effectiveness for the dual ramsey theorem · 2018. 11. 10. · n+2 set in the borel hierarchy, then...

EFFECTIVENESS FOR THE DUAL RAMSEY THEOREM

DAMIR D. DZHAFAROVSTEPHEN FLOOD

REED SOLOMON

LINDA BROWN WESTRICK

Abstract. We analyze the Dual Ramsey Theorem for k partitions and ` col-

ors (DRTk` ) in the context of reverse math, effective analysis, and strong re-

ductions. Over RCA0, the Dual Ramsey Theorem stated for Baire coloringsBaire-DRTk` is equivalent to the statement for clopen colorings ODRTk` and to

a purely combinatorial theorem CDRTk` .

When the theorem is stated for Borel colorings and k ≥ 3, the resulting

principles are essentially relativizations of CDRTk` . For each α, there is a

computable Borel code for a ∆0α coloring such that any partition homogeneous

for it computes ∅(α) or ∅(α−1) depending on whether α is infinite or finite.For k = 2, we present partial results giving bounds on the effective content

of the principle. A weaker version for ∆0n reduced colorings is equivalent to

Dn2 over RCA0 + IΣ0n−1 and in the sense of strong Weihrauch reductions.

1. Introduction

The Dual Ramsey Theorem states that for every sufficiently nice coloring of thek-block partitions of ω using ` colors, there is a partition of ω into infinitely manyblocks such that every coarsening of it down to exactly k blocks has the same color.The theorem was proved for Borel colorings by Carlson and Simpson [3] (who alsoshow it is not true for arbitrary colorings by a straightforward choice argument)and was extended to colorings with the Baire property by Promel and Voigt [10].

Dual Ramsey Theorem ([3], [10]). For all finite k, ` ≥ 1, if (ω)k = C0∪· · ·∪C`−1,where each Ci is Borel (or more generally has the Baire property), then there existp ∈ (ω)ω and i < ` such that (p)k ⊆ Ci.

In this statement, (ω)k is the set of partitions of ω into k nonempty pieces, (ω)ω

is the set of partitions of ω into infinitely many nonempty pieces and (p)k is theset of coarsenings of p down to exactly k many blocks. The partition p in the DualRamsey Theorem is said to be homogeneous for the coloring. Typically, we thinkof the colors Ci being disjoint although they do not have to be. Throughout thisarticle, when talking about versions of the Dual Ramsey Theorem with parametersk and `, we will assume k, ` ≥ 2.

To study the Dual Ramsey Theorem in computability theory or reverse math-ematics, we must choose a method to code the coloring of k-partitions. Previouswork in these areas focused on ODRTk` (requiring each color to be open), or avoidedcoding by considering variants of the Carlson-Simpson Lemma CSL(k, `) (the main

Dzhafarov was supported in part by NSF Grant DMS-1400267. The authors thank Jose Mijaresand Ludovic Patey for useful comments and discussions during the preparation of this paper.

1

2 DAMIR D. DZHAFAROV STEPHEN FLOOD REED SOLOMON LINDA BROWN WESTRICK

combinatorial lemma in [3], defined below in Section 3) such as the Variable andOrdered Variable Word principles VW(k, `) and OVW(k, `).

Miller and Solomon [9] defined the principle ODRTk` to be the restriction ofthe Dual Ramsey Theorem for k partitions and ` colors in which each color isrepresented by a code for an open set (and hence each color is clopen). Formalizing

arguments from [3], they showed that ODRTk+1` implies RTk` over RCA0, and hence

when k ≥ 4, ODRTk` implies ACA0.For the variants of Carlson-Simpson Lemma, Erhard [6] proved that COH does

not imply VW(2, 2) and that SRT22 does not imply OVW(2, 2), while Solomon and

Miller [9] showed that WKL0 does not imply VW(2, 2). However, the methods usedin these proofs do not extend to the more general Carlson-Simpson Lemma and itremains an open question whether CSL(2, 2) is computably true.

In the present paper, we consider a broader range of representations for the col-orings in the Dual Ramsey Theorem. After fixing notation in Section 2, we specifyfour version of the Dual Ramsey Theorem in Section 3. Three of the versions aredirectly related to the formal method of coding the coloring: Borel-DRTk` uses Borel

codes, ODRTk` uses codes for open sets and Baire-DRTk` uses a Baire approximation

to the coloring. The fourth variant, CDRTk` is a combinatorial statement similar tothe Carlson-Simpson Lemma (with a slight shift in parameters) but is more closely

tied to the topological versions. In particular, CDRTk` implies the Carlson-SimpsonLemma (with appropriate parameters) and it follows from transfinitely many nestedapplications of the Carlson-Simpson Lemma (again, with appropriate parameters).

In Section 3.1, we show that Baire-DRTk` , ODRTk` and CDRTk` are equivalent overRCA0, and that when k = 2, these principles are provable in RCA0, and hence arecomputably true. The coding issues for Borel-DRTk` are more subtle. We discuss the

connections between Borel-DRTk` , Baire-DRTk` and ATR0 in Section 3.2 but delaythe formal reverse mathematics proofs until Section 6.

Simpson noted a connection between CSL(2, `) and Hindman’s Theorem (see [3,page 268]), and we thank Ludovic Patey for showing us a proof of CSL(2, `) fromHindman’s Theorem. With minor modifications, we adapt this proof in Section 3.4to show that Hindman’s Theorem for ` colorings implies CDRT3

` and hence ACA+0

implies CDRT3` by Blass, Hirst and Simpson [2]. We also show that the method

does not generalize for k > 3.The earliest claim we are aware of for a proof of CDRTk` is in [10] where a

generalization of CDRTk` labeled Theorem A is attributed to a preprint of Voigttitled “Parameter words, trees and vector spaces”. However, as far as we can tell,this paper never appeared in print. Another proof of CDRTk` can be found in[14] where it comes as a corollary of a larger theory. Therefore, in Section 3.5, we

present a self contained classical proof of CDRTk` for k ≥ 3 (since CDRT2` is provable

in RCA0) in which the only non-constructive steps are ω ·(k−2) nested applicationsof the Carlson-Simpson Lemma.

In Section 4, we consider Borel-DRTk2 for k ≥ 3 from the perspective of effectivecombinatorics rather than reverse mathematics. For each ordinal 0 < α < ωCK1 ,there is a ∅(α)-computable clopen coloring (ω)k = R∪R for which any homogeneousinfinite partition p computes ∅(α). We pull the ∅(α) description of the open set Rdown to a computable code for R at the expense of describing R as a topologically∆0α or ∆0

α+1 set in the Borel hierarchy depending on whether α is infinite or finite.

EFFECTIVENESS FOR THE DUAL RAMSEY THEOREM 3

Figure 1. Implications over RCA0 between variants of the DualRamsey Theorem considered in this paper and some related prin-ciples. The parameter k ≥ 4 is arbitrary.

Π11-CA0

++ttATR0

��

Borel-DRTk2

��ACA+

0

��

CDRTk2 ↔ Baire-DRTk2 ↔ ODRTk2

��vvss

HT2

**��ACA0

��

CDRT32 ↔ Baire-DRT3

2 ↔ ODRT32

tt ++

CSL(k − 1, 2)

��RT2

2

��

CSL(2, 2)

vv

SRT22 ↔ ∆0

2-rDRT22

**RCA0 ↔ CDRT2

2 ↔ Baire-DRT22 ↔ ODRT2

2

Therefore, for each 0 < α < ωCK1 , there is a computable Borel code for a set Ras a topologically ∆0

α set (although R is classically open) such that every infinitehomogeneous partition for the coloring (ω)k = R ∪ R computes ∅(α−1) or ∅(α)

depending on whether α is finite or infinite.We interpret the results in Sections 3 and 4 as indicating that Baire codes are a

more natural representation than Borel codes for studying computational propertiesof the Dual Ramsey Theorem and that the Borel version of the Dual RamseyTheorem can be thought of as a relativization of the Baire version.

In Section 5, we study Borel-DRT22 and give upper bounds on the complexity

of finding an infinite homogeneous partition for colorings (ω)2 = R ∪ R where Ris coded as a set at a finite level of the Borel hierarchy. If R is a computableopen set, then there is a computable infinite homogeneous partition, although theconstruction of this partition is necessarily non-uniform. If R has a computablecode as a Σ0

n+2 set in the Borel hierarchy, then there is either a ∅(n)-computable

homogeneous partition for R or a ∅(n+1)-computable homogeneous partition for R.Because of the non-uniformity in these results, we end Section 5 by character-

izing a restriction of Borel-DRT22 under strong Weihrauch reducibility. For this

reducibility, we think of Borel-DRT22 as an instance-solution problem. Such a prob-

lem consists of a collection of subsets of ω called the instances of this problem, andfor each instance, a collection of subsets of ω called the solutions to this instance(for this problem). A problem P is strongly Weihrauch reducible to a problem Qif there are fixed Turing functionals Φ and Ψ such that given any instance A ofP, ΦA is an instance of Q, and given any solution B to ΦA in Q, ΨB is a solution


to A in P. There are a number of variations on this reducibility and we refer tothe reader to [5] and [8] for background on these reductions and for connections toreverse mathematics. In this paper, we will only be interested in problems arisingout of Π1

2 statements of second order arithmetic. Any such statement can be putin the form ∀X(ϕ(X) → ∃Y ψ(X,Y )), where φ and ψ are arithmetical. We canthen regard this as a problem, with instances being all X such that ϕ(X), andthe solutions to X being all Y such that ψ(X,Y ). Note that while the choice ofϕ and ψ is not unique, we always have a fixed such choice in mind for a given Π1

2

statement, and so also a fixed assignment of instances and solutions.We formulate a version of Borel-DRT2

2, denoted ∆0n-rDRT2

2, for which the in-stances are reduced colorings (ω)2 = R ∪ R where R and R are given by Borelcodes for Σ0

n sets in the Borel hierarchy and the solutions are homogeneous infinitepartitions. (We define a reduced coloring in Section 3.) We show this problemis strong Wiehrauch equivalent (and equivalent over RCA0 + IΣ0

n−1) to Dn2 whichstates that every stable coloring c : [ω]n → 2 has an infinite limit-homogeneousset. In particular, since D2

2 is equivalent to SRT22 over RCA0 by Chong, Lempp and

Yang [4], it follows that ∆02-rDRT2

2 is equivalent to SRT22 over RCA0.

In Section 6, we present a number of technical results in reverse mathematicsconnected to Borel and Baire codes. In particular, we show that ATR0 is equivalentto the statement that every Borel code has a Baire approximation and to thestatement that for every Borel code B, there is some point x such that x ∈ B orx 6∈ B. The proofs use a method of effective transfinite recursion, ETR, which isavailable in ACA0 (and possibly in weaker systems). Variations of these results areknown in some branches of effective mathematics. For example, Ash and Knight[1] prove similar results in the context of computable fragments of Lω1,ω ratherthan Borel codes. Greenberg and Montalban [7] use ETR to establish equivalencesof ATR0 and claim that ETR is provable in RCA0. However, their proof of ETRoverlooks an application of Σ0

1 transfinite induction, and in general, transfiniteinduction for Σ0

1 formulas does not hold in RCA0. While the main results in [7]continue to hold because Greenberg and Montalban show the classified theoremsimply ACA0 without reference to ETR (and hence can use ETR in ACA0 to completethe equivalence with ATR0), we have included a proof of ETR to make explicit theuse of transfinite induction.

We end this section with two comments on notation. First, we use ω to denotethe natural numbers, which in subsystems of Z2 is the set {x : x = x}, oftendenoted by N in the literature. Despite this notation, we do not restrict ourselvesto ω-models. Second, when we refer to the parameters k and ` in versions of theDual Ramsey Theorem, we assume k and ` are arbitrary standard numbers withk, ` ≥ 2. By a statement such as “RCA0 proves Borel-DRTk` implies Baire-DRTk` ”,

we mean, for all k, ` ≥ 2, RCA0 ` Borel-DRTk` → Baire-DRTk` . For many results,the quantification over k and ` can be pulled inside the formal system. However,in some cases, issues of induction arise and we wish to set those aside in this work.

2. Notation

For k ≤ ω, let k<ω denote the set of finite strings over k and let kω denote the setof functions f : ω → k. As noted above, unless explicitly stated otherwise, we willalways assume that k ≥ 2. For σ ∈ k<ω, |σ| denotes the length of σ, and if |σ| > 0,σ(0), . . . , σ(|σ| − 1) denote the entries of σ in order. For p ∈ kω and σ ∈ k<ω, we


write σ ≺ p if σ is an initial segment of p. Similarly, if σ, τ ∈ k<ω, we write σ � τif σ is an initial segment of τ and σ ≺ τ if σ is a proper initial segment of τ . Wewrite p � n to denote the string obtained by restricting the domain of p to n. Thestandard (product) topology on kω is generated by basic clopen sets of the form

[σ] = {p ∈ kω : σ ≺ p}for σ ∈ k<ω.

Informally, for k ≤ ω, a k-partition p of ω is a collection of k many pairwisedisjoint nonempty sets Bpi ⊆ ω (called blocks or p-blocks) such that ∪i<kBpi = ω.When the partition is clear from context, we may drop the superscript p. We denotethe least element of Bpi by µp(i) or simply µ(i). To fix a unique representation foreach k-partition, we assume the blocks are indexed such that µp(i) < µp(i + 1).With this convention, each k-partition is represented by a unique surjective functionp : ω → k with p−1(i) = Bpi .

More formally, we say f ∈ kω is ordered if

∀n ∀i < k(f(n) = i→ ∀j < i∃m < n (f(m) = j)

)and we say that σ ∈ k<ω is ordered if it satisfies the analogous condition for alln < |σ|. We let (ω)k ⊆ kω denote the set of ordered surjective functions f ∈ kω. Insecond order arithmetic, we view the notation p ∈ (ω)k as shorthand for the formalstatement that p : ω → k is an ordered surjective function. Similar comments applyto many of the sets defined below. In RCA0, we define a k-partition as follows.

Definition 2.1. Let k ≤ ω. A k-partition of ω is a function p ∈ (ω)k. If k < ω, wesay that p is a finite partition and if k = ω, we say that p is an infinite partition.

For finite k, (ω)kfin denotes the set of ordered σ ∈ k<ω in which all i < k appear.That is, σ ∈ (ω)kfin represents a k-partition of an initial segment of ω. The set (ω)k

inherits the subspace topology from kω with basic open sets of the form [σ] ∩ (ω)k

for σ ∈ k<ω. The space (ω)k is not compact since, for example, the collectionof open sets [0n1] for n ≥ 1 cover (ω)2 but this collection has no finite subcover.However, if σ ∈ (ω)kfin, then [σ] ⊆ (ω)k and [σ] is a compact clopen subset of (ω)k.To generate the topology on (ω)k, it suffices to restrict to the basic clopen sets ofthe form [σ] with σ ∈ (ω)kfin. Although the notation [σ] is ambiguous about whetherthe ambient space is kω or (ω)k (or `ω or (ω)` for some ` > k), the meaning will beclear from context.

Definition 2.2. Let p ∈ (ω)ω and k ≤ ω. We say q is a k-coarsening of p ifq ∈ (ω)k and for all n,m ∈ ω, if p(n) = p(m), then q(n) = q(m). In other words, qis a k-partition and each p-block is contained in a q-block. We let (p)k denote theset of all k-coarsenings of p.

Similarly, for τ ∈ (ω)hfin with h ∈ ω, σ ∈ (ω)kfin is a k-coarsening of τ if k ≤ h,|σ| = |τ | and for all n,m < |τ | such that τ(n) = τ(m), we have σ(n) = σ(m).

As with (ω)k, (p)k inherits the subspace topology from kω. For k < ω, we let(p)kfin denote the set of all σ ∈ (ω)kfin which are coarsenings of p � µp(n) for somen ≥ k. The topology on (p)k is generated by [σ] for σ ∈ (p)kfin.

Coarsenings have a natural composition operation. Let p ∈ (ω)ω, k ∈ ω andr ∈ (ω)k. Viewing p and r as functions, the composition r ◦ p : ω → k is an orderedsurjective map with r ◦ p ∈ (p)k. Intuitively, the partition coded by r ◦ p uses r todescribe how to combine the p-blocks. If r(m) = r(n) = i, then the p-blocks Bpn


and Bpm will be contained in the (r ◦ p)-block Br◦pi . The map from (ω)k to (p)k

defined by r 7→ r ◦ p gives a canonical homeomorphism between these sets.We can also compose elements of (ω)kfin and (ω)hfin under the right conditions.

Let k < h, σ ∈ (ω)kfin, and τ ∈ (ω)hfin. If |σ| ≥ h, then σ ◦ τ : |τ | → k is ordered.Moreover, if |σ| ≥ h > µσ(k − 1), then σ ◦ τ ∈ (ω)kfin because the least element ofthe last σ-block appears in the range of τ and thus σ ◦ τ is onto k.

We will use this compositional structure in two ways. First, we will use effectiveversions of the following lemma which states that an open coloring ∪i<Òi = (p)k

of the k-coarsenings of a fixed partition p ∈ (ω)ω can be turned into a coloring

∪i<Òi = (ω)k of the k-coarsenings of ω such that the set of all k-coarsenings of

q ∈ (ω)ω are contained in Oi if and only if the set of all k-coarsenings of q ◦ p arecontained in Oi.

