Ramsey-like theorems and moduli of computation
Ludovic Patey
Institut Camille Jordan
Université Claude Bernard Lyon 1
43 boulevard du 11 novembre 1918
F-69622 Villeurbanne Cedex
[email protected]
Abstract.
Ramsey’s theorem asserts that every k-coloring of [ω]n admits an infinite monochromatic set. Whenever n≥3, there exists a computable k-coloring of [ω]n whose solutions compute the halting set. On the other hand, for every computable k-coloring of [ω]2 and every non-computable set C, there is an infinite monochromatic set H such that C≤TH. The latter property is known as cone avoidance.
In this article, we design a natural class of Ramsey-like theorems encompassing many statements studied in reverse mathematics. We prove that this class admits a maximal statement satisfying cone avoidance and use it as a criterion to re-obtain many existing proofs of cone avoidance.
This maximal statement asserts the existence, for every k-coloring of [ω]n, of an infinite subdomain H⊆ω over which the coloring depends only on the sparsity of its elements. This confirms the intuition that Ramsey-like theorems compute Turing degrees only through the sparsity of its solutions.
Ramsey’s theorem asserts that every k-coloring of [ω]n admits an infinite monochromatic set, where [X]n denotes the set of the unordered n-tuples over X. This theorem plays an central role in reverse mathematics, as Ramsey’s theorem for pairs is historically the first example of a theorem escaping the structural phenomenon known as the “Big Five” phenomenon. See Simpson [25] for a reference on the early reverse mathematics. In his celebrated theorem, Seetapun [24] proved that Ramsey’s theorem for pairs admits cone avoidance, that is, for any fixed non-computable set C, every computable k-coloring of [ω]2 admits an infinite monochromatic set H which does not compute C. Since then, many consequences of Ramsey’s theorem have been studied from a computability-theoretic viewpoint, including the Erdős-Moser theorem [1], the Ascending Descending sequence principle [12], the free set and thin set theorems [3] and the rainbow Ramsey theorem [6]. Seeing these statements as problems, in terms of instances and solutions, the community studied basis theorems for various computability-theoretic properties, including cone avoidance for computable instances, but also for arbitrary instances. This latter property is known as strong cone avoidance.
In this article, we generalize the above analysis by designing a general class of Ramsey-like statements encompassing the above examples, and providing general criteria to decide whether any such statement admits (strong) cone avoidance. We start with a short survey on cone avoidance for Ramsey’s theorem in Section 1. Then, we define in Section 2 the class of Ramsey-like statements to be those of the form “For every coloring f:[ω]n→k, there is an infinite set H⊆ω avoiding some set of forbidden patterns relative to f.” This class contains Ramsey’s theorem, but also the Erdős-Moser theorem. We prove that this class contains a maximal statement admitting strong cone avoidance (SCA\mbox−RTkn) and cone avoidance (CA\mbox−RTkn) and characterize the statements admitting strong cone avoidance and cone avoidance as those identically reducible to SCA\mbox−RTkn and CA\mbox−RTkn, respectively. In Section 3, we define the class of promise Ramsey-like statements which generalizes the class of Ramsey-like statements by restricting the instances to those satisfying some property. These statements as then of the form “For every coloring f:[ω]n→k such that ω avoids some set of forbidden patterns relative to f, there is an infinite set H⊆ω avoiding some other set of forbidden patterns relative to f.” This enables us to express statements about other structures, such as linear orders and partial orders, including the Ascending Descending sequence and the Chain Antichain principle, but also statements about ω-colorings over [ω]n, such as the free set or the rainbow Ramsey theorem. In Section 4, we apply the previous analysis to reprove many existing theorems, including cone avoidance of Ramsey’s theorem for pairs [24], strong cone avoidance of Ramsey’s theorem for singletons [8], strong cone avoidance of the Erdős-Moser theorem [20], strong cone avoidance of the thin set and free set theorems [27], among others. Last, in Section 5, we state some remaining open questions and suggest further developments.
Put aside the practical application of this general framework to obtain forcing-free proofs of cone and strong cone avoidance, the actual statements of SCA\mbox−RTkn and CA\mbox−RTkn and the resulting decidability criteria provide some further insights on the nature of computation of Ramsey-like statements. One can see a function μ:ω→ω as a measure of largeness of the intervals over ω, by saying that [x,y] is μ-large if μ(x)≤y. The statements SCA\mbox−RTkn and CA\mbox−RTkn are both of the form “For every coloring f:[ω]n→k, there is a function μ:ω→ω and an infinite set H⊆ω such that for every D∈[H]n, f(D) depends only on the μ-largeness analysis over D.” This gives further evidence that Ramsey-like theorems get their computational power out of the sparsity of the solutions, and more generally that Ramsey-like theorems compute through moduli.
1. A short survey on encodability by Ramsey’s theorem
For the sake of clarity, we will adopt a thematic presentation, independently of the historical aspects of these discoveries.
Ramsey’s theorem is a combinatorial theorem at the foundation of Ramsey’s theory. This theory studies the conditions under which given a sufficiently large amount of data, one can see the emergence of some structure.
A k-coloring of [ω]n is a function f:[ω]n→k. A set H⊆ω is f-homogeneous if f is constant on [H]n.
Statement 1.1** (Ramsey’s theorem).**
RTkn: Every coloring f:[ω]n→k admits an infinite f-homogeneous set.
Ramsey’s theorem plays a central role in reverse mathematics. It is historically the first theorem which does not belong to the empirical structural observation of mathematics.
This motivated the computability-theoretic analysis of Ramsey’s theorem and its consequences.
One can see Ramsey’s theorem as a mathematical problem, expressed in terms of instances and solutions. Here, an instance of RTkn is a coloring f:[ω]n→k. A solution to an RTkn-instance f is an infinite f-homogeneous set. The computable analysis of Ramsey’s theorem consists of, given an instance of RTkn, studying the complexity of its solutions from a computable and a proof-theoretic viewpoint. This study started with Jockusch [13], who proved that every computable instance of RTkn admits an arithmetical solution.
In this article, we are interested in the ability of Ramsey’s theorem to compute Turing degrees, that is, the existence of (computable or not) instances of RTkn such that every solution computes a fixed Turing degree.
1.1. Encodability by Ramsey’s theorem
Given a problem P with instances and solutions, we say that a set A is encodable by P, or P-encodable, if there is an (arbitrary) instance of P such that every solution computes A.
The study of RTkn-encodable sets is closely bound to the ability of Ramsey’s theorem to compute fast-growing functions.
Suppose a set A admits a modulus, that is, a function μ:ω→ω such that every function dominating μ computes A. Then, define the RT22-instance f:[ω]2→2 for each x<y by f(x,y)=1 if and only if y≥μ(x), that is, if the interval [x,y] is sufficiently large. Every infinite f-homogeneous set H must be of color 1, and its principal function pH:ω→ω, which on n associates the nth element of H, dominates μ and therefore computes A.
Such an argument shows that if a set A admits a modulus, then it is RTkn-encodable, for n≥2 and k≥2. Moreover, this encodability is witnessed by constructing an instance of RTkn whose solutions are sparse enough so that their principal functions are fast-growing.
The sets admitting a modulus have been studied by Groszek and Slaman [10], who proved that these are precisely the hyperarithmetical sets.
On the other hand, Solovay [26] proved that no other degree can be encoded by Ramsey’s theorem, using the notion of computable encodability. Given a set X⊆ω, we let [X]ω be the collection of all the infinite subsets of X.
Definition 1.2**.**
A set A is computably encodable if for every set X∈[ω]ω, there is a set Y∈[X]ω such that Y≥TA.
Suppose that a set A is computed by an instance f:[ω]n→k of RTkn, in other words, every infinite f-homogeneous set computes A.
In particular, since for every set X∈[ω]ω, there is an infinite f-homogeneous set H⊆X, then the set A is computably encodable. The following equivalence proves that the only Turing degrees which can be computed by an instance of Ramsey’s theorem are the ones which admit a modulus, hence those who can be computed using fast-growing functions.
Theorem 1.3** (Solovay [26], Groszek and Slaman [10]).**
Given a set A, the following are equivalent
A* is computably encodable*
A* is hyperarithmetic*
A* admits a modulus*
One case however remains, namely, Ramsey’s theorem for singletons. Intuitively, one cannot build an instance of RT21 whose solutions are sparse everywhere, so that their principal functions are sufficiently fast-growing. Indeed, given a RT21-instance f:ω→2, in order to ensure that the f-homogeneous sets for color 0 are all sparse, one has to set f(x)=1 for many x∈ω. But then, one can construct non-sparse f-homogeneous sets for color 1. Actually, Ramsey’s theorem for singletons has no encodability power, as formalized by the notion of strong cone avoidance.
Definition 1.4**.**
A problem P admits strong cone avoidance if for every set Z, every C≤TZ and every P-instance X, there is a P-solution Y to X such that C≤TZ⊕Y.
Building on the work of Seetapun and Slaman [24] and of Cholak, Jockusch and Slaman [4], Dzhafarov and Jocksuch [8] proved that Ramsey’s theorem for singletons (RTk1) admits strong cone avoidance, thereby completing the picture of which sets are encodable by Ramsey’s theorem. To summarize, a set is encodable by RTkn with n≥2 if and only if it is hyperarithmetic, and is encodable by RTk1 if and only if it is computable.
1.2. Encodability by computable instances
One may naturally want to refine the previous analysis, and study the RTkn-encodable sets with respect to the computational complexity of its instances. Jockusch [13] proved that every computable instance of RTkn admits an arithmetical solution. On the lower bound side, he proved that for every n≥3, there is computable instance of RT2n such that every solution computes ∅(n−2). The proof is a simple effectivization of the previous section.
On the other direction, Seetapun and Slaman [24] proved that Ramsey’s theorem for pairs (RTk2) has no encodability power when restricted to computable instances, in the following sense.
Definition 1.5**.**
A problem P admits cone avoidance if for every set Z, every C≤TZ and every Z-computable P-instance X, there is a P-solution Y to X such that C≤TZ⊕Y.
While strong cone avoidance expresses the combinatorial failure of P to encode any non-computable set, cone avoidance only expresses the computational weakness of P. There is a deep link between the combinatorial features of RTkn and the computational features of RTkn+1, as expressed by Cholak and Patey [2, Theorem 1.5]. One can deduce cone avoidance of RTk2 from strong cone avoidance of RTk1, although historically, Seetapun and Slaman [4] first proved cone avoidance of RTk2.
Cholak, Jockusch and Slaman [4, Theorem 12.2] (proof fixed in [11, Appendix A]), relativized cone avoidance of RTk2 to prove that for every n≥2, if a set A is not Δn−10, then every computable instance of RTkn admits a solution H such that A≤TH. This completes the study of the sets encodable by computable instances of RTkn. Indeed, for every n≥2, a set A is encodable by a computable instance of RTkn if and only if A is Δn−10.
1.3. Encodability by the thin set theorems
The previous sections give a complete picture of which sets are encodable by Ramsey’s theorem, with or without restricting the complexity of the instances. This could be the end of the story. However, Wang [27] surprisingly showed that by weakening the notion of homogeneity to allow sufficiently many colors in the solutions, one obtains strong cone avoidance.
Statement 1.6** (Thin set theorem).**
RT<∞,ℓn: For every coloring f:[ω]n→k, there is an infinite set H⊆ω such that ∣f[H]n∣≤ℓ.
Wang [27] proved that for every n≥1 and every sufficiently large ℓ∈ω, RT<∞,ℓn admits strong cone avoidance. On the other hand, Cholak and Patey [2, Theorem 3.2], adapting Dorais et al. [7, Proposition 5.5], proved that if ℓ<2n−1, RT<∞,ℓn can still compute arbitrarily fast-growing functions, and therefore computes all the hyperarithmetic sets. More precisely.
Theorem 1.7** (Cholak and Patey).**
Fix n≥1.
For every function μ:ω→ω, there is an instance of RT<∞,2n−1−1n such that every solution computes a function dominating μ.
If A is not arithmetical, then every instance of RT<∞,2n−1n has a solution which does not compute A.
Using the previous sections, whenever a set A is not hyperarithmetical, it is not RT2n-encodable, hence not RT<∞,ℓn-encodable for any ℓ≥1. Whenever A is hyperarithmetical, but not arithmetical, then it is RT<∞,ℓn-encodable if and only if ℓ<2n−1. The case of the the arithmetical sets must be treated independently.
Consider the halting set ∅′. The standard modulus of ∅′ is defined by letting μ(n) be the smallest time t at which for every e<n, if Φe(e)↓, then Φe(e) halts before stage t. One can in particular computably approximate the function μ from below. This is the notion of left-c.e. function.
Definition 1.8**.**
A function μ:ω→ω is left-c.e. if there is a uniformly computable sequence of functions μ0,μ1,… with μs:ω→ω such that for every s∈ω, μs≤μs+1, and for every x∈ω, limsμs(x)=μ(x).
The notion relativizes, and we say that a function is left-X-c.e. if the sequence of functions is uniformly X-computable. When mentioning a left-c.e. function, we will always assume that the sequence of its approximations is specified. This is why we will sometime talk about relative left-c.e. function simply to say that a sequence of lower approximations of the function is fixed, no matter the effectiveness of the sequence.
Cholak and Patey [2] studied the threshold of ℓ under which, given a left-c.e. function μ, there is an instance of RT<∞,ℓn such that every solution computes a function dominating μ. This happens to be exactly the Catalan sequence, inductively defined by C0=1 and
[TABLE]
In particular, C0=1, C1=1, C2=2, C3=5, C4=14, C5=42, C6=132, C7=429, … Note that this sequence corresponds to the OEIS sequence A000108.
Theorem 1.9** (Cholak and Patey [2]).**
Fix n≥1.
For every left-c.e. function μ:ω→ω, there is an instance of RT<∞,Cn−1n such that every solution computes a function dominating μ.
If A is not computable, then every instance of RT<∞,Cnn has a solution which does not compute A.
By iterating the notion of left-c.e. function, one can RT<∞,Cn−1n-encode all the arithmetical sets for every n≥2.
This completes the picture of the RT<∞,ℓn-encodable sets, depending on the value of n and ℓ. There are actually three classes of RT<∞,ℓn-encodable sets: the computable, arithmetical, and hyperarithmetical sets. One can also deduce which sets are encodable by computable instances of RT<∞,ℓn, using the bridge between the combinatorics of RT<∞,ℓn and the computations of RT<∞,ℓn+1. See Cholak and Patey [2] for this analysis.
2. Ramsey-like theorems
As explained, RT2n does not admit strong cone avoidance for every n≥2. However, there exist some weakenings of Ramsey’s theorem which admit strong cone avoidance. The thin set theorem is an example, but also the Erdos-Moser theorem. Given a coloring f:[ω]2→2, the Erdos-Moser theorem (EM) asserts the existence of an infinite set H⊆ω over which f is transitive, that is, for every x<y<z and i<2, if f(x,y)=i and f(y,z)=i then f(x,z)=i. The author [20] proved that EM admits strong cone avoidance. Ramsey’s theorem, the thin set theorem and the Erdős-Moser theorem are all of the form “For every coloring f:[ω]n→k, there exists an infinite set H⊆ω avoiding some set of forbidden patterns relative to f.” In this section, we design a general class of Ramsey-like theorems, and provide a criterion to decide which statements admit strong cone avoidance.
