongs to I. Thrfor (ω, I) is a modl of RCA 0 + ADS, but is not a modl of SCAC. 3 EM background In this sction, w prsnt proofs of Thorms 1.9 and 1.10, which ar rstatd blow for convninc. W bgin with som basic proprtis of infinit transitiv tournamnts and thir transitiv substs. W rgard vry tournamnt T (including finit subtournamnts) as containing lmnts and with th proprty that T (, x) and T (x, ) hold for vry x T. If T is a transitiv tournamnt, thn th T rlation dfins a linar ordr on th domain of T with as th last lmnt and as th gratst lmnt. W will dnot 17

18 this ordr by T, or by F if F is a finit transitiv subst of som ambint (nontransitiv) tournamnt T. Dfinition 3.1. Lt T b an infinit tournamnt and lt a, b T b such that T (a, b) holds. Th intrval (a, b) is th st of all x T such that both T (a, x) and T (x, b) hold. That is, (a, b) is th st of points btwn a and b in T. Dfinition 3.2. Lt F T b a finit transitiv subst of an infinit tournamnt T. For a, b F such that T (a, b) holds (i.. a F b), w say (a, b) is a minimal intrval of F if thr is no c F such that T (a, c) and T (c, b) both hold (i.. b is th succssor of a in F ). In th contxt of Dfinition 3.2, (a, b) is an intrval in T wll as in F. Howvr, th fact that (a, b) is a minimal intrval of F is a proprty of this intrval in F. Dfinition 3.3. Lt T b an infinit tournamnt and F T b a finit transitiv st. F is xtndabl if F is a subst of som solution to T. A on point transitiv xtnsion of F is a transitiv st F {a} such that a F. Lmma 3.4. Lt T b an infinit transitiv tournamnt and F T b a finit transitiv st. F is xtndabl if and only if F has infinitly many on point transitiv xtnsions. Proof. If F is xtndabl, thn it clarly has infinitly many on point xtnsions. Convrsly, suppos F has infinitly many on point xtnsions. Lt T b th st of all a T \ F such that F {a} is transitiv. Sinc F is transitiv, w can list F in F ordr as < F x 0 < F x 1 < F < F x k < F Bcaus T is infinit and thr ar finitly many minimal intrvals in F, thr must b a minimal intrval (a, b) of F such that (a, b) T is infinit. (Not that a could b, if thr ar infinitly many lmnts a T such that T (a, x 0 ) holds. Similarly, b could b.) Fix such a minimal intrval (a, b) in F and lt T = T (a, b). T is an infinit subtournamnt of T and hnc (viwing T as an infinit tournamnt), T contains an infinit transitiv tournamnt T. Sinc T is containd in a minimal intrval of F, T F is transitiv, and hnc is a solution to T containing F. Lmma 3.5. Lt T b an infinit tournamnt. containing a. For any a T, thr is a solution to T Proof. Fix a T and lt F = {a}. For all b T, {a, b} is a transitiv, so F has infinitly many on point transitiv xtnsions. By Lmma 3.4, F is xtndabl. Lmma 3.6. Lt T b an infinit transitiv tournamnt and lt F T b a finit transitiv xtndibl st. Cofinitly many of th on point transitiv xtnsions of F ar xtndabl. Proof. Suppos for a contradiction that thr ar infinitly many x T \ F such that F {x} is transitiv but not xtndabl. Lt T b th st of all such x. As in th proof of Lmma 3.4, thr must b a minimal intrval (a, b) of F such that T (a, b) is infinit. Fix such an intrval (a, b) and lt T = T (a, b). T is an infinit subtournamnt of T, so thr is an infinit transitiv st T T. F T is a solution to T containing F as wll as infinitly many point from T giving th dsird contradiction. 18

