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The Reverse Mathematics of Non-Decreasing Subsequences Ludovic Patey The reverse mathematics of non-decreasing subsequences Ludovic Patey To cite this version: Ludovic Patey. The reverse mathematics of non-decreasing subsequences. Archive for Mathematical Logic, Springer Verlag, 2017, 56 (5-6), pp.491 - 506. 10.1007/s00153-017-0536-9. hal-01888765 HAL Id: hal-01888765 https://hal.archives-ouvertes.fr/hal-01888765 Submitted on 5 Oct 2018 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. THE REVERSE MATHEMATICS OF NON-DECREASING SUBSEQUENCES LUDOVIC PATEY Abstract. Every function over the natural numbers has an infinite subdo- main on which the function is non-decreasing. Motivated by a question of Dzhafarov and Schweber, we study the reverse mathematics of variants of this statement. It turns out that this statement restricted to computably bounded functions is computationally weak and does not imply the existence of the halting set. On the other hand, we prove that it is not a consequence of Ramsey's theorem for pairs. This statement can therefore be seen as an arguably natural principle between the arithmetic comprehension axiom and stable Ramsey's theorem for pairs. 1. Introduction A non-decreasing subsequence for a function f : N ! N is a set X ⊆ N such that f(x) ≤ f(y) for every pair x < y 2 X. Every function over N ! N admits an infinite non-decreasing subsequence. Moreover, such a sequence can be computably, but not uniformly obtained from the function f. Indeed, given f : N ! N, either there is a value y 2 N for which the set Sy = fx 2 N : f(x) = yg is infinite, or for every y 2 N, there is a threshold t 2 N after which f(x) > y for every x > t. In the former case, the set Sy is an infinite f-computable non- decreasing subsequence for f, while in the latter case, one can f-computably thin out the set N to obtain an infinite, strictly increasing subsequence. This non-uniform argument can be shown to be necessary by Weihrauch reducing the limited principle of omniscience (LPO) to this statement [1]. In this paper, we study the reverse mathematics of variants of this state- ment by considering various classes of non-computable functions over N ! N. This study is motivated by the following question of Dzhafarov and Schweber in MathOverflow [6] and taken up by Hirschfeldt in his open questions paper [7]. A function f : N × N ! N is stable if for every x 2 N, limy f(x; y) exists. A set X is a limit non-decreasing subsequence for a stable function f : N × N ! N if it is a non-decreasing subsequence of its limit function f~ : N ! N defined by f~(x) = lims f(x; s). Let f : N × N ! N be a computable function such that f(x; s + 1) ≤ f(x; s) for every x; s 2 N. Let X be an infinite limit non- decreasing subsequence for f. How complicated must such an X be? In particular, can it avoid computing the halting set? Let LNS be the statement asserting the existence of an infinite limit non- decreasing subsequence for any such function f : N × N ! N. Liang Yu noticed that LNS implies the existence of a diagonally non-computable function, that is, a function h : N ! N such that h(e) 6= Φe(e) for every e 2 N. We identify a nat- ural strengthening of LNS that we call CNS, standing for \computably bounded 1 2 LUDOVIC PATEY non-decreasing subsequence". We prove that every computable instance of CNS admits low2 solutions using the first jump control of Cholak, Jockusch and Sla- man [2], and show that CNS is a computationally weak statement by proving that it does not imply weak weak K¨onig'slemma (WWKL). On the other hand, 2 CNS is not a consequence of Ramsey's theorem for pairs (RT2) and implies sta- 2 ble Ramsey's theorem for pairs (SRT2). Finally, we separate LNS from CNS by proving that the former does not imply the stable ascending descending sequence principle (SADS) in reverse mathematics. 