Definition 2.1**.**
Fix a countable collection of variables x0,x1,…
An RTkn-pattern P is a finite conjunction of formulas of the form f({xi:i∈D})=v for some D∈[ω]n and v<k, where f is a function symbol of type [ω]n→k.
Given an actual coloring f:[ω]n→k, a set of integers E={n0<n1<⋯<nr−1} f-satisfies an RTkn-pattern P≡f({xi:i∈D0})=v0∧⋯∧f({xi:i∈Dℓ−1})=vℓ−1 if for every s<ℓ, letting Es={ni:i∈Ds}, f(Es)=vs.
A set H⊆ω f-meets P if H contains an finite subset f-satisfying P. Otherwise, H f-avoids P.
For example, f(x5,x6)=0∧f(x6,x7)=1∧f(x5,x7)=0 is an RT22-pattern.
Definition 2.2**.**
Given a collection W of RTkn-patterns, the RTkn-like problem RTkn(W) is the problem whose instances are colorings f:[ω]n→k. An RTkn(W)-solution to an instance f is an infinite set H⊆ω f-avoiding every pattern in W.
In particular, RT22 is the RT22-like problem RT22(WRT22) with WRT22={f(x0,x1)=0∧f(x2,x3)=1,f(x0,x1)=1∧f(x2,x3)=0,f(x0,x2)=0∧f(x1,x3)=1,f(x0,x2)=1∧f(x1,x3)=0}. Similarly, EM is the RT22-like problem RT22(WEM) with WEM={f(x0,x1)=0∧f(x1,x2)=0∧f(x0,x2)=1,f(x0,x1)=1∧f(x1,x2)=1∧f(x0,x2)=0}.
2.1. True Ramsey-like problems
We say that a problem is true if every instance has a solution. Intuitively, Ramsey’s theorem is the strongest structural property we can get out of a k-coloring of [X]n, where X is an abstract set. We formalize this intuition by proving that Ramsey’s theorem is the maximal true Ramsey-like theorem.
Definition 2.3**.**
Let P and Q be two problems with dom(P)⊆dom(Q).
We say that P is identically reducible to Q (written P≤idQ)
if for every I∈dom(P), every Q-solution to I is a P-solution to I.
Theorem 2.4**.**
Fix n,k≥1 and a collection W of RTkn-patterns.
Then RTkn(W) is true if and only if RTkn(W)≤idRTkn.
Proof.
⇒: Suppose RTkn(W)≤idRTkn. Let f:[ω]n→k be an instance of RTkn(W) and let H be an infinite f-homogeneous set for some color i<2 which f-meets some P∈W. Then P is of the form f({xj:j∈D0})=i∧⋯∧f({xj:j∈Dℓ−1})=i. Then, letting g:[ω]n→k be defined for every D∈[ω]n by g(D)=i, g has no RTkn(W)-solution, therefore RTkn(W) is not true.
⇐: Suppose RTkn(W)≤idRTkn. Given an instance f:[ω]n→k of RTkn(W), by the classical Ramsey theorem, there is an infinite f-homogeneous set H. In particular, H is an RTkn(W)-solution to f, so f admits an RTkn(W)-solution and RTkn(W) is true.
∎
Before starting the analysis of cone avoidance for Ramsey-like theorems, let us introduce an important concept which will be implicitly used all over the article.
Definition 2.5**.**
A problem P is zoomable if for every P-instance I and every infinite set X={x0<x1<…}, there is a P-instance IX such that for every P-solution Y to IX, {xn:n∈Y} is a P-solution to I.
One can see zoomable problems as saying that given an instance I and an infinite set X, we can zoom in on the set X and consider it as the new set ω by renaming the elements of X. Then, after having built a solution within X seen as ω, we can zoom out and see it as a subset of X. For example, Ramsey’s theorem is a zoomable problem, while Hindman’s theorem is not, since the zoom operation changes the semantics of the addition.
The following lemma which is implicitly used everywhere asserts that whenever a zoomable problem P admits strong cone avoidance, then one can apply this strong cone avoidance within the scope of a cone avoiding reservoir.
Lemma 2.6**.**
Let P be a zoomable problem which admits strong cone avoidance.
For every set Z, every set C≤TZ, every infinite Z-computable set X and every P-instance I, there is a P-solution Y⊆X such that C≤TZ⊕Y.
Proof.
Fix Z, C and X. Since P is zoomable, there is a P-instance IX such that for every P-solution Y to IX, {xn:n∈Y} is a P-solution to I. By strong cone avoidance of P applied to IX, there is a P-solution Y to IX such that C≤TZ⊕Y. The set YX={xn:n∈Y} is a Z⊕Y-computable P-solution to I. In particular, YX⊆X and C≤TZ⊕YX.
∎
Similar lemmas can be proven for cone avoidance and strong cone avoidance for non-arithmetical cones. We will use these lemmas without any further mention.
2.2. Strongly avoiding non-arithmetical cones
Recall that given a modulus μ:ω→ω of a set A, one can define a coloring f:[ω]2→2 by f(x,y)=1 if and only if μ(x)≤y. Any infinite f-homogeneous set computes a function dominating μ. Cholak and Patey [2] generalized this coding in the following sense.
Definition 2.7**.**
A graph G=({0,…,n−1},E) of size n is a vector graph if
E⊆{{i,i+1}:i<n−1}.
A vector graph of size n is just a representation of a {0,1}-value vector of size n−1. The choice of a graph representation is for uniformity with the later sections. Given n≥1, we let Vn be the set of all vector graphs of size n. In particular, ∣Vn∣=2n−1. We shall actually consider functions μ:ω→ω+, where ω+ is the successor ordinal of ω.
Definition 2.8**.**
Let μ:ω→ω+ be a function. For every n≥1 and D={x0<⋯<xn−1}∈[ω]n, let Vn(μ,D) be the graph G=({0,…,n−1},E) such that for each i<n−1, {i,i+1}∈E if and only if μ(xi)≤xi+1.
An interval [x,y] is μ-large if μ(x)≤y. Otherwise, it is μ-small.
Given a finite set D={x0<⋯<xn−1}, Vn(μ,D) is supposed to code the whole μ-largeness information over D. Actually, Vn(μ,D) contains only the information about the adjacent intervals, namely, intervals of the form [xi,xi+1]. What about the largeness information about non-adjacent ones? For example, if [x,y] and [y,z] are both μ-small, is [x,z] μ-small or μ-large? We shall see through the notion of μ-transitivity that we can always restrict ourselves to sets over which the whole information of μ-largeness is already fully specified by the μ-largeness information on the adjacent intervals.
A set H⊆ω is μ-transitive if for every x<y<z∈H, μ(x)>y and μ(y)>z if and only if μ(x)>z.
Lemma 2.9**.**
Fix μ:ω→ω+ and ρ:ω→ω+.
Given n≥1, let E={x0<⋯<xn−1} be a μ-transitive set, and F={y0<⋯<yn−1} be a ρ-transitive set such that Vn(μ,E)=Vn(ρ,F). Then for every i<j<n, [xi,xj] is μ-large if and only if [yi,yj] is ρ-large.
Proof.
We prove by induction over m≥1 that for every i, [xi,xi+m] is μ-large if and only if [yi,yi+m] is ρ-large. The base case m=1 is the lemma hypothesis. Suppose it holds up to m. Since E is μ-transitive, then [xi,xi+m+1] is μ-large if and only if either [xi,xi+m] or [xi+m,xi+m+1] is μ-large. By induction hypothesis, this holds if and only if either [yi,yi+m] or [yi+m,yi+m+1] is ρ-large. Since F is ρ-transitive, this holds if and only if [yi,yi+m+1] is ρ-large.
∎
Cholak and Patey [2, Theorem 3.2] proved the following theorem. Note that μ ranges over ω and not ω+.
Theorem 2.10** (Cholak and Patey [2]).**
Let μ:ω→ω be a function. For every n≥1, define fn:D↦Vn(μ,D). For every infinite set H⊆ω such that Vn⊆fn[H]n, H computes a function dominating μ.
In particular, this proves that for every hyperarithmetical set A, there is an RT<∞,2n−1−1n-instance such that every solution computes A. Moreover, this coding technique optimal for non-arithmetical sets from the viewpoint of the number of colors, in the sense that if A is non-arithmetical, then every RT<∞,2n−1n-instance admits a solution which does not compute A.
We now prove that this optimality is not only about the number of colors, but also on the nature of the coding, by proving that for every coloring f:[ω]n→k and every non-arithmetical set A, then there exists an infinite subdomain H⊆ω such that A≤TH, and over which f↾[H]n behaves exactly like our function D↦Vn(μ,D), up to a renaming of the colors.
Statement 2.11**.**
ARITH\mbox−SCA\mbox−RTkn: For every function f:[ω]n→k,
there is a function μ:ω→ω+, an infinite μ-transitive set H⊆ω, and a coloring χ:Vn→k such that for every D∈[H]n, f(D)=χ(Vn(μ,D)).
For n=1, there is only one vector graph of size 1, namely G=({0},∅}. Therefore ∣V1∣=1, and ARITH\mbox−SCA\mbox−RTk1 states the existence, for every function f:ω→k, of a unique color i<k such that for every x∈H, f(x)=i. Thus ARITH\mbox−SCA\mbox−RTk1 is RTk1.
The case n=2 yields a new principle.
Statement 2.12**.**
LARGEk: For every coloring f:[ω]2→k, there are some colors is,iℓ<k and an infinite set H⊆ω such that f[H]2⊆{is,iℓ} and for every x<y<z∈H, f(x,y)=f(y,z)=is if and only if f(x,z)=is.
Intuitively, is and iℓ are the colors of small and large intervals, respectively.
Lemma 2.13**.**
ARITH\mbox−SCA\mbox−RTk2* is the statement LARGEk.*
Proof.
We first prove that LARGEk≤idARITH\mbox−SCA\mbox−RTk2.
Let f:[ω]2→k be a coloring, and let H be an infinite ARITH\mbox−SCA\mbox−RTk2-solution. By definition of ARITH\mbox−SCA\mbox−RTk2, there is some function μ:ω→ω+ and a coloring χ:V2→k such that H is μ-transitive, and for every x<y∈H, f(x,y)=χ(V2(μ,{x,y})). Let G0=({0,1},∅} and G1=({0,1},{{0,1}}). In particular, V2={G0,G1}. Let is=χ(G0) and iℓ=χ(G1).
Since for every x<y∈H, f(x,y)=χ(V2(μ,{x,y})), f[H]2⊆{is,iℓ}.
We claim that for every x<y<z∈H, f(x,y)=f(y,z)=is if and only if f(x,z)=is. If is=iℓ, then H is f-homogeneous for color is, and satisfies the property, so suppose is=iℓ. In other words, χ is one-to-one.
Fix x<y<z∈H. Then f(x,y)=f(y,z)=is if and only if V2(μ,{x,y})=V2(μ,{y,z})=G0, if and only if μ(x)>y and μ(y)>z. By μ-transitivity of H, this holds if and only if μ(x)>z, hence V2(μ,{x,z})=G0. Since χ is one-to-one, this holds if and only if f(x,z)=χ(V2(μ,{x,z}))=χ(G0)=is.
We now prove that ARITH\mbox−SCA\mbox−RTk2≤idLARGEk.
Let f:[ω]2→k be a coloring, and let H⊆ω and is,iℓ<2 be such that f[H]2⊆{is,iℓ} and for every x<y<z∈H, f(x,y)=f(y,z)=is if and only if f(x,z)=is. Let χ(G0)=is and χ(G1)=iℓ.
For every x∈ω, let μ(x) be either min{y>x:y∈H∧f(x,y)=iℓ} if it exists, and μ(x)=ω otherwise.
We first claim that H is μ-transitive. Indeed, for every x<y<z∈H, μ(x)>y and μ(y)>z if and only if f(x,y)=f(y,z)=is. Since H is a P-solution to f, this holds if and only if f(x,z)=is, hence if and only if μ(x)>z.
Last, we claim that H is an ARITH\mbox−SCA\mbox−RTk2-solution to f with witness χ and μ. Fix x<y∈H. Then f(x,y)=is if and only if μ(x)>y, if and only if V2(μ,{x,y})=G0, if and only if χ(V2(μ,{x,y}))=χ(G0)=is. Thus f(x,y)=χ(V2(μ,{x,y})).
∎
By compactness, if we don’t consider μ and χ as part of the solution , then ARITH\mbox−SCA\mbox−RTkn can be seen as an RTkn-like problem.
Lemma 2.14**.**
There is a c.e. set of RTkn-like patterns W such that ARITH\mbox−SCA\mbox−RTkn is the problem RTkn(W).
Proof.
Fix a coloring f:[ω]n→k.
By compactness, a set H⊆ω is an ARITH\mbox−SCA\mbox−RTkn-solution if and only if for every finite set F⊆H, there is a function μ:F→ω+ and a coloring χ:Vn→k such that F is μ-transitive and for every D∈[F]n, f(D)=χ(Vn(μ,D)).
Let W be the set of all RTkn-patterns such that the above property does not hold. Then ARITH\mbox−SCA\mbox−RTkn is the statement RTkn(W).
∎
We say that a problem P admits strong cone avoidance for non-arithmetical cones if for every set Z, every non-Z-arithmetical set C, and every P-instance X, there is a P-solution Y such that C≤TZ⊕Y.
Theorem 2.15**.**
ARITH\mbox−SCA\mbox−RTkn* admits strong cone avoidance for non-arithmetical cones.*
Proof.
Fix a set Z, a non-Z-arithmetical set C and a coloring f:[ω]n→k.
Suppose first that C is not Z-hyperarithmetical. By Solovay [26], C is not computably encodable relative to Z. Since for every infinite set X⊆ω, there is an infinite ARITH\mbox−SCA\mbox−RTkn-solution Y⊆X to f, there is an ARITH\mbox−SCA\mbox−RTkn-solution H to f such that C≤TZ⊕H.
Suppose now that C is Z-hyperarithmetical. By Groszek and Slaman [10], there is a modulus μ:ω→ω relative to Z, that is, for every function g dominating μ, C≤TZ⊕g.
Let f1 be defined for each D∈[ω]n by f1(D)=⟨f(D),Vn(μ,D)⟩.
By strong cone avoidance of RT<∞,2n−1n for non-arithmetical cones (see Cholak and Patey [2, Theorem 4.15]), there is an infinite set H⊆X such that C≤TZ⊕H and ∣f1[H]n∣≤2n−1. In particular, H⊕Z does not compute a function dominating μ, so by Theorem 2.10, for every G∈Vn, there is some i<k and some D∈[H]n such that f1(D)=⟨i,G⟩. Since ∣Vn∣=2n−1, this i is unique. For each G∈Vn, let χ(G) be this unique i.
We claim that for every D∈[H]n, f(D)=χ(Vn(μ,D)).
By definition of χ, f1(D)=⟨f(D),Vn(μ,D)⟩=⟨χ(Vn(μ,D)),Vn(μ,D)⟩. It follows that f(D)=χ(Vn(μ,D)). By Cholak and Patey [2, Corollary 5.5], there is an infinite μ-transitive subset H1⊆H such that C≤TZ⊕H1. Therefore, H1 is an ARITH\mbox−SCA\mbox−RTkn-solution to f.
∎
The following technical lemma will be useful for Theorem 2.17, Lemma 3.3 and Lemma 3.4. Note that μ0 ranges over ω+ while μ1 ranges over ω.
Lemma 2.16**.**
Fix χ:Vn→k, and let f0:[ω]n→k and f1:[ω]n→k be two colorings.