19 Thorm 3.7 (Kach, Lrman, Solomon and Wbr). Thr is a computabl infinit tournamnt T with no infinit Σ 0 2 transitiv subtournamnts. Proof. Sinc vry infinit Σ 0 2 st contains an infinit 0 2 subst, it suffics to construct an infinit computabl tournamnt T with no infinit 0 2 transitiv subtournamnts. W build T in stags to mt th following rquirmnts. R : If D (x) = lim s ϕ (x, s) xists for vry x, thn D is finit or D is not transitiv. As stag s, w dtrmin whthr T (x, s) or T (s, x) holds for ach x < s by acting in substags < s. At substag, R chooss th last lmnts x 0 < x 1 < < x 2+1 (lss than s) that ϕ currntly claims ar in D. (If thr ar not many such lmnts, thn w procd to th nxt substag.) Lt x i and x j b th last from this st which hav not bn chosn as witnsss by a highr priority rquirmnt at this stag and assum that T (x i, x j ) holds. Dclar that T (s, x i ) and T (x j, s) hold so that {x i, x j, s} is not transitiv. Procd to th nxt substag. Whn all substags ar complt, dclar T (x, s) for any x < s for which w hav not dclard ithr T (x, s) or T (s, x). This nds stag s. It is clar that this procss dfins a computabl infinit tournamnt T. To s that R is mt, assum that D (x) is dfind for all x. Lt x 0 < x 1 < < x 2+1 b last such that D (x i ) = 1 and lt s b such that ϕ claims that ach x i is in D for all t s. For vry t s, R chooss a pair of lmnt from {x 0,..., x 2+1 } to mak a cycl with t. Thrfor, {x 0,..., x 2+1 } has only finitly many on point transitiv xtnsions and hnc is not a subst of any infinit transitiv subtournamnt. Thorm 3.8 (Dzhafarov, Kach, Lrman and Solomon). Thr is a computabl infinit tournamnt T with no infinit hyprimmun-fr transitiv subtournamnts. Proof. W build T in stags to mt, for ach, th rquirmnt R that if {D ϕ(x) x N} is a strong array, thn thr ar x 0 < x 1 such that for all y 0 D ϕ(x0 ) and all y 1 D ϕ(x1 ), th st {y 0, y 1 } is not xtndibl. Th stratgy to mt an individual rquirmnt R in isolation is straightforward. W wait for ϕ (x 0 ) to convrg for som x 0, and start dfining T (y, s) for all y D ϕ(x0 ) and all s. If {D ϕ : ω} is a strong array, w must vntually find an x 1 such that ϕ (x 1 ) convrgs with T (y 0, y 1 ) for all y 0 D ϕ(x0 ) and all y 1 D ϕ(x1 ). W thn start dfining T (s, y) for all y D ϕ(x0 ) and all s, and T (y, s) for all y D ϕ(x1 ) and all s. Thus nsurs that R is mt. Sorting out compting rquirmnts can b handld via a standard finit injury priority argumnt, as w now show. At stag s, w dfin T (x, s) or T (s, x) for all x < s. W procd by substags s. At substag, w act as follows, braking into thr cass. Cas 1: R has no witnsss. Lt x 0 b th last x < s, if it xists, such that (1) ϕ,s (x) ; (2) D ϕ(x) and ach lmnt of D ϕ(x) is < s; (3) for all i < and any witnss y of R i, x > y and D ϕ(x) is disjoint from D ϕi (y). 19

20 If thr is no such x 0, thn do nothing and procd to th nxt substag. If thr is such an x 0, thn call x 0 th first witnss of R, dfin T (y, s) for all y D ϕ(x 0 ), cancl th witnsss of ach R i with i > and procd to th nxt substag. Cas 2: R has xactly on witnss. Call this first witnss x 0. Lt x 1 b th last x < s, if it xists, that satisfis conditions (1) (3) abov, as wll as (4) T (y 0, y 1 ) for all y 0 D ϕ(x 0 ) and all y 1 D ϕ(x). If thr is no such x 1, th dfin T (y, s) for all y D ϕ(x 0 ) and procd to th nxt substag. If thr is such a witnss x 1, thn call x 1 th scond witnss of R, dfin T (s, y) for all y D ϕ(x 0 ) and T (y, s) for all y D ϕ(x 1 ), cancl th witnsss of ach R i with i > and procd to th nxt substag. Cas 3: R has two witnsss. Lt x 0 b th first witnss and x 1 b th scond witnss. Dfin T (s, y) for all y D ϕ(x 0 ) and T (y, s) for all y D ϕ(x 1 ). Procd to th nxt substag. Whn all substags < s ar complt, dfin T (x, s) for any x < s for which nithr T (x, s) nor T (s, x) has bn dfind. This complts th dscription for th construction. It is clar that T is a computabl tournamnt on N. To vrify that rquirmnt R is mt, suppos {D ϕ(x) : x N} is a strong array. By induction, support furthr that ach R i, i <, is satisfid. Sinc ach rquirmnt R i has at most two witnsss at any stag, and sinc it can los ths witnsss only for th sak of som R i, i < i, bing assignd a witnss, w lt s b th last stag such that no R i, i <, is assignd a witnss at any stag s s. By minimality of s, it must b that R has no witnsss at stag s. Sinc {D ϕ(x) : x N} is a strong array, w lt s 0 s b th last stag such that som x < s 0 satisfis conditions (1) (3) of th construction. Thn th last such x