Lemma 2.3. Fix p ∈ (ω)ω and k, ` ∈ ω. Let Oi, i < `, be open sets in (p)k such

that (p)k = ∪i<Òi. There are open sets Oi, i < `, in (ω)k such that ∪i<Òi = (ω)k

and for any q ∈ (ω)ω and i < `, (q)k ⊆ Oi if and only if (q ◦ p)k ⊆ Oi.Proof. This follows from the continuity of the canonical homeomorphism φ : (ω)k →(p)k, where φ(r) = r ◦ p. Letting Oi = φ−1(Oi), it is straightforward to check thatthese are as required. �

Later we will need to use the fact that this lemma holds in RCA0. For thatpurpose it is useful to express Oi as a union of basic open sets and describe theinverse image of each. Let Si ⊆ (p)kfin be such that Oi =

⋃σ∈Si [σ]. For each

σ ∈ (p)kfin, let n be such that σ is a coarsening of p � µp(n) and define τσ ∈ (ω)kfin tobe the string such that |τσ| = n and τσ(i) = σ(µp(i)) for all i < n. This definitionensures that τσ ◦ p = σ. We have for any r ∈ (ω)k, r ∈ [τσ] if and only if r ◦ p ∈ [σ].

Therefore, φ−1([σ]) = [τσ] and Oi =⋃σ∈Si [τσ].

Second, we will use the compositional structure to describe the k-coarsenings ofa given τ ∈ (ω)s+1

fin .

Lemma 2.4. Let s ∈ ω and τ ∈ (ω)s+1fin . For k ≤ s+ 1, the k-coarsenings of τ are

exactly the strings σ ◦ τ where σ ∈ (ω)kfin with |σ| = s+ 1.

Proof. Fix s, τ and k ≤ s + 1. Let σ ∈ (ω)kfin be such that |σ| = s + 1. Because

τ ∈ (ω)s+1fin and |σ| = s + 1 > µσ(k − 1), we have σ ◦ τ ∈ (ω)kfin by the comments

preceding Lemma 2.3. Therefore, σ ◦ τ is a k-coarsening of τ .Conversely, let τ ′ be a k-coarsening of τ . Define σ ∈ (ω)kfin with |σ| = s + 1 by

σ(i) = τ ′(µτ (i)) for all i < s + 1. By calculations similar to those in the proof ofLemma 2.3, we have that for all n < |τ | = |τ ′|, σ(τ(n)) = τ ′(n) as required. �

3. The Dual Ramsey Theorem

3.1. Four versions of the Dual Ramsey Theorem. We formulate four versionsof the Dual Ramsey Theorem in second order arithmetic and examine how they arerelated in reverse mathematics.

Definition 3.1 (RCA0). A code for an open set in (ω)k is a set O ⊆ ω × (ω)kfin.We say that a partition p ∈ (ω)k is in the open set coded by O (or just in O andwrite p ∈ O) if there is a pair 〈n, σ〉 ∈ O such that p ∈ [σ].

A code for an closed set in (ω)k is also a set V ⊆ ω× (ω)kfin. In this case, we sayp ∈ (ω)k is in V (and write p ∈ V ) if for all pairs 〈n, σ〉 ∈ V , p 6∈ [σ].


We code a sequence of open sets {Oi}i<ω by O ⊆ ω × ω × (ω)kfin with p ∈ Oi ifthere is a triple 〈i, n, σ〉 ∈ O such that p ∈ [σ]. The codes in Definition 3.1 can begeneralized to Borel sets which we describe in Section 3.2, although we delay theformal definition until Section 6 when we prove the results about Borel codes.

It will be useful to consider not only open colorings of (ω)k but also open coloringsof (p)k for p ∈ (ω)ω. Modifying Definition 3.1, a code for an open set in (p)k isa set O ⊆ ω × (p)kfin and we write x ∈ O if there is a pair 〈n, σ〉 ∈ O such thatσ ≺ x. With these definitions, the proof of Lemma 2.3 goes through in RCA0 with

the modification that O = {〈n, τσ〉 : 〈n, σ〉 ∈ O}.Coding colorings or sets with the Baire property in second order arithmetic is

complicated by the fact that there are 2c (where c = 2ℵ0) many subsets of (ω)k or kω

with the Baire property. Our definition for Baire codes (given below) is motivatedby considering how facts about sets with the Baire property are typically proved.

Definition 3.2 (RCA0). An open set O ⊆ (ω)k is dense if for all τ ∈ (ω)kfin,[τ ] ∩ O 6= ∅. That is, for all τ , there is a pair 〈n, σ〉 ∈ O such that σ and τ arecomparable as strings.

RCA0 suffices to prove the Baire Category Theorem: if {Dn}n<ω is a sequence ofdense open sets, then ∩n<ωDn is dense. Classically, if a coloring ∪i<`Ci = (ω)k hasthe Baire property, then it has a comeager approximation given by sequences of opensets {Oi}i<` and {Dn}n<ω such that each Dn is dense and for each p ∈ ∩n<ωDn,p ∈ Ci if and only if p ∈ Oi. The fact that ∪i<`Ci = (ω)k implies that ∪i<Òi isdense. Often, a classical proof about colorings or sets with the Baire property willstart by fixing a comeager approximation and will proceed by working exclusivelywith this approximation. This classical observation motivates our definition of acode for a Baire coloring.

Definition 3.3 (RCA0). A code for a Baire `-coloring of (ω)k is a sequence ofdense open sets {Dn}n<ω together with a sequence of open sets {Oi}i<` such that⋃i<Òi is dense in (ω)k.

In Definition 3.3, the code consists of a comeager approximation to the intendedcoloring and thus avoids the difficulties of explicitly describing the coloring in secondorder arithmetic. Note that if we define (classically) Ci = Oi ∩

⋂n<ωDn, then

∪i<`Ci will differ from (ω)k on a meager set. Thus, a single code for a Bairecoloring will represent many different classical colorings, each of which admits thesame comeager approximation.

We abuse terminology and refer to the Baire code as a Baire `-coloring of (ω)k.Similarly, an open (or Borel) `-coloring is a coloring (ω)k = ∪i<`Ci in which eachCi is given by an open (or Borel, respectively) code.

Definition 3.4. For each (standard) k, ` ≥ 2, we define Borel-DRTk` , Baire-DRTk` ,

ODRTk` and CDRTk` in RCA0.

(1) Borel-DRTk` : For every Borel `-coloring (ω)k = ∪i<`Ci, there is a partitionp ∈ (ω)ω and a color i < ` such that for all x ∈ (p)k, x ∈ Ci.

(2) Baire-DRTk` : For every Baire `-coloring {Oi}i<` and {Dn}n<ω of (ω)k, thereis a partition p ∈ (ω)ω and a color i < ` such that for all x ∈ (p)k,x ∈ Oi ∩

⋂nDn.

(3) ODRTk` : For every open `-coloring (ω)k = ∪i<Òi, there is a partitionp ∈ (ω)ω and a color i < ` such that for all x ∈ (p)k, x ∈ Oi.


(4) CDRTk` : For every coloring (ω)k−1fin = ∪i<`Ci, there is a partition p ∈ (ω)ω

and a color i < ` such that for all x ∈ (p)k, x � µx(k − 1) ∈ Ci.

In the statement of CDRTk` , each color Ci ⊆ (ω)k−1fin is a set of strings in second

order arithmetic and there is no assumption that the colors are disjoint. However, in

RCA0, given a coloring (ω)k−1fin = ∪i<`Ci, we can define a partition (ω)k−1

fin = ∪i<`Ciwhere σ ∈ Ci if and only if i is the least index such that σ ∈ Ci. Thus, it is equivalentover RCA0 to state CDRTk` in terms of colorings given as functions c : (ω)kfin → `.Similarly, in RCA0, we can replace open sets Oi, i < `, by pairwise disjoint open

sets Oi such that Oi ⊆ Oi and ∪i<Òi = ∪i<Òi. Therefore, in RCA0, we canassume without loss of generality that the individual colors in Baire-DRTk` , ODRTkànd CDRTk` are pairwise disjoint.

Our first goal is to show that the instances of CDRTk` are in one-to-one canonical

correspondence with those instances of ODRTkl for which the coloring of (ω)k isreduced. We define a reduced coloring without considering the coding method andnote that any reduced coloring is classically open. In RCA0, we will use the notionof a reduced coloring only in the context of an open coloring.

Definition 3.5. Let y ∈ (ω)ω and m < k. A coloring of (y)k is m-reduced ifwhenever p, q ∈ (y)k and p � µp(m) = q � µq(m), p and q have the same color. Acoloring of (y)k is reduced if it is (k − 1)-reduced.

Proposition 3.6 (RCA0). The following are equivalent.

(1) CDRTk` .(2) For every y ∈ (ω)ω and open reduced coloring (y)k = ∪i<Òi, there are

p ∈ (y)ω and i < ` such that for all x ∈ (p)k, x ∈ Oi.

Proof. Assume (2) and fix c : (ω)k−1fin → `. Let y ∈ (ω)ω be the trivial partition

with blocks {0}, {1}, . . . and note that (y)k = (ω)k. For each i < `, let

Oi = {〈0, σa(k − 1)〉 : σ ∈ (ω)k−1fin and c(σ) = i}

be an open code for the union of clopen sets [σa(k − 1)] such that c(σ) = i.(ω)k = ∪i<Òi is an open reduced coloring of (ω)k such that any p ∈ (ω)ω which ishomogeneous for this coloring is also homogeneous for c.

For the other direction, assume CDRTk` . Fix y ∈ (ω)ω and a reduced open

coloring (y)k = ∪i<Òi. By Lemma 2.3, let ∪i<Òi = (ω)k be an open coloring such

that for any q ∈ (ω)ω, (q)k ⊆ Oi if and only if (q ◦ y)k ⊆ Oi. It is straightforward

to check that the coloring (ω)k = ∪i<Òi is also reduced.

We claim that for each σ ∈ (ω)k−1fin , there is a triple 〈n, τ, i〉 ∈ ω× (ω)kfin× ` such

that 〈n, τ〉 ∈ Oi and σa(k − 1) � τ . To see why, let p ∈ (ω)k be any partition

extending σa(k − 1). Because (ω)k = ∪i<Òi, there is a color i < ` and a pair

〈n, τ〉 ∈ Oi such that τ ≺ p. Since τ ∈ (ω)kfin, we have σa(k − 1) � τ , proving the

existence of the triple 〈n, τ, i〉. Because the coloring {Oi}i<` is reduced, for any

x ∈ (ω)k, if x � µx(k − 1) = σ = p � µp(k − 1), then x ∈ Oi as well.

For each σ ∈ (ω)k−1fin , let 〈nσ, τσ, iσ〉 be the least triple satisfying the conditions

in the previous paragraph. Define c : (ω)k−1fin → ` by c(σ) = iσ. It follows that for

each x ∈ (ω)k and each i < `, if c(x � µ(k− 1)) = i, then x ∈ Oi. Applying CDRTk`to c, there are i < ` and q ∈ (ω)ω such that for all x ∈ (q)k, c(x � µ(k − 1)) = i.

Therefore, (q)k ⊆ Oi. Setting p = q ◦ y ∈ (y)ω, we have (p)k = (q ◦ y)k ⊆ Oi. �


It is now routine to show that the number of colors does not matter.

Proposition 3.7 (RCA0). CDRTk` and CDRTk2 are equivalent.

Proof. Collapse colors and iterate CDRTk2 finitely many times using the canonicalcorrespondence in Lemma 2.3. �

Lemma 3.8 (RCA0). For any Baire `-coloring of (ω)k given by {O}i<` and{Dn}n∈ω, there is a function f : (ω)kfin × ω → (ω)kfin × ` such that for all τ ∈ (ω)kfin

and s ∈ ω, f(τ, s) = 〈δ, i〉 where τ ≺ δ and [δ] ⊆ Oi ∩⋂n<sDn.

Proof. The function f is defined in a straightforward way by primitive recursionand minimization using the density of each Dn and ∪i<`Oi. �

The next proof is essentially an effective version of an argument in [10].

Theorem 3.9 (RCA0). Baire-DRTk` , ODRTk` and CDRTk` are equivalent.

Proof. By setting Dn = (ω)k in Baire-DRTk` , ODRTk` is a special case of Baire-DRTk` ,

and by Proposition 3.6, CDRTk` is a special case of ODRTk` . It remains to prove in

RCA0 that CDRTk` implies Baire-DRTk` .Let {Oi}i<`, {Dn}n<ω be a Baire `-coloring of (ω)k for which the open sets Oi

are pairwise disjoint. We construct a partition y ∈ (ω)ω such that (y)k ⊆ ∩nDn

and ∪iOi restricted to (y)k is reduced. By Proposition 3.6 and CDRTk` , there is ahomogeneous z ∈ (y)ω for this open reduced coloring. Since (z)k ⊆ (y)k ⊆ ∩nDn,this partition z is homogeneous for the original Baire coloring.

To build y ∈ (ω)ω, we construct a sequence τk−1 ≺ τk ≺ · · · of strings τs ∈ (ω)sfin

in stages starting with τk−1 for notational convenience and set y = ∪s≥k−1τs. Wedefine τk−1 by |τk−1| = k − 1 and τk−1(n) = n for n < k − 1, so τk−1 correspondsto the trivial partition {0}, {1}, . . . , {k − 1} of k.

At stage s+ 1, assume we have defined τs ∈ (ω)sfin with |τs| = ms. We extend τsfinitely many times to obtain τs+1. For the initial extension, let τ0

s = τas s ∈ (ω)s+1fin

which ensures that µx(s) = ms.For the remaining extensions, consider the ways to coarsen a finite partition with

(s+ 1) many blocks down to a partition with k many blocks. By Lemma 2.4, thesecoarsenings correspond to composing with strings σ ∈ (ω)kfin of length s + 1. LetMs denote the number of σ ∈ (ω)kfin with |σ| = s + 1 and let σ0

s , . . . , σMs−1s list

these strings. We define extensions τ js for 1 ≤ j ≤Ms (with τ0s defined above) and

set τs+1 = τMss .

Assume τ js has been defined. By Lemma 3.8, let δjs and ijs be such that σjs◦τ js ≺ δjsand [δjs] ⊆ Oijs ∩

⋂n<s+1Dn. That is, we coarsen τ js by the j-th canonical way to

coarsen the (s+ 1) many blocks to k blocks and then we take an extension of thiscoarsening that lies in Oijs ∩

⋂n<s+1Dn for some ijs < `.

To define τ j+1s , we want to “uncollapse” δjs by reversing the coarsening done to

τ js by σjs. Define τ j+1s with |τ j+1

s | = |δjs| by considering each n < |δjs|. If n < |τ js | setτ j+1s (n) = τ js (n), guaranteeing that τ js ≺ τ j+1

s . For n ≥ |τ js |, let τ j+1s (n) = τ js (m)

where m is least such that δjs(m) = δjs(n). That is, m = µδjs(δjs(n)). (Below we

show that m ≤ ms and σjs ◦ τ j+1s = δjs.) This completes the definition of τs+1 and

hence the construction of y in RCA0 since the initial segments of y are defined byprimitive recursion using the function f from Lemma 3.8.

Claim. In the definition of τ j+1s (n) when n ≥ |τ js |, the number m satisfies m ≤ ms.

10DAMIR D. DZHAFAROV STEPHEN FLOOD REED SOLOMON LINDA BROWN WESTRICK

Since τ0s � τ js , if m is the least element of a τ js -block, then m ≤ ms. Because

σj ◦ τ js is a coarsening of τ js , it follows that if m is the least element of a (σj ◦ τ js )-block, then m ≤ ms. Therefore, since σjs ◦ τ js ≺ δjs and m is the least element of aδjs-block, we have m ≤ ms.

Claim. For all s ≥ k and j < Ms, σjs ◦ τ j+1

s = δjs.

We show that for all n < |δjs|, σjs(τ j+1s (n)) = δjs(n). For n < |τ js |,

σjs(τj+1s (n)) = σjs(τ

js (n)) = δjs(n)

where the last equality holds because σjs ◦ τ js ≺ δjs. For n ≥ |τ js |, τ j+1s (n) = τ js (m)

for an m ≤ ms such that δjs(m) = δjs(n). Therefore,

σjs(τj+1s (n)) = σjs(τ

js (m)) = δjs(m) = δjs(n)

where the second equality holds because m ≤ ms < |τ js | and σjs ◦ τ js ≺ δjs.

Claim. (y)k ⊆ ∪i<Òi and the coloring ∪i<Òi restricted to (y)k is reduced.

Let x ∈ (y)k. We need to show that there is an i such that x ∈ Oi. (Since theopen sets Oi are pairwise disjoint, there is at most one such index i.) Furthermore,

for any x′ ∈ (y)k with x′ � µx′(k−1) = x � µx(k−1), we need to show that x′ ∈ Oi.

Let m = µx(k − 1). Since m is the least element of an x-block, it must be theleast element of a y-block. Fix s such that m = µy(s) and note that s ≥ k − 1.In the notation of the construction, m = ms and y(m) = s was first decided atstage s+ 1. In particular, τ0

s = τas s is the initial segment of y ending with the leastelement of Bys . Similarly, x � (µx(k− 1) + 1) is the initial segment of x ending withthe least element of Bxk−1.

Since Bys is the least y-block collapsed into Bxk−1, we can fix the index j < Ms

from stage s + 1 such that σjs ∈ (ω)kfin with |σjs| = s + 1 satisfies σjs ◦ τ0s = x �

(µx(k− 1) + 1). That is, σjs described how x collapses the first (s+ 1)-many blocksof y into the k-many blocks of x. For any initial segment γ of y with τ0

s � γ(so γ contains elements from each of the first (s + 1)-many y-blocks) and withγ � τs+1 (so γ only containing elements from the first (s + 1)-many y-blocks),σjs ◦ γ will collapse the (s + 1)-many γ-blocks as x specifies and so will satisfyσjs ◦ γ ≺ x. In particular, since τ0

s � τ j+1s ≺ y and |τ j+1

s | ≤ |τs+1| = µy(s + 1),we have σjs ◦ τ j+1

s = δjs ≺ x. However, [δjs] ⊆ Oijs by construction, so x ∈ Oijs . If

x′ ∈ (y)k satisfies x′ � µx′(k−1) = x � µx(k−1), then x′ determines the same string

σjs ∈ (ω)kfin and hence x′ ∈ Oijs as well. This completes the proof of the claim.

