Samuel Braunfeld

David Chodounský

Noé De Rancourt
email: nderancour@univ-lille.fr

Jan Hubička
email: hubicka@kam.mff.cuni.cz

AND MAT ĚJ KONE Jamal Kawach
email: jkawach@math.toronto.edu

Matěj Konečný

Ramsey 2023 Big

Hal

NO É DE RANCOURT David Chodounsk Ý

Jan Hubi Čka

Big Ramsey degrees and infinite languages

Keywords: Homogeneous structures, Ramsey theory, big Ramsey degrees, Milliken tree theorem

à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Introduction

Given L-structures A, B and C, we write C -→ (B) A k,ℓ to denote the following statement:

For every colouring χ : C A → k there exists an embedding f : B → C such that χ takes at most ℓ values on f [B] A . For a countably infinite structure B and its finite induced substructure A, the big Ramsey degree of A in B is the least number ℓ ∈ ω + 1 such that B -→ (B) A k,ℓ for every k ∈ ω; see [START_REF] Kechris | Fraïssé limits, Ramsey theory, and topological dynamics of automorphism groups[END_REF]. A countably infinite structure B has finite big Ramsey degrees if the big Ramsey degree of A in B is finite for every finite substructure A of B.

The study of big Ramsey degrees dates back to Ramsey's theorem itself which can be stated as (ω) -→ (ω) n k,1 for every n, k ∈ ω, where we understand the ordinals ω and n as structures with their natural linear orders (using the standard set-theoretic convention that n = {0, 1, . . . , n -1} and ω = {0, 1, . . .}). However, the real origin of this project lies in the work of Galvin [START_REF] Galvin | Partition theorems for the real line[END_REF][START_REF] Galvin | Errata to "Partition theorems for the real line[END_REF] who proved that the big Ramsey degree of pairs in the order of the rationals, denoted by (Q, ≤), is equal to 2. Subsequently, Laver in late 1969 proved that (Q, ≤) has in fact finite big Ramsey degrees, see [START_REF] Erdős | Unsolved and solved problems in set theory[END_REF][START_REF] Laver | Products of infinitely many perfect trees[END_REF], and Devlin determined the exact values of ℓ [Dev79, Page 73]. In 2006 Sauer [START_REF] Sauer | Coloring subgraphs of the Rado graph[END_REF] proved that the Rado graph has finite big Ramsey degrees and Laflamme, Sauer, Vuksanovic [START_REF] Laflamme | Canonical partitions of universal structures[END_REF] obtained their exact values. Behind both this result and the result for (Q, ≤) was Milliken's tree theorem for a single binary tree. This was refined to unconstrained structures in binary languages [START_REF] Laflamme | Canonical partitions of universal structures[END_REF] and additional special classes [DLS16, NVT09, LNVTS10]. Milliken's tree theorem remained the key in all results in the area (here we consider Ramsey's Theorem as a special case of Milliken's tree theorem for the unary tree). See also [START_REF] Dobrinen | Ramsey theory on infinite structures and the method of strong coding trees[END_REF] for a recent survey.

Recently, Balko, Chodounský, Hubička, Konečný, Nešetřil and Vena [BCH + 22, BCH + 19] applied the product version of Milliken's tree theorem to prove that the generic countable 3-uniform hypergraph has finite big Ramsey degrees. The method could be extended to prove big Ramsey degrees of the generic countable k-uniform hypergraph for an arbitrary k, and in this paper we further extend these results and prove the following theorem (the definition of an unrestricted structure is given later, see Definition 2.1):

Theorem 1.1. Let L be a relational language with finitely many relations of every arity greater than one and with finitely or countably many unary relations and let H be an unrestricted L-structure where all relations are injective. Then H has finite big Ramsey degrees.

We believe that this result is the limit of how far Milliken's tree theorem can be pushed in this area (at least by using the passing number representation and its generalisations). In Section 6 we give evidence for this and discuss infinite lower bounds. In fact, we conjecture that this theorem is tight when there are only finitely many unary relations (see Section 7.1 and Conjecture 7.8).

