Phase transitions related to the pigeonhole principle

Transcription

1 Phase transitions relate to the pigeonhole principle Michiel De Smet an Anreas Weiermann Smisestraat 92 B 9000 Ghent Belgium ms@michielesmet.eu 2 Department of Mathematics Ghent University Builing S22 Krijgslaan 28 B 9000 Ghent Belgium Anreas.Weiermann@UGent.be Abstract. Since Jeff Paris introuce them in the late seventies [Par78], ensities turne out to be useful for stuying inepenence results. Motivate by their simplicity an surprising strength we investigate the combinatorial complexity of two such ensities which are strongly relate to the pigeonhole principle. The aim is to miniaturise Ramsey s Theorem for -tuples. The first principle uses an unlimite amount of colours, whereas the secon has a fixe number of two colours. We show that these principles give rise to Ackermannian growth. After parameterising these statements with respect to a function f : N N, we investigate for which functions f Ackermannian growth is still preserve. Keywors Ackermann function, pigeonhole principle, Ramsey theory, phase transitions. Introuction The pigeonhole principle is one of the most well-know combinatorial principles, ue to both its simplicity an usefulness. The principle is also known as the chest-of-rawers principle or Schubfachprinzip an is attribute to Dirichlet in 834. The pigeonhole principle can also be consiere as a finite instance of Ramsey s theorem for -tuples. So, if RT n k stans for Ramsey s Theorem for n imensions an k colours, i.e. RT n k For every G: [N] n k there exists an infinite set H such that G [H] n is constant, Research supporte in part by Fons Wetenschappelijk Onerzoek (FWO) - Flaners Research supporte in part by(fwo) an the John Templeton Founation

2 then the pigeonhole principle is a finite instance of RT < = k RT k. In this paper we will investigate miniaturisations of the statements RT < an RT 2. Let us recall some results from reverse mathematics: for any fixe natural number k, RCA 0 RT k, whereas WKL 0 RT <. Both results are ue to Hirst (see [Hir87], Theorem 6.3 an Theorem 6.5). In aition, there it is also prove that RT < oes not imply ACA 0 over RCA 0. As it oes not fit nicely into the programme of reverse mathematics, one might be tempte to think that RT < is of little importance. However, it pops up every now an then in the literature. It is, for instance, equivalent to Rao s Lemma over RCA 0 (see [Hir87], Theorem 6.6). For miniaturising RT < an RT 2 we efine two notions of ensity, n-ensity an (α, 2)-ensity, which are parametrise by a function f : N N. Using these notions we efine two first orer assertions an stuy their provability with respect to IΣ, the first-orer part of RCA 0. We show which f give rise to Ackermannian growth an etermine the exact phase transition. In case of n-ensity Ackermannian growth is obtaine for f(i) = A ω A i (i), whereas for f(i) = i (i) it is not. Here A enotes the -th branch of the Ackermann function A ω. Our proof will show that in these results A ω (A ) coul be replace by any non ecreasing unboune non primitive recursive function (resp. by any non ecreasing unboune primitive recursive function). In the case of (α, 2)-ensity we restrict ourselves to only two colours an strength isappears, as expecte. Surprisingly, iterating up to ω 2 suffices to gain proof theoretic strength again. It turns out that f(i) = log(i) gives rise to A (i) no more than primitive recursive growth, but f(i) = A ω (i) log(i) oes. Our proof will show that also in these results A ω (A ) coul be replace by any non ecreasing unboune non primitive recursive function (resp. by any non ecreasing unboune primitive recursive function). We woul like to mention that the n-ensity threshol functions are exactly the same as those for the parameterise Kanamori-McAloon principle, whereas the (ω 2, 2)-ensity functions equal those for the parameterise Paris-Harrington principle [KLOW08,WVH202]. It is our hope that by investigating miniaturisations of RT < an RT 2 2 one coul obtain insights into the seemingly ifficult question whether RT 2 2 oes or oes not prove the totality of the Ackermann function (the so calle Ramsey for pairs problem). For relate work we also we also refer to [DSW08] an the unpublishe PhD thesis of the first author [DS]. 2 n-density Henceforth, let f : N N be any (elementary recursive function), such that f(x) x, for x large enough. We efine the functions F f,k an F f, epening