Finally, we show that if x ∈ (y)k, then x ∈ ∩n<ωDn. Fix x and let m and s beas in the proof of the previous claim. Since δjs ≺ x and [δjs] ⊆ ∩n<s+1Dn, we havex ∈ ∩n<s+1Dn. The following claim will complete the proof that x ∈ ∩n<ωDn.

Claim. For all t > s, by the end of stage t+ 1, it is forced that x ∈ Dt.

The proof of this claim is almost identical to the proof that the coloring ∪i<Òirestricted to (y)k is reduced. Fix a stage t + 1 for t > s. The string τ0

t = τat tis the initial segment of y ending with the least element of Byt . Because t > s, x

coarsens τ0t down to k-many blocks. Fix the index j for the string σjt ∈ (ω)kfin with

|σjt | = t + 1 that describes this coarsening, so σjt ◦ τ0t = x � (mt + 1). As before,

σjt ◦ τj+1t = δjt ≺ x. Since [δjt ] ⊆ ∩n<t+1Dn, we have x ∈ Dt completing the proof

of the claim and the fact that x ∈ ∩n<ωDn. Because (y)k ⊆ ∩nDn and ∪i<Òi


restricted to (y)k is an open reduced coloring, we have also completed the proof

that CDRTk` implies Baire-DRTk` . �

Since ODRTk+1` implies RTk` over RCA0, we have the following corollary.

Corollary 3.10 (RCA0). CDRTk+1` implies RTk` .

Proposition 3.11. For any ` ≥ 2, RCA0 proves CDRT2` and hence also ODRT2

` .

Proof. Let c : (ω)1fin → `. Since (ω)1

fin = {0n : n ≥ 1}, c can be viewed as an

`-coloring of ω. By RT1` , there is a color i and an infinite set X such that for every

n ∈ X, c(0n) = i. Let z be the partition which has a block of the form {n} for eachn ∈ X and puts all the other numbers in Bz0 . Then z is homogeneous for c. �

3.2. Relationships with Borel-DRTk` . We give the formal definition of a Borelcode in Section 6. Informally, a Borel code B for a subset of (ω)k is a well foundedtree in ω<ω in which each leaf codes a clopen set and the interior nodes code eitheran intersection or a union. Given a point x ∈ (ω)k, an evaluation map for B at xis a function f : B → {0, 1} such that f(σ) = 1 for a leaf σ if x is in the clopenset coded by σ and f correctly propagates down the tree respecting unions andintersections. We say x ∈ B if there is an evaluation function with value 1 at theroot, and we say x 6∈ B if there is an evaluation function with value 0 at the root.Therefore, both x ∈ B and x 6∈ B are Σ1

1 statements, and in general, ATR0 isrequired to show that evaluation maps exists. Similarly, (ω)k = B0 ∪ . . . ∪ B`−1 isthe Π1

2 statement that for every x ∈ (ω)k and i < `, there is an evaluation map forBi at x and for some i < `, x ∈ Bi.

Proposition 3.12 (RCA0). Borel-DRTk` implies Baire-DRTk` .

Proof. In Section 6, we show in RCA0 that if (ω)k = O0 ∪ · · · ∪ O`−1 is an opencoloring, then each Oi has an equivalent Borel code Bi such that (ω)k = B0 ∪ · · · ∪B`−1. It follows that Borel-DRTk` implies ODRTk` and hence implies Baire-DRTk` . �

Definition 3.13 (RCA0). Let B be a Borel (or open or closed) code for subset of(ω)k. A Baire code for B consists of open sets U and V and a sequence 〈Dn : n ∈ ω〉of dense open sets such that U ∪ V is dense and for every x ∈ ∩n∈ωDn, if p ∈ Uthen p ∈ B and if p ∈ V then p 6∈ B.

In Section 6, we will prove the following theorem.

Theorem 3.14 (RCA0). The following are equivalent.

(1) ATR0.(2) For every Borel code B for a subset of (ω)k, there is an x ∈ (ω)k such that

x ∈ B or x 6∈ B.(3) Every Borel code B for a subset of (ω)k has a Baire code.

Definition 3.15 (RCA0). A Baire code for a Borel coloring (ω)k = C0 ∪ · · · ∪C`−1

consists of open sets Oi, i < `, and a sequence 〈Dn : n ∈ ω〉 of dense open sets suchthat ∪i<`Oi is dense and for every p ∈ ∩n∈ωDn and i < `, if p ∈ Oi then p ∈ Ci.

Proposition 3.16 (ATR0). Every Borel coloring (ω)k = C0∪· · ·∪C`−1 has a Bairecode.


Proof. By Theorem 3.14, fix Baire codes Ui, Vi and Dn,i for each Ci. We claimthat the open sets Ui for i < ` and the sequence of dense open sets Dn,i for i < `and n < ω form a Baire code for this coloring. Note that if i < ` and x ∈ ∩n,iDn,i,then x ∈ Ui implies x ∈ Ci. Therefore, it suffices to show that ∪i<`Ui is dense.

Suppose not. Then there is τ such that [τ ] ∩ Ui = ∅ for all i. Because each setUi∪Vi is open and dense, by the Baire Category Theorem there is x ∈ [τ ] such thatx ∈ ∩n∈ω,i<`Dn,i and x ∈ ∩i<`(Ui∪Vi). Since x is not in any Ui, we have x ∈ Vi foreach i. Therefore, for each i, x 6∈ Ci, contradicting that (ω)k = C0 ∪ · · · ∪C`−1. �

Proposition 3.17 (ATR0). Baire-DRTk` implies Borel-DRTk` .

Proof. By Proposition 3.16, each Borel coloring has a Baire code. Baire-DRTk`guarantees a homogeneous partition for the coloring given by this Baire code andthis partition is homogeneous for the Borel coloring. �

3.3. Alternate coding methods for two complementary colors. While webelieve our formal statement of the Borel Dual Ramsey Theorem is the most natural,there is an alternate formal version of this theorem for colorings (classically) of theform (ω)k = B ∪B that avoids explicitly stating that every partition p ∈ (ω)k hasan evaluation map. The material in this subsection is somewhat tangential to ourmain story and nothing from it is used later in the paper.

Theorem 3.18 (RCA0). The statement “for every Borel code B for a subset of(ω)k, there is a partition p ∈ (ω)ω such that either ∀x ∈ (p)k(x ∈ B) or ∀x ∈(p)k(x 6∈ B)” implies ATR0.

Proof. Fix a Borel code B. The given statement implies there is an x ∈ (ω)k suchthat x ∈ B or x 6∈ B. (Let x be any coarsening of p down to k-blocks.) By Theorem3.14, this suffices to prove ATR0. �

Note that this argument does not suffice to prove an implication from Borel-DRTk2to ATR0 because the hypotheses of Borel-DRTk2 include that (ω)k = C0 ∪C1 whichrequires the existence of an evaluation map for every x ∈ (ω)k witnessing x ∈ C0

or x ∈ C1.There is an analogous variant of the Dual Ramsey Theorem for colorings (ω)k =

O ∪ O where O is a code for an open set. Here, O = O as sets but O is viewedas a code for a closed set. That is, for p ∈ (ω)k, p ∈ O if for every 〈s, σ〉 ∈ O,σ 6≺ p. Therefore O is a code for the complement of O. (Note that this version

differs significantly from ODRTk2 because one of the colors is closed.)For 0 < a < b, let Oa,b = {x ∈ (ω)3 : µx(1) = a ∧ µx(2) = b}. Oa,b is a

finite union of basic open sets [σ] with σ ∈ (ω)3fin and |σ| = b + 1. For notational

convenience, we write σ ∈ Oa,b if |σ| = b+ 1, µσ(1) = a and µσ(2) = b.The use of exponent (at least) 3 in the following theorem is important. In

Theorem 5.1, we will see that RCA0 suffices to prove that for every open set O in(ω)2, there is an infinite homogeneous partition for (ω)2 = O ∪O.

Theorem 3.19 (RCA0). The statement “for every open set O in (ω)3, there is aninfinite homogeneous partition for the coloring (ω)3 = O ∪O” implies ACA0.

Proof. Fix a 1-to-1 function g and we show the range of g exists. Define O by

〈s, σ〉 ∈ O ⇔ ∃ 0 < a < b < s(σ ∈ Oa,b ∧ ∃u ≤ a ∃b < t ≤ s (g(t) = u)

).


Note that for all 0 < a < b, either Oa,b ⊆ O (when ∃u ≤ a∃t > b(g(t) = u))

or Oa,b ⊆ O (when ∀u ≤ a¬∃t > b(g(t) = u)). Let p ∈ (ω)ω be a homogeneous

partition for (ω)3 = O ∪O. We claim that p is homogeneous for O.We prove the claim by constructing x ∈ (p)3 ∩O. Let n = g(0), let m1 be such

that n < µp(m1) and let a = µp(m1). Since g is 1-to-1, the set C = {t : g(t) ≤ a}is finite. Let m2 > m1 be such that µp(m2) > max(C) and let b = µp(m2).Define x ∈ (p)3 by Bx0 = ∪i<m1B

pi , Bx1 = ∪m1≤i<m2B

pi and Bx2 = ∪i≥m2B

pi . By

definition, µx(1) = a and µx(2) = b, so x ∈ Oa,b. By our choice of a and b,

∀u ≤ a¬∃t ≥ b (g(t) = u) and therefore x ∈ Oa,b ⊆ O as required.Since the function µp is strictly increasing, we can define the function f(n) =

the least m such that µp(m) > n. We claim that n ∈ range(g) if and only if∃t ≤ µp(f(n) + 1) (g(t) = n). Suppose n = g(t) but t > µp(f(n) + 1). In this case,〈t, Oµp(f(n)),µp(f(n)+1)〉 ∈ O. But, collapsing p as above, there is an x ∈ (p)3 suchthat µx(1) = µp(f(n)) and µx(2) = µp(f(n) + 1) and hence x ∈ O contradictingthe fact that p is homogeneous for O. �

We have seen that obtaining Baire codes for Borel colorings codes a significantamount of information. The next theorem shows that even obtaining Baire codesfor open colorings codes a non-trivial amount of information.

Theorem 3.20 (RCA0). The following are equivalent.

(1) ACA0.(2) Every closed subset of (ω)k has a Baire code.(3) Every open subset of (ω)k has a Baire code.

Proof. (2) and (3) are equivalent by trading the roles of U and V in their respectiveBaire codes. To see (1) implies (2), fix a closed set C, so C is a set of pairs 〈s, σ〉 andx ∈ C if for all 〈s, σ〉 ∈ C, σ 6≺ x. To define a Baire code for C, set V = C as sets,but view V as a code for the open set C. That is, x ∈ V if there is 〈s, σ〉 ∈ V = Csuch that σ ≺ x. Let U = {〈0, τ〉 : ¬∃〈s, σ〉 ∈ V (σ � τ∨τ ≺ σ)} and set Dn = (ω)k

for all n ∈ ω. It is straightforward to check that U ∪ V is dense.Suppose x ∈ ∩nDn = (ω)k. If x ∈ V , then by definition, x ∈ C. On the other

hand, suppose x ∈ U and fix 〈0, τ〉 ∈ U with τ ≺ x. For every 〈s, σ〉 ∈ V = C, τ isincomparable with σ and hence σ 6≺ x. Therefore, x ∈ C as required.

We show (3) implies (1) for the case when k = 2. The proof is similar for othervalues of k. Fix a 1-to-1 function g. Let O = {〈n, 0m+11〉 : g(n) = m}. LetU, V, {Dn}n∈ω be a Baire code for O. Then the range of g has a ∆0

1 definition:

∃n[g(n) = m] ⇐⇒ ∃σ[0m+11 ≺ σ and σ ∈ U ] ⇐⇒ ¬(∃σ[0m+11 ≺ σ and σ ∈ V ]).

The complementarity of the above Σ01 formulas does not require any induction. For

each m, either [0m+11] ⊆ O or [0m+11]∩O = ∅. The density of U ∪V implies thereis a σ ∈ U ∪ V with 0m+1 ≺ σ. If [0m+11] ⊆ O then σ ∈ U , and if [0m+11]∩O = ∅,then σ ∈ V . �

3.4. Connections to Hindman’s theorem. In this section, we show that Hind-man’s Theorem for `-colorings implies CDRT3

` .

Definition 3.21 (RCA0). Let Pfin(ω) denote the set of (codes for) all non-emptyfinite subsets of ω. X ⊆ Pfin(ω) is an IP set if X is closed under finite unions andcontains an infinite sequence of pairwise disjoint sets.


Theorem 3.22 (Hindman’s theorem for `-colorings). For every c : Pfin(ω) → `there is an IP set X and a color i < ` such that c(F ) = i for all F ∈ X.

Proposition 3.23 (RCA0). Hindman’s theorem for `-colorings implies CDRT3` . In

particular, CDRT3` is provable in ACA+

0 .

Proof. Fix ` ≥ 2 and assume Hindman’s Theorem for `-colorings. Since Hindman’sTheorem for 2-colorings implies ACA0, we reason in ACA0. By Proposition 3.6 andLemma 2.3, it suffices to fix an open reduced coloring (ω)3 = ∪i<`Oi and producep ∈ (ω)ω and i < ` such that for all x ∈ (p)3, x ∈ Oi. We write the coloring asc : (ω)3 → ` with the understanding that c(x) = i is shorthand for x ∈ Oi.

For a nonempty finite set F ⊆ ω with 0 /∈ F and a number n > maxF , we letxF,n ∈ (ω)3 be the following partition.

xF,n(k) =

0 if k /∈ F and k 6= n

1 if k ∈ F2 if k = n

Thus, BxF,n0 = ω − (F ∪ {n}), BxF,n1 = F and B

xF,n2 = {n}. Note that we can

determine the color c(xF,n) as a function of F and n and that since c is reduced, ifx ∈ (ω)3 and x � µx(2) = xF,n � n, then c(x) = c(xF,n).

The remainder of the proof is most naturally presented as a forcing construction.After giving a classical description of this construction, we indicate how to carryout the construction in ACA0. The forcing conditions are pairs (F, I) such that

• F is a non-empty finite set such that 0 /∈ F ,• I is an infinite set such that maxF < min I, and• for every nonempty subset U of F there is an i < ` such that c(xU,n) = i

for all n ∈ F ∪ I with maxU < n.

Extension of conditions is defined as for Mathias forcing: (F , I) ≤ (F, I) if F ⊆F ⊆ F ∪ I and I ⊆ I.

By the pigeonhole principle, there is an i < l such that c(x{1},n) = i for infinitelymany n > 1. For any such i, the pair ({1}, {n ∈ ω : n > 1 and c(x{1},n) = i}) is

a condition. More generally, given a condition (F, I) there is an infinite set I ⊆ I

such that (F ∪ {min I}, I) is also a condition. To see this, let U0, . . . , Us−1 bethe nonempty subsets of F ∪ {min I} containing min I. By arithmetic induction,for each positive k ≤ s, there exist colors i0, . . . , ik−1 < ` such that there areinfinitely many n ∈ I with c(xUj ,n) = ij for all j < k. (If not, fix the least k forwhich the fact fails, and apply the pigeonhole principle to obtain a contradiction.)

Let i0, . . . , is−1 be the colors corresponding to k = s and let I be the infinite set{n ∈ I : ∀j < s (c(xUj ,n) = ij)}.

Fix a sequence of conditions (F1, I1) > (F2, I2) > · · · with |Fk| = k and letG =

⋃k Fk. To complete the proof, we use G to define a coloring d : Pfin(ω) → `

to which we can apply Hindman’s Theorem. However, first we indicate why we canform G in ACA0.

The conditions (F, I) used to form G can be specified by the finite set F , thenumber m = min I and the finite sequence δ ∈ `M where M = 2|F | − 1 such thatif F0, . . . , FM−1 is a canonical listing of the nonempty subsets of F , then I = {n ≥m : ∀j < M (c(xFj ,n) = δ(j))}. The extension procedure above can be captured byan arithmetically definable function f(F,m, δ) = 〈F ∪ {m},m′, δ′〉 where F ∪ {m},


m′ and δ′ describe the extension (F ∪ {m}, I). Because the properties of thisextension where verified using arithmetic induction and the pigeonhole principle,both of which are available in ACA0, we can define f in ACA0 and form a sequenceof conditions (F1,m1, δ1) > (F2,m2, δ2) > · · · giving G =

⋃k Fk.

It remains to use G = {g0 < g1 < · · · } to complete the proof. By construction,for each non-empty finite subset U of G, there is color iU < ` such that c(xU,n) = iUfor all n ∈ G with n > maxU . Define d : Pfin(ω) → ` by d(F ) = i{gm:m∈F}. Weapply Hindman’s theorem to d to obtain an IP set X and a color i < `. Since Xcontains an infinite sequence of pairwise disjoint members, we can find a sequenceE1, E2, . . . of members of X such that maxEk < minEk+1. Define p ∈ (ω)ω to bethe partition whose blocks are Bp0 = ω −

⋃k{gm : m ∈ Ek} and, for each k ≥ 1,

Bpk = {gm : m ∈ Ek}. Note that for all k ≥ 1,

maxBpk = max{gm : m ∈ Ek} < minBpk+1 = min{gm : m ∈ Ek+1}.

It remains to verify that p and i have the desired properties. Consider any x ∈(p)3; we must show that c(x) = i. Let U = Bx1 � µx(2) and let n = µx(2) = minBx2 .Then n = µxU,n(2) and x � n = xU,n � n, so since c is reduced, c(x) = c(xU,n).Therefore, it suffices to show c(xU,n) = i.