is assignd as a first witnss x 0 of R, and this witnss is nvr canclld. If, at any latr stag s 1 > s 0, w assign a scond witnss x 1 for R, thn R will b satisfid. (Bcaus x 1 will nvr b cancld, w hav T (y 0, y 1 ), T (s, y 0 ) and T (y 1, s) for all s > s 1, all y 0 D ϕ(x 0 ) and all y 1 D ϕ(x 1 ). Thrfor, {y 0, y 1 } is not xtndibl.) So suppos w nvr find a scond witnss x 1. Thn by construction, w dfin T (y, s) for all s s 0 and all y D ϕ(x 0 ). But if s is larg nough that for som x < s, ϕ,s (x) and all lmnts of D ϕ(x) li btwn s 0 and s, thn x will satisfy conditions (1) (4) of th construction. Th last such x is assignd as a scond witnss x 1 of R for th dsird contradiction. 4 EM dos not imply SRT 2 2 Bfor giving th proof of Thorm 1.15 in a styl similar to th proof of Thorm 1.7, w prsnt som motivating idas for th forcing construction. Fix an indx. W sktch a stratgy to mt a singl diagonalization rquirmnt towards constructing a stabl coloring c such that if Φ c is th charactristic function for an infinit tournamnt T c givn by such that c S dos not comput a solution to Φ c, thn thr is a solution S to T c c. A singl diagonalization rquirmnts has th form R i : Φ c S i is not a solution to c. 20

21 To approximat c w us a tripls (c, A, B ) (calld partial stabl colorings) such that c is a 2-coloring of th two lmnt substs of a finit domain [0, c ], and A and B ar disjoint substs of this domain. W say (c α, A α, B α) xtnds (c β, A β, B β ) if c β c α, A β A α, B β B α, if a A β and c β < x c α, thn c α (a, x) = 0, and if b B β and c β < x c α, thn c α (b, x) = 1. In th full construction, ths partial stabl colorings will b our ground forcing conditions, and w can forc statmnts such as F is a finit transitiv subtournamnt of T c or I is a minimal intrval in F which is infinit in T c in a standard mannr. For xampl, th st of (c, A, B ) such that i A B is obviously dns, so a gnric coloring c will b stabl. Givn α = (c α, A α, Bα), w lt C α dnot th st of suitably gnric infinit stabl colorings xtnding α. To approximat a solution S to T c, w augmnt a partial stabl coloring α by adding a finit transitiv subtournamnt F α of T cα and a minimal intrval I α of F α such that I α is infinit in vry tournamnt T c for c C α. F α dnots th part of S spcifid so far and I α witnsss th fact that no mattr how c α is (gnrically) xtndd to c, F α is xtndibl in T c. Thus, a condition for th purposs of this sktch has th form α = (c α, A α, Bα, F α, I α ). W say α xtnds β if th partial colorings xtnd as abov, F β F α, I α is a subintrval of I β and for ach x F α \ F β, x > max(f β ) and x I β. Givn a condition α, w would lik to mt R i by xtnding c α to c β and F α to F β so that Φ c β F β i (y) = 1 for som larg y A α Bα. Assuming w can do this without xpanding A α Bα, w ar fr to add y to ithr A α or Bα. Thrfor, if w can prform such an xpansion twic, w will arriv at a condition γ such that a A γ b Bγ (Φ cγ Fγ i (a) = Φ cγ Fγ i (b) = 1) and hnc will hav succssfully diagonalizd. Th obvious difficulty is that w hav to maintain that F γ is xtndibl in T c for all c C γ. W us following partition thorm to hlp addrss this problm. Lmma 4.1. Lt T b an infinit tournamnt, F b a finit transitiv st and (a, b) b a minimal intrval of F which is infinit in T. For any finit st J (a, b) such that F J is transitiv, thr is a partition P Q = J such that both F P and F Q ar xtndibl and contain minimal intrvals in (a, b) which ar infinit in T. Givn a condition α, w ask our main qustion: is thr a coloring c C α xtnding c α, an infinit transitiv st S in T c containd in I α with F α < S, and a finit initial sgmnt J of S such that for all partitions P Q = J, thr is a transitiv F P or F Q for which Φ c (Fα F ) i (y) = 1 for som y A α Bα? Suppos th answr to this qustion is ys. W collct a finit st Y disjoint from A α Bα such that for ach partition P Q = J, thr is som F P,Q P or F P,Q Q and som y Y 21