3 on f, by F f,0 (n) := n + F f,k+ (n) := F f,k (... (F f,k (n))...) := F f(n) f,k }{{} (n) f(n) times F f (n) := F f,n (n), for every k, n N. If it is clear which f we are working with, we leave out the subscript f an simply write F k an F, instea of F f,k an F f, respectively. We also consier functions f with non integer values. It is then unerstoo that we roun a value f(i) own to f(i), the biggest natural number below f(i). Moreover we assume that f has always values at least as big as. It is easy to verify that the functions F f,k an F f are strictly monotonic increasing if the parameter function f is non ecreasing. In case of f(i) = i we write A ω for F f an A for F f,. A ω is a stanar choice for the well known Ackermann function, which is a recursive but not a primitive recursive function. The function A is calle the -th branch of the Ackermann function. Every function A is primitive recursive. In [OW09] a classification is given of those functions f for which F f is primitive or non primitive recursive. Let us efine n-ensity, the first ensity notion relate to the pigeonhole principle. In this case the number of colours epens on the minimum of X an the function f. Definition X is calle 0-ense(f) if X max{f(min X), 3}. X is calle (n + )-ense(f) if for all G : X f(min X), there exists Y X, such that Y is homogeneous for G an Y is n-ense(f). Lemma Assume that k l an that X [k, l] an that f, g : [k, l] N are two functions such that f(i) g(i) for all i [k, l]. If X is n-ense(g) then X is n-ense(f). Proof. One verifies the claim easily by inuction on n. 2. Upper boun Lemma 2 Let f be non ecreasing. Let n N an X N be a finite set. If X is n-ense(f), then max X F f,n (min X). Proof. Being of no importance for the proof itself, we leave out the subscript f. Henceforth, let x 0 = min X an c = f(x 0 ). The proof goes by inuction on n. If X is 0-ense(f), then X max{f(x 0 ), 3}. Thus, max X x F 0 (x 0 ). Seconly, assume the statement is proven for n an X is (n + )-ense(f). Consier the following partition of X = 0 i<c Y i, where Y i is efine by Y i = {x X F i n(x 0 ) x < F i+ n (x 0 )}

4 for 0 i < c an Y c = {x X Fn c (x 0 ) x}. Now, efine G : X c, as follows G(x) := i, for x Y i. Since X is (n + )-ense(f), there exists a subset Y of X, such that Y is n-ense(f) an homogeneous for G. By contraiction assume Y Y i0 for some i 0 with 0 i 0 < c. The n-ensity of Y an the monotonicity of F n yiel F i0+ n (x 0 ) max Y i0 max Y F n (min Y ) F n (min Y i0 ) F n (F i0 n (x 0 )) = F i0+ n (x 0 ), a contraiction. So Y Y c, which implies max X = max Y c max Y F n (min Y ) F n (min Y c ) F n (Fn c (x 0 )) = Fn(x c 0 ) = Fn f(x0) (x 0 ) = F n+ (x 0 ), by the n-ensity of Y. This conclues the inuction argument. Definition 2 Define PHP f : N N by PHP f (n) := min{n N [n, n ] is n-ense(f)}. Let f(i) = i A ω (i), where A ω enotes the Ackermann fuction. Then F f is Ackermannian, ue to Theorem in [OW09]. If f woul be non ecreasing then Lemma 2 woul yiel PHP f (n) F f,n (n) = F f (n), for all n N, hence also PHP f woul Ackermannian. Since the provably total functions of IΣ are exactly the primitive recursive functions, we woul immeiately obtain that PHP f woul not be provably recursive in IΣ. We now show how to overcome this problem. Theorem If f(i) = i A ω (i), then IΣ ( n)( a)( b)([a, b] is n-ense(f)). Proof. Let p(n) := n+ + (n + ) n+ an let f k (i) := i k. It suffices to show that PHP f (p(n)) > A ω (n) Assume that PHP f (p(n)) A ω (n). Then for i A ω (n) one has that A ω (i) n which yiels f(i) f n (i) for all i PHP f (p(n)). The proof of Claim 2.2 from [KLOW08] yiels F fn+,n+n 2 +4n+5(p(n) > A ω (n). Together with Lemma this yiels PHP f (p(n)) PHP fn (p(n)) F fn+,n+n 2 +4n+5(p(n)) > A ω (n) which is a contraiction.