Besides Milliken's tree theorem, other partition theorems have been used in the area, such as the Carlson-Simpson theorem [Hub20, BCD + 21, BCH + 21] or various custom theorem proved using forcing [START_REF] Coulson | Fraisse structures with SDAP+, Part II: Simply characterized big Ramsey structures[END_REF][START_REF] Dobrinen | The Ramsey theory of the universal homogeneous triangle-free graph[END_REF][START_REF] Dobrinen | The Ramsey theory of Henson graphs[END_REF][START_REF] Zucker | On big Ramsey degrees for binary free amalgamation classes[END_REF]. While we only use Milliken's theorem here, the product tree structure we develop is, to a large degree, inherent to the problem, not to the method. For this reason we believe that our development of the concept of valuation trees (extending [BCH + 22]) and k-enveloping embeddings will serve as an important basis for future big Ramsey theorems for structures with relations of arity greater than two.

Preliminaries

A relational language L is a collection of symbols, each having an associated arity, denoted by a(R) ∈ ω. An L-structure A consists of a vertex set A and an interpretation of every R ∈ L, which is R A ⊆ A a(R) . We say that a relation R is injective in A if every tuple x ∈ R A is injective (i.e. contains no repeated occurrences of vertices) and it is symmetric if whenever x ∈ R A and ȳ is a permutation of x then ȳ ∈ R A . Equivalently, we can consider a symmetric relation to be a subset of A a(R) . An L-hypergraph is an L-structure where all relations are injective and symmetric and every tuple is in at most one relation (so it can be seen as an edge-colored hypergraph with number of colours for each arity given by L).

We adopt the standard model-theoretic notions of embeddings etc. with one exception: Unless explicitly stated otherwise, every structure in this paper will be implicitly equipped with an enumeration (i.e. a linear order for finite structures and an ω-type order for countably infinite structures) and all embeddings will be monotone with respect to the enumerations. When we do not explicitly describe the enumeration, one can pick an arbitrary one.

Since all structures will be at most countable, we can assume without loss of generality that the vertex set of every structure is either some natural number n or ω, the symbol ≤ will always denote the standard order of natural numbers and all embeddings will be monotone with respect to ≤. We will always assume that ≤ / ∈ L. If A and B are L-structures, the symbol B A denotes the set of all embeddings A → B. (Remember that these are monotone with respect to ≤.)

Given a class of finite and countably infinite structures C, a structure A ∈ C is universal if for every B ∈ C there exists an embedding B → A. Examples of universal structures come, for example, from Fraïssé theory which produces special (homogeneous) unenumerated structures which retain their universality (for all enumerations of members of the respective class) even when enumerated. Often we will say that A is a universal structure of some kind (e.g. A is a universal F-free structure). This will mean that A is a structure of the given kind and is universal for all countable structures of that kind.

Definition 2.1. An L-structure F is covered by a relation if there is some relation R ∈ L and a tuple x containing all vertices of F such that x ∈ R F . If F is a collection of L-structures and A is an L-structure such that there is no F ∈ F with an embedding F → A, we say that A is F-free. An L-structure is unrestricted if there is a family F containing only finite L-structures which are covered by a relation such that A is a universal F-free structure.

Note that, in particular, every unrestricted structure is countable (by our definition of universality).

2.1. Milliken's tree theorem. Our argument will make use of the vector (or product) form of Milliken's tree theorem. All definitions and results in this section are taken from [START_REF] Dodos | Ramsey theory for product spaces[END_REF]. Given an integer ℓ, we use both the combinatorial notion [ℓ] = {1, . . . , ℓ} and the set-theoretical convention ℓ = {0, 1, . . . , ℓ -1}.

A tree is a (possibly empty) partially ordered set (T, < T ) such that, for every t ∈ T , the set {s ∈ T : s < T t} is finite and linearly ordered by < T . All trees considered in this paper will be finite or countable. All nonempty trees we consider are rooted, that is, they have a unique minimal element called the root of the tree. An element t ∈ T of a tree T is called a node of T and its level, denoted by |t| T , is the size of the set {s ∈ T : s < T t}. Note that the root has level 0. For D ⊆ T , we write L T (D) = {|t| T : t ∈ D} for the level set of D in T . We use T (n) to denote the set of all nodes of T at level n, and by T (<n) the set {t ∈ T : |t| T < n}. The height of T is the smallest natural number h such that T (h) = ∅. If there is no such number h, then we say that the height of T is ω. We denote the height of T by h(T ).