2. The weakness of non-decreasing subsequences First, note that the general statement of the existence of a non-decreasing sequence for any function over N ! N implies the existence of the halting set. Indeed, let µ be the modulus function of ;0, that is, µ(x) is the minimal stage s 0 0 such that ;sx = ; x, and let f : N ! N be the function defined by f(x) = µ(n) − k − 1, where n and k are the unique solution to the equation x = k + P j<n µ(j) satisfying k < µ(n). µ(2) µ(1) µ(0) The above argument uses finite decreasing sequences to ensure a sufficient amount of sparsity in the non-decreasing subsequences for f, to compute fast- growing functions. At first sight, such an argument does not seem to be applica- ble to LNS since the value of f~(x) = lims f(x; s) is bounded by f(x; 0) for every instance f of LNS. Therefore, one cannot force the solutions to have a hole of size more than f(x; 0) starting from x. Actually, this computable bounding of the function f~ is the essential feature of the weakness of the LNS statement. A function f : N ! N is computably bounded if it is dominated by a computable 0 function. Let CNS be the statement \Every computably bounded ∆2 function admits an infinite non-decreasing subsequence." In particular, CNS generalizes 0 LNS, in that every instance of LNS can be seen as the ∆2 approximation of a computably bounded function. As a warm-up, we prove that CNS admits cone avoidance. Theorem 2.1 Fix a set C and a set A 6≤T C. For every C-computably bounded function f : N ! N, there is an infinite set G non-decreasing for f such that A 6≤T G ⊕ C. Proof. Let b : N ! N be a C-computable function bounding f. Assume that there is no infinite non-decreasing subsequence G such that A 6≤T G ⊕ C, oth- erwise we are done. We will construct the set G using a variant of Mathias forcing. Our forcing condition are pairs (F; X) where F is a finite set of integers non- decreasing for f, X is an infinite set such that max F < min X and such that THE REVERSE MATHEMATICS OF NON-DECREASING SUBSEQUENCES 3 f(x) ≤ f(y) for every x 2 F and y 2 X. We furthermore require that A 6≤T X ⊕ C. A condition d = (E; Y ) extends c = (F; X) if d Mathias extends c, that is, E ⊇ F , Y ⊆ X and E r F ⊆ X. A set G satisfies a condition (F; X) if F ⊆ G ⊆ F [ X. We start by proving that every sufficiently generic filter for this notion of forcing yields an infinite set. Lemma 2.2 For every condition c = (F; X), there is an extension d = (E; Y ) of c such that jEj > jF j. Proof. Pick any x 2 X. By strong cone avoidance of the infinite pigeonhole principle [5], there is an infinite set Y ⊆ X r [0; x] such that either f is constant over Y for some value smaller than f(x), or f(y) ≥ f(x) for every y 2 Y . In the former case, Y is an infinite non-decreasing subsequence for f such that A 6≤T Y ⊕ C, contradicting our assumption. In the latter case, the condition d = (F [ fxg;Y ) is the desired extension of c. A condition c forces a formula '(G) if '(G) holds for every infinite set G satisfying c. We now prove that A 6≤T G ⊕ C for every set G yielded by a sufficiently generic filter. Lemma 2.3 For every condition c = (F; X) and every Turing functional Γ, there is an extension d of c forcing ΓG⊕C 6= A. 0;X⊕C Proof. For every x 2 N and i < 2, let Fx;i be the Π1 class of all functions g : N ! N dominated by b such that for every set E ⊂ X non-decreasing for g, Γ(F [E)⊕C (x) " or Γ(F [E)⊕C (n) #6= i Also define P = f(x; i): Fx;i = ;g. We have three outcomes. In the first case, Fx;1−A(x) 2 P for some x 2 N. In other words, Fx;1−A(x) = ;. In particular, f 62 Fx;1−A(x), so there is a finite set E ⊆ X non-decreasing for f, such that Γ(F [E)⊕C (x) #= 1−A(x). Apply strong cone avoidance avoidance of the infinite pigeonhole principle as in Lemma 2.2 to obtain an infinite set Y ⊆ X such that d = (F [ E; Y ) is a valid extension of c forcing ΓG⊕C (x) #6= A(x). In the second case, there is some x 2 N such that Fx;A(x) 62 P . By the cone avoidance basis theorem [10], there is some g 2 Fx;A(x) such that A 6≤T g⊕X⊕C.
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