5 2.2 Lower boun A Let f(i) = i (i), where A enotes the -th branch of the Ackermann function A ω. This function is not weakly increasing on its omain. For i [A (k), A (k + ) ] one has that A (i) = k an on such intervals f will be non ecreasing. In intervals of the form i [A (k), A (k)] the function A jumps from k to k. But since the intervals of the form [A (k), A (k + ) ] are rather long it is very easy to fin enough points b such that f(b) f(i) for all i b. One simply has to choose b so large that f majorizes f(c) where c is the initial point of a last jump interval which comes before b. With this caveat we can consier f basically as non ecreasing function although it in fact is not. Theorem 2 If f(i) = i A (i), then IΣ ( n)( a)( b)([a, b] is n-ense(f)). Proof. Assume that n an a are given. Put b := 2 A (a2 n+ )2 n+. Then f(i) f(b) for all i b. We claim that any Y [a, b] with Y > 2 A (a2 n+ )2 k is k-ense(f). To prove the claim we procee by inuction on k. Assume the claim hols for k an consier Y [a, b] with Y > 2 A (a2 n+ )2 k. Since 2 A (a2 n+ )2 n+ > A (2 n+ ), we have f(min Y ) < f(b) = (2 A (a2 n+ )2 n+ ) (2 A (a2 n+ )2 n+ ) A (2A (a2n+ )2 n+ ) A (A (2n+ )) = (2 A (a2 n+ )2 n+ ) 2 n+ = 2 A (a2 n+) 2 A (a2 n+ )2 k. Let c = f(min Y ) an G : Y c be any function. Consier the partition of Y inuce by G, i.e. Y = 0 i<c Y i, with Y i = {y Y G(y) = i}. By contraiction, assume that Y i 2 A (a2 n+ )2 k for every 0 i < c. Then 2 A (a2 n+ )2 k < Y c 2 A (a2 n+ )2 k = f(min Y ) 2 A (a2 n+ )2 k < 2 A (a2 n+ )2 k 2 A (a2 n+ )2 k = 2 A (a2 n+ )2 k +A (a2 n+ )2 k = 2 A (a2 n+ )2 k, a contraiction. Thus, there exists an inex i 0 {0,..., c }, such that Y i0 > 2 A (a2 n+ )2 k. The inuction hypothesis yiels that Y i0 is (k )-ense(f) an by efinition Y i0 is homogeneous for G, so Y is k-ense(f). If k = 0 then Y > 2 A (a2 n+) > f(min Y ), which completes the inuction argument an proves the claim. Now return to [a, b]. [a, b] is n-ense(f) since [a, b] 2 A (a2 n+ )2 n+ a 2 A (a2 n+ )2 n. Remarking that the function E : N N N, efine by E(a, n) = 2 A (a2 n+ )2 n+ is primitive recursive, completes the proof. Let PHP f stan for ( n)( a)( b)([a, b] is n-ense(f)). Then we obtain the following picture.

6 IΣ PHP f f(i) = i A ω (i) threshol region f(i) = i A (i) IΣ PHP f Fig.. Phase transition for PHP f. 3 (α, 2)-Density In this section we work with a fixe number of colours, namely two. For a limit orinal not exceeing ω 2 we efine the funamental sequence as follows. We put ω (k + )[n] = ω k + n an we set ω 2 [n] = ω n. Definition 3 X is calle (0, 2)-ense(f) if X max{f(min X), 3}. X is calle (α +, 2)-ense(f) if for all G : X 2 there exists Y X, such that Y is (α, 2)-ense(f) an Y is homogeneous for G. If λ is a limit orinal, then X is calle (λ, 2)-ense(f) if for all G : X 2 there exists Y X, such that Y is (λ[min X], 2)-ense(f) an Y is homogeneous for G. Lemma 3 Let k l an f, g : [k, l] N be non ecreasing such that f(i) g(i) for all i [k, l]. If X [k, l] is (n, 2)-ense(g) then X is is (n, 2)-ense(g). 3. Upper boun In this section we will use another hierarchy which we call B f,α an which turns out to be relate to F f,k. We efine B α, epening on f, by B f,0 (n) := n + B f,α+ (n) := B f,α (B f,α (n)) := B 2 f,α(n) B f,λ (n) := B f,λ[f(n)] (n), for all n N an orinals α an λ, with the latter a limit orinal. As for F f, we leave out the subscript f an write B α if it is clear which f we are working with. We first show a simple lemma concerning the relation between the two hierarchies efine in this paper. This lemma might be consiere as folklore. Lemma 4 Let k, l an m be natural numbers. Then B f,ω k+l (m) = F 2l 2 f,k (m). Proof. We procee by main inuction on k an subsiiary inuction on l. If k equals l equals zero, we have B f,0 (m) = m + = F 2 f,0(m).