We claim U is a finite union of p-blocks. Because x is a coarsening of p, Bx1 isa (possibly infinite) union of p-blocks Bpj1 ∪ B

pj2∪ · · · with 0 < j1 < j2 < · · · and

n = µx(2) = minBx2 = minBpb for some b ≥ 2. Let ja be the largest index such thatja < b. Since the p-blocks are finite and increasing, U = Bx1 � µx(2) = Bpj1∪· · ·∪B

pja

.

Note that n ∈ G (because Bpb 6= Bp0) and maxU < n.It follows that U = {gm | m ∈ F} where F = Ej1 ∪ · · · ∪ Eja . Since our fixed

IP set X is closed under finite unions, F ∈ X and therefore d(F ) = i. By thedefinition of d, d(F ) = i{gm|m∈F} = iU , so i = iU . Finally, U is a finite subset ofG, n ∈ G and maxU < n, so c(xU,n) = iU = i as required. �

Observe that this proof of CDRT3` from HT produces a homogeneous p with a

special property: maxBpi < minBpi+1 for all i > 0. We show that this strengthened

“ordered finite block” version of CDRT3` is equivalent to HT. However, there is no

finite block version of CDRTk` for k > 3.

Proposition 3.24 (RCA0). If for every `-coloring of (ω)2fin there is an infinite

homogeneous partition p with maxBi < minBi+1 for all i > 0, then Hindman’sTheorem for `-colorings holds.

Proof. Given c : Pfin(ω)→ `, define c : (ω)2fin → ` by c(σ) = c({i < |σ| : σ(i) = 1}).

Let p be a homogeneous partition for c with maxBpi < minBpi+1 for all i > 0.The set of all finite unions of the blocks Bpi for i > 0 satisfies the conclusion ofHindman’s Theorem. �

Proposition 3.25. There is a 2-coloring of (ω)3fin such that any infinite homoge-

neous partition p has Bpi infinite for all i > 0.

Proof. For σ ∈ (ω)3fin, set c(σ) = 1 if σ contains more 1’s than 2’s and set c(σ) = 0

otherwise. Let p be homogeneous for this coloring. Suppose for contradiction thati > 0 is such that Bpi is finite. Let N = i+ 2 + |Bpi | and let x be the coarsening ofp with nonzero blocks

Bx1 = Bpi , Bx2 = ∪Nj=i+1B

pj and Bx3 = BpN+1.


Since |Bx2 | > |Bx1 |, c(x � µx(3)) = 0. Now coarsen in a different way: let h ∈[i + 1, N ] be chosen so that the size of Bph ∩ [0, µx(3)] is minimized. Let y be thecoarsening of p whose nonzero blocks are

By1 = ∪Nj=iBpi \B

ph, B

y2 = Bph and By3 = BpN+1.

Since at least one p-block has moved from Bx2 to By1 and since By2 contains onlythe smallest p-block from Bx2 , c(y � µy(3)) = 1. So p was not homogeneous. �

3.5. CDRT and the Carlson-Simpson Lemma. The Carlson-Simpson Lemmais the main technical tool in the original proof of the Borel version of the DualRamsey Theorem. The principle is usually stated in the framework of variablewords, but it can also be understood as a special case of the Combinatorial DualRamsey Theorem.

Carlson-Simpson Lemma (CSL(m, `)). For every coloring (ω)mfin = ∪i<`Ci, thereis a partition p ∈ (ω)ω and a color i such that for all x ∈ (p)m+1, if Bpj ⊆ Bxj for

each j < m, then x � µx(m) ∈ Ci.

The condition Bpj ⊆ Bxj for j < m captures those x ∈ (p)k which keep the first m

many blocks of p distinct in x. Therefore, CSL(m, `) is a special case of CDRTm+1` .

Proposition 3.26 (RCA0). CDRTm+1` implies CSL(m, `).

Note that we will use CSL(m, `) to denote this form of the Carlson-SimpsonLemma, despite its close connection to CDRTm+1

` , as this convention is establishedin the literature. As with the Combinatorial Dual Ramsey Theorem, we can assumethe coloring in CSL(m, `) is given in the form c : (ω)mfin → `. In Proposition 3.27,we give three equivalent versions of the Carlson-Simpson Lemma. The version inProposition 3.27(2) is (up to minor notational changes which are easily translatedin RCA0) the statement from Carlson and Simpson [3]. After proving Proposition3.27, we will use the statement in Proposition 3.27(3) to give a classical proof of

CDRTk` using transfinitely many applications of the Carlson-Simpson Lemma.

Proposition 3.27 (RCA0). The following are equivalent.

(1) CSL(m, `).(2) For each coloring (ω)mfin = ∪i<`Ci, there is a partition p ∈ (ω)ω and a color

i such that for all a < m, a ∈ Bpa and for all x ∈ (p)m+1, if Bpj ⊆ Bxj for

each j < m, then x � µx(m) ∈ Ci.(3) For each y ∈ (ω)ω and open reduced coloring (y)m+1 = ∪i<`Oi, there is a

partition p ∈ (y)ω and a color i such that for all a < m, Bya ⊆ Bpa and forall x ∈ (p)m+1, if Bpj ⊆ Bxj for each j < m, then x ∈ Oi.

Proof. (2) implies (1) because CSL(m, `) is a special case of (2). The extra conditionin (2) that a ∈ Bpa for a < m says that the partition p does not collapse anyof the first m-many blocks of the trivial partition defined by Bn = {n}. Theequivalence between (2) and (3) is proved in a similar way to Proposition 3.6 usingthe transformation in Lemma 2.3.

It remains to prove (1) implies (2). Fix an `-coloring c : (ω)mfin → `. Definec : (ω)mfin → ` by c(σ) = c(0a1a · · ·a (m − 1)aσ). Apply CSL(m, `) to c to get

p ∈ (ω)ω and i < ` such that for all x ∈ (p)m+1, if Bpj ⊆ Bxj for all j < m, then

c(x � µx(m)) = i. We treat p as an infinite string 〈p(0), p(1), · · · 〉 with entries in ω.


Let p ∈ (ω)ω be the partition corresponding to the infinite string p =0a1a · · ·a (m−1)ap. We claim that p satisfies the conditions in (2) for the coloringc with the fixed color i. By the definition of p, a ∈ Bpa for all a < m.

Fix x ∈ (p)m+1 such that Bpj ⊆ Bxj for all j < m. We need to show that

c(x � µx(m)) = i. Since x does not collapse any of the firstm-many p-blocks, a ∈ Bxafor all a < m and x (as an infinite string) has the form x = 0a1a · · ·a (m − 1)axsuch that the infinite string x is an ordered function from [m,∞) onto m + 1.

Letting x(n) = x(n+m), we obtain a partition x ∈ (p)m+1 such that Bpj ⊆ Bxj for

all j < m. Therefore, c(x � µx(m)) = i. Translating back through x to x, we haveµx(m) = µx(m) + m and x � µx(m) = 0a1a · · ·a (m − 1)ax � µx(m). Translatingfrom c to c, we have

i = c(x � µx(m)) = c(0a1a · · ·a (m− 1)ax � µx(m)) = c(x � µx(m))

as required to complete the proof that (1) implies (2). �

We will also use a variant of Lemma 2.4. Let p ∈ (ω)ω and s ≥ m. Consider theways to collapse the first s-many blocks Bp0 , . . . , B

ps−1 of p to exactly m-many blocks

while leaving the remaining p-blocks unchanged. Collapsing s-many blocks to m-many blocks is described by a string σ ∈ (ω)mfin with |σ| = s. To leave the remainingp-blocks unchanged, we extend σ to σ∗ ∈ (ω)ω to renumber the blocks Bpa for a ≥ sstarting with index m. Formally, σ∗(n) = σ(n) for n < s and σ∗(n) = n− (s−m)for n ≥ s. An argument similar to the proof of Lemma 2.4 gives the next lemma.

Lemma 3.28. Fix s ≥ m ≥ 2. Let σs,0, . . . , σs,Ms−1 list the strings σs,j ∈ (ω)mfin

with |σs,j | = s. For any p ∈ (ω)ω, the coarsenings of p which collapse the firsts-many blocks of p to m-many blocks and leave the remaining p-blocks unchangedare σ∗s,0 ◦ p, . . . , σ∗s,Ms−1 ◦ p.

Let y ∈ (ω)ω and (y)k = ∪i<`Ci be an m-reduced coloring for some 1 < m < k.

We define the induced coloring (y)m+1 = ∪i<`Ci as follows. For q ∈ (y)m+1,

q ∈ Ci if and only if q ∈ Ci for some (or equivalently all) q ∈ (y)k such thatq � µq(m) = q � µq(m). This induced coloring is a reduced coloring of (y)m+1 andtherefore we can apply CSL(m, `) to it.

Lemma 3.29. Let 1 < m < k, y ∈ (ω)ω and (y)k = ∪i<`Ci be an m-reduced

coloring. Let (y)m+1 = ∪i<`Ci be the induced coloring and let z ∈ (y)ω and i < `be obtained by applying CSL(m, `) as in Proposition 3.27(3) to the induced coloring.If x ∈ (z)k with Bza ⊆ Bxa for a < m, then x ∈ Ci.

Proof. Given x ∈ (z)k as in the lemma, let x ∈ (x)m+1 be the coarsening of x withblocks Bxa = Bxa for a < m and Bxm = ∪m≤a<kBxa . By definition, x ∈ (z)m+1 with

Bza ⊆ Bxa for a < m, and therefore x ∈ Ci. Since x � µx(m) = x � µx(m), x ∈ Ciby the definition of the induced coloring. �

Our proof of CDRTk` from the Carlson-Simpson Lemma will use repeated appli-cations of the following lemma which is proved using ω many nested applicationsof CSL(m, `).

Lemma 3.30. Fix 1 < m < k and y ∈ (ω)ω. Let (y)k = ∪i<`Ci be an m-reducedcoloring. There is an x ∈ (y)ω such that the coloring restricted to (x)k is (m− 1)-reduced.


Proof. Fix an m-reduced coloring (y)k = ∪i<`Ci. We define a sequence of infinitepartitions xm, xm+1, · · · starting with index m such that xm = y and xs+1 is acoarsening of xs for which Bxsa ⊆ B

xs+1a for all a < s. That is, we do not collapse

any of the first s-many blocks of the partition xs when we coarsen it to xs+1.This property guarantees that the sequence has a well-defined limit x ∈ (ω)ω. Weshow this limiting partition x satisfies the conclusion of the lemma. The process ofpassing from xs to xs+1 will use finitely many nested applications of CSL(m, `).

Assume xs has been defined for a fixed s ≥ m and we construct xs+1. Letσ∗s,0, . . . , σ

∗s,Ms−1 be the infinite partitions from Lemma 3.28. Set x0

s = xs. We

define a sequence of coarsenings x1s, . . . , x

Mss and set xs+1 = xMs

s . The definition ofxj+1s from xjs will use one application of CSL(m, `).

Assume that xjs has been defined. Let wjs = σ∗s,j ◦ xjs be the result of collapsing

the first s-many blocks of xjs into m-many blocks in the j-th possible way andleaving the remaining blocks of xjs unchanged. Since wjs is a coarsening of y, thecoloring (y)k = ∪i<`Ci restricts to an m-reduced coloring of (wjs)

k which induces areduced coloring of (wjs)

m+1 as described above. Let zjs be the result of applyingCSL(m, l) as stated in Proposition 3.27(3) to this reduced coloring of (wjs)

m+1.To define xj+1

s , we want to “uncollapse” the first m-many blocks of zjs to reversethe action of σ∗s,j in defining wjs. Since wjs collapsed the first s-blocks of xjs to

m-many blocks and since zjs is a coarsening of wjs, if xjs(u) < s, then zjs(u) < m.We define xj+1

s by cases as follows.

(1) If xjs(u) < s, then xj+1s (u) = xjs(u).

(2) If xjs(u) ≥ s and zjs(u) = a < m, then xj+1s (u) = xjs(µ

zjs (a)).(3) If zjs(u) ≥ m, then xj+1

s (u) = zjs(u) + (s−m).

Below we verify that xj+1s is an infinite partition coarsening xjs which does not

collapse any of the first s-many blocks of xjs. This completes the construction ofxj+1s and hence of xs+1 and x.

We verify the required properties of xj+1s . By (1), B

xjsa ⊆ Bx

j+1sa for all a < s, so

we do not collapse any of the first s-many blocks of xjs in xj+1s . There is no conflict

between (1) and (3) because xjs(u) < s implies zjs(u) < m. Furthermore, (3)renumbers the zjs-blocks starting with index m to xj+1

s -blocks starting with indexs without changing any of these blocks. Therefore, xj+1

s is an infinite partition.In (2), we handle the case when the xjs-block containing u is not changed by

wjs (except to renumber its index) but is collapsed by zjs into one of the first m-

many zjs-blocks. In this case, µzjs (a) = µx

js(b) for some b < s and we have set

xj+1s (u) = b. It is straightforward to check (as in the proof of Theorem 3.9) thatxj+1s is a coarsening of xjs and that σ∗s,j ◦ xj+1

s = zjs .

To complete the proof, we verify that the restriction of ∪i<`Ci to (x)k is (m−1)-reduced. Fix p ∈ (x)k and we show the color of p depends only on p � µp(m− 1).

Fix s ∈ ω such that µx(s − 1) = µp(m − 1). The partition p collapses the firsts-many x-blocks into the first m-many p-blocks. The string σ ∈ (ω)mfin with |σ| = sdefined by σ(a) = b if Bxa ⊆ Bpb describes this collapse. Fix an index j such thatσ = σs,j in our fixed enumeration of such strings in Lemma 3.28. Note that σs,j isdetermined by p � µp(m− 1) and that p is a coarsening of σ∗s,j ◦ x.

Consider how xj+1s was defined from xjs in the construction. We set wjs = σ∗s,j ◦xjs

and applied CSL(m, `) to the induced coloring of (wjs)m+1 to get zjs . Although we


did not use it in the construction, this application of CSL(m, `) also determined ahomogeneous color ijs < ` which we will use below.

Since x is a coarsening of xj+1s for which B

xj+1sa ⊆ Bxa for all a < s, σ∗s,j ◦ x is a

coarsening of σ∗s,j ◦ xj+1s = zjs . Because p is a coarsening of σ∗s,j ◦ x, it follows that

p is a coarsening of zjs .We claim that the first m-many blocks of zjs remain distinct in p. That is, if

zjs(u) < m, then p(u) = zjs(u). To see why, assume zjs(u) < m. It follows thatxj+1s (u) < s and hence that x(u) = xj+1

s (u) < s because x does not collapse any ofthe first s-many blocks of xj+1

s . Therefore, σs,j(x(u)) is defined and we have

p(u) = σs,j(x(u)) = σs,j(xj+1s (u)) = zjs(u)

as required.We obtained zjs and the color ijs < ` by applying CSL(m, `) to the induced

coloring of (wjs)m+1. Since p ∈ (zjs)

k satisfies Bzjsa ⊆ Bpa for a < m, we can apply

Lemma 3.29 to conclude that p ∈ Cijs . This completes the proof that the restriction

of ∪i<`Ci to (x)k is (m−1)-reduced because the indices s and j in zjs are determinedby p � µp(m− 1) and the color of p is equal to the homogeneous color ijs obtainedwhen we applied CSL(m, `) to obtain zjs . �

We end this section using Lemma 3.30 to prove CDRTk` for k ≥ 3. (For k = 2,

CDRTk` follows from the pigeonhole principle as in Proposition 3.11.) Consider

CDRTk` is the form given in Proposition 3.6. Let y ∈ (ω)ω and (y)k = ∪i<`Oibe an open reduced coloring. These satisfy the assumptions of Lemma 3.30 withm = k−1. After k−2 applications of Lemma 3.30, we obtain x ∈ (y)ω such that therestriction of ∪i<`Oi to (x)k is 1-reduced and hence the color of p ∈ (x)k dependsonly on p � µp(1). Since the numbers n < µp(1) must lie in Bp0 , the color of p isdetermined by the value of µp(1). By the pigeonhole principle, there is an infiniteset X ⊆ {µx(a) : a ≥ 1} and a color i such that for all p ∈ (x)k, if µp(1) ∈ X,then p ∈ Ci. It follows that for any z ∈ (x)ω such that µz(a) ∈ X for all a ≥ 1,(z)k ⊆ Ci as required. It is interesting to note that the only non-constructive stepsin this proof are the ω · (k− 2) nested applications of the Carlson-Simpson Lemma.

4. The Borel Dual Ramsey Theorem for k ≥ 3

In the next two sections we consider the Borel Dual Ramsey Theorem from theperspective of effective mathematics. For continuity with Section 3, we define acode for an open set in (ω)k to be a set O ⊆ ω × (ω)kfin. We say O is a ∆0

n codeif O is ∆0

n as a set of natural numbers. Equivalently, a ∆0n code for an open set is

a subset of (ω)kfin which is c.e. relative to ∅(n−1), or by replacing elements of O asthey are enumerated with sets of sufficiently long strings, is a subset of (ω)kfin which

is computable in ∅(n−1). We will shift between these coding methods in Sections 4and 5.

We define Borel codes for topologically Σ0α subsets of (ω)k by induction on the

ordinals below ω1. This definition gives another method of coding an open (topo-logically Σ0

1) set which is easily translated into the codes described above. Let Bn,

n ∈ ω, be an effective listing of the clopen sets ∅, (ω)k and [σ] and [σ] for σ ∈ (ω)kfin.

Definition 4.1. We define a Borel code for a (topologically) Σ0α or Π0

α set.