Let H0 be an infinite ARITH\mbox−SCA\mbox−RTkn-solution to f0 with witnesses χ and μ0:ω→ω+, and let H1 be an infinite ARITH\mbox−SCA\mbox−RTkn-solution to f1 with witnesses χ and μ1:ω→ω.
If H0 f0-meets some RTkn-pattern P but H1 f1-avoids P, then H1 computes a function dominating μ1.
Proof.
Since H0 f0-meets some RTkn-pattern P, there is a finite set E0={x0<⋯<xr−1}⊆H0 which f0-meets P.
Suppose for the sake of contradiction that H1 does not compute a function dominating μ1.
By Theorem 2.10, there is a finite set E1={y0<⋯<yr−1}⊆H1 of size r such that Vr(μ0,E0)=Vr(μ1,E1). By Lemma 2.9, since H0 is μ0-transitive and H1 is μ1-transitive, for every I∈[r]n, letting D0={xi:i∈I} and D1={yi:i∈I}, Vn(μ0,D0)=Vn(μ1,D1). Since f0(D0)=χ(Vn(μ0,D0)) and f1(D1)=χ(Vn(μ1,D1)), then f0(D0)=f1(D1). Thus f0↾[E0]n and f1↾[E1]n have the same function graph.
It follows that E1 f1-meets P, so H1 f1-meets P. Contradiction.
∎
Theorem 2.17**.**
Let W be a collection of RTkn-patterns such that RTkn(W)≤idARITH\mbox−SCA\mbox−RTkn. Then for every function μ:ω→ω, there is an RTkn(W)-instance such that every solution computes a function dominating μ.
Proof.
Since RTkn(W)≤idARITH\mbox−SCA\mbox−RTkn, there is a coloring ffail:[ω]n→k and an ARITH\mbox−SCA\mbox−RTkn-solution Hfail to ffail witnessed by a function μfail:ω→ω+ and a coloring χ:Vn→k, and such that Hfail meets some RTkn-pattern P∈W.
Let μ:ω→ω be a function. Let f:[ω]n→k be an instance of RTkn(W) defined by f(D)=χ(Vn(μ,D)).
We claim that every RTkn(W)-solution H⊆ω to f computes a function dominating μ. Let H be an RTkn(W)-solution to f. In particular, H f-avoids P. If H does not compute a function dominating μ, then by Cholak and Patey [2, Theorem 5.11], there is an infinite μ-transitive subset H1⊆H such that H1 does not compute a function dominating μ.
In particular, H1 is an infinite ARITH\mbox−SCA\mbox−RTkn-solution to f with witnesses χ and μ, and such that H1 f-avoids P. By Lemma 2.16, H computes a function dominating μ, contradiction.
∎
Actually, ARITH\mbox−SCA\mbox−RTkn is the strongest RTkn-like problem which admits this avoidance property. The following theorem therefore provides a simple criterion to decide whether an RTkn-like problem admits strong cone avoidance for non-arithmetical cones.
Theorem 2.18**.**
A problem RTkn(W) admits strong cone avoidance for non-arithmetical cones if and only if RTkn(W)≤idARITH\mbox−SCA\mbox−RTkn.
Proof.
⇐: Suppose RTkn(W)≤idARITH\mbox−SCA\mbox−RTkn. Fix a set Z, a non-Z-arithmetical set C and a coloring f:[ω]n→k. By Theorem 2.15, there is an ARITH\mbox−SCA\mbox−RTkn-solution H to f such that C≤TZ⊕H. In particular, H is an RTkn(W)-solution to f.
⇒: Suppose RTkn(W)≤idARITH\mbox−SCA\mbox−RTkn. Let μ:ω→ω be a modulus of some non-arithmetical set C. By Theorem 2.17, there is a RTkn(W)-instance such that every solution computes a function dominating μ, hence computes C. Therefore RTkn(W) does not admit strong cone avoidance for non-arithmetical cones.
∎
2.3. Strongly avoiding non-computable cones
Whenever the modulus μ:ω→ω is left-c.e., that is, there is a uniformly computable sequence of functions μ0≤μ1≤… pointwise limiting to μ, one can exploit more information to compute functions dominating μ. For example, let f:[ω]3→2 be defined by f(x,y,z)=⟨b0,b1,b2⟩, where b0=1 if and only if μ(x)≤y, b1=1 if and only if μ(y)≤z, and b2=1 if and only if μz(x)≤y. Cholak and Patey [2, Theorem 3.17] proved that every infinite set H⊆ω such that ∣f[H]3∣≤4 computes a function dominating μ.
We refine again their analysis to obtain a maximal RTkn-like principle admitting strong cone avoidance.
Definition 2.19** (Cholak and Patey [2]).**
A largeness graph of size n is a graph ({0,…,n−1},E) such that
If {i,i+1}∈E, then for every j>i+1, {i,j}∈E
If i<j<n, {i,i+1}∈E and {j,j+1}∈E, then {i,j+1}∈E
If i+1<j<n−1 and {i,j}∈E, then {i,j+1}∈E
If i+1<j<k<n and {i,j}∈E but {i,k}∈E, then {j−1,k}∈E
Let us explain the intuition behind the definition of a largeness graph. Given a left-c.e. function μ:ω→ω and a finite μ-transitive set D={x0<x1<⋯<xn−1}, one can define a graph G=({0,…,n−1},E) which will represent the whole information about the largeness of the intervals [xi,xj] for each i<j<n. Since D is μ-transitive, the information about μ-largeness of the intervals [xi,xj] for every i<j is already fully specified by the information about μ-largeness of [xi,xi+1]. Therefore, this information is coded only into the adjacent edges. We then set {i,i+1}∈E if and only if [xi,xi+1] is μ-large.
Whenever [xi,xi+1] is μ-small, this is witnessed after a finite approximation stage μs, that is, [xi,xi+1] is μs-small for all but finitely many s∈ω. We can therefore code in G the information whether some xj (with j>i+1) is large enough to witness this fact, in the sense that μxj(xi)>xi+1. This information is coded into the edges {i,j} with i+1<j. We then set {i,j}∈E if [xi,xi+1] is μxj-small. Although this coding does not seem consistent with the adjacent edges since the existence of an adjacent edge gives a largeness information, one should really see μxj-smallness of [xi,xi+1] as an information of how big the number xj is and not on how small the interval [xi,xi+1] is.
Let us now look at the properties (a) to (d), successively. Property (a) says that if the interval [xi,xi+1] is μ-large, then it is never μxj-small. This property is ensured by construction.
Property (b) says that if [xi,xi+1] is μ-small and [xj,xj+1] is μ-large, then the approximation time xj+1 is large enough to witness the smallness of [xi,xi+1]. In other words, [xi,xi+1] is gxj+1-small. This property is not structurally ensured, and must be obtained by some extra assumptions on the function μ.
Property (c) simply says that if [xi,xi+1] is μxj-small, then it is μxj+1-small. This property is structurally ensured by the fact that μxj≤μxj+1, which is true of all left-c.e. approximations.
Property (d) says that if xk is large enough to witness the μ-smallness of [xi,xi+1], but that xj was not large enough to witness it, then not only by property (b), [xj−1,xj] cannot be μ-large, but furthermore xk is large enough to witness the smallness of [xj−1,xj]. In some sense, Property (d) is a refinement of property (b). This property must be also ensured by some extra assumptions on μ.
Let Ln be the set of all largeness graphs of size n. Recall that Cn is the nth Catalan number. Cholak and Patey [2, Lemma 3.16] proved that for every n≥1, ∣Ln∣=Cn.
Definition 2.20**.**
Given a relative left-c.e. function μ:ω→ω+ with approximations μ0,μ1,… and a set D={x0,…,xn−1}⊆ω,
let Ln(μ,D) be the graph ({0,…,n−1},E) where
E={{p,q}:μxq(xp)>xp+1∧p+1<q<n}∪{{p,p+1}:μ(xp)≤xp+1}.
Note that Ln(μ,D) is not a largeness graph in general, because the properties (b) and (d) are not structurally satisfied. We give a sufficient property to ensure Ln(μ,D)∈Ln.
Definition 2.21**.**
A relative left-c.e. function μ:ω→ω is strongly increasing
if for every s∈ω and x<y∈ω, μs(x)≤μs(y), and if μs+1(x)>μs(x) then μs+1(y)>s.
Lemma 2.22**.**
Fix a strongly increasing relative left-c.e. function μ:ω→ω+.
For every n≥1 and D∈[ω]n, Ln(μ,D)∈Ln.
Proof.
Claim 1: For every w<x<y<z, if μy(w)≤x and μz(w)>x then μz(x)>y. Let s∈[y,z) be such that μs(w)≤x and μs+1>x. Then since μ is strongly increasing, μs+1(x)>s≥y. In particular, μz(x)≥y. This proves Claim 1.
Claim 2: For every w<x<y, if μ(w)>x and μ(x)≤y, then μy(w)>x.
Suppose that Claim 2 does not hold. Then there is some w<x<y such that
μ(w)>x, μ(x)≤y and μy(w)≤x. Then there is a stage z>y such that μz(w)>x and μz(x)≤y. Since μy(w)≤x and μz(w)>x, but μz(x)≤y, we contradict Claim 1. This proves Claim 2.
Fix n≥1 and D={x0<⋯<xn−1}∈[ω]n.
We check properties (a−d) of Definition 2.19 for Ln(μ,D)=({0,…,n−1},E).
(a): If {i,i+1}∈E, then μ(xi)≤xi+1. In particular, for every j>i+1, since μxj≤μ, μxj(xi)≤xi+1, so {i,j}∈E.
(b): If i<j<n and {i,i+1}∈E and {j,j+1}∈E. Then μ(xi)>xi+1 and μ(xj)≤xj+1. Since μ is increasing, μ(xi+1)≤xj+1. Then by Claim 2, μxj+1(xi)>xi+1. Thus {i,j+1}∈E.
(c): If i+1<j<n−1 and {i,j}∈E, then μxj(xi)>xi+1. Since μxj≤μxj+1, then μxj+1(xi)>xi+1, so {i,j+1}∈E.
(d): If i+1<j<k<n and {i,j}∈E but {i,k}∈E. Then μxj(xi)≤xi+1 but μxk(xi)>xi+1. By Claim 1, μxk(xi+1)>xj. Since μxk is increasing, μxk(xj−1)>xj. Thus {j−1,k}∈E.
∎
We now prove that whenever we choose a left-c.e. modulus for a set, we can always assume that it is strongly increasing, without loss of generality.
Lemma 2.23**.**
Every left-c.e. function μ:ω→ω
is dominated by a strongly increasing left-c.e. function g:ω→ω.
Proof.
Fix μ:ω→ω with approximations μ0,μ1,….
We define a uniformly computable sequence of functions g0≤g1≤⋯:ω→ω pointwisely, that is, at stage x, we define gs(x) for every s∈ω. During the construction, for each approximation time s and value x, we associate a moving threshold ts,x∈ω which can only increase, starting with ts,x=0. We will ensure that gs(x)≥ts,x.
At stage x, suppose we have defined gs(y) for every y<x and s∈ω. We define gs(x) for every s∈ω.
At approximation time s, having defined gu(x) for every u<s, we define gs(x) to be the maximum value among {gs(y):y<x}, {gu(x):u<s}, ts,x and μs(x).
If s>0 and gs(x)>gs−1(x), then set ts,y=max(ts,y,s) for every y>x.
Then go to the next approximation time s+1.
Once all the approximations ⟨gs(x):s∈ω⟩ are defined, then go to the next stage x+1.
Claim 1: g is a strongly increasing left-c.e. function.
By making gs(x) larger than {gs(y):y<x}, we have ensured that the function gs is non-decreasing, that is, for every x<y, gs(x)≤gs(y).
By making gs(x) larger than {gu(y):u<s}, we have ensured that gs≤gs+1, hence that g0,g1,… are left-c.e. approximations.
By making gs(x) larger than ts,x, we have ensured that if there is some y<x such that gs−1(y)<gs(y), then gs(x)>s−1. This proves Claim 1.
By making gs(x) larger than μs(x), we have ensured that g(x)=limsgs(x)≥limsμs(x)=μ(x). So g dominates μ.
Claim 2: For every x, g(x)<ω. We prove it by induction over x. Indeed, by induction hypothesis, there is a time v∈ω such that for every y<x and s>v, gs(y)=gv(y).
For every s>v, the maximum value among {gs(y):y<x} is bounded by
maximum value among {g(y):y<x}, which is a finite number by induction hypothesis. Moreover, ts,x=0 since only stages y<x at time s can change the threshold ts,x. Moreover, μs(x) is bounded by μ(x) which is again a finite number. Let m be the maximum value among {g(y):y<x}, {gu(x):u<v} and μ(x). Then gs(x)≤m for every s>v. It follows that limsgs(x)≤m.
This completes the proof of Lemma 2.23.
∎
By Lemma 2.22, taking a strongly increasing relative left-c.e. modulus ensures that Ln(μ,D) is a largeness graph (Definition 2.19). As seen in Lemma 2.9, μ-transitivity ensures that μ-largeness of the intervals over a set D is fully specified by the μ-largeness of its adjacent intervals. However, some information still seems to be missing in Ln(μ,D). Indeed, suppose that [xi,xi+1] and [xi+1,xi+2] are both μxj-small, that is, {i,j} and {i+1,j} are both edges on Ln(μ,D). By μ-transitivity, we know that [xi,xi+2] is μ-small. However, is it μxj-small? Thanks to a stronger notion of μ-transitivity, we can ensure that it will always be the case.
A set H⊆ω is strongly μ-transitive if it is μ-transitive, and for every w<x<y<z∈H such that μz(w)>x and μz(x)>y, then μz(w)>y. In other words, if z witnesses μ-smallness of both [w,x] and [x,y], then it witnesses μ-smallness of [w,y].
Lemma 2.24**.**
Fix two relative left-c.e. functions μ:ω→ω+ and ρ:ω→ω+.
Given n≥1, let E={x0<⋯<xn−1} be a strongly μ-transitive set, and F={y0<⋯<yn−1} be a strongly ρ-transitive set such that Ln(μ,E)=Ln(ρ,F). Then for every i<j<k<n, [xi,xj] is μk-large if and only if [yi,yj] is ρk-large.
Proof.
We prove by induction over m≥1 that for every i,k such that i+m<k, [xi,xi+m] is μk-large if and only if [yi,yi+m] is ρk-large. The base case m=1 is the lemma hypothesis. Suppose it holds up to m. Since E is strongly μ-transitive, then [xi,xi+m+1] is μk-large if and only if either [xi,xi+m] or [xi+m,xi+m+1] is μk-large. By induction hypothesis, this holds if and only if either [yi,yi+m] or [yi+m,yi+m+1] is ρk-large. Since F is strongly ρ-transitive, this holds if and only if [yi,yi+m+1] is ρk-large.
∎
We now prove that we can computably thin out an infinite μ-transitive set to obtain a strongly μ-transitive infinite subset.
Lemma 2.25**.**
Let μ:ω→ω+ be a strongly increasing left-c.e. function.
Every infinite μ-transitive set X⊆ω has an infinite X-computable strongly μ-transitive subset Y⊆X.
Proof.