7 Assume the statement is proven for k, we will prove it for k by subsiiary inuction on l. If l = 0, then the main inuction hypothesis yiels B f,ω (k ) (m) = F 2 f,k (m). Assume the claim is proven for l. We have B f,ω (k )+l (m) = B f,ω (k )+l (B f,ω (k )+l (m)) = F 2l 2l 2l 2 f,k (F2 f,k (m)) = F2 f,k (m), which proves the statement for k an every l. Using this fact, we obtain B f,ω k (m) = B f,ω (k )+ω[f(m)] (m)) = B f,ω (k )+f(m) (m)) = F 2f(m) 2 f,k (m) = F 2 f,k(m), which conclues the main inuction an proves the statement. Lemma 5 Let f be non ecreasing. Let n be a natural number an α be any orinal. If X N is (α, 2)-ense(f), then max X B f,α (min X). Proof. Being of no importance for the proof itself, we leave out the subscript f. Henceforth, let x 0 = min X. The proof goes by transfinite inuction on α. If X is (0, 2)-ense(f), then X max{f(x 0 ), 3}. Thus, max X x > x 0 + = B 0 (x 0 ). Assume the statement is proven for α an X is (α +, 2)-ense(f). Define G : X 2 as follows { 0 if x 0 x < B α (x 0 ) G(x) :=, if B α (x 0 ) x for all x X. Since X is (α +, 2)-ense(f), there exists a subset Y of X, such that Y is (α, 2)-ense(f) an Y is homogeneous with respect to G. By contraiction, assume G takes colour 0 on Y. Then, by the inuction hypothesis, B α (x 0 ) max Y B α (min Y ) = B α (x 0 ), a contraiction. So, the colour nees to be, which implies The inuction hypothesis yiels Y {x X B α (x 0 ) x}. max X max Y B α (min Y ) = B α (B α (x 0 )) = B α+ (x 0 ). Finally, assume the statement is proven for all α < λ, with λ a limit orinal, an X is (λ, 2)-ense(f). There exists a subset Y which is (λ[f(x 0 )], 2)-ense(f). We obtain by the inuction hypothesis max X max Y B λ[f(x0)](min Y ) B λ[f(x0)](x 0 ) = B λ (x 0 ). This completes the proof.

8 Definition 4 Define PHP2 f : N N by Fix f(i) = PHP2 f (n) := min{n N [n, n ] is (ω 2, 2)-ense(f)}. A ω (i) log(i) for the rest of this subsection. Lemma 6 Let f n (i) := n log 2(i). Then PHP2 fn (2 n2 ) F 2 fn (n) hols for every n N. Proof. Let X = [2 n2, PHP2 f (2 n2 )]. Define G : X 2 by G(x) = 0 for every x X. Since X is (ω 2, 2)-ense(f n ) there exists Y X, such that Y is (ω 2 [f n (min X)], 2)-PHP2-ense(f n ), i.e. (ω f n (2 n2 ), 2)-ense(f n ). Lemma 4 an Lemma 5 yiel PHP2(2 n2 ) max Y B fn,ω f(min Y )(min Y ) B fn,ω f(2 n2 ) (2n2 ) since f n (2 n2 ) = n. Corollary If f(i) = = F 2 fn,f n(2 n2 ) (2n2 ) F 2 fn,n(n) = F 2 fn (n), A ω (i) log(i), then IΣ ( a)( b)([a, b] is (ω 2, 2)-ense(f)). Proof. Let p(n) = 4+3 n+ +(n+) n+ an f k (i) := k log 2(i). It suffices to show that PHP2 f (2 p(n)2 ) > A ω (n). Assume for a contraiction that PHP2 f (2 p(n)2 ) A ω (n). For i A ω (n) one has A ω (i) n hence f(i) f n (i) for all i PHP2 f (2 p(n)2 ). This yiels PHP2 f (2 p(n)2 ) PHP2 fn (2 p(n)2 ) F 2 fn (p(n) > A ω (n). Contraiction! 3.2 Lower boun As in Section 2.2 let f(i) = log(i), where A A (i) enotes the th branch of the Ackermann function A ω. Recall from Section 2.2 that f is almost non ecreasing an that it is easy to ientify the jumps for f. Theorem 3 If f(i) = log(i), then A (i) IΣ ( a)( b)([a, b] is (ω 2, 2)-ense(f)). Proof. Assume that a is given. Put b := 2 A (2 a+2 )2 a+. Then f(i) f(b) for all i b. We claim that any Y [a, b], with Y > 2 A (2 a+2 )2 k is (ω k, 2)-ense(f). The proof goes by inuction on k. Let k = 0. Since 2 A (2 a+2 )2 a+ > A (2 a+2 ), we have f(min Y ) < f(b) = A (2A (2 a+2 )2 a+ ) log(2a (2 a+2 )2 a+ ) < 2 a+2 A (2 a+2 )2 a+ < A (2 a+2 ),