• A Borel code for a Σ00 or a Π0

0 set is a labelled tree T consisting of just aroot λ in which the root is labeled by a clopen set Bnλ . The code representsthe set Bnλ .• For α ≥ 1, a Borel code for a Σ0

α set is a labelled tree with a root labelledby ∪ and attached subtrees at level 1, each of which is a Borel code for aΣ0βn

or Π0βn

set An for some βn < α. The code represents the set ∪nAn.

• For α ≥ 1, a Borel code for a Π0α set is the same, except the root is labelled

∩. The code represents the set ∩nAn.

For α ≥ 1, a Borel code for a ∆0α set is a pair of labelled trees which encode the

same set, where one encodes it as a Σ0α set and the other encodes it as a Π0

α set.

The codes are faithful to the Borel hierarchy in the sense that every code for aΣ0α set represents a Σ0

α set and every Σ0α set is represented by a Borel code for a

Σ0α set. There is a uniform procedure to transform a Borel code B for a Σ0

α set Ainto a Borel code B for a Π0

α set A: leave the underlying tree structure the same,swap the ∪ and ∩ labels and replace the leaf labels by their complements.

If a Borel set A has a computable code (i.e. the labeled subtree of ω<ω is com-putable), then the Turing machine Φe giving the computable labelled tree is acomputable Borel code for A.

We recall some notation from hyperarithmetic theory. Let O denote Kleene’sset of computable ordinal notations. The ordinal represented by a ∈ O is denoted|a|O, with |1|O = 0, |2a|O = |a|O + 1, and |3 · 5e|O = supj |ϕe(j)|O. The H-sets aredefined by effective transfinite recursion on O as follows: H1 = ∅, H2a = H ′a andH3·5a = {〈i, j〉 | i ∈ Hϕa(j)}. The reader referred to Sacks [11] for more details. As

usual, ωCK1 denotes the least noncomputable ordinal.It is well-known that an open set of high hyperarithmetic complexity can be

represented by a computable Borel code for a Σ0α set, where α is an appropriate

computable ordinal. In the following proposition, we use a standard technique tomake this correspondence explicit. Let

height(a) =

{|a|O if |a|O < ω

|a|O−1 if |a|O ≥ ω.

where α−1 = α if α is a limit and α − 1 otherwise. Note that for |a|O < ω,height(2a) = |a|O + 1 and for |a|O ≥ ω, height(2a) = |a|O. Fix an effective 1-to-1enumeration τn for the strings τ ∈ (ω)kfin.

Proposition 4.2. There is a partial computable function p(x, y) such that p(a, e) isdefined for all a ∈ O and e ∈ ω and such that if a ∈ O and R =

⋃{[τn] : n ∈WHa

e },then Φp(a,e) is a computable Borel code for R as a Σ0

height(2a) set.

Proof. We define p(a, e) for all e by effective transfinite recursion on a ∈ O. SinceH1 = ∅, let Φp(1,e) be a Borel code for the open set R =

⋃{[τn] : n ∈We}.

For the successor step, consider R =⋃{[τn] : n ∈ WH2a

e }. Each set which is Σ01

in H2a is Σ02 in Ha and for such sets, we can effectively pass from a Σ0,H2a

1 index

to a Σ0,Ha2 description. Specifically, uniformly in e, we compute an index e′ such

that for all oracles X, ΦXe′ (x, y) is a total {0, 1}-valued function and

n ∈WX′

e if and only if ∃t∀s ≥ t (ΦXe′ (n, s) = 1).


Let Rt =⋃{[τn] : ∃s ≥ t (ΦHae′ (n, s) = 0)}. R0 ⊇ R1 ⊇ · · · is a decreasing sequence

of sets such that x 6∈ R if and only if ∀t (x ∈ Rt). Therefore, R = ∪tRt. Each set Rtcan be represented as Rt =

⋃{[τn] : n ∈ WHa

et }, where et is uniformly computablefrom e and t. Applying the induction hypothesis, we define p(2a, e) to encode atree whose root is labelled by a union and whose t-th subtree at level 1 is the Borelcode representing the complement of Φp(a,et).

For the limit step, consider R =⋃{[τn] : n ∈ WH3·5a

e }. Uniformly in e, weconstruct a sequence of indices et for t ∈ ω such that for all oracles X, ΦXet(x)

converges if and only if ΦXe (x) converges and only asks oracle questions about

numbers in the first t many columns of X. Let Rt =⋃{[τn] : n ∈ W⊕i≤tHϕa(i)

et }and note that R = ∪tRt. We can effectively pass to a sequence of indices e′tsuch that Rt =

⋃{[τn] : n ∈ W

Hϕa(t)

e′t}. By induction, each p(ϕa(t), e′t) is the

index for a computable Borel code for Rt as a Σ0height(2ϕa(t))

set, so we may define

p(3 · 5a, e) to be the index of a tree which has Φp(ϕa(t),e′t)as its subtrees. Since

height(2ϕa(t)) < |3 · 5a|O = height(23·5a) for all t, the resulting Borel code has therequired height. �

To force the Dual Ramsey Theorem to output computationally powerful homo-geneous sets, we use the following definition and classical result of Solovay [13].

Definition 4.3. For functions f, g : ω → ω, we say g dominates f , and write g � f ,if f(n) ≤ g(n) for all but finitely many n.

Theorem 4.4 (Solovay). For each a ∈ O, there is a function fa such that fa ≡T Ha

and for every g � fa, we have Ha ≤T g.

In Theorem 4.7, we use these functions fa to show that for every a ∈ O, there isa computable Borel code for a set R ⊆ (ω)3 such that any homogeneous partitionp ∈ (ω)ω for the coloring (ω)3 = R ∪R computes Ha.

Theorem 4.5. Let A be a set and fA be a function such that A ≡T fA and for everyg � fA, we have A ≤T g. There is an A-computable clopen coloring (ω)3 = R ∪ Rfor which every homogeneous partition p satisfies p ≥T A.

Proof. Fix A and fA as in the statement of the theorem. Without loss of generality,we assume that if n < m, then fA(n) < fA(m). For x ∈ (ω)3, let ax = µx(1) andbx = µx(2). As in the proof of Theorem 3.19, let Oa,b = {x ∈ (ω)3 : ax = a∧bx = b}.Set R = {x ∈ (ω)3 : fA(ax) ≤ bx}. Since R =

⋃{On,m | fA(n) ≤ m} and

R =⋃{On,m | fA(n) > m} both R and R are A-computable open sets.

Claim. If p ∈ (ω)ω is homogeneous, then (p)3 ⊆ R.

It suffices to show that there is an x ∈ (p)3 with x ∈ R. Let u = µp(1). Becausep has infinitely many blocks, there must be a p-block V 6= Bp1 with least elementv ≥ f(u). Consider the partition x ∈ (p)3 with Bx0 = ω \ (Bp1 ∪ V ), Bx1 = Bp1 andBx2 = V . Since ax = u and bx = v, we have x ∈ (p)3 with f(ax) ≤ bx, so x ∈ R.

Claim. If p ∈ (ω)ω is homogeneous, then A ≤T p.

Fix p and let g(n) = µp(n + 2). Since g is p-computable, it suffices to showg � fA. Because n < µp(n+1) and fA is increasing, we have fA(n) < fA(µp(n+1)).Therefore, to show g � fA, it suffices to show fA(µp(n+ 1)) ≤ µp(n+ 2) = g(n).


Let xn ∈ (p)3 be the coarsening with blocks ω \ (Bpn+1 ∪Bpn+2), Bpn+1 and Bpn+2.

Note that axn = µp(n+ 1) and bxn = µp(n+ 2). By the previous claim, xn ∈ R, sofA(axn) ≤ bxn . In other words, fA(µp(n+ 1)) ≤ µp(n+ 2) as required. �

Corollary 4.6. For each k ≥ 3 and each a ∈ O, there is an Ha-computable clopenset R ⊆ (ω)k such that if p ∈ (ω)ω is homogeneous for (ω)k = R∪R, then Ha ≤T p.

Proof. For k = 3, this corollary follows from Theorems 4.4 and 4.5. For k > 3, usesimiliar definitions for R and R ignoring what happens after the first three blocksof the partition. �

Theorem 4.7. For every recursive α > 0, and every k ≥ 3, there is a computableBorel code for a ∆0

α set R ⊆ (ω)k such that every p ∈ (ω)ω homogeneous for thecoloring (ω)k = R ∪R computes ∅(α−1) if α < ω and computes ∅(α) if α ≥ ω.

Proof. Given a ∈ O with height(2a) = α, let R,R be Ha-computable clopen setsfrom the previous corollary. By Proposition 4.2, both R and R have computableBorel codes as Σ0

height(2a) subsets of (ω)k. Therefore, R has a computable Borel

code as ∆0α set. By the previous corollary, if p is homogeneous for (ω)k = R ∪ R,

then p ≥T Ha, which corresponds to the indicated number of Turing jumps. �

For α = 2, Theorem 4.7 says there is a classically clopen set R ⊆ (ω)3 suchthat R and R have computable Borel codes as Σ0

2 sets (and hence as ∆02 sets) and

any homogeneous partition for (ω)3 = R ∪ R computes ∅′. Theorem 3.19 givesan analogous result at a slightly better coding level in the sense that the coloring(ω)3 = O ∪O is given in terms of an open code and a closed code.

5. The Borel Dual Ramsey Theorem for k = 2

5.1. Effective Analysis. We consider the complexity of finding infinite homoge-neous partitions for colorings (ω)2 = R∪R when R is a computable code for a set ata finite level of the Borel hierarchy. We begin by showing that if R is a computableopen set, there is a computable homogeneous partition.

Theorem 5.1. Let R be a computable code for an open set in (ω)2. There is acomputable p ∈ (ω)ω such that (p)2 ⊆ R or (p)2 ⊆ R.

Proof. If there is an n ≥ 1 such that [0n] ∩ R = ∅, then the partition x ∈ (ω)ω

with blocks {0, 1, . . . , n}, {n + 1}, {n + 2}, . . . satisfies (x)2 ⊆ R. Otherwise, forarbitrarily large n there are τ � 0n1 with [τ ] ⊆ R, and hence there is a computablesequence τ1, τ2, . . . of such τ with 0i ≺ τi. Computably thin this sequence so thatfor each i, 0|τi| ≺ τi+1. The partition x with blocks Bxi = {j : τi(j) = 1} for i > 0satisfies (x)2 ⊆ R. �

To extend to sets coded at higher finite levels of the Borel hierarchy, we will needthe following generalization of the previous result.

Theorem 5.2. Let R be a computable code for an open set in (ω)2 such thatR ∩ [0n] 6= ∅ for all n. Let {Di}i<ω be a uniform sequence of computable codes foropen sets such that each Di is dense in R. There is a computable x ∈ (ω)ω suchthat (x)2 ⊆ R ∩ (∩iDi).


Proof. We build x as the limit of an effective sequence τ0 ≺ τ1 ≺ · · · with τs ∈(ω)s+1

fin . We define the strings τs in stages starting with τ0 = 〈0〉 which puts 0 ∈ Bx0 .For s ≥ 1, we ensure that at the start of stage s + 1, we have [σ ◦ τs] ⊆ R for allσ ∈ (ω)2

fin with |σ| = s+ 1. That is, the open sets in (ω)2 determined by each wayof coarsening the s+ 1 many blocks of τs to two blocks is contained in R.

At stage s + 1, assume we have defined τs ∈ (ω)s+1fin . If s ≥ 1, assume that for

all σ ∈ (ω)2fin with |σ| = s + 1, [σ ◦ τs] ⊆ R. Let σ0, . . . , σMs−1 list the strings

σ ∈ (ω)2fin such that |σ| = s + 2. We define a sequence of strings τ0

s , . . . , τMss and

set τs+1 = τMss .

We define τ0s to start the block Bxs+1. Since [0|τs|] ∩R 6= ∅, we effectively search

for γs ∈ (ω)2fin such that 0|τs| ≺ γs and [γs] ⊆ R. Since γs ∈ (ω)2

fin, there is at least

one m < |γs| such that γs(m) = 1. Define τ0s ∈ (ω)s+2

fin with |τ0s | = |γs| by

τ0s (m) =

τs(m) if m < |τs|s+ 1 if γs(m) = 1 (and hence m ≥ |τs|)0 if m ≥ |τs| and γs(m) = 0.

Note that τs ≺ τ0s . Intuitively, τ0

s partitions {0, . . . , |τ0s | − 1} into (s + 2) many

blocks as follows. It leaves the blocks Bτs1 , . . . , Bτss unchanged, starts a new block

Bτ0ss+1 = Bγs1 and puts the remaining elements in B

τ0s

0 .

Before proceeding, we claim that if σ ∈ (ω)2fin with |σ| = s+2, then [σ ◦ τ0

s ] ⊆ R.First, suppose s = 0. In this case, τ0

s ∈ (ω)2fin and τ0

s = γs because s + 1 = 1.Therefore, [τ0

s ] = [γs] ⊆ R. Furthermore, the only string σ ∈ (ω)2fin with |σ| = 2 is

σ = 〈0, 1〉. Therefore, σ◦τ0s = τ0

s = γs and the claim follows. Second, suppose s ≥ 1and let j be the least number such that σ(j) = 1. If j = s+ 1, then σ ◦ τ0

s = γs andthe claim follows. If j < s+1, then let σ′ ≺ σ with |σ′| = s+1. Since σ′ ∈ (ω)2

fin, wehave by induction that [σ′ ◦ τs] ⊆ R and since σ′ ◦ τs ≺ σ ◦ τ0

s , we have [σ ◦ τ0s ] ⊆ R.

We continue to define the τ js strings by induction. Assume that τ js has beendefined and consider the j-th string σj enumerated above describing how to collapse(s + 1) many blocks into 2 blocks. Since τ0

s � τ js , we have σj ◦ τ0s � σj ◦ τ js and

hence [σj ◦ τ js ] ⊆ R. Because ∩n<s+1Dn is dense in R, we can effectively search fora string δjs ∈ (ω)2

fin such that σj ◦ τ js � δjs and [δjs] ⊆ ∩n<s+1Dn. To define τ j+1s ,

we uncollapse δjs. Let j∗ be the least number such that σj(j∗) = 1. Define

τ j+1s (m) =

τ js (m) if m < |τ js |j∗ if m ≥ |τ js | and δjs(m) = 10 if m ≥ |τ js | and δjs(m) = 0

It is straightforward to check that τ js � τ j+1s and that σj◦τ j+1

s = δjs. This completesthe construction of the sequence τ0

s � · · · � τMss and of the computable partition

x. It remains to show that if p ∈ (x)2, then p ∈ R and p ∈ ∩n∈ωDn. Fix p ∈ (x)2

and let s0 be the least number such that Bxs0+1 is not collapsed into Bp0 .

Claim. p ∈ R.

Let σ ∈ (ω)2fin with |σ| = s0 + 2 be the sequence defined by σ(m) = 0 for all

m < s0 + 1 and σ(s0 + 1) = 1. Thus σ describes how the blocks Bx0 , . . . , Bxs0+1

are collapsed in p. At stage s0 + 1, we defined τ0s0 ≺ x with the property that

[σ ◦ τ0s0 ] ⊆ R. Since σ ◦ τ0

s ≺ p, we have p ∈ R.

Claim. p ∈ ∩n<ωDn.


Fix k ∈ ω and we show p ∈ Dk. Let s = max{k, s0}. Consider the actionduring stage s + 1 of the construction. Let σj ∈ (ω)2

fin with |σj | = s + 2 describehow p collapses Bx0 , . . . , B

xs+1 into Bp0 and Bp1 . We defined δjs and τ j+1

s such that

σj ◦ τ j+1s = δjs and [δjs] ⊆ ∩n<s+1Dn, so in particular, [δsj ] ⊆ Dk. Since τ j+1

s ≺ x

contains the least elements of the first (s + 2) many x-blocks, we have δjs = σj ◦τ j+1s ≺ p, so p ∈ Dk as required. �

The next proposition is standard, but we present the proof because some detailswill be relevant to Theorem 5.4. In the proof, we use codes for open sets as inDefinition 3.1 and we equate a partition p ∈ (ω)2 (namely, a surjection p : ω → 2)with the set for which p is the characteristic function.

Proposition 5.3. Let n ∈ ω and let A ⊆ 2ω be defined by a Σ0n+1 predicate. There

is a ∆0n+1 code U for an open set in (ω)2, a ∆0

n+2 code V for an open set in (ω)2

and a uniformly ∆0n+1 sequence 〈Di : i ∈ ω〉 of codes for dense open sets such that

U ∪ V is dense and for all p ∈ ∩i∈ωDi, if p ∈ U , then p ∈ A and if p ∈ V thenp 6∈ A. Furthermore, the ∆0

n+1 and ∆0n+2 indices for U , V and 〈Di : i ∈ ω〉 can be

obtained uniformly from a Σ0n+1 index for A.

Proof. We proceed by induction on n. Throughout this proof, σ, τ , ρ and δ denoteelements of (ω)2

fin. In addition to the properties stated in the proposition, we ensurethat if 〈m,σ〉 ∈ U (or V ) and τ � σ, then there is a k such that 〈k, τ〉 ∈ U (or Vrespectively). Thus, if U ∩ [σ] 6= ∅, then there is 〈k, τ〉 ∈ U with σ � τ .

For n = 0, we have X ∈ A⇔ ∃k ∃mP (m,X � k) where P (x, y) is a Π00 predicate.

Without loss of generality, we assume that if P (m,X � k) holds, then P (m′, Y � k′)holds for all k′ ≥ k, m′ ≥ m and Y ∈ 2ω such that Y � k = X � k. LetU = {〈n, σ〉 : P (σ, n)}, V = {〈0, σ〉 : ∀x∀τ � σ (¬P (τ, x))} and Di = (ω)2

fin fori ∈ ω. It is straightforward to check these codes have the required properties.