Fix μ and X. We build an infinite subsequence y0<y1<⋯∈X as follows. At stage [math], let y0=minX. At stage s>0, suppose we have defined a strongly μ-transitive finite sequence y0<⋯<ys−1. Let ys>ys−1 be the least element of X such that for every i<j<k<s, if [yi,yj] and [yj,yk] are μys-small, then [yi,yk] is μys-small. Such ys must be found, since given a fixed tuple i<j<k<s if [yi,yj] and [yj,yk] are μ-small, then [yi,yk] is μ-small, and therefore for all but finitely many y, [yi,yk] is μk-small. Taking a value in X bigger than the max of the thresholds for each tuple i<j<k<s, we obtain a ys with the desired property.
∎
In the following theorem, note again that μ ranges over ω and not ω+.
Theorem 2.26**.**
Let μ:ω→ω be a strongly increasing left-c.e. function. For every n≥1, define fn:D↦Ln(μ,D). For every infinite set H⊆ω such that Ln⊆fn[H]n, H computes a function dominating μ.
Proof.
Fix μ, n≥1 and H.
By Lemma 2.22, fn[H]n⊆Ln. By Cholak and Patey [2, Lemma 3.16], ∣Ln∣=Cn. If Ln⊆fn[H]n, then ∣fn[H]n∣<Cn. Then by Cholak and Patey [2, Theorem 3.17], H computes a function dominating μ.
∎
In the following statement, note that we state the existence of a relative left-c.e. function, which is another way of stating the existence of a sequence of functions μ0,μ1,… which satisfy the properties of a left-c.e. function independently of its effectiveness.
Statement 2.27**.**
SCA\mbox−RTkn: For every function f:[ω]n→k,
there is a strongly increasing relative left-c.e. function μ:ω→ω+, an infinite strongly μ-transitive set H⊆ω and a coloring χ:Ln→k such that for every D∈[H]n, f(D)=χ(Ln(μ,D)).
Again, for n=1, ∣Ln∣=1 and SCA\mbox−RTk1 is the statement RTk1. Similarly, for n=2, L2=V2, and SCA\mbox−RTk2 is the same statement as ARITH\mbox−SCA\mbox−RTk2, or equivalently LARGEk, that is, the statement “For every coloring f:[ω]2→k, there is an infinite set H⊆ω and two colors is,iℓ<k such that f[H]2⊆{is,iℓ} and for every x<y<z∈H, f(x,y)=f(y,z)=is if and only if f(x,z)=is.”
Theorem 2.28**.**
SCA\mbox−RTkn* admits strong cone avoidance.*
Proof.
Fix two sets Z and C with C≤TZ and let f:[ω]n→k be an instance of SCA\mbox−RTkn.
By Lerman [14, 4.18], there is a set Z1≥TZ such that C is Δ20(Z1) but C≤TZ1. Since C is Δ20(Z1), there is a left Z1-c.e. modulus μ:ω→ω for C. By Lemma 2.23, we can assume that μ is strongly increasing.
Let f1:[ω]n→k×Ln be defined for each D∈[ω]n by f1(D)=⟨f(D),Ln(μ,D)⟩.
By strong cone avoidance of RT<∞,Cnn (Cholak and Patey [2, Theorem 4.18]), there is an infinite set H⊆ω such that C≤TZ1⊕H and ∣f1[H]n∣≤Cn. In particular, Z1⊕H does not compute a function dominating μ, so by Theorem 2.26, for every G∈Ln, there is some i<k and some D∈[H]n such that f1(D)=⟨i,G⟩. Since ∣Ln∣=Cn (Cholak and Patey [2, Lemma 3.16]), this i is unique. For each G, let χ(G) be such an i.
We claim that for every D∈[H]n, f(D)=χ(Ln(μ,D)).
By definition of χ, f1(D)=⟨f(D),Ln(μ,D)⟩=⟨χ(Ln(μ,D)),Ln(μ,D)⟩. It follows that f(D)=χ(Ln(μ,D)). By Cholak and Patey [2, Corollary 5.5], there is an infinite μ-transitive subset H1⊆H such that C≤TZ⊕H1. By Lemma 2.25, there is an infinite strongly μ-transitive subset H2⊆H1 such that C≤TZ⊕H2. Therefore, H2 is a SCA\mbox−RTkn-solution to f.
∎
The following technical lemma will be useful for Theorem 2.30, Lemma 3.8 and Lemma 3.9.
Lemma 2.29**.**
Fix χ:Ln→k, and let f0:[ω]n→k and f1:[ω]n→k be two colorings.
Let H0 be an infinite SCA\mbox−RTkn-solution to f0 with witnesses χ and μ0:ω→ω+, and let H1 be an infinite SCA\mbox−RTkn-solution to f1 with witnesses χ and μ1:ω→ω.
If H0 f0-meets some RTkn-pattern P but H1 f1-avoids P and μ1 is left-c.e. then H1 computes a function dominating μ1.
Proof.
Since H0 f0-meets some RTkn-pattern P, there is a finite set E0={x0<⋯<xr−1}⊆H0 which f0-meets P.
Suppose for the sake of contradiction that H1 does not compute a function dominating μ1.
By Theorem 2.26, there is a finite set E1={y0<⋯<yr−1}⊆H1 of size r such that Lr(μ0,E0)=Lr(μ1,E1). By Lemma 2.9 and Lemma 2.24, since H0 is strongly μ0-transitive and H1 is strongly μ1-transitive, for every I∈[r]n, letting D0={xi:i∈I} and D1={yi:i∈I}, Ln(μ0,D0)=Ln(μ1,D1). Since f0(D0)=χ(Ln(μ0,D0)) and f1(D1)=χ(Ln(μ1,D1)), then f0(D0)=f1(D1). Thus f0↾[E0]n and f1↾[E1]n have the same function graph.
It follows that E1 f1-meets P, so H1 f1-meets P. Contradiction.
∎
Theorem 2.30**.**
Let W be a collection of RTkn patterns such that RTkn(W)≤idSCA\mbox−RTkn. Then for every left-c.e. function μ:ω→ω, there is an RTkn(W)-instance such that every solution computes a function dominating μ.
Proof.
Since RTkn(W)≤idSCA\mbox−RTkn, there is a coloring ffail:[ω]n→k and a SCA\mbox−RTkn-solution Hfail to ffail witnessed by a strongly increasing relative left-c.e. function μfail and a coloring χ:Ln→k, such that Hfail is strongly μfail-transitive, and such that Hfail meets some RTkn-pattern P∈W.
Let μ:ω→ω be a left-c.e. function. By Lemma 2.23, there is a strongly increasing left-c.e. function g:ω→ω dominating μ. Let f:[X]n→k be an instance of RTkn(W) defined by f(D)=χ(Ln(g,D)).
We claim that every RTkn(W)-solution H⊆X to f computes a function dominating g, hence dominating μ.
Fix H and suppose for the contradiction that H does not compute a function dominating g.
By Cholak and Patey [2, Theorem 5.11] and Lemma 2.25, there is an infinite strongly g-transitive subset H1⊆H such that H1 does not compute a function dominating g.
In particular, H1 is an infinite SCA\mbox−RTkn-solution to f with witnesses χ and g, and such that H1 f-avoids P. By Lemma 2.29, H computes a function dominating g, contradiction.
∎
Theorem 2.31**.**
A problem RTkn(W) admits strong cone avoidance if and only if RTkn(W)≤idSCA\mbox−RTkn.
Proof.
⇐: Suppose RTkn(W)≤idSCA\mbox−RTkn. Fix a set Z, a non-Z-computable set C and a coloring f:[ω]n→k. By Theorem 2.28, there is an SCA\mbox−RTkn-solution H to f such that C≤TZ⊕H. In particular, H is an RTkn(W)-solution to f.
⇒: Suppose RTkn(W)≤idSCA\mbox−RTkn. Let μ:ω→ω be a left-c.e. modulus of ∅′. By Theorem 2.30, there is a RTkn(W)-instance such that every solution computes a function dominating μ, hence computes ∅′. Therefore RTkn(W) does not admit strong cone avoidance.
∎
Interestingly, Theorem 2.31 admits multiple abstract consequences of which one might expect to have a more direct proof. However, there does not seem to be any simpler proof of these facts than proving Theorem 2.31.
The following corollary states that the question of strong cone avoidance of a collection of patterns can be reduced to strong cone avoidance of each pattern individually.
Corollary 2.32**.**
If RTkn({P}) admits strong cone avoidance for every P∈W,
then RTkn(W) admits strong cone avoidance.
Proof.
If RTkn({P}) admits strong cone avoidance for every P∈W,
then by Theorem 2.31, RTkn({P})≤idSCA\mbox−RTkn for every P∈W. It follows that RTkn(W)≤idSCA\mbox−RTkn, so again by Theorem 2.31, RTkn(W) admits strong cone avoidance.
∎
The following corollary gives an external definition of the maximal problem which admits strong cone avoidance.
Corollary 2.33**.**
Let Wn,k=⋃{W:RTkn(W)\mboxadmitsstrongconeavoidance}. Then RTkn(Wn,k) admits strong cone avoidance.
Proof.
By Corollary 2.32, it suffices to prove that RTkn({P}) admits strong cone avoidance for every P∈Wn,k.
Fix some P∈Wn,k. By definition, P∈W for some collection W such that RTkn(W) admits strong cone avoidance. In particular, RTkn({P}) admits strong cone avoidance.
∎
2.4. Avoiding non-computable cones
As explained in Section 1.2, there is a deep relation between the combinatorial features of the colorings over [ω]n and the computational features the colorings over [ω]n+1. This link is formalized in the case of cone avoidance by Cholak and Patey in [2, Theorem 1.5]. We prove the existence of a maximal Ramsey-like principle which admits cone avoidance. For this, we must restrict ourselves to a particular type of largeness graphs, namely, packed graphs.
Definition 2.34** (Cholak and Patey [2]).**
A largeness graph G=({0,…,n−1},E) is packed
if for every i<n−2, {i,i+1}∈E.
Let Pn be the set of all packed largeness graphs of size n.
The definition of Ln(μ,D), and more precisely the adjacent edges, codes some μ-largeness information. However, in a computable setting, we only have access to the left-c.e. approximations of μ. This is why we must restrict ourselves to μ-largeness approximations Pn(μ,D), which can be computably coded.
Definition 2.35**.**
Given a relative left-c.e. function μ:ω→ω+ with approximations μ0,μ1,… and a set D={x0,…,xn−1}⊆ω,
let Pn(μ,D) be the graph ({0,…,n−1},E) where
E={{p,q}:μxq(xp)>xp+1∧p+1<q<n}.
The following two lemmas are obtained by the exact same proof as Lemma 2.22 and Lemma 2.24, respectively.
Lemma 2.36**.**
Fix a strongly increasing relative left-c.e. function μ:ω→ω+.
For every n≥1 and D∈[ω]n, Pn(μ,D)∈Pn.
Lemma 2.37**.**
Fix μ:ω→ω+ and ρ:ω→ω+.
Given n≥1, let E={x0<⋯<xn−1} be a strongly μ-transitive set, and F={y0<⋯<yn−1} be a strongly ρ-transitive set such that Pn(μ,E)=Pn(ρ,F). Then for every i<j<k<n, [xi,xj] is μk-large if and only if [yi,yj] is ρk-large.
A coloring f:[ω]n+1→k is stable if for every D∈[ω]n, limyf(D∪{y}) exists.
Given a set D={x0<⋯<xn−1}, if we take some y∈ω sufficiently large, then μy and μ will coincide over [D]2.
Then, looking at the packed largeness graph Pn+1(μ,D∪{y}), this graph codes exactly the information of the packed largeness graph Pn(μ,D) and the μy-largeness information over [D]2, which is by assumption the μ-largeness information over [D]2. These two kind of informations are exactly the ones coded by Ln(μ,D). Then, Pn+1(μ,D∪{y}) and Ln(μ,D) are in one-to-one correspondance.
Let us define explicitly the one-to-one mapping.
Given a packed largeness graph G=({0,…,n},E) of size n+1, let Ln(G) be the largeness graph of size n ({0,…,n−1},E1) where E1={{i,j}∈E:j<n}∪{{i,i+1}:{i,n}∈E}.
The following lemma uses this one-to-one correspondance to relate stable colorings over [ω]n+1 to colorings over [ω]n.
Lemma 2.38**.**
Let μ:ω→ω+ be a strongly increasing relative left-c.e. function. Fix n≥1 and define fn+1:D↦Pn+1(μ,D).
Let H be an infinite strongly μ-transitive set over which fn+1 is stable. For every D∈[H]n, letting G=limy∈HPn+1(μ,D∪{y}), Ln(μ,D)=Ln(G).
Proof.
Fix D={x0<⋯<xn−1}∈[H]n, and let xn∈H be sufficiently large so that Pn+1(μ,D∪{xn})=limy∈HPn+1(μ,D∪{y}). Let G=({0,…,n},E) be the packed largeness graph Pn+1(μ,D∪{xn}) and ({0,…,n−1},E1)=Ln(G).
We first prove that the adjacent edges in Ln(G) are exactly the adjacent edges in Ln(μ,D).
For every i<n−1, by definition of Ln(G), {i,i+1}∈E1 if and only if {i,n}∈E. By definition of Pn+1(μ,D∪{xn}), {i,n}∈E if and only if [xi,xi+1] is μy-large, hence if and only if [xi,xi+1] is μ-large.
We now prove that the non-adjacent adjacent edges in Ln(G) are exactly the non-adjacent edges in Ln(μ,D).
For every i+1<j<n, {i,j}∈E1 if and only if {i,j}∈E,
hence if and only if [xi,xi+1] is μxj-large.
Therefore Ln(G)=Ln(μ,D).
∎
Theorem 2.39**.**
Let μ:ω→ω be a strongly increasing left-c.e. function. For every n≥1, define fn:D↦Pn(μ,D). For every infinite set H⊆ω such that Pn⊆fn[H]n, H computes a function dominating μ.
Proof.
Fix μ, n≥1 and H.
For n=1, then ∣P1∣=1. For every infinite set H⊆ω, ∣f1[H]1∣=1, so P1⊆f1[H]1, and the property vacuously holds.
For n>1. Suppose for the contradiction that H does not compute a function dominating μ. By Patey [23, Theorem 12], there is an infinite subset H1⊆H over which fn is stable and such that H1 does not compute a function dominating μ. Let f~:[ω]n−1→Pn be defined by f~(D)=limy∈H1fn(D∪{y})=limy∈H1Pn(μ,∪{y}). By Lemma 2.38, for every D∈[H1]n−1, Ln−1(μ,D)=Ln(f~(D)).
By Lemma 2.36, fn[H1]n⊆Pn. By Cholak and Patey [2, Lemma 3.15,Lemma 3.16], ∣Pn∣=Cn−1. Since Pn⊆fn[H]n, then ∣fn[H]n∣<Cn−1. It follows that, letting gn−1:D↦Ln−1(μ,D), ∣gn−1[H]n−1∣=∣fn[H]n∣<Cn−1.
By Cholak and Patey [2, Lemma 3.16], ∣L−n−1∣=Cn−1, therefore, Ln−1⊆gn−1[H]n−1, so by Theorem 2.26, H1 computes a function dominating μ. Contradiction.
∎
Statement 2.40**.**
CA\mbox−RTkn: For every function f:[ω]n→k,
there is a strongly increasing relative left-c.e. function μ:ω→ω+, an infinite strongly μ-transitive set H⊆ω and a coloring χ:Pn→k such that for every D∈[H]n, f(D)=χ(Pn(μ,D)).