9 so, Y > 2 A (2 a+2) > max{f(min Y ), 3}, i.e. Y is (0, 2)-ense(f). Assume the assertion hols for k an consier Y [a, b] with Y > 2 A (2 a+2 )2 k. We claim that if Z Y an Z > 2 A (2 a+2 )2 k +l, then Z is (ω (k ) + l, 2)-ense(f). The proof goes by subsiiary inuction on l. If l = 0, then the claim follows by the main inuction hypothesis. Assume the claim hols for l an Z > 2 A (2 a+2 )2 k +l. Let G : Z 2 be any function. Consier the partition of Z inuce by G, i.e. Z = Z 0 Z, with Z i = {z Z G(z) = i}. By contraiction, assume that for i = 0,. Then Z i 2 A (2 a+2 )2 k +l, 2 A (2 a+2 )2 k +l < Z 2 2 A (2 a+2 )2 k +l = 2 A (2 a+2 )2 k +l, a contraiction. Thus, there exists an inex i 0 {0, }, such that Z i0 > 2 A (2 a+2 )2 k +l. The inuction hypothesis yiels Z i0 is (ω (k ) + l, 2)- ense(f), an so Z is (ω (k ) + l, 2)-ense(f), since Z i0 is homogeneous for G. This proves the latter claim. Now return to Y. Let G : Y 2 be any function. Consier the partition of Y inuce by G, i.e. Y = Y 0 Y, with Y i = {y Y G(y) = i}. In the same way as above, one can prove there exists an inex i 0 {0, }, such that Since Y i0 > 2 A (2 a+2 )2 k = 2 A (2 a+2 )2 k +A (2 a+2 )2 k. A (2 a+2 )2 k A (2 a+2 ) f(min Y ) +, we have Y i0 > 2 A (2 a+2 )2 k +f(min Y ). The latter claim yiels Y i0 is (ω (k ) + f(min Y ), 2)-ense(f), i.e. (ω k[f(min Y )], 2)-ense(f). Thus Y is (ω k, 2)- ense(f), since Y i0 is homogeneous for G. We finally prove that [a, b] is (ω 2, 2)-ense(f). Let G : [a, b] 2 be any function an consier the partition of [a, b] inuce by G, i.e. [a, b] = Y 0 Y, with Y i = {y [a, b] G(y) = i}. Remark that [a, b] > 2 A (2 a+2 )2 a+ a 2 A (2 a+2 )2 a +. Similarly as before, there exists an inex i 0 {0, }, such that Y i0 > 2 A (2 a+2 )2 a 2 A (2 a+2 )2 f(a). The main claim yiels Y is (ω f(a), 2)-ense(f), i.e. (ω 2 [f(a)], 2)-ense(f). In combination with Y i0 being homogeneous for G, this implies [a, b] is (ω 2, 2)- ense(f).

10 IΣ PHP2 f f(i) = A ω (i) log(i) threshol region f(i) = log(i) A (i) IΣ PHP2 f Fig. 2. Phase transition for PHP2 f. Let PHP2 f stan for ( n)( a)( b)([a, b] is (ω 2, 2)-ense(f)). Once again, we obtain the following picture. In accorance with the referees (for which we are grateful for valuable comments) we expect that it will be not too har to show that for natural choices of f the principles PHP log f an PHP2 f are equivalent over IΣ. References [DSW08] Michiel De Smet an Anreas Weiermann: Phase Transitions for Weakly Increasing Sequences. Proceeings of CiE 2008, (2008) [DS] Michiel De Smet : Unprovability an phase transitions in Ramsey theory. PhD thesis, Ghent 20. [Hir87] Jeffry Lynn Hirst. Combinatorics in subsystems of secon-orer arithmetic. PhD thesis, Pennsylvania State University, 987. [KLOW08] Menachem Kojman, Gyesik Lee, Eran Omri an Anreas Weiermann: Sharp threshols for the phase transition between primitive recursive an Ackermannian Ramsey numbers JCT A. 5(6). (2008), p [OW09] Eran Omri an Anreas Weiermann. Classifying the phase transition threshol for Ackermannian functions. Ann. Pure Appl. Logic, 58(3):56 62, [WVH202] Anreas Weiermann an Wim Van Hoof: Sharp phase transition threshols for the Paris Harrington Ramsey numbers for a fixe imension. Proc. Amer. Math. Soc. 40 (202), no. 8, [Par78] Jeff B. Paris. Some inepenence results for Peano arithmetic. J. Symbolic Logic, 43(4):725 73, 978.