For the induction case, let A ⊆ 2ω be defined by a Σ0n+2 predicate, so X ∈ A⇔

∃kP (X, k) where P is a Π0n+1 predicate. For k ∈ ω, let Ak = {X : ¬P (X, k)}.

Apply the induction hypothesis to Ak to fix indices (uniformly in k) for the ∆0n+1

codes Uk and 〈Di,k : i ∈ ω〉 and for the ∆0n+2 code Vk so that if p ∈ ∩i∈ωDi,k, then

p ∈ Uk implies ¬P (k, p) and p ∈ Vk implies P (k, p). Let

U = {〈〈k,m〉, σ〉 : 〈m,σ〉 ∈ Vk} and

V = {〈0, σ〉 : ∀k ∀τ � σ ∃m∃ρ � τ 〈m, ρ〉 ∈ Uk}.

U is a ∆0n+2 code for ∪kVk, and V is a ∆0

n+3 code such that 〈m,σ〉 ∈ V if andonly if every Uk is dense in [σ]. We claim that U ∪ V is dense. Fix σ and assumeU ∩ [σ] = ∅, so Vk ∩ [σ] = ∅ for all k. Since Uk ∪ Vk is dense, Uk ∩ [τ ] 6= ∅ for allτ � σ and all k, so 〈0, σ〉 ∈ V .

For i = 〈ai, bi〉, define Di = Dai,bi ∩ (Ui ∪ Vi). Di has a ∆0n+2 code as a dense

open set and the index can be uniformly computed from the indices for Ui, Vi andDai,bi . Furthermore, if p ∈ ∩iDi then p ∈ ∩i,kDi,k and p ∈ ∩k(Uk ∪ Vk).

Assume that p ∈ ∩iDi. First, we show that if p ∈ U , then p ∈ A. Supposep ∈ U = ∪kVk and fix k such that p ∈ Vk. Since p ∈ ∩iDi,k for this fixed k, p 6∈ Akby the induction hypothesis. Therefore, P (k, p) holds and hence p ∈ A.

Second, we show that if p ∈ V then p 6∈ A. Assume p ∈ V and fix 〈0, σ〉 ∈ Vsuch that σ ≺ p. It suffices to show ¬P (k, p) holds for an arbitrary k ∈ ω. Sincep ∈ ∩iDi, we have p ∈ Uk ∪ Vk and p ∈ ∩iDi,k. If p ∈ Uk, then ¬P (k, p) holds byinduction and we are done. Therefore, suppose for a contradiction that p ∈ Vk. Fix


〈0, τ〉 ∈ Vk such that σ � τ and τ ≺ p. Since 〈0, σ〉 ∈ V and σ � τ , there are ρ � τand m such that 〈m, ρ〉 ∈ Uk, and therefore [ρ] ⊆ Uk ∩ Vk. This containment is thedesired contradiction because q ∈ [ρ]∩∩iDi,k would satisfy q ∈ Ak and q 6∈ Ak. �

Theorem 5.4. For every coloring (ω)2 = R ∪R such that R is a computable codefor a Σ0

n+2 set, there is either a ∅(n)-computable x ∈ (ω)ω which is homogeneous

for R or a ∅(n+1)-computable x ∈ (ω)ω which is homogeneous for R.

Proof. Fix R and fix a Π0n+1 predicate P (k, y) such that for y ∈ (ω)2, y ∈ R ⇔

∃k P (k, y). Let Uk, Vk and 〈Di,k : i ∈ ω〉 be the codes from Proposition 5.3 forRk = {y : ¬P (y, k)}. Let U = ∪kVk, V = ∪{[σ] : ∀k Uk is dense in [σ]} and Di,i ∈ ω, be the corresponding codes for R. We split non-uniformly into cases.

Case 1: Assume V is dense in [0`] for some fixed `. We make two observations.First, U is disjoint from [0`]. Therefore, each Vk is disjoint from [0`] and hence eachUk is dense in [0`]. Second, suppose y ∈ (

⋂i,kDi,k) ∩ (

⋂k Uk). For each k we have

y ∈ ∩iDi,k and y ∈ Uk, so ∀k¬P (k, y) holds and hence y ∈ R.

We apply Theorem 5.2 relativized to ∅(n) to the computable open set O = [0`](which has nonempty intersection with [0j ] for every j) and the ∅(n)-computablesequence of codes Di,k and Uk for i, k < ω. By the first observation, each coded

set in this sequence is dense in O. Therefore, there is a ∅(n)-computable x ∈ (ω)ω

such that (x)2 ⊆ [0`] ∩ (⋂i,kDi,k) ∩ (

⋂k Uk). By the second observation, (x)2 ⊆ R

as required.Case 2: Assume V is not dense in [0m] for any m. In this case, since U ∪ V is

dense, we have U ∩ [0m] 6= ∅ for all m. We apply Theorem 5.2 relativized to ∅(n+1)

to the ∅(n+1)-computable open set U and the ∅(n+1)-computable sequence of densesets Di for i ∈ ω to obtain an ∅(n+1)-computable x with (x)2 ⊆ U ∩ (

⋂iDi) ⊆ R

as required. �

We end this section by showing that the non-uniformity in the proof of Theorem5.1 is necessary.

Theorem 5.5. For every Turing functional ∆, there are computable codes R0

and R1 for complementary open sets in (ω)2 such that ∆R0⊕R1 is not an infinitehomogeneous partition for the reduced coloring (ω)2 = R0 ∪R1.

Proof. Fix ∆. We define R0 and R1 in stages as R0,s and R1,s. Our construction

proceeds in a basic module while we wait for ∆R0,s⊕R1,ss to provide appropriate com-

putations. If these computations appear, we immediately diagonalize and completethe construction.

For the basic module at stage s, put 02s+11 ∈ R0,s and 02s+21 ∈ R1,s. Check

whether there is a 0 < k < s such that ∆R0,s⊕R1,ss (i) = 0 for all i < k and

∆R0,s⊕R1,ss (k) = 1. If there is no such k, then we proceed to stage s + 1 and

continue with the basic module.If there is such a k, then we stop the basic module and fix i < 2 such that

0k1 ∈ Ri,s. (Since k < s, we have already enumerated 0k1 into one of B0,s or B1,s

depending on whether k is even or odd.) We end the construction at this stage anddefine Ri = Ri,s and R1−i = R1−i,s ∪ {0t1 | 2s+ 2 < t}.

This completes the construction. It is clear that R0 and R1 are computablecodes for complementary open sets and (ω)2 = R0∪R1 is a reduced coloring. If theconstruction never finds an appropriate value k, then ∆R0⊕R1 is not an element of


(ω)ω and we are done. Therefore, assume we find an appropriate value k at stages in the construction. Fix i such that 0k1 ∈ Ri,s and assume that p = ∆R0⊕R1 isa element of (ω)ω. We show p is not homogeneous by giving elements q0, q1 ∈ (p)2

such that q0 ∈ Ri and q1 ∈ R1−i.By construction, [0, k) ⊆ Bp0 and k = µp(1) ∈ Bp1 . Let q0 ∈ (p)2 be any

coarsening such that µq0(1) = k. Since [0, k) ⊆ Bq00 and k ∈ Bq01 , we have 0k1 ≺ q0

and so q0 ∈ [0k1] ⊆ Ri as required.On the other hand, since p ∈ (ω)ω, there are infinitely many p-blocks. Let Bpn

be the first p-block with µp(n) > 2s + 2. Let q1 ∈ (p)2 be any coarsening forwhich µq1(1) = µp(n) and hence q1 ∈ [0µ

p(n)1]. Since µp(n) > 2s + 2, we put0µ

p(n)1 ∈ R1−i, so q1 ∈ R1−i as required. �

5.2. Strong reductions for reduced colorings. A reduced coloring (ω)2 = R0∪R1 is classically open and the color of p ∈ (ω)2 depends only on µp(1). When R0

and R1 are codes for open sets, there is a homogeneous partition computable inR0 ⊕R1, although by Theorem 5.5, not uniformly. We consider the case when theopen sets R0 and R1 are represented by Borel codes for Σ0

n sets with n ≥ 2.∆0n-rDRT2

2 is the statement that for each reduced coloring (ω)2 = R0∪R1 whereR0 and R1 are Borel codes for Σ0

n sets, there exists an x ∈ (ω)ω and an i < 2 suchthat (x)2 ⊆ Ri. In effective algebra, this statement is clear, but in RCA0, we needto specify how to handle these codes.

A Borel code for a Σ0n set is a labelled subtree of ω<n+1 which, in this section,

we write as (B, `) to specify the labeling function `. The labels come from the setS = {∪,∩}∪L where ∪ is the label for an interior node to denote a union, ∩ is thelabel for an interior node to denote an intersection and L is the set of labels for theleaves, namely our fixed codes for the clopen sets ∅, (ω)2 [τ ] and [τ ] for τ ∈ (ω)2

fin.For a leaf σ and a partition p, we write p ∈ `(σ) if p is an element of the clopen setcoded by `(σ). Similarly, we write `(σ) = [τ ] to avoid specifying a coding scheme.Since this code is for a Σ0

n set, we require that `(λ) = ∪.We construct a Σ0

n formula η(B, `, p) such that if (B, `) is a Borel code for a Σ0n

set and p ∈ (ω)2, then η(B, `, p) says p is in the set coded by (B, `). We begin bydefining formulas βk(σ,B, `, p) for 1 ≤ k ≤ n by downward induction on k. Forσ ∈ B with |σ| = k, βk(σ,B, `, p) says that p is in the set coded by the labeledsubtree of (B, `) above σ. Since any σ ∈ B with |σ| = n is a leaf, βn(σ,B, `, p) isthe formula p ∈ `(σ). For 1 ≤ k < n, βk(σ,B, `, p) is the formula

(`(σ) = ∪ → α∪k ) ∧ (`(σ) = ∩ → α∩k ) ∧ (`(σ) ∈ L→ αLk ), where

α∪k (σ,B, `, p) is ∃τ ∈ B(σ ≺ τ ∧ |τ | = k + 1 ∧ βk+1(τ,B, `, p)

)α∩k (σ,B, `, p) is ∀τ ∈ B

((σ ≺ τ ∧ |τ | = k + 1)→ βk+1(τ,B, `, p)

)and αLk (σ,B, `, p) is p ∈ `(σ).

The formula η(B, `, p) is ∃σ ∈ B (|σ| = 1∧ β1(σ,B, `, p)). In RCA0, we write p ∈ Bfor η(B, `, p). The statement ∆0

n-rDRT22 now has the obvious translation in RCA0.

A Borel code (B, `) for a Σ0n set is in normal form if B = ω<n+1 and for every

σ with |σ| < n, if |σ| is even, then `(σ) = ∪, and if |σ| is odd, then `(σ) = ∩. In

RCA0, for every (B, `), there is a (B, ) in normal form such that for all p ∈ (ω)2,

p ∈ B if and only if p ∈ B. Moreover, the transformation from (B, `) to (B, ) isuniformly computable in (B, `). We describe the transformation when (B, `) is aBorel code for a Σ0

2 set. The case for a Σ0n set is similar.


Let (B, `) be a Borel code for a Σ02 set. By definition, λ ∈ B with `(λ) = ∪.

Each σ ∈ B with |σ| = 1 is the root of a subtree coding a Σ00 set (if `(σ) ∈ L), a

Σ01 set (if `(σ) = ∪) or a Π0

1 set (if `(σ) = ∩). Consider the following sequence oftransformations.

• To form (B1, `1), for each σ ∈ B with |σ| = 1 and `(σ) = ∪, remove thesubtree of B above σ (including σ). For each τ ∈ B with τ � σ, add a newnode τ ′ to B1 with |τ ′| = 1 and `1(τ ′) = `(τ) ∈ L.• To form (B2, `2), for each leaf σ ∈ B1 with |σ| = 1, relabel σ by `2(σ) = ∩

and add a new successor τ to σ with label `2(τ) = `1(σ).• To form (B3, `3), for each σ ∈ B2 with |σ| = 1, let τσ ∈ B1 be the first

successor of σ. Add infinite many new nodes δ � σ to B3 with `3(δ) =`2(τσ).• To form (B4, `4), let σ be the first node of B3 at level 1. Add infinitely

many copies of the subtree above σ to B4 with the same labels as in B3.

In (B4, `4), the leaves are at level 2, every interior node is infinitely branchingand `4(σ) = ∩ when |σ| = 1. There is a uniform procedure to define a bijection

f : B4 → ω<3. We define (B, ) by B = ω<3 and (σ) = `4(f−1(σ)). In RCA0, for

all p ∈ (ω)2, η(B, `, p) holds if and only if η(B, , p) holds.When (B, `) is a Borel code for a Σ0

n set in normal form, η(B, `, p) is equivalentto ∃x0 ∀x1 · · ·Qn−1xn−1 (p ∈ `(〈x0, x1, . . . , xn−1〉)) where Qn−1 is ∀ or ∃ dependingon whether n− 1 is odd or even. We have analogous definitions for Borel codes forΠ0n sets in normal form.To define Dn2 , let [ω]n denote the set of n element subsets of ω. We view the

elements of [ω]n as strictly increasing sequences s0 < s1 < · · · < sn−1.

Definition 5.6. A coloring c : [ω]n → 2 is stable if for all k, the limit

lims1· · · lim

sn−1

c(k, s1, . . . , sn−1)

exists. L ⊆ ω is limit-homogeneous for a stable coloring c if there is an i < 2 suchthat for each k ∈ L,

lims1· · · lim

sn−1

c(k, s1, . . . , sn−1) = i.

Dn2 is the statement that each stable coloring c : [ω]n → 2 has an infinite limit-homogeneous set.

Below, the proof of Theorem 5.7(2) is a formalization of the proof of Theorem5.7(1), and the additional induction used is a consequence of this formalization.We do not know if its use is necessary; that is, we do now if RCA0 + IΣ0

n−1 can bereplaced simply by RCA0 when n > 2.

Theorem 5.7. Fix n ≥ 2.

(1) ∆0n-rDRT2

2 ≡sW Dn2 .(2) Over RCA0 + IΣ0

n−1, ∆0n-rDRT2

2 is equivalent to Dn2 .

Corollary 5.8. ∆02-rDRT2

2 is equivalent to SRT22 over RCA0.

Proof. D22 is equivalent to SRT2

2 over RCA0 by Chong, Lempp, and Yang [4]. �

Corollary 5.9. ∆02-rDRT2

2 <sW SRT22.

Proof. D22 <sW SRT2

2 by Dzhafarov [5, Corollary 3.3]. (It also follows immediatelythat ∆0

2-rDRT22 ≡W D2

2 <W SRT22.) �


Proof of Theorem 5.7. We prove the two parts simultaneously, remarking, whereneeded, how to formalize the argument in RCA0 + IΣ0

n−1.

To show that ∆0n-rDRT2

2 ≤sW Dn2 , and that ∆0n-rDRT2

2 is implied by Dn2 overRCA0 + IΣ0

n−1, fix an instance (ω)2 = R0 ∪ R1 of ∆0n-rDRT2

2 where each Ri is aBorel code for a Σ0

n set. Without loss of generality, R0 and R1 are in normal form.For each k ≥ 1, fix the partition pk defined by Bpk0 = ω − {k} and Bpk1 = {k}.

For m < n, we let Ri(t0, . . . , tm) denote the Borel set coded by the subtree of Riabove 〈t0, . . . , tm〉. Since 〈t0, . . . , tn−1〉 is a leaf, Ri(t0, . . . , tn−1) is the clopen set`i(〈t0, . . . , tn−1〉). If m < n− 1, then Ri(t0, . . . , tm) is a code for a Σ0

n−(m+1) set (if

m is odd) or a Π0n−(m+1) set (if m is even) in normal form.

We define a coloring c : [ω]n → 2 as follows. Let c(0, s1, . . . , sn−1) = 0 for alls1 < · · · < sn−1. For m ≤ n, let Qm stand for ∃ or ∀, depending as m is even orodd, respectively. Given 1 ≤ k < s1 < . . . < sn−1, define

c(k, s1, . . . , sn−1) = 1

if and only if there is a t0 ≤ s1 such that

(∀t1 ≤ s1) · · · (Qmtm ≤ sm) · · · (Qn−1tn−1 ≤ sn−1) pk ∈ `0(〈t0, . . . , tn−1〉)and for which there is no u0 < t0 such that

(∀u1 ≤ s1) · · · (Qmum ≤ sm) · · · (Qn−1un−1 ≤ sn−1) pk ∈ `1(〈u0, . . . un−1〉).(Note that s1 bounds t0, t1 and u1, whereas the other sm bound only tm and um.)The coloring c is uniformly computable in (R0, `0) and (R1, `1) and is definable inRCA0 as a total function since all the quantification is bounded.

We claim that for each k ≥ 1,

lims1· · · lim

sn−1

c(k, s1, . . . , sn−1)

exists. Furthermore, if this limit equals 1, then pk ∈ R0, and if this limit equals 0,then pk ∈ R1. We break this claim into two halves.

First, for 1 ≤ m ≤ n− 1, we claim that for all fixed 1 ≤ k < s1 < . . . < sm,

limsm+1

· · · limsn−1

c(k, s1, . . . , sm, sm+1, . . . , sn−1)

exists, and the limit equals 1 if and only if there is a t0 ≤ s1 such that

(1) (∀t1 ≤ s1) · · · (Qmtm ≤ sm) pk ∈ R0(t0, . . . , tm)

and there is no u0 < t0 such that

(2) (∀u1 ≤ s1) · · · (Qmum ≤ sm) pk ∈ R1(u0, . . . , um).

The proof is by downward induction on m. (In RCA0, the induction is performedexternally, so we do not need to consider its complexity.) For m = n− 1, there areno limits involved and the values of c are correct by definition.