In the cases n=1 and n=2, there is exactly one packed largeness graph of size n, namely, the graph with no edges. Therefore ∣Pn∣=1, and CA\mbox−RTk1 and CA\mbox−RTk2 are exactly RTk1 and RTk2, respectively. The case n=3 yields a new principle.
Statement 2.41**.**
PACKEDk: “For every coloring f:[ω]3→k, there are two colors is,iℓ<k and an infinite set H⊆ω such that f[H]3⊆{is,iℓ} and for every w<x<y<z∈H,
f(w,x,z)=f(x,y,z)=is if and only if f(w,y,z)=is
if f(w,x,y)=is then f(w,x,z)=is
if f(w,x,y)=iℓ and f(w,x,z)=is then f(x,y,z)=is.”
Informally, f(x,y,z)=is if [x,y] is μz-small, and f(x,y,z)=iℓ otherwise. We now prove that CA\mbox−RTk3 and PACKEDk are the same statement by bi-reduction.
Lemma 2.42**.**
PACKEDk≤idCA\mbox−RTk3.
Proof.
Let f:[ω]3→k be a coloring, and let H be an infinite CA\mbox−RTk3-solution to f. By definition of CA\mbox−RTk3, there is a strongly increasing relative left-c.e. modulus μ:ω→ω+ and a function χ:P3→k such that H is strongly μ-transitive, and for every D∈[H]3, f(D)=χ(P3(μ,D)). Let G0=({0,1,2},∅) and G1=({0,1,2},{{0,2}}). In particular, P3={G0,G1}. Let is=χ(G1) and iℓ=χ(G0). We have f[H]3⊆{is,iℓ}. If is=iℓ, then H is f-homogeneous, and H is an PACKEDk-solution to f, so suppose is=iℓ. In particular, χ is one-to-one. We now prove properties (a-c) of Statement 2.41. Fix w<x<y<z∈H.
(a): f(w,x,z)=f(x,y,z)=is if and only if χ(P3(μ,{w,x,z}))=χ(P3(μ,{x,y,z}))=is if and only if P3(μ,{w,x,z})=P3(μ,{x,y,z})=G1 if and only if [w,x] and [x,y] are μz-small. Since H is strongly μ-transitive, this holds if and only if [w,y] is μz-small, if and only if P3(μ,{w,y,z})=G1 if and only if f(w,y,z)=χ(P3(μ,{w,y,z}))=χ(G1)=is.
(b): If f(w,x,y)=is, then χ(P3(μ,{w,x,y}))=is, so P3(μ,{w,x,y})=G1. In particular, [w,x] is μy-small, hence is μz-small. Therefore, P3(μ,{w,x,z})=G1, so f(w,x,z)=χ(P3(μ,{w,x,z}))=χ(G1)=is.
(c): If f(w,x,y)=iℓ and f(w,x,z)=is, then χ(P3(μ,{w,x,y}))=iℓ and χ(P3(μ,{w,x,z}))=is. Therefore, P3(μ,{w,x,y})=G0 and P3(μ,{w,x,z})=G1. It follows that [w,x] is μy-large but μz-small. By Claim 1 in the proof of Lemma 2.22, since μ is strongly increasing, [x,y] is μz-small. Thus P3(μ,{x,y,z})=G1, so f(x,y,z)=χ(P3(μ,{x,y,z}))=χ(G1)=is.
∎
Lemma 2.43**.**
CA\mbox−RTk3≤idPACKEDk.
Proof.
Let f:[ω]3→k be a coloring, and let H be an infinite PACKEDk-solution, that is, there are some colors is,iℓ<k such that f[H]3⊆{is,iℓ} and properties (a-c) of Statement 2.41 hold.
For every x,z∈ω, let x0 and z0 be the least elements of H∩[x,∞) and H∩[z,∞), respectively. Let μz(x) be the least element y0 of H∩(x0,z0) such that f(x0,y0,z0)=iℓ if it exists. Otherwise μz(x)=z.
Claim 1: For every x,z∈ω, μz(x)≤z.
Indeed, let x0 and z0 be the least elements of H∩[x,∞) and H∩[z,∞), respectively. If there is a least element y0 of H∩(x0,z0) such that f(x0,y0,z0)=iℓ, then y0<z0 and by definition of z0, y0<z. It follows that μz(x)=y0<z. If there is no such y0, then μz(x)=z≤z. This proves Claim 1.
Claim 2: For every x∈ω and u≤v∈ω, μu(x)≤μv(x).
Let x0, u0 and v0 be the least elements of H∩[x,∞), H∩[u,∞) and H∩[v,∞), respectively. In particular, u0≤v0. Case 1: there is no yu∈H∩(x0,u0) such that f(x0,yu,u0)=iℓ. Then μu(x)=u. If there is no yv∈H∩(x0,v0) such that f(x0,yv,v0)=iℓ, then μv(x)=v≥u=μu(x), and we are done. So suppose there is such a yv∈H∩(x0,v0). If yv≥u0, then μv(x)=yv≥u0≥u=μu(u) and we are done. If yv<u0, then f(x0,yv,v0)=iℓ and since yu does not exist, f(x0,yv,u0)=is. This contradicts property (b) of Statement 2.41. Case 2: yu exists. Then yu<u0, so yu<u and for every y∈H∩(x0,yu), f(x0,y,u0)=is. By property (b) of Statement 2.41, for every y∈H∩(x0,yu), f(x0,y,v0)=is. Therefore if yv exists, then yv≥yu and μu(x)≤μv(x). If yv does not exist, then μv(x)=v≥u>yu=μu(x). This proves Claim 2.
Claim 3: For every z∈ω and u≤v, μz(u)≤μz(v).
Let u0,v0 and z0 be the least elements of H∩[u,∞), H∩[v,∞) and H∩[z,∞), respectively. In particular, u0≤v0. Case 1: there is no yv∈H∩(v0,z0) such that f(v0,yv,z0)=iℓ. Then μv(x)=v. Since by Claim 1, μu(x)≤u≤v=μv(x), we are done. Case 2: there is such a yv. In particular, f(v0,yv,z0)=iℓ. By property (a) of Statement 2.41, f(u0,yv,z0)=iℓ, so there is a least yu∈H∩(u0,z0) such that f(u0,yu,z0)=iℓ and yu≤yv. Then μz(u)=yu≤yv=μz(v). This proves Claim 3.
Claim 4: For every z∈ω and u<v∈ω, if μz+1(u)>μz(u), then μz+1(v)>z.
Let z0,z1,u0,v0 be the least elements of H∩[z,∞), H∩[z+1,∞), H∩[u,∞) and H∩[v,∞), respectively.
If z0=z1, we are done, so suppose z0<z1. In particular, z1 is the immediate successor of z0 in H. If there is no least yv∈H∩(v0,z1) such that f(v0,yv,z1)=iℓ, then μz+1(v)=z+1 and we are done as well. So suppose there is such a yv. By property (a) of Statement 2.41, f(u0,yv,z1)=iℓ, so there is a least yu∈H∩(u0,z1) such that f(u0,yu,z1)=iℓ. Since z1 is the immediate successor of z0 in H, either yu=z0, or yu<z0. Case 1: yu=z0. By Claim 3, μz+1(v)≥μz+1(u)=yu=z0. If z0>z, we are done, so suppose yu=z0=z. If μz(u)=z, then Claim 4 is vacuously satisfied, so suppose μz(u)<z. It follows that there is a least y∈H∩(u0,z0) such that f(u0,y,z0)=iℓ. By minimality of y0, f(u0,y,z1)=is. Then by property (c) of Statement 2.41, f(y,z0,z1)=is. Since f(u0,yu,z1)=iℓ, and yu=z0=z, then f(y,z,z1)=is and f(u0,z,z1)=iℓ. So by property (a) of Statement 2.41, f(u0,y,z1)=iℓ, contradiction.
Case 2: yu<z0. Then there is some some y∈H∩(u0,z0) such that f(u0,y,z0)=iℓ, with y<yu. By definition of yu being a least element, f(u0,y,z1)=is. Then by property (c) of Statement 2.41, f(y,z0,z1)=is, and since y<yu<z0, and f(u0,yu,z1)=iℓ, then by property (a) of Statement 2.41, f(u0,y,z1)=iℓ or f(y,yu,z1)=iℓ. The former case does not hold, so f(y,yu,z1)=iℓ. Again by property (a) of Statement 2.41, f(y,z0,z1)=iℓ. Contradiction.
Claim 5: For w<x<y<z∈H such that μz(w)>x and μz(x)>y, then μz(w)>y. By definition of μ, f(w,x,z)=f(x,y,z)=is. By property (a) of Statement 2.41, f(w,y,z)=is. Therefore, μz(w)>y. This proves Claim 5.
By Claim 2, 3 and 4, μ is a strongly increasing left-c.e. function.
Moreover, for every x<y<z∈H, f(x,y,z)=iℓ if and only if μz(x)≤y, so f(x,y,z)=χ(P3(μ,{x,y,z})).
By Claim 5, H is strongly μ-transitive. Therefore H is a CA\mbox−RTk3-solution to f with witnesses χ and μ.
∎
Theorem 2.44**.**
CA\mbox−RTkn* admits cone avoidance.*
Proof.
Fix two sets Z and C with C≤TZ and let f:[ω]n→k be a Z-computable instance of CA\mbox−RTkn.
By Lerman [14, 4.18], there is a set Z1≥TZ such that C is Δ20(Z1) but C≤TZ1. Since C is Δ20(Z1), there is a left Z1-c.e. modulus μ:ω→ω for C. By Lemma 2.23, we can assume that μ is strongly increasing.
Let f1:[ω]n→k×Pn be defined for each D∈[ω]n by f1(D)=⟨f(D),Pn(μ,D)⟩.
By cone avoidance of RT<∞,Cn−1n (Cholak and Patey [2, Corollary 4.19]), there is an infinite set H⊆ω such that C≤TZ1⊕H and ∣f1[H]n∣≤Cn−1. In particular, Z1⊕H does not compute a function dominating μ, so by Theorem 2.26, for every G∈Pn, there is some i<k and some D∈[H]n such that f1(D)=⟨i,G⟩. Since ∣Pn∣=Cn−1 (Cholak and Patey [2, Lemma 3.15,Lemma 3.16]), this i is unique. For each G, let χ(G) be such an i.
We claim that for every D∈[H]n, f(D)=χ(Pn(μ,D)).
By definition of χ, f1(D)=⟨f(D),Pn(μ,D)⟩=⟨χ(Pn(μ,D)),Pn(μ,D)⟩. It follows that f(D)=χ(Pn(μ,D)). By Cholak and Patey [2, Corollary 5.5], there is an infinite μ-transitive subset H1⊆H such that C≤TZ⊕H1. By Lemma 2.25, there is an infinite strongly μ-transitive subset H2⊆H1 such that C≤TZ⊕H2. Therefore, H2 is a CA\mbox−RTkn-solution to f.
∎
The following technical lemma will be useful for Theorem 2.46, Lemma 3.13 and Lemma 3.14.
Lemma 2.45**.**
Fix χ:Pn→k, and let f0:[ω]n→k and f1:[ω]n→k be two colorings such that f1 is computable.
Let H0 be an infinite CA\mbox−RTkn-solution to f0 with witnesses χ and μ0:ω→ω, and let H1 be an infinite CA\mbox−RTkn-solution to f1 with witnesses χ and μ1:ω→ω.
If H0 f0-meets some RTkn-pattern P but H1 f1-avoids P and μ1 is left-c.e., then H1 computes a function dominating μ1.
Proof.
Since H0 f0-meets some RTkn-pattern P, there is a finite set E0={x0<⋯<xr−1}⊆H0 which f0-meets P.
Suppose for the sake of contradiction that H1 does not compute a function dominating μ1.
By Theorem 2.39, there is a finite set E1={y0<⋯<yr−1}⊆H1 of size r such that Pr(μ0,E0)=Pr(μ1,E1). By Lemma 2.37, since H0 is strongly μ0-transitive and H1 is strongly μ1-transitive, for every I∈[r]n, letting D0={xi:i∈I} and D1={yi:i∈I}, Pn(μ0,D0)=Pn(μ1,D1). Since f0(D0)=χ(Pn(μ0,D0)) and f1(D1)=χ(Pn(μ1,D1)), then f0(D0)=f1(D1). Thus f0↾[E0]n and f1↾[E1]n have the same function graph.
It follows that E1 f1-meets P, so H1 f1-meets P. Contradiction.
∎
Theorem 2.46**.**
Let W be a collection of RTkn-patterns such that RTkn(W)≤idCA\mbox−RTkn. Then for every left-c.e. function μ:ω→ω, there is a computable RTkn(W)-instance such that every solution computes a function dominating μ.
Proof.
Since RTkn(W)≤idCA\mbox−RTkn, there is a coloring ffail:[ω]n→k and a CA\mbox−RTkn-solution Hfail to ffail witnessed by strongly increasing left-c.e. function μfail:ω→ω+ and a coloring χ:Pn→k, such that Hfail is strongly μfail-transitive, and Hfail meets some RTkn-pattern P∈W.
Let μ:ω→ω be a left-c.e. function. By Lemma 2.23, there is a strongly increasing left-c.e. function g:ω→ω dominating μ. Let f:[X]n→k be a computable instance of RTkn(W) defined by f(D)=χ(Pn(g,D)).
We claim that every RTkn(W)-solution H to f computes a function dominating g, hence dominating μ.
Fix H and suppose for the contradiction that H does not compute a function dominating g. By Cholak and Patey [2, Theorem 5.11] and Lemma 2.25, there is an infinite strongly g-transitive subset H1⊆H such that H1 does not compute a function dominating g.
In particular, H1 is an infinite CA\mbox−RTkn-solution to f with witnesses χ and g, and such that H1 f-avoids P. By Lemma 2.45, H computes a function dominating g, contradiction.
∎
Theorem 2.47**.**
A problem RTkn(W) admits cone avoidance if and only if RTkn(W)≤idCA\mbox−RTkn.
Proof.
⇐: Suppose RTkn(W)≤idCA\mbox−RTkn. Fix a set Z, a non-Z-computable set C and a Z-computable coloring f:[ω]n→k. By Theorem 2.44, there is a CA\mbox−RTkn-solution H to f such that C≤TZ⊕H. In particular, H is an RTkn(W)-solution to f.
⇒: Suppose RTkn(W)≤idCA\mbox−RTkn. Let μ:ω→ω be a left-c.e. modulus of ∅′. By Theorem 2.46, there is a computable RTkn(W)-instance such that every solution computes a function dominating μ, hence computes ∅′. Therefore RTkn(W) does not admit cone avoidance.
∎
3. Promise Ramsey-like theorems
The class of Ramsey-like problems encompasses Ramsey’s theorem and the Erdős-Moser theorem [1] since both statements are of the form “For every coloring f:[ω]n→k, there is an infinite set H which avoids some set of patterns.” There are however two kind of consequences of Ramsey’s theorem which do not fit within this framework.
First, some statements have a restricted class of instances. For example, the Ascending Descending Sequence [12] (ADS) principle asserts that every infinite linear order admits an infinite ascending or descending sequence. A linear order L=(ω,≺L) can be formalized as a coloring f:[ω]2→2 such that for every x<y, f(x,y)=1 if and only if x≺Ly. Such a coloring is called transitive, since for every x<y<z and i<2, if f(x,y)=i and f(y,z)=i then f(x,z)=i. ADS can therefore be formulated as the statement “For every transitive coloring f:[ω]2→2, there exists an infinite f-homogeneous set.”
Second, some consequences of Ramsey’s theorem such as the Free Set theorem [3] involve ω-colorings of [ω]n. The free set theorem for n-tuples is the statement “For every coloring f:[ω]n→ω, there is an infinite set H such that for every x∈H, x∈f[H∖{x}]n.” Suppose that for every D∈[ω]n, f(D)≤minD.
We can define a coloring g:[ω]n+1→2 by g(x0,x1,…,xn)=1 if and only if f(x1,…,xn)=x0.
Every g-homogeneous set must be of color 0, and therefore be a free set solution to f.
Then one can formulate this restricted version of the free set theorem as the statement “For every coloring g:[ω]n+1→ω such that for every x∈ω, there is at most one D∈[ω]n such that g({x}∪D)=1, there is an infinite g-homogeneous set.” The full free set theorem can also fit within this framework with additional technicalities.
In both examples, the problems can be formulated as statements “For every coloring f:[ω]n→k which avoids some set of patterns, there is an infinite set H which avoids another set of patterns.” In computational complexity, problems whose class of instances is restricted by some properties are known as promise problems. This motivates the following definition.
Definition 3.1**.**
Given two collections V and W of RTkn-patterns, the promise RTkn-like problem RTkn(V,W) is the problem whose instances are colorings f:[ω]n→k such that ω f-avoids every pattern in V. An RTkn(V,W)-solution to an instance f is an infinite set H⊆ω f-avoiding every pattern in W.
For example, ADS is the promise RT22-like problem RT22(WADS,WRT22) with WADS={f(x0,x1)=1∧f(x1,x2)=1∧f(x0,x2)=0,f(x0,x1)=0∧f(x1,x2)=0∧f(x0,x2)=1}. We interpret a function f:[ω]2→2 such that ω f-avoids the pattern WADS as a linear order ≺ defined by x≺y if and only if x<Ny and f(x,y)=1 or x>Ny and f(x,y)=0.
Similarly, CAC is the promise RT32-like problem RT32(WCAC,WRT22) with WCAC=WADS.
We interpret a function f:[ω]2→3 such that ω f-avoids the pattern WADS as a partial order ≺ defined by x≺y if and only if x<Ny and f(x,y)=1 or x>Ny and f(x,y)=0.
3.1. Strongly avoiding non-arithmetical cones
We now extend our analysis of strong cone avoidance for non-arithmetical cones to the class of promise Ramsey-like problems. For this, we need to define a restricted version of ARITH\mbox−SCA\mbox−RTkn, in which the coloring χ:Vn→k must belong to a predefined set of colorings.
Statement 3.2**.**
Let R be a set of functions of type χ:Vn→k.
R\mbox−ARITH\mbox−SCA\mbox−RTkn: For every function f:[ω]n→k,
there is a function μ:ω→ω+, an infinite μ-transitive set H⊆ω, and a coloring χ∈R such that for every D∈[H]n, f(D)=χ(Vn(μ,D)).
One particular case of interest is whenever R is a singleton {χ}. The following notion will be very useful in our case analysis.
Given some n,k∈ω and a collection W of RTkn-patterns,
let
[TABLE]
In particular, if RTkn(W) is a true statement, then every constant function χ:Vn→k belongs to Rkn(W).
Lemma 3.3**.**
Let V and W be collections of RTkn-patterns such that
RTkn(W)≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn.
Then RTkn(V,W) admits strong cone avoidance for non-arithmetical cones.
Proof.
Fix a set Z, a non-Z-arithmetical set C and a coloring f:[ω]n→k.
Suppose first that C is not Z-hyperarithmetical. By Solovay [26], C is not computably encodable relative to Z. Since for every infinite set X⊆ω, there is an infinite ARITH\mbox−SCA\mbox−RTkn-solution Y⊆X to f, there is an ARITH\mbox−SCA\mbox−RTkn-solution H to f such that C≤TZ⊕H.
Suppose now that C is Z-hyperarithmetical. By Groszek and Slaman [10], there is a modulus μ:ω→ω relative to Z, that is, for every function g dominating μ, C≤TZ⊕g.
Let f1 be defined for each D∈[ω]n by f1(D)=⟨f(D),Vn(μ,D)⟩.
By strong cone avoidance of RT<∞,2n−1n for non-arithmetical cones (see Cholak and Patey [2, Theorem 4.15]), there is an infinite set H such that C≤TZ⊕H and ∣f1[H]n∣≤2n−1. In particular, Z⊕H does not compute a function dominating μ, so by Theorem 2.10, for every G∈Vn, there is some i<k and some D∈[H]n such that f1(D)=⟨i,G⟩. Since ∣Vn∣=2n−1, this i is unique. For each G∈Vn, let χ(G) be such an i.
In particular, for every D∈[H]n, f(D)=χ(Vn(μ,D)).
By Cholak and Patey [2, Corollary 5.5], there is an infinite μ-transitive subset H1⊆H such that C≤TZ⊕H1. Therefore, H1 is a ARITH\mbox−SCA\mbox−RTkn-solution to f.
If χ∈Rkn(V), then since RTkn(W)≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn, H2 is an RTkn(V,W)-solution to f, and we are done. So suppose χ∈Rkn(V). Unfolding the definition, RTkn(V)≤id{χ}\mbox−ARITH\mbox−SCA\mbox−RTkn, so there is a coloring fχ:[ω]n→k and some infinite ARITH\mbox−SCA\mbox−RTkn-solution Hχ to fχ with witnesses μχ:ω→ω+ and χ:Vn→k, that meets some RTkn-pattern Pχ∈V.
However, H2 f-avoids Pχ since f is an instance of RTkn(V,W). By Lemma 2.16, H2⊕Z computes a function dominating μ, hence C≤TZ⊕H2, contradiction. This completes the proof of Lemma 3.3.
∎
Lemma 3.4**.**
Let V and W be collections of RTkn-patterns such that
RTkn(W)≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn.
For every function μ:ω→ω, there is an RTkn(V,W)-instance such that every solution computes a function dominating μ.
Proof.
Since RTkn(W)≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn, there is some coloring ffail:[ω]n→k and some infinite ARITH\mbox−SCA\mbox−RTkn-solution H to f with some witness μfail:ω→ω and χ∈Rkn(V) that meets some RTkn-pattern P∈W.
Define f:[ω]n→k by f(D)=χ(Vn(μ,D)).
Suppose μ is not dominated by any computable function, otherwise we are done.
By Cholak and Patey [2, Theorem 5.11], there is an infinite μ-transitive set H which does not compute a function dominating μ.
Therefore, H is an ARITH\mbox−SCA\mbox−RTkn-solution to f with witnesses μ and χ∈Rkn(V). By definition of Rkn(V), RTkn(V)≤id{χ}\mbox−ARITH\mbox−SCA\mbox−RTkn and H is an RTkn(V)-solution to f, so f:[H]n→k is an instance of RTkn(V,W).
We claim that for every RTkn(V,W)-solution G⊆H to f, G computes a function dominating μ. In particular, G f-avoids P, so by Lemma 2.16, G computes a function dominating μ. This completes the proof of Lemma 3.4.
∎
Theorem 3.5**.**
Let V and W be two collections of RTkn-patterns.
RTkn(V,W) admits strong cone avoidance for non-arithmetical cones if and only if RTkn(W)≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn.
Proof.
⇐: This is Lemma 3.3. ⇒: We prove the contrapositive. Suppose RTkn(W)≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn Let C be a non-arithmetical set with modulus μ. By Lemma 3.4, there is a RTkn(V,W)-instance such that every solution computes a function dominating μ, hence computes C.
∎
In the case V=∅, then Rkn(V) is the set of all functions χ:Vn→k and therefore Rkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn is ARITH\mbox−SCA\mbox−RTkn. We obtain Theorem 2.18, that is, RTkn(∅,W) admits strong cone avoidance for non-arithmetical cones if and only if RTkn(W)≤idARITH\mbox−SCA\mbox−RTkn.
More interestingly, we consider the case where W is the set WRTkn of patterns forbidding non-homogeneous sets.
Corollary 3.6**.**
Let V be a collection of RTkn-patterns.
RTkn(V,WRTkn) admits strong cone avoidance for non-arithmetical cones if and only if Rkn(V) contains only constant functions.
Proof.
By Theorem 3.5,
RTkn(V,WRTkn) admits strong cone avoidance for non-arithmetical cones if and only if RTkn≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn.
Case 1: Rkn(V) contains only constant functions. Then for any f:[ω]n→k, any Rkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn-solution to f is f-homogeneous, hence RTkn≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn, so RTkn(V,WRTkn) admits strong cone avoidance for non-arithmetical cones.
Case 2: Rkn(V) contains a non-constant function χ:Vn→k.
Let μ:ω→ω be a modulus of a non-arithmetical set C.
Let f(D)=χ(Vn(μ,D)). By Cholak and Patey [2, Corollary 5.5], there is an infinite μ-transitive set H which does not compute a function dominating μ. Therefore, H is an Rkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn-solution to f with witnesses μ and χ. We claim that H is not f-homogeneous. Indeed, by Theorem 2.10, for every G∈Vn, there is some D∈[H]n such that G=Vn(μ,D). Since χ is not constant on Vn, H is not f-homogeneous. Thus RTkn≤idRkn(V)\mbox−ARITH\mbox−SCA\mbox−RTkn and RTkn(V,WRTkn) does not admit strong cone avoidance for non-arithmetical cones.
∎
3.2. Strongly avoiding non-computable cones
The strong cone avoidance analysis is very similar to the one for non-arithmetical cones, mutatis mutandis. We start again by defining a restricted version of SCA\mbox−RTkn.
Statement 3.7**.**
Let S be a set of functions of type χ:Ln→k.
S\mbox−SCA\mbox−RTkn: For every function f:[ω]n→k,
there is a strongly increasing relative left-c.e. function μ:ω→ω+, an infinite strongly μ-transitive set H⊆ω, and a coloring χ∈S such that for every D∈[H]n, f(D)=χ(Ln(μ,D)).
Given a collection of RTkn-patterns, we define a class Skn(W) for SCA\mbox−RTkn is a similar way as the class Rkn(W) for ARITH\mbox−SCA\mbox−RTkn, that is, given some n,k∈ω and a collection W of RTkn-patterns,
let
[TABLE]
In particular, if RTkn(W) is a true statement, then every constant function χ:Ln→k belongs to Skn(W).
Lemma 3.8**.**
Let V and W be collections of RTkn-patterns such that
RTkn(W)≤idSkn(V)\mbox−SCA\mbox−RTkn.
Then RTkn(V,W) admits strong cone avoidance.
Proof.
Fix two sets Z and C with C≤TZ and let f:[ω]n→k be an instance of SCA\mbox−RTkn.
By Lerman [14, 4.18], there is a set Z1≥TZ such that C is Δ20(Z1) but C≤TZ1. Since C is Δ20(Z1), there is a left Z1-c.e. modulus μ:ω→ω for C. By Lemma 2.23, we can assume that μ is strongly increasing.
Let f:[ω]n→k be an instance of RTkn(V,W).
Let f1:[ω]n→k×Ln be defined for each D∈[ω]n by f1(D)=⟨f(D),Ln(μ,D)⟩.
By strong cone avoidance of RT<∞,Cnn (Cholak and Patey [2, Theorem 4.18]), there is an infinite set H⊆ω such that C≤TZ1⊕H and ∣f1[H]n∣≤Cn. In particular, Z1⊕H does not compute a function dominating μ, so by Theorem 2.26, for every G∈Ln, there is some i<k and some D∈[H]n such that f1(D)=⟨i,G⟩. Since ∣Ln∣=Cn (Cholak and Patey [2, Lemma 3.16]), this i is unique. For each G, let χ(G) be such an i. Then for every D∈[H]n, f(D)=χ(Ln(μ,D)). By Cholak and Patey [2, Corollary 5.5], there is an infinite μ-transitive subset H1⊆H such that C≤TZ⊕H1. By Lemma 2.25, there is an infinite strongly μ-transitive subset H2⊆H1 such that C≤TZ⊕H2. Therefore, H2 is a SCA\mbox−RTkn-solution to f.
If χ∈Skn(V), then since RTkn(W)≤idSkn(V)\mbox−SCA\mbox−RTkn, H2 is an RTkn(V,W)-solution to f, and we are done. So suppose χ∈Skn(V). Unfolding the definition, RTkn(V)≤id{χ}\mbox−SCA\mbox−RTkn, so there is a coloring fχ:[ω]n→k and some infinite SCA\mbox−RTkn-solution Hχ to fχ with witnesses μχ:ω→ω+ and χ:Ln→k, that meets some RTkn-pattern Pχ∈V.
However, H2 f-avoids Pχ since f is an instance of RTkn(V,W). By Lemma 2.29, H2⊕Z computes a function dominating μ, hence C≤TZ⊕H2, contradiction. This completes the proof of Lemma 3.8.
∎
Lemma 3.9**.**
Let V and W be collections of RTkn-patterns such that
RTkn(W)≤idSkn(V)\mbox−SCA\mbox−RTkn.
For every strongly increasing left-c.e. function μ:ω→ω, there is an RTkn(V,W)-instance such that every solution computes a function dominating μ.
Proof.
Since RTkn(W)≤idSkn(V)\mbox−SCA\mbox−RTkn, there is some coloring ffail:[ω]n→k and some infinite SCA\mbox−RTkn-solution Hfail to ffail with some witness μfail:ω→ω and χ∈Skn(V) that meets some RTkn-pattern P∈W.
Define f:[ω]n→k by f(D)=χ(Ln(μ,D)).
Suppose that μ is not dominated by a computable function.
By Cholak and Patey [2, Theorem 5.11] and Lemma 2.25, there is an infinite strongly μ-transitive set H which does not compute a function dominating μ. Therefore, H is a SCA\mbox−RTkn-solution to f with witnesses μ and χ∈Skn(V). By definition of Skn(V), RTkn(V)≤id{χ}\mbox−SCA\mbox−RTkn and H is an RTkn(V)-solution to f, so f:[H]n→k is an instance of RTkn(V,W).
We claim that for every RTkn(V,W)-solution G⊆H to f, G computes a function dominating μ. In particular, G f-avoids P, so by Lemma 2.29, G computes a function dominating μ. This completes the proof of Lemma 3.9.
∎
Theorem 3.10**.**
Let V and W be two collections of RTkn-patterns.
RTkn(V,W) admits strong cone avoidance if and only if RTkn(W)≤idSkn(V)\mbox−SCA\mbox−RTkn.
Proof.
⇐: This is Lemma 3.8. ⇒: We prove the contrapositive. Suppose RTkn(W)≤idSkn(V)\mbox−SCA\mbox−RTkn. Let μ be a strongly increasing left-c.e. modulus of ∅′. By Lemma 3.9, there is a RTkn(V,W)-instance such that every solution computes a function dominating μ, hence computes ∅′.
∎
Corollary 3.11**.**
Let V be a collection of RTkn-patterns.
RTkn(V,WRTkn) admits strong cone avoidance if and only if Skn(V) contains only constant functions.
Proof.
By Theorem 3.10,
RTkn(V,WRTkn) admits strong cone avoidance if and only if RTkn≤idSkn(V)\mbox−SCA\mbox−RTkn.