Assume the result is true for m+ 1 and we show it remains true for m. By thedefinition of R0(t0, . . . , tm), t0 satisfies (1) if and only if

(∀t1 ≤ s1) · · · (Qmtm ≤ sm)(Qm+1tm+1) pk ∈ R0(t0, . . . , tm, tm+1),

which in turn holds if and only if there is a bound v such that for all sm+1 ≥ v,

(∀t1 ≤ s1) · · · (Qmtm ≤ sm)(Qm+1tm+1 ≤ sm+1) pk ∈ R0(t0, . . . , tm, tm+1).

If Qm+1 is ∃, then over RCA0, this equivalence requires a bounding principle. Sincepk ∈ R0(t0, . . . , tm+1) is a Π0

n−(m+2) predicate and m + 2 ≥ 3, we need at most


BΠ0n−3 which follows from IΣ0

n−1. An analogous analysis applies to numbers u0

satisfying (2). Thus, we can fix a common bound v that works for all t0 ≤ s1 in (1)and all u0 < t0 ≤ s1 in (2).

Suppose there is a t0 ≤ s1 satisfying (1) for which there is no u0 < t0 satisfying(2). Then, for all sm+1 ≥ v, t0 satisfies the version of (1) for m + 1, and there isno u0 < t0 satisfying the version of (2) for m+ 1. Therefore, by induction

∃v∀sm+1 ≥ v(

limsm+2

· · · limsn−1

c(k, s1, . . . , sn−1) = 1)

and hence limsm+1· · · limsn−1

c(k, s1, . . . , sn−1) = 1 as required.On the other hand, suppose that there is no t0 ≤ s1 satisfying (1), or that

for every t0 ≤ s1 satisfying (1), there is a u0 < t0 satsifying (2). Then, for allsm+1 ≥ v, we have the analogous condition for m+ 1 and the induction hypothesisgives limsm+1 · · · limsn−1 c(k, s1, . . . , sn−1) = 0. This completes the first part of theclaim.

We can now prove the rest of the claim. For each k ≥ 1, we have pk ∈ R0 orpk ∈ R1. Let t0 be least such that pk ∈ R0(t0) or pk ∈ R1(t0). Since pk ∈ Ri(t) isa Π0

n−1 statement, we use IΣ0n−1 to fix this value in RCA0.

Suppose pk ∈ R0(t0), so for all u0 < t0, it is not the case that pk ∈ R1(u0). Bythe first half of the claim with m = 1, we have for every s1 ≥ t0

lims2· · · lim

sn−1

c(k, s1, s2, . . . , sn−1) = 1,

and therefore lims1 · · · limsn−1 c(k, s1, . . . , sn−1) = 1.Suppose pk 6∈ R0(t0), and hence pk ∈ R1(t0). Again, by the first half of the

claim with m = 1, we have for every s1 ≥ t0lims2· · · lim

sn−1

c(k, s1, s2, . . . , sn−1) = 0,

so lims1 · · · limsn−1 c(k, s1, . . . , sn−1) = 0. This completes the proof of the claim.Since c is an instance of Dn2 , fix i < 2 and an infinite limit-homogeneous set L

for c with color i. By the claim, pk ∈ R1−i for all k ∈ L. List the non-zero elementsof L as k0 < k1 < · · · , and let p ∈ (ω)ω be the partition whose blocks are [0, k0)and [km, km+1) for m ∈ ω. Each x ∈ (p)2 satisfies µx(1) = km for some m. SinceR0 ∪R1 is a reduced coloring, x and pkm have the same color, which is R1−i. Sincex was arbitrary, (p)2 ⊆ R1−i as required to complete this half of the theorem.

Next, we show that Dn2 ≤sW ∆0n-rDRT2

2, and that Dn2 is implied by ∆0n-rDRT2

2

over RCA0. (No extra induction is necessary for this implication.) Fix an instancec : [ω]n → 2 of Dn2 , and define a partition R0 ∪R1 of (ω)2 as follows. For x ∈ (ω)2

with µx(1) = k, x ∈ Ri for the unique i such that

lims1· · · lim

sn−1

c(k, s1, . . . , sn−1) = i.

Since each of the iterated limits is assumed to exist over what follows on the right,we may express these limits by alternating Σ0

2 and Π02 definitions, as

(∃t1∀s1 ≥ t1)(∀t2 ≥ s1∃s2 ≥ t2) · · · c(k, s1, . . . , sn−1) = i.

Thus, R0 and R1 are Σ0n-definable open subsets of (ω)2. By standard techniques,

there are Borel codes for R0 and R1 as Σ0n sets uniformly computable in c and in

RCA0. (Below, we illustrate this process for D32.)

By definition, (ω)2 = R0∪R1 is a reduced coloring and hence is an instance of ∆0n-

rDRT22. Let p ∈ (ω)ω be a solution to this instance, say with color i < 2. Thus, for


every x ∈ (p)2, the limit color of k = µx(1) is i. Define L = {µp(m) : m ≥ 1}. Sincefor each k ∈ L, there is an x ∈ (p)2 such that µx(1) = k, L is limit-homogeneousfor c with color i.

We end this proof by illustrating how to define the Borel codes for R0 and R1

as Σ03 sets from a stable coloring c(k, s1, s2). In this case, we have

lims1

lims2c(k, s1, s2) = i⇔ ∃t1(∀s1 ≥ t1 ∀t2 ≥ s1)(∃s2 ≥ t2) c(k, s1, s2) = i.

The nodes in each Ri are the initial segments of the strings 〈〈k, t1〉, 〈s1, t2〉, s2〉 fork ≤ t1 < s1 ≤ t2 < s2 and the labeling functions are ì(σ) = ∪ if |σ| ∈ {0, 2},ì(σ) = ∩ if |σ| = 1 and ì(〈〈k, t1〉, 〈s1, t2〉, s2〉) = [0k1] if c(k, s1, s2) = i and isequal to ∅ if c(k, s1, s2) = 1 − i. It is straightforward to check in RCA0 that Rirepresents the union of clopen sets [0k1] such that the limit color of k is i. �

6. Reverse math and Borel codes

In this section, we define Borel codes in second order arithmetic and prove Theo-rem 3.14. Although we give definitions specific to the setting of (ω)k, we assume thereader is familiar with ATR0 as well as the Turing jump and the hyperarithmetichierarchy in second order arithmetic from Simpson [12] Chapters V and VIII. Webegin by defining a Borel code for a subset of (ω)k in RCA0. Although the treestructure is similar to Definition 4.1, these codes are not defined inductively andthe label for each node is coded by its last numerical entry. Let τ0, τ1, · · · be a fixedenumeration of (ω)kfin and let λ denote the empty string.

Definition 6.1 (RCA0). A Borel code for a subset of (ω)k is a tree B ⊆ ω<ω withno infinite path such that there is exactly one m ∈ ω (denoted mB) with 〈m〉 ∈ B.A Borel code is trivial if σ(|σ| − 1) ∈ {0, 1} for every leaf σ ∈ B.

Definition 6.2 (RCA0). Let B be a Borel code for a subset of (ω)k and p ∈ (ω)k.An evaluation map for B at p is a function f : B → {0, 1} such that for all σ ∈ Band n = |σ| − 1

• if σ is a leaf, then

f(σ) =

σ(n) if σ(n) ∈ {0, 1}1 if σ(n) = 2m+ 2 and τm ≺ p1 if σ(n) = 2m+ 3 and τm 6≺ p0 otherwise

• if σ 6= λ is not a leaf, then

f(σ) =

1 if σ(n) is even and ∃m (σam ∈ B ∧ f(σam) = 1)1 if σ(n) is odd and ∀m (σam ∈ B → f(σam) = 1)0 otherwise

• and f(λ) = f(〈mB〉).We write p ∈ B if there is an evaluation map for B at p with f(λ) = 1 and p 6∈ Bif there is an evaluation map f for B at p with f(λ) = 0.

The leaf nodes of a Borel code B code the basic clopen sets ∅, (ω)k, [τm] or [τm]depending on the last entry in σ. The interior nodes of B code either a union (ifthe last entry is even) or an intersection (if the last entry is odd). A trivial Borelcode represents a set built from ∅ and (ω)k using unions and intersections. SinceACA0 suffices to prove that the Kleene-Brouwer order on B is a well order and to


prove arithmetic transfinite induction, ACA0 proves that an evaluation map for Bat p is unique, provided it exists.

For a binary string σ, let σ be defined by |σ| = |σ|, σ(n) = 2m if σ(n) = 2m+ 1,and σ(n) = 2m+ 1 if σ(n) = 2m. For a Borel code B, let B = {σ : σ ∈ B}. RCA0

proves that if B is a Borel code, then B is a Borel code for the complement of Bin the sense that for every p ∈ (ω)k, f is an evaluation map for B at p if and onlyif f(σ) = 1− f(σ) is an evaluation map for B at p. In particular, f(λ) = 1− f(λ).

Lemma 6.3 (RCA0). For every code O for an open set, there is a Borel code Bsuch that (ω)k = B ∪B and for all x ∈ (ω)k, x ∈ B if and only if x ∈ O.

Proof. Fix O. Let B contain λ, 〈0〉 and, for all 〈s, τm〉 ∈ O, both 〈0, 〈s, τm〉〉 and〈0, 〈s, τm〉, 2m+ 2〉. We claim that for every x ∈ (ω)k, there is a unique evaluationmap f for B at x, and f(λ) = 1 if and only if x ∈ O. To prove this claim, we definetwo potential evaluation functions, f0 and f1, and show that one of them is correct.

For each i < 2 and leaf τ = 〈0, 〈s, τm〉, 2m+ 2〉, let fi(〈0, 〈s, τm〉〉) = fi(τ) = 1 ifτm ≺ x and have value 0 otherwise. Note that fi(〈0, 〈s, τm〉〉) is correctly defined be-cause τ is the unique successor of 〈0, 〈s, τm〉〉 and therefore fi(〈0, 〈s, τm〉〉) = fi(τ)regardless of whether 〈s, τm〉 is coded by an even or odd number. Set fi(λ) =fi(〈0〉) = i. If there is a pair 〈s, τm〉 ∈ O with τm ≺ x, then f1 satisfies the condi-tions for an evaluation function and hence x ∈ B. Otherwise, f0 is an evaluationfunction and x 6∈ B. In either case, the corresponding fi is the unique evaluationfunction for B at x and it agrees with whether x ∈ O or x ∈ O. �

If B is a trivial Borel code, then an evaluation map for B at p is independent of p,so we can refer to an evaluation map f for B. Below, we show the statement “everytrivial Borel code has an evaluation map” implies ACA0 over RCA0. We prove aform of effective transfinite recursion in ACA0 and use this recursion method toshow “every trivial Borel code has an evaluation map” implies ATR0. The mainideas in the effective transfinite recursion are similar to those in Section 7.7 of Ashand Knight [1]. Since “for every Borel code B, there is a p such that p ∈ B orp 6∈ B” implies “every trivial Borel code has an evaluation map” these results show(2) implies (1) in Theorem 3.14. Because we work with trivial Borel codes, theunderlying topological space does not matter as long as Borel codes are defined ina manner similar to Definitions 6.1 and 6.2. For example, Theorem 3.14 holds forBorel codes of subsets of 2ω or ωω as defined in Simpson [12].

Proposition 6.4 (RCA0). The statement “every trivial Borel code has an evalua-tion map” implies ACA0.

Proof. Fix g : ω → ω and we show range(g) exists. Let B be the trivial Borel codeconsisting of the initial segments of 〈0, 2n,m, 1〉 for g(m) = n and 〈0, 2n,m, 0〉 forg(m) 6= n. Let f be an evaluation function for B.

Assume g(m) = n and we show f(〈0, 2n〉) = 1. By definition, 〈0, 2n,m, 1〉 ∈ Bis a leaf and f(〈0, 2n,m, 1〉) = 1. Since 〈0, 2n,m〉 has only one successor in B,f(〈0, 2n,m〉) = 1 regardless of whether m is even or odd. Since 2n is even, itfollows that f(〈0, 2n〉) = 1.

Similarly, if n 6∈ range(g), then f(〈0, 2n〉) = 0 because all the leaves extending〈0, 2n〉 have the form 〈0, 2n,m, 0〉 and f(〈0, 2n,m, 0〉) = 0. Therefore, range(g) ={n : f(〈0, 2n〉) = 1}. �


Let LO(X) and WO(X) be the standard formulas in second order arithmeticsaying X is a linear order and X is a well order. We abuse notation and writex ∈ X in place of x ∈ field(X). For a formula ϕ(n,X), Hϕ(X,Y ) is the formulastating LO(X) and Y = {〈n, j〉 : j ∈ X ∧ ϕ(n, Y j)} where Y j = {〈m, a〉 : a <Xj ∧ 〈m, a〉 ∈ Y }. When ϕ is arithmetic, Hϕ(X,Y ) is arithmetic and ACA0 provesthat if WO(X), then there is at most one Y such that Hϕ(X,Y ). We define ourformal version of effective transfinite recursion.

Definition 6.5. ETR is the axiom scheme

∀X[(WO(X) ∧ ∀Y ∀n (ϕ(n, Y )↔ ¬ψ(n, Y ))

)→ ∃Y Hϕ(X,Y )

]where ϕ and ψ range over Σ0

1 formulas.

We show that ETR is provable in ACA0. Following Simpson [12], we avoid usingthe recursion theorem and note that the only place the proof goes beyond RCA0

is in the use of transfinite induction for Π02 formulas, which holds is ACA0 and is

equivalent to transfinite induction for Σ01 formulas. Greenberg and Montalban [7]

point out that ETR can also be proved using the recursion theorem, although thisproof also uses Σ0

1 transfinite induction.

Proposition 6.6. ETR is provable in ACA0.

Proof. Fix a well order X and Σ01 formulas ϕ and ψ. Throughout this proof, we

let f , g and h be variables denoting finite partial functions from ω to {0, 1} codedin the canonical way as finite sets of ordered pairs. We write f � g (or f ≺ X)if f ⊆ g (or f ⊆ χX) as sets of ordered pairs. By the usual normal form results(e.g. Theorem II.2.7 in Simpson), we fix a Σ0

0 formula ϕ0 such that

∀Y ∀n(ϕ(n, Y )↔ ∃f (f ≺ Y ∧ ϕ0(n, f))

)and such that if ϕ0(n, f) and f ≺ g, then ϕ0(n, g). We fix a formula ψ0 related toψ in the same manner. Since ϕ(n, Y ) ↔ ¬ψ(n, Y ), we cannot have compatible fand g such that ϕ0(n, f) and ψ0(n, g).

Our goal is to use partial functions f as approximations to a set Y such thatHϕ(X,Y ). Therefore, we view dom(f) as consisting of coded pairs 〈n, a〉. For f tobe a suitable approximation to Y , we need that if 〈n, a〉 ∈ dom(f) and a 6∈ X, thenf(〈n, a〉) = 0. Similarly, if f is an approximation to Y j , we need that f(〈n, a〉) = 0whenever 〈n, a〉 ∈ dom(f) and a ≥X j. These observations motivate the followingdefinitions.

Let f be a finite partial function and let i ∈ X. We define

f i = f � {〈n, a〉 : n ∈ ω ∧ a <X i}.We say g � f is an i-extension of f if for all 〈n, a〉 ∈ dom(g)−dom(f), g(〈n, a〉) = 0and either a 6∈ X or i ≤X a.

For j ∈ X, f is a j-approximation if the following conditions hold.

• If 〈n, a〉 ∈ dom(f) with a 6∈ X or j ≤X a, then f(〈n, a〉) = 0.• If 〈n, a〉 ∈ dom(f) and a <X j, then

– if f(〈n, a〉) = 1, then there is an a-extension h of fa such that ϕ0(n, h),and

– if f(〈n, a〉) = 0, then there is an a-extension h of fa such that ψ0(n, h).

Note that if f is a j-approximation and i <X j, then f i is an i-approximation. Also,if f is a j-approximation and g is a j-extension of f , then g is a j-approximation.


Claim. For all j ∈ X, there do not exist m ∈ ω and j-approximations f and g suchthat ϕ0(m, f) and ψ0(m, g).

The proof is by transfinite induction on j. Fix the least j ∈ X for which thisproperty fails and fix witnesses m, f and g. To derive a contradiction, it suffices toshow that f and g are compatible. Fix 〈k, a〉 such that both f(〈k, a〉) and g(〈k, a〉)are defined. If a 6∈ X or j ≤X a, then f(〈k, a〉) = g(〈k, a〉) = 0.

Suppose for a contradiction that a <X j and f(〈k, a〉) 6= g(〈k, a〉). Withoutloss of generality, f(〈k, a〉) = 1 and g(〈k, a〉) = 0. Fix a-extensions h and h′ of fa

and ga respectively such that ϕ0(k, h) and ψ0(k, h′). Since f is a j-approximation,fa is an a-approximation, and since h is an a-extension of fa, h is also an a-approximation. Similarly, h′ is an a-approximation. Therefore, we have k ∈ ω,a <X j and a-approximation h and h′ such that ϕ0(k, h) and ψ0(k, h′) contradictingthe minimality of j.

Claim. For any j-approximation f and any m ∈ ω, there is a j-approximationg � f such that either ϕ0(m, g) or ψ0(m, g).

The proof is again by transfinite induction on j. Fix the least j for which thisproperty fails and fix witnesses f and m. Let 〈ns, is〉 enumerate the pairs not in thedomain of f . Below, we define a sequence f = f0 � f1 � · · · of j-approximationssuch that fs+1(〈ns, is〉) is defined. Let Y be the set with χY = ∪sfs. Eitherϕ(m,Y ) or ψ(m,Y ) holds, and so there is a g ≺ Y such that ϕ0(m, g) or ψ0(m, g)holds. Fixing s such that g � fs shows that either ϕ0(m, fs) or ψ0(m, fs) holds forthe desired contradiction.