Case 1: Skn(V) contains only constant functions. Then for any f:[ω]n→k, any Skn(V)\mbox−SCA\mbox−RTkn-solution to f is f-homogeneous, hence RTkn≤idSkn(V)\mbox−SCA\mbox−RTkn, so RTkn(V,WRTkn) admits strong cone avoidance.
Case 2: Skn(V) contains a non-constant function χ:Ln→k.
Let μ:ω→ω be a strongly increasing left-c.e modulus of a non-computable set C.
Let f(D)=χ(Ln(μ,D)). By Cholak and Patey [2, Corollary 5.5] and Lemma 2.25, there is an infinite strongly μ-transitive set H such that C≤TH. Therefore, H is an Skn(V)\mbox−SCA\mbox−RTkn-solution to f with witnesses μ and χ. We claim that H is not f-homogeneous. Since C≤TH, then in particular, H does not compute a function dominating μ, so by Theorem 2.26, for every G∈Ln, there is some D∈[H]n such that G=Ln(μ,D). Since χ is not constant on Ln, H is not f-homogeneous. Thus RTkn≤idSkn(V)\mbox−SCA\mbox−RTkn and RTkn(V,WRTkn) does not admit strong cone avoidance.
∎
3.3. Avoiding non-computable cones
Last, we complete our analysis of this extended class of promise Ramsey-like theorems with cone avoidance. The analysis follows the same scheme.
Statement 3.12**.**
Let T be a set of functions of type χ:Pn→k.
T\mbox−CA\mbox−RTkn: For every function f:[ω]n→k,
there is a strongly increasing relative left-c.e. function μ:ω→ω+, an infinite strongly μ-transitive set H⊆ω, and a coloring χ∈T such that for every D∈[H]n, f(D)=χ(Pn(μ,D)).
We now define Tkn(W) the same way we defined Rkn(W) and Skn(W) for ARITH\mbox−SCA\mbox−RTkn and SCA\mbox−RTkn, respectively.
Given some n,k∈ω and a collection W of RTkn-patterns,
let
[TABLE]
In particular, if RTkn(W) is a true statement, then every constant function χ:Pn→k belongs to Tkn(W).
Lemma 3.13**.**
Let V and W be collections of RTkn-patterns such that RTkn(W)≤idTkn(V)\mbox−CA\mbox−RTkn.
Then RTkn(V,W) admits cone avoidance.
Proof.
Fix two sets Z and C with C≤TZ and let f:[ω]n→k be a Z-computable instance of CA\mbox−RTkn.
By Lerman [14, 4.18], there is a set Z1≥TZ such that C is Δ20(Z1) but C≤TZ1. Since C is Δ20(Z1), there is a left Z1-c.e. modulus μ:ω→ω for C. By Lemma 2.23, we can assume that μ is strongly increasing.
Let f1:[ω]n→k×Pn be defined for each D∈[ω]n by f1(D)=⟨f(D),Pn(μ,D)⟩.
By cone avoidance of RT<∞,Cn−1n (Cholak and Patey [2, Corollary 4.19]), there is an infinite set H⊆ω such that C≤TZ1⊕H and ∣f1[H]n∣≤Cn−1. In particular, Z1⊕H does not compute a function dominating μ, so by Theorem 2.26, for every G∈Pn, there is some i<k and some D∈[H]n such that f1(D)=⟨i,G⟩. Since ∣Pn∣=Cn−1 (Cholak and Patey [2, Lemma 3.15,Lemma 3.16]), this i is unique. For each G, let χ(G) be such an i.
For every D∈[H]n, f(D)=χ(Pn(μ,D)). By Cholak and Patey [2, Corollary 5.5], there is an infinite μ-transitive subset H1⊆H such that C≤TZ⊕H1. By Lemma 2.25, there is an infinite strongly μ-transitive subset H2⊆H1 such that C≤TZ⊕H2. Therefore, H2 is a CA\mbox−RTkn-solution to f.
If χ∈Tkn(V), then since RTkn(W)≤idTkn(V)\mbox−CA\mbox−RTkn, H2 is an RTkn(V,W)-solution to f, and we are done. So suppose χ∈Tkn(V). Unfolding the definition, RTkn(V)≤id{χ}\mbox−CA\mbox−RTkn, so there is a coloring fχ:[ω]n→k and some infinite CA\mbox−RTkn-solution Hχ to fχ with witnesses μχ:ω→ω+ and χ:Pn→k, that meets some RTkn-pattern Pχ∈V.
However, H2 f-avoids Pχ since f is an instance of RTkn(V,W). By Lemma 2.45, H2⊕Z computes a function dominating μ, hence C≤TZ⊕H2, contradiction. This completes the proof of Lemma 3.13.
∎
Lemma 3.14**.**
Let V and W be collections of RTkn-patterns such that
RTkn(W)≤idTkn(V)\mbox−CA\mbox−RTkn.
For every strongly increasing left-c.e. function μ:ω→ω, there is a set Z which does not compute a function dominating μ, and a Z-computable RTkn(V,W)-instance such that every solution Z-computes a function dominating μ.
Proof.
Since RTkn(W)≤idTkn(V)\mbox−CA\mbox−RTkn, there is some coloring ffail:[ω]n→k and some infinite CA\mbox−RTkn-solution H to f with some witness μfail:ω→ω+ and χ∈Tkn(V) that meets some RTkn-pattern P∈W.
Define f:[ω]n→k by f(D)=χ(Pn(μ,D)). Suppose that μ is not dominated by a computable function.
By Cholak and Patey [2, Theorem 5.11] and Lemma 2.25, there is an infinite strongly μ-transitive set H which does not compute a function dominating μ. Therefore, H is a CA\mbox−RTkn-solution to f with witnesses μ and χ∈Tkn(V). By definition of Tkn(V), RTkn(V)≤id{χ}\mbox−CA\mbox−RTkn and H is an RTkn(V)-solution to f, so f:[H]n→k is an H-computable instance of RTkn(V,W) such that H does not compute a function dominating μ.
We claim that for every RTkn(V,W)-solution G⊆H to f, G⊕H computes a function dominating μ. In particular, G f-avoids P, so by Lemma 2.45, G⊕H computes a function dominating μ. This completes the proof of Lemma 3.14.
∎
Theorem 3.15**.**
Let V and W be two collections of RTkn-patterns.
RTkn(V,W) admits cone avoidance if and only if RTkn(W)≤idTkn(V)\mbox−CA\mbox−RTkn.
Proof.
⇐: This is Lemma 3.13. ⇒: We prove the contrapositive. Suppose RTkn(W)≤idTkn(V)\mbox−CA\mbox−RTkn. Let μ be a strongly increasing left-c.e. modulus of ∅′. By Lemma 3.14, there is a set Z which does not compute a function dominating μ and a Z-computable RTkn(V,W)-instance such that every solution Z computes a function dominating μ, hence computes ∅′. Therefore RTkn(W) does not admit cone avoidance.
∎
Corollary 3.16**.**
Let V be a collection of RTkn-patterns.
RTkn(V,WRTkn) admits cone avoidance if and only if Tkn(V) contains only constant functions.
Proof.
By Theorem 3.15,
RTkn(V,WRTkn) admits cone avoidance if and only if RTkn≤idTkn(V)\mbox−CA\mbox−RTkn.
Case 1: Tkn(V) contains only constant functions. Then for any f:[ω]n→k, any Tkn(V)\mbox−CA\mbox−RTkn-solution to f is f-homogeneous, hence RTkn≤idTkn(V)\mbox−CA\mbox−RTkn, so RTkn(V,WRTkn) admits cone avoidance.
Case 2: Tkn(V) contains a non-constant function χ:Pn→k.
Let μ:ω→ω be a strongly increasing left-c.e modulus of a non-computable set C.
Let f(D)=χ(Pn(μ,D)). By Cholak and Patey [2, Corollary 5.5] and Lemma 2.25, there is an infinite strongly μ-transitive set H which does not compute C. Therefore, H is a Tkn(V)\mbox−CA\mbox−RTkn-solution to f with witnesses μ and χ. We claim that H is not f-homogeneous. Indeed, since C≤TH, then H does not compute a function dominating μ, so by Theorem 2.39, for every G∈Pn, there is some D∈[H]n such that G=Pn(μ,D). Since χ is not constant on Pn, H is not f-homogeneous. Thus RTkn≤idTkn(V)\mbox−CA\mbox−RTkn and RTkn(V,WRTkn) does not admit cone avoidance.
∎
4. Applications
In this section, we exemplify the use of this framework by reproving existing theorems in reverse mathematics without involving any forcing argument.
4.1. Ramsey’s theorem
Ramsey’s theorem for pairs is one of the most famous theorems studied in reverse mathematics as it is historically the first one which is not equivalent (and not even linearly ordered with) the five systems of axioms known as the Big Five [25].
The first theorem that we reprove with our framework is the celebrated Seetapun theorem [24, Theorem 2.1], answering negatively the long-standing question whether RT22 is equivalent to the Arithmetical Comprehension Axiom (ACA) over RCA0.
Theorem 4.1** (Seetapun and Slaman [24]).**
For every k≥1, RTk2 admits cone avoidance.
Proof.
By Theorem 2.44, CA\mbox−RTk2 admits cone avoidance.
As explained, CA\mbox−RTk2 is nothing but the statement RTk2. Indeed, since ∣P2∣=1, the function χ:P2→k is constant, and CA\mbox−RTk2 asserts the existence of an infinite set H over which f belongs to the range of χ. In particular H is f-homogeneous.
∎
Later, Jockusch and Dzhafarov [13, Lemma 3.2] adapted the proof of Seetapun’s theorem to obtain strong cone avoidance of RTk1.
Theorem 4.2** (Jockusch and Dzhafarov [13]).**
For every k≥1, RTk1 admits strong cone avoidance.
Proof.
By Theorem 2.28, SCA\mbox−RTk1 admits strong cone avoidance.
Here again, SCA\mbox−RTk1 is nothing but the statement RTk1, since ∣L1∣=1.
∎
Wang [27, Theorem 3.1] surprisingly proved that for every n, whenever ℓ is sufficiently large, then RT<∞,ℓn admits strong cone avoidance. Cholak and Patey [2, Theorem 4.18] improved his bound to Catalan’s sequence, and proved the tightness of the result. The following theorem was used all over the article, and we can actually get a reversal.
Theorem 4.3** (Cholak and Patey [2]).**
For every n≥1, RT<∞,Cnn admits strong cone avoidance.
Proof.
By Theorem 2.28, SCA\mbox−RTkn admits strong cone avoidance.
In particular, SCA\mbox−RTkn asserts the existence, for every coloring f:[ω]n→k, of an infinite set H and a function χ:Ln→k such that f[H]n⊆χ(Ln). Since ∣Ln∣=Cn, it follows that ∣f[H]n∣≤Cn and therefore that H is an RT<∞,Cnn-solution to f.
∎
4.2. The Erdős-Moser theorem and the Ascending Descending sequence
A tournament T is a directed graph such that any two vertices has exactly one arrow. We consider the predicate T(x,y) to be true if there is an arrow from x to y. A set H is T-transitive if for every x,y,z∈H such that T(x,y) and T(y,z) both hold, then T(x,z) holds.
Statement 4.4** (Erdős-Moser).**
EM: Every infinite tournament admits an infinite transitive set.
A tournament T can be represented by a coloring f:[ω]2→2 such that for every x<y, f(x,y)=1 if and only if T(x,y) holds. A set H is f-transitive if for every x<y<z∈H and i<2, if f(x,y)=f(y,z)=i then f(x,z)=i. One can see the Erdős-Moser theorem as the Ramsey-like statement “For every coloring f:[ω]2→2, there is an infinite f-transitive set H.”
The following ADS principle was introduced and studied by Hirschfeldt and Shore [12] in the context of reverse mathematics.
Statement 4.5** (Ascending Descending sequence).**
ADS: Every infinite linear order has an infinite ascending or descending sequence.
As explained, a linear order (ω,≺L) can be represented by a coloring f:[ω]2→2 such that for every x<y∈ω, f(x,y)=1 if and only if x≺Ly. In particular, ω is f-transitive.
The Ascending Descending sequence principle can be seen as the promise Ramsey-like statement “For every coloring f:[ω]2→2 such that ω is f-transitive, there is an infinite f-homogeneous set.”
The EM principle was introduced by Bovykin and Weiermann [1] as a way to decompose the proof of RT22 into two steps. Indeed, given a coloring f:[ω]2→2, by EM, there is an infinite set X over which f is transitive, and by ADS, there is an infinite f-homogeneous subset Y⊆X. The author [20] proved that EM admits strong cone avoidance.
Theorem 4.6** (Patey [20]).**
EM* admits strong cone avoidance.*
Proof.
By Theorem 2.31, it suffices to prove that EM≤idSCA\mbox−RT22. However, SCA\mbox−RT22 is the statement LARGE2 saying that for every coloring f:[ω]2→2, there is some i<2 and an infinite set H such that for every x<y<z∈H, f(x,y)=f(y,z)=i if and only if f(x,z)=i. In particular H is f-transitive.
∎
The author [20] also deduced that ADS does not admit strong cone avoidance. Indeed, if ADS and EM both admit strong cone avoidance, then RT22 does, which is known not to be the case. One can however reprove it directly from our criterion for promise Ramsey-like theorems.
Theorem 4.7** (Patey [20]).**
ADS* does not admit strong cone avoidance.*
Proof.
Let VADS be the set of RT22-patterns forbidding non-transitive colorings. ADS is the promise Ramsey-like statement RT22(VADS,WRT22). By Corollary 3.11, ADS admits strong cone avoidance if and only if S22(VADS) contains only constant functions. Since RT22(VADS)≤idLARGE2 and LARGE2 is the statement SCA\mbox−RT22, then S22(VADS) contains all the functions χ:L2→2. Since ∣L2∣=2, ADS does not admit strong cone avoidance.
∎
Dorais et al [7] studied combinatorial principles, including Ramsey’s theorem, under the Weihrauch reduction. They introduced for this some new consequences of Ramsey’s theorem for pairs, such as SHER. Given a coloring f:[ω]2→k, a set H⊆ω is f-semi-hereditary if for all i<k except possibly one, whenever x<y<z∈H and f(x,z)=f(y,z)=i, then f(x,y)=i.
Statement 4.8** (Semi-hereditary).**
SHERk: Every coloring f:[ω]2→k such that ω is f-semi-hereditary has an infinite f-homogeneous set.
The restriction SHER2 was studied by Dorais (unpublished), who showed that it follows from ADS. Thanks to our general criterion for promise Ramsey-like principles, we can prove the following theorem.
Theorem 4.9**.**
For every k≥2, SHERk does not admit strong cone avoidance.
Proof.
Let VSHERk be the set of RTk2-patterns forbidding non-semi-hereditary colorings. SHERk is the promise Ramsey-like statement RT22(VSHERk,WRTk2). By Corollary 3.11, SHERk admits strong cone avoidance if and only if Sk2(VSHERk) contains only constant functions.
We now prove that SHERk≤idLARGEk.
Fix a coloring f:[ω]2→k and let H be an infinite LARGEk-solution to f. In particular, there are two colors is,iℓ<k such that f[H]2⊆{is,iℓ} and for every x<y<z∈H, f(x,y)=f(y,z)=is if and only if f(x,z)=is. Therefore, for all colors i<k except possibly iℓ, whenever f(x,z)=f(y,z)=i, then f(x,y)=i.