To define fs+1, we need to extend fs to a j-approximation fs+1 with 〈ns, is〉 ∈dom(fs+1). We break into several cases. If fs(〈ns, is〉) is already defined, let fs+1 =fs. Otherwise, if is 6∈ X or j ≤X is, set fs+1(〈ns, is〉) = 0 and leave the remainingvalues as in fs. In both cases, it is clear that fs+1 is a j-approximation.

Finally, if is <X j and fs(〈ns, is〉) is undefined, we apply the induction hypothe-sis to the is-approximation f iss to get an is-approximation g � f iss such that eitherϕ0(ns, g) holds or ψ0(ns, g) holds. Define fs+1 as follows.

• For 〈m, a〉 ∈ dom(g) with a <X is, set fs+1(〈m, a〉) = g(〈m, a〉).• For 〈m, a〉 ∈ dom(fs) with is ≤X a or a 6∈ X, set fs+1(〈m, a〉) = fs(〈m, a〉).• Set fs+1(〈ns, is〉) = 1 if ϕ0(ns, g) holds and fs+1(〈ns, is〉) = 0 if ψ0(ns, g)

holds.

It is straightforward to verify that fs ≺ fs+1, g is an is-extension of f iss+1 and fs+1

is a j-approximation, completing the proof of the claim.We define the set Y for which we will show Hϕ(X,Y ) holds by 〈m, j〉 ∈ Y if

and only if j ∈ X and there is a j-approximation f such that ϕ0(m, f). It followsfrom the claims above that 〈m, j〉 6∈ Y if and only if either j 6∈ X or there is aj-approximation f such that ψ0(m, f). Therefore, Y has a ∆0

1 definition. The nexttwo claims show that Hϕ(X,Y ) holds, completing our proof.

Claim. If f is a j-approximation, then f ≺ Y j .

Consider 〈m, a〉 ∈ dom(f). If a 6∈ X or j ≤X a, then f(〈m, a〉) = Y j(〈m, a〉) = 0.Suppose a <X j. If f(〈m, a〉) = 1, then there is an a-extension g of fa such thatϕ0(m, g). Since fa is an a-approximation and g is an a-extension of fa, g is ana-approximation. Therefore, 〈m, a〉 ∈ Y by definition and hence 〈m, a〉 ∈ Y j . Bysimilar reasoning, if f(〈m, a〉) = 0, then 〈m, a〉 6∈ Y and hence 〈m, a〉 6∈ Y j .


Claim. 〈m, j〉 ∈ Y if and only if ϕ(m,Y j).

Assume that 〈m, j〉 ∈ Y and fix a j-approximation f such that ϕ0(m, f). Sincef ≺ Y j , ϕ(m,Y j). For the other direction, assume that ϕ(m,Y j). Fix a j-approximation f such that either ϕ0(m, f) or ψ0(m, f). Since f ≺ Y j and ϕ(m,Y j),we must have ϕ0(m, f) and therefore 〈m, j〉 ∈ Y by definition. �

We recall some notation and facts from Simpson [12] to state the equivalence ofATR0 we will prove. We let TJ(X) denote the Turing jump in ACA0 given by fixinga universal Π0

1 formula. We use the standard recursion theoretic notations ΦXe andΦXe,s with the understanding that they are defined by this fixed universal formula.O+(a,X) is the arithmetic statement that a = 〈e, i〉, e is an X-recursive index

of an X-recursive linear order ≤Xe and i ∈ field(≤Xe ). OX+ = {a : O+(a,X)} exists

in ACA0. For a, b ∈ OX+ , we write b <XO a if a = 〈e, i〉, b = 〈e, j〉 and j <Xe i. For

a ∈ OX+ , the set {b : b <XO a} exists in ACA0.

O(a,X) is the Π11 statement O+(a,X)∧WO({b : b <XO a}). Intuitively, O(a,X)

says that a = 〈e, i〉 is an X-recursive ordinal notation for the well ordering givenby the restriction of ≤Xe to {j : j <Xe i}. In ATR0, if O(a,X), then the set

HXa = {〈y, 0〉 : y ∈ X} ∪ {〈y, b+ 1〉 : b <XO a ∧ y ∈ TJ(HX

b )}exists. In ACA0, there is an arithmetic formula H(a,X, Y ) which, under the as-sumption that O(a,X), holds if and only if Y = HX

a .By Theorem VIII.3.15 in Simpson [12], ATR0 is equivalent over ACA0 to

∀X ∀a (O(a,X)→ HXa exists).

If O(a,X) with a = 〈e, i〉, then we can assume without loss of generality that thereare a′ and a′′ such that O(a′, X), O(a′′, X) and a <XO a′ <XO a′′ by adding two newsuccessors of i in ≤Xe if necessary. Therefore, to prove ATR0, it suffices to fix a andX such that O(a,X) and prove ∀c <XO b (HX

c exists) for each b <XO a.

Theorem 6.7 (ACA0). The statement “every trivial Borel code has an evaluationfunction” implies ATR0.

Proof. Fix a and X such that O(a,X), so the restriction of <XO to {b : b <XO a} isa well order. Using ETR, we define trivial Borel codes Bx,b for x ∈ ω by transfiniterecursion on b <XO a. We explain the intuitive construction before the formaldefinition.

Let b <XO a and x ∈ ω. We want to define a trivial Borel code Bx,b such that iff is an evaluation map for Bx,b, then f(λ) = 1 if and only if x ∈ TJ(HX

b ). We put〈0〉 ∈ Bx,b, so Bx,b codes a union of sets. For each binary string σ such that Φσx,|σ|(x)

converges, we add a successor 〈0, σ〉 with a unique extension 〈0, σ, 1〉. Therefore,regardless of whether the code for σ is even or odd, we have f(〈0, σ〉) = f(〈0, σ, 1〉).It follows that f(λ) = f(〈0〉) = 1 if and only if there is a σ such that Φσx,|σ|(x)

converges and f(〈0, σ, 1〉) = 1. (If Φσx,|σ|(x) always diverges, then 〈0〉 is a leaf. In

this case, f(λ) = f(〈0〉) = 0 and x 6∈ TJ(HXb ) which is what we want.)

Next, we want to ensure f(〈0, σ, 1〉) = 1 if and only if σ ≺ HXb . For each k < |σ|,

we add a successor 〈0, σ, 1, k〉. Since 〈0, σ, 1〉 codes the intersection of the sets codedby 〈0, σ, 1, k〉, we want f(〈0, σ, 1, k〉) = 1 if and only if σ(k) = HX

b (k). We breakinto cases to determine the extensions of 〈0, σ, 1, k〉.

For the first case, suppose k = 〈y, 0〉. We want f(〈0, σ, 1, k〉) = 1 if and onlyif y ∈ X. If σ(k) = X(y), we add 〈0, σ, 1, k, 1〉 to Bx,b as the unique successor of


〈0, σ, 1, k〉, and if σ(k) 6= X(y), we add 〈0, σ, 1, k, 0〉 as the unique successor. Ineither case, the successor nodes will be leaves so we have f(〈0, σ, 1, k〉) = 1 if andonly if k ∈ HX

b .For the second case, suppose k = 〈y, c + 1〉 and c <XO b. By the induction

hypothesis, we have defined the trivial Borel code By,c already. If σ(k) = 1, thenwe extend 〈0, σ, 1, k〉 by a copy of By,c, treating 〈0, σ, 1, k〉 as the root of this tree.The map f restricted to the subtree above 〈0, σ, 1, k〉 is an evaluation map for By,cand hence by the inductive hypothesis

f(〈0, σ, 1, k〉) = 1⇔ y ∈ TJ(HXc )⇔ k ∈ HX

b ⇔ σ(k) = HXb (k).

On the other hand, if σ(k) = 0, then we extend 〈0, σ, 1, k〉 by a copy of By,c. Bythe inductive hypothesis, we have

f(〈0, σ, 1, k〉) = 1⇔ y 6∈ TJ(HXc )⇔ k 6∈ HX

b ⇔ σ(k) = HXb (k).

For the third case, suppose that k = 〈y, c + 1〉 and c 6<XO b. In this case, weknow HX

b (k) = 0. If σ(k) = 0, then we add a unique successor 〈0, σ, 1, k, 1〉 to〈0, σ, 1, k〉 as a leaf. We have f(〈0, σ, 1, k, 1〉) = 1 (since this node is a leaf) andhence f(〈0, σ, 1, k〉) = 1 which is what we want since σ(k) = HX

b (k) = 0. On theother hand, if σ(k) = 1, then we add a unique successor 〈0, σ, 1, k, 0〉 as a leaf, sof(〈0, σ, 1, k〉) = f(〈0, σ, 1, k, 0〉) = 0. Since 〈0, σ, 1〉 codes an intersection, we getf(〈0, σ, 1〉) = 0 which is what we want since σ(k) 6= HX

b (k) and hence σ 6≺ HXb .

The formal construction follows this outline. To simplify the notation, for atrivial Borel code B, we let B1 = B and B0 = B. Since “Φσx,|σ|(x) converges” is

a bounded quantifier statement and c <XO b is a ∆01 statement with parameter X,

the following recursion on b <XO a can be done with ETR. For each x ∈ ω, we putλ and 〈0〉 in Bx,b. For each σ such that Φσx,|σ|(x) converges, we put 〈0, σ〉, 〈0, σ, 1〉and 〈0, σ, 1, k〉 in Bx,b for all k < |σ|. We extend 〈0, σ, 1, k〉 as follows.

• For k = 〈y, 0〉: if σ(k) = X(y), then 〈0, σ, 1, k, 1〉 ∈ Bx,b and if σ(k) 6= X(y),then 〈0, σ, 1, k, 0〉 ∈ Bx,b.

• For k = 〈y, c+ 1〉 with c <XO b, 〈0, σ, 1, k〉aτ ∈ Bx,b for all τ ∈ Bσ(k)y,c .

• For k = 〈y, c+ 1〉 with c 6<XO b, 〈0, σ, 1, k, 1− σ(k)〉 ∈ Bx,b.This completes the construction of the trivial Borel codes Bx,b for b <XO a by

ETR. To complete the proof, we fix an arbitrary b <XO a and show that ∀c <XOb (HX

c exists).Fix an index x and s ∈ ω such that Φ1s

x,s(x) converges. Let N be the least value

of s witnessing this convergence so Φ1s

x,s(x) converges for all s ≥ N . Let f be anevaluation map for Bx,b.

For c <XO b and y ∈ ω, let σ = 1N+k where k = 〈y, c + 1〉. Define fy,c(τ) =f(〈0, σ, 1, k〉aτ). We claim fy,c is an evaluation map for By,c. By the choice of x,Φσx,|σ|(x) converges. Since c <XO b and σ(k) = 1, we have 〈0, σ, 1, k〉aτ ∈ Bx,b if and

only if τ ∈ By,c. Therefore, fy,c is defined on By,c and it satisfies the conditions foran evaluation map because f does.

Recall that H(x,X, Y ) is a fixed arithmetic formula such that if O(x,X), thenH(x,X, Y ) holds if and only if Y = HX

x . Define

Z = {〈y, 0〉 : y ∈ X} ∪ {k : k = 〈y, c+ 1〉 ∧ c <XO b ∧ f(〈0, 1N+k, 1, k〉) = 1}.For c <XO b, let Zc = {〈y, r〉 ∈ Z : r = 0 ∨ r − 1 <XO c}. We show the followingproperties by simultaneous arithmetic induction on c <XO b.


(1) H(c,X,Zc) holds. That is, Zc = HXc .

(2) For all y, fy,c(λ) = 1 if and only if y ∈ TJ(Zc) = TJ(HXc ).

These properties imply ∀c <XO b (HXc exists) completing our proof.

Fix c <XO b and assume (1) and (2) hold for d <XO c. To see (1) holds for c, fix k.If k = 〈y, 0〉, then k ∈ Zc ⇔ y ∈ X ⇔ k ∈ HX

c . Suppose k = 〈y, d+ 1〉. If d 6<XO c,then k 6∈ HX

c and k 6∈ Zc. If d <XO c, then

k ∈ Zc ⇔ f(〈0, 1N+k, 1, k〉) = 1⇔ fy,d(λ) = 1.

By the induction hypothesis, k ∈ Zc if and only if y ∈ TJ(Zd) = TJ(HXd ), which

holds if and only if k ∈ HXc , completing the proof of (1).

To prove (2), fix y and let k = 〈y, c+ 1〉. By definition,

k ∈ Zc ⇔ fy,c(λ) = f(〈0, 1N+k, 1, k〉) = 1,

and y ∈ TJ(Zc) = TJ(HXc ) if and only if there is a σ such that Φσy,|σ|(y) converges

and σ ≺ Zc = HXc .

Suppose there are no σ such that Φσy,|σ|(y) converges. In this case, y 6∈ TJ(HXc )

and (since By,c consists of λ and 〈0〉) fy,c(λ) = 0. Therefore fy,c(λ) = 1 if and onlyif y ∈ TJ(HX

c ) as required.Suppose Φσy,|σ|(y) converges for some σ. For any such σ, 〈0, σ, 1, `〉 ∈ By,c for

all ` < |σ|. By the induction hypothesis and the case analysis in the intuitiveexplanation of the construction, we have fy,c(〈0, σ〉) = fy,c(〈0, σ, 1〉) = 1 if andonly if σ ≺ HX

c = Zc, and therefore, fy,c(λ) = 1 if and only if there is a σsuch that Φσy,|σ|(y) converges and σ ≺ HX

c , completing the proof of (2) and of the

theorem. �

Proposition 6.8 (ATR0). Every Borel code for a subset of (ω)k has a Baire code.

Proof. Fix a Borel code B. For σ ∈ B, let Bσ = {τ ∈ B : τ is comparable to σ}.Bσ is a Borel code for the set coded coded by the subtree of B above σ in thefollowing sense. Let f be an evaluation map for B at x. The function g : Bσ → 2defined by g(τ) = f(τ) for τ � σ and g(τ) = f(σ) for τ ≺ σ is an evaluationfunction for Bσ at x which witnesses x ∈ Bσ if and only if f(σ) = 1. We denotethis function g by fσ,x.

Formally, our proof proceeds in two steps. First, by arithmetic transfinite re-cursion on the Kleene-Brouwer order KB(B), we construct open sets Uσ, Vσ andDn,σ, n ∈ ω, which are intended to form a Baire code for Bσ. This constructionis essentially identical to the proof of Proposition 5.3. Second, for any x ∈ (ω)k

and evaluation map f for B at x, we show by arithmetic transfinite induction onKB(B) that if x ∈ ∩n∈ωDn,σ, then x ∈ Uσ implies x ∈ Bσ via fσ,x and x ∈ Vσimplies x 6∈ Bσ via fσ,x. For ease of presentation, we combine these two steps.Since ATR0 suffices to construct evaluation maps, we treat Borel codes as sets in anaive manner and suppress explicit mention of the evaluation maps.

If σ is a leaf coding a basic clopen set [τ ], we set Uσ = [τ ], Vσ = [τ ] and

Dn,σ = (ω)k. Similarly, if σ codes [τ ], we switch the values of Uσ and Vσ. In eithercase, it is clear that these open sets form a Baire code for Bσ.

Suppose σ is an internal node coding a union, so Bσ is the union of Bσak forσak ∈ B. We define Uσ to be the union of Uσak for σak ∈ B and Vσ to be theunion of [τ ] such that Vσak is dense in [τ ] for all σak ∈ B. The sequence Dn,σ is


the sequence of all open sets of the form Dn,σak ∩ (Uσak ∪ Vσak) for n ∈ ω and

σak ∈ B. As in the proof of Proposition 5.3, Uσ ∪ Vσ and each Dn,σ are dense.Let x ∈ ∩n∈ωDn,σ. Suppose x ∈ Uσ and we show x ∈ Bσ. By the definition of

Uσ, fix σak ∈ B such that x ∈ Uσak. Since x ∈ ∩n∈ωDn,σak, we have by inductionthat x ∈ Bσak and hence x ∈ Bσ. On the other hand, suppose x ∈ Vσ and weshow x 6∈ Bσ. Fix τ such that τ ≺ x and [τ ] ⊆ Vσ, and fix k such that σak ∈ B.Since x ∈ ∩n∈ωDn,σ, x ∈ Uσak ∪ Vσak. However, Vσak is dense in [τ ]. Therefore,x 6∈ Uσak (because Uσak and Vσak must be disjoint as in the proof of Proposition5.3), so x ∈ Vσak. Since x ∈ ∩n∈ωDn,σak, we have by induction that x 6∈ Bσak.

Because this holds for every σak ∈ B, it follows that x 6∈ Bσ, completing the casefor unions.

The case for an interior node coding an intersection is similar with the roles ofUσ and Vσ switched. Finally, the Baire codes for the unique 〈m〉 ∈ B satisfy theconditions to be Baire codes for B itself. �

We conclude with a proof of Theorem 3.14.

Proof. Lemma V.3.3 in Simpson [12] shows (1) implies (2) in the space 2ω and theproof translates immediately to (ω)k. By Proposition 6.8, (1) implies (3). It followsfrom Theorem 6.7 that (2) implies (1). We show (3) implies (2). Let B be a Borelcode. Fix a Baire code U , V and Dn for B. Since each Dn and U ∪ V is a denseopen set, there is an x ∈ (U ∪ V ) ∩ ∩n∈ωDn. If x ∈ U , then by the definition ofa Baire code, x ∈ B, and similarly, if x ∈ V , then x 6∈ B. Therefore, we have apartition x such that x ∈ B or x 6∈ B as required. �

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effectiveness for the dual ramsey theorem · 2018. 11. 10. · n+2 set in the borel hierarchy, then...

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