Since LARGEk is the statement SCA\mbox−RTk2, then Sk2(VSHERk) contains all the functions χ:L2→k. Since ∣L2∣=2, SHERk does not admit strong cone avoidance
∎
4.3. The free set theorem
We now provide an example of application to a statement which involves ω-colorings of [ω]n. The free set theorem was introduced by Friedman [9] and then studied by Cholak et al [3] in the framework of reverse mathematics. Given a coloring f:[ω]n→ω, a set H⊆ω is f-free if for every D∈[H]n, whenever f(D)∈H then f(D)∈D. Equivalently, H is f-free if for every x∈H, x∈f[H∖{x}]n.
Statement 4.10** (Free set theorem).**
FSn: Every coloring f:[ω]n→ω has an infinite f-free set.
Given a coloring f:[ω]n→ω, for every D={x0<⋯<xn}∈[ω]n+1, let g(D) be the function gD:[n+1]n→2 defined for every E∈[n+1]n by gD(E)=1 if and only if f({xi:i∈E})=xj where j is the unique value in D∖E. Since there are 2n+1 functions of type [n+1]n→2, one can see g as a function of type [ω]n+1→2n+1.
Moreover, every g-homogeneous set H must be for color the 0-constant function of type [n+1]n→2, and H is then f-free.
We say that an infinite set H is g-functional if for every D1={x0<⋯<xn},D2={y0<⋯<yn}∈[ω]n+1 and E∈[n+1]n, if g(D1)(E)=g(D2)(E)=1, then {xi:i∈E}={yi:i∈E}. In particular, ω is g-functional for the coloring g:[ω]n+1→2n+1 defined above.
We can therefore see FSn as the promise Ramsey-like statement “For every coloring g:[ω]n+1→2n+1 such that ω is g-functional, there is an infinite g-homogeneous set.”
Wang [27, Theorem 4.1] proved the following theorem.
Theorem 4.11** (Wang [27]).**
For every n≥1, FSn admits strong cone avoidance.
Proof.
Let VFSn be the set of RT2n+1n+1-patterns which ensures that ω is g-functional. By Corollary 3.11, FSn admits strong cone avoidance if and only if S2n+1n+1(VFSn) contains only constant functions.
Let χ:Ln+1→2n+1 be a non-constant function. In particular, there is some G∈Ln+1 such that χ(G):[n+1]n→2 is different from the 0-constant function of type [n+1]n→2.
Let μ:ω→ω be a strongly increasing left-c.e. modulus of ∅′. Let ρ:ω→ω be the left-c.e. function defined for every n,x∈ω by ρ2n(2x)=2μn(x), ρ2n(2x+1)=ρ2n(2x) and ρ2n+1=ρ2n. Informally, ρ is obtained from μ by considering the integers of μ as even numbers for ρ, and interleaving odd numbers which do not change the value of ρ. We claim that ρ is strongly increasing. First of all for every n, ρ2n is non-decreasing, and since ρ2n+1=ρ2n, neither is ρ2n+1. We need to check that for every x<y∈ω and s∈ω, if ρs+1(x)>ρs(x) then ρs+1(y)>s. By construction of ρ, if ρs+1(x)>ρs(x) then s is of the form 2s1+1. Let x1=⌊x/2⌋ and y1=⌊y/2⌋. In particular μs1+1(x1)>μs1(x1), and since μ is strongly increasing, μs1+1(y1)>s1 so ρs+1(y)=2μs1+1(y1)>2s1 so ρs+1(y)>s. Thus ρ is a strongly increasing left-c.e. function. Moreover, every function dominating ρ computes ∅′.
Let f:[ω]n+1→2n+1 be defined by f(D)=χ(Ln+1(ρ,D)). By Cholak and Patey [2, Corollary 5.5] and Lemma 2.25, there is an infinite strongly ρ-transitive set Heven⊆{2n:n∈ω} which does not compute ∅′, hence does not compute a function dominating ρ. We claim that H={x,x+1:x∈Heven} is strongly ρ-transitive. Let w<x<y<z∈H be such that ρz(w)>x and ρz(x)>y. Let w1,x1,y1 and z1 be the largest even value smaller or equal to w, x, y and z, respectively.
Then ρz1(w1)=ρz(w)>x≥x1 and ρz1(x1)=ρz(x)>y≥y1. By strong ρ-transitivity of Heven, ρz1(w1)>y1. In particular, ρz(w)=ρz1(w1)>y1, and since ρz(w) is even, ρz(w)>y.
Therefore, H is a χ\mbox−SCA\mbox−RT2n+1n+1-solution to f with witness ρ.
We claim that H is not an RT2n+1n+1(VFSn)-solution to f. Let G∈Ln+1 be such that χ(G) is not the 0-constant function of type [n+1]n→2. In particular, there is some E∈[n+1]n such that χ(G)(E)=1.
Since Heven does not compute ∅′, hence does not compute a function dominating ρ, by Theorem 2.26, there is some Deven={x0<⋯<xn}∈[Heven]n+1 such that G=Ln+1(ρ,Deven).
Let t be the unique element of {0,…,n+1}∖E
and let D={xi:i∈E}∪{xt+1}. In particular, D∈[H]n+1. By construction of ρ, Ln+1(ρ,Deven)=Ln+1(ρ,D)=G, so f(D)=f(Deven)=χ(G).
Then D and Deven∈[H]n+1 witness that H is not f-functional, and therefore that H is not an RT2n+1n+1(VFSn)-solution to f.
∎
4.4. The canonical Ramsey theorem and the rainbow Ramsey theorem
We conclude this section by providing a more general translation scheme from principles involving ω-colorings of [ω]n into promise Ramsey-like statements.
As shown by Theorem 2.4, Ramsey’s theorem is the maximal true Ramsey-like theorem for finite colorings of [ω]n.
The following canonical Ramsey theorem can be seen as the maximal true Ramsey-like theorem for ω-colorings from [ω]n. Given a coloring f:[ω]n→ω, a set H is f-canonical if there is a set U⊆{0,…,n−1} such that for every D0={x0<⋯<xn−1}∈[H]n and D1={y0<⋯<yn−1}∈[H]n, f(D0)=f(D1) if and only if for every i∈U, xi=yi.
Statement 4.12** (Canonical Ramsey theorem).**
CRTn: Every coloring f:[ω]n→ω has an infinite f-canonical set.
The canonical Ramsey theorem was first studied by Mileti [16] from a computability-theoretic viewpoint. He proved in particular that CRT2 does not admit cone avoidance. We now explain how to express the canonical Ramsey theorem as a promise Ramsey-like statement.
Given a function f:[ω]n→ω, for every D={x0<⋯<x2n−1}∈[ω]2n, let g(D) be the function gD:[2n]n×[2n]n→2 defined for every E0,E1∈[2n]n by
gD(E0,E1)=1 if and only if f({xi:i∈E0})=f({xi:i∈E1}).
Lemma 4.13**.**
An infinite set H is g-homogeneous if and only if it is f-canonical.
Proof.
⇒: Suppose first that H is g-homogeneous, say for color c:[2n]n×[2n]n→2.
Claim 1: There is a finite set U⊆{0,…,n−1} such that for every E0={a0<⋯<an−1}∈[2n]n and E1={b0<⋯<bn−1}∈[2n]n, c(E0,E1)=1 if and only if for every i∈U, ai=bi.
Fix E0 and E1.
By the classical canonical Ramsey theorem, there is an infinite subset H1⊆H which is f-canonical with some witness set U⊆{0,…,n−1}. Note that c(E0,E1)=1 if and only if for every F={z0<⋯<z2n−1}∈[H1]2n, letting D0={zi:i∈E0} and D1={zi:i∈E1}, f(D0)=f(D1). Since H1 is f-canonical with witness U, then f(D0)=f(D1) if and only if for every i∈U, the ith element of D0 equals the ith element of D1, if for every i∈U, ai=bi. This proves Claim 1. From now on, fix the set U.
Claim 2: H is f-canonical with witness U. Fix some D0,D1∈[H]n, and let F={z0<⋯<z2n−1}∈[H]2n be some set such that D0∪D1⊆F. Let E0={i<2n:zi∈D0} and E1={i<2n:zi∈D1}. By definition of g, f(D0)=f(D1) if and only if c(E0,E1)=1. By Claim 1, c(E0,E1)=1 if and only if for every i∈U, the ith element of E0 equals the ith element of E1. Therefore, f(D0)=f(D1) if and only if for every i∈U, the ith element of D0 equals the ith element of D1. This proves Claim 2.
⇐: Suppose now that H is f-canonical, with witness U⊆{0,…,n−1}. We claim that H is g-homogeneous. Fix some D={z0<⋯<z2n−1}∈[H]2n and let c=f(D), with c:[2n]n×[2n]n→2. We claim that c is fully specified by U. Fix some E0,E1∈[2n]n and let D0={zi:i∈E0} and D1={zi:i∈E1}. Then c(E0,E1)=1 if and only if f(D0)=f(D1), and by f-canonicity of H, this holds if and only if for every i∈U, the ith element of D0 equals the ith element of D1, which again holds if and only if the ith element of E0 equals the ith element of E1.
This property depends only on E0, E1 and U. Therefore c is unique.
∎
We say that a set H is g-comparing if g behaves as the coding of some function f, that is, for every F1,2={x0<⋯<x2n−1}∈[H]2n, F1,3={y0<⋯<y2n−1}∈[H]2n and F2,3={z0<⋯<z2n−1}∈[H]2n and E1,21,E1,31,E1,22,E2,32,E1,33,E2,33 such that {xi:i∈E1,21}={yi:i∈E1,31}, {xi:i∈E1,22}={zi:i∈E2,32} and {yi:i∈E1,33}={zi:i∈E2,33}, if g(F1,2)(E1,21,E1,22)=g(F2,3)(E2,32,E2,33)=1 then g(F1,3)(E1,31,E1,33)=1. Moreover g(F1,2)(E1,21,E1,22)=1 and if g(F1,2)(E1,21,E1,22)=1 then g(F1,2)(E1,22,E1,21)=1.
Let VCRTn be the set of RTℓ2n-patterns (where ℓ is the number of functions of type [2n]n×[2n]n→2) forbidding the sets which are non g-comparing. Then CRTn can be seen as the promise Ramsey-like statement RTℓ2n(VCRTn,WRTℓ2n).
A function f:[ω]n→ω is k-bounded if for every c∈ω, ∣f−1(c)∣≤k, that is, each color appears at most k times. A set H⊆ω is an f-rainbow if f is injective over [H]n.
Statement 4.14** (Rainbow Ramsey theorem).**
RRTkn: Every k-bounded coloring f:[ω]n→ω has an infinite f-rainbow.
Cisma and Mileti [5] first studied the rainbow Ramsey theorem in the context of reverse mathematics. Wang [27, Theorem 4.2] proved that RRT2n follows directly from FSn, thus that RRT2n admits strong cone avoidance for every n≥1. Later, the author [19, Theorem 4.6] proved that FSn follows from RRT22n+1.
Given a k-bounded function f:[ω]n→ω, we can define g:[ω]2n→[2n]n×[2n]n→2 as for the canonical Ramsey theorem. An infinite set H is g-homogeneous if and only if it is an f-rainbow. Then letting VRRTkn be the set of RTℓ2n-patterns which force the sets to be g-comparing and to code a k-bounded function, the rainbow Ramsey theorem can be seen as the promise Ramsey-like statement RTℓ2n(VRRTkn,WRTℓ2n).
5. Open questions
This article provides an extensive analysis of cone avoidance for Ramsey-like and promise Ramsey-like theorems. Other weakness notions have been proven to be very useful in reverse mathematics. It would be interesting to extend this analysis to these notions. We detail some of the remaining questions.
5.1. PA avoidance
Among the five main subsystems of second-order arithmetics studied in reverse mathematics, weak König’s lemma (WKL) captures compactness arguments. Weak König’s lemma asserts that every infinite binary tree admits an infinite path. The question whether RT22 implies WKL was a long-standing open question, until Liu [15] answered it negatively using the notion of PA avoidance. A Turing degree d is PA relative to X if every infinite X-computable binary tree has an infinite path bounded by d.
Definition 5.1** (PA avoidance).**
A problem P admits PA avoidance if for every set Z of non-PA degree and every Z-computable P-instance X, there is a P-solution Y to X such that Z⊕Y is of non-PA degree.
The notion of strong PA avoidance is defined accordingly. Liu [15] proved that RT21 admits strong PA avoidance and deduced that RT22 admits PA avoidance.
Question 5.2*.*
What Ramsey-like statements admit PA and strong PA avoidance, respectively?
The cone avoidance analysis for Ramsey-like statements strongly relies on finding the exact bounds for which the thin set theorems admits cone avoidance. The author [20] proved that for every n≥1, there is some ℓ∈ω such that RT<∞,ℓn admits strong PA avoidance.
5.2. Preservation of hyperimmunities
A very important and successful computability-theoretic notion to separate statements in reverse mathematics is simultaneous preservation of hyperimmunities. A function f:ω→ω is X-hyperimmune if it is not dominated by any X-computable function.
Definition 5.3** (Preservation of hyperimmunities).**
A problem P admits preservation of k hyperimmunities if for every set Z, every k-tuple of Z-hyperimmune functions f0,…,fk−1 and every Z-computable P-instance X, there is a P-solution Y such that all the functions f0,…,fk−1 are Z⊕Y-hyperimmune.
Again, the notion of strong preservation of k hyperimmunities is defined accordingly. The analysis of Section 2 actually shows that whenever RTkn(W)≤idSCA\mbox−RTkn, then RTkn(W) does not admit strong preservation of 1 hyperimmunity. Actually, the proof of Cholak and Patey [2] that RT<∞,Cnn admits strong cone avoidance can be adapted to prove that RT<∞,Cnn admits strong preservation of 1 hyperimmunity. By a similar analysis, we can prove that SCA\mbox−RTkn admits strong preservation of 1 hyperimmunity, and thus that the Ramsey-like statements which admit strong preservation of 1 hyperimmunity and strong cone avoidance coincide. The situation becomes different when preserving 2 hyperimmunities. Indeed, the author [18, Lemma 25 and Lemma 27] proved that RT21 does not admit strong preservation of 2 hyperimmunities, while Dzhafarov and Jockusch [8] proved that RT21 can strongly avoid multiple cones simultaneously.
Question 5.4*.*
What Ramsey-like statements admit preservation and strong preservation of k hyperimmunities, respectively?
The author [22, Theorem 8.4.1] proved that for every k∈ω and n≥1, there is some ℓ∈ω such that RT<∞,ℓn admits preservation of k hyperimmunities.
5.3. Jump cone avoidance
The question of the relation between stable Ramsey’s theorem for pairs and cohesiveness [21] motivated the study of jump computation and yielded the notion of jump cone avoidance which is similar to the notion of cone avoidance, but for jump computation.
Definition 5.5** (Jump cone avoidance).**
A problem P admits jump cone avoidance if for every set Z, every non-Δ20(Z) set C, and every Z-computable P-instance X, there is a P-solution Y such that C is not Δ20(Z⊕Y).
Once again, the notion of strong jump cone avoidance is defined accordingly by dropping the effectiveness restraint on the P-instance.
Recently, Monin and Patey [17, Theorem 4.1] proved that RT21 admits strong jump cone avoidance.
It is however currently unknown whether for every n≥1, there is some ℓ∈ω such that RT<∞,ℓn admits strong jump cone avoidance.
Question 5.6*.*
What Ramsey-like statements admit jump cone and strong jump cone avoidance, respectively?