Given a tree T and nodes s, t ∈ T we say that s is a successor of t in T if t ≤ T s. The node s is an immediate successor of t in T if t < T s and there is no s ′ ∈ T such that t < T s ′ < T s. We denote the set of all successors of t in T by Succ T (t) and the set of immediate successors of t in T by ImmSucc T (t). We say that the tree T is finitely branching if ImmSucc T (t) is finite for every t ∈ T .

For s, t ∈ T , the meet of s and t, denoted by s ∧ T t, is the largest s ′ ∈ T such that s ′ ≤ T s and s ′ ≤ T t. A node t ∈ T is maximal in T if it has no successors in T . The tree T is balanced if it either has infinite height and no maximal nodes, or all its maximal nodes are in T (h -1), where h is the height of T .

A subtree of a tree T is a subset T ′ of T viewed as a tree equipped with the induced partial ordering such that s ∧ T ′ t = s ∧ T t for each s, t ∈ T ′ . Note that our notion of a subtree differs from the standard terminology, since we require the additional condition about preserving meets. Definition 2.2. A subtree S of a tree T is a strong subtree of T if either S is empty, or S is nonempty and satisfies the following three conditions.

(1) The tree S is rooted and balanced.

(2) Every level of S is a subset of some level of T , that is, for every n < h(S) there exists m ∈ ω such that S(n) ⊆ T (m). (3) For every non-maximal node s ∈ S and every t ∈ ImmSucc T (s) the set ImmSucc S (s) ∩ Succ T (t) is a singleton.

Observation 2.1. If E is a subtree of a balanced tree T , then there exists a strong subtree S ⊇ E of T such that L T (E) = L T (S). □

A vector tree of dimension d ∈ ω + 1 (often also called product tree) is a sequence T = (T i : i ∈ d) of trees having the same height h(T i ) for all i ∈ d. This common height is the height of T and is denoted by h(T). A vector tree T = (T i : i ∈ d) is balanced if the tree T i is balanced for every i ∈ d.

If T = (T i : i ∈ d) is a vector tree, then a vector subset of T is a sequence

D = (D i : i ∈ d) such that D i ⊆ T i for every i ∈ d. We say that D is level compatible if there exists L ⊆ ω such that L Ti (D i ) = L for every i ∈ d. This (unique) set L is denoted by L T (D) and is called the level set of D in T. If k ∈ d, we denote T ↾ k = (T i : i ∈ k) and T (k) = (T i : k ≤ i < d).
Definition 2.3. Let T = (T i : i ∈ d) be a vector tree. A strong vector subtree (or strong product subtree) of T is a level compatible vector subset S = (S i : i ∈ d) of T such that S i is a strong subtree of T i for every i ∈ d.

For every k ∈ ω + 1 with k ≤ h(T), we use Str k (T) to denote the set of all strong vector subtrees of T of height k. We also use Str ≤k (T) to denote the set of all strong subtrees of T of height at most k.

Theorem 2.2 (Milliken [START_REF] Keith | A Ramsey theorem for trees[END_REF]). For every rooted, balanced and finitely branching vector tree T of infinite height and finite dimension, every non-negative integer k and every finite colouring of Str k (T) there is S ∈ Str ω (T) such that the set Str k (S) is monochromatic.

Valuation trees

Given n ∈ ω + 1 and 0 < ℓ < ω, we denote I n ℓ = {(i 0 , ..., i ℓ-1 ) :

n > i 0 > • • • > i ℓ-1 ≥ 0}. If n, ℓ ∈ ω + 1, we put I n <ℓ = 1≤k<ℓ I n k .
Let σ = (σ 1 , σ 2 , . . .) be an infinite sequence of positive natural numbers, we will call it a signature. Let f : I n <ω → ω be a function such that if x ∈ I n ℓ then f (x) < σ ℓ . We call such f a valuation function of level n and signature σ (or a σ-valuation function of level n, or just a valuation function of level n if σ is clear from the context), and write |f | = n. Given i ∈ ω we denote by σ (i) the i-shift of σ defined by σ (i) j = σ j+i . Given a signature σ, let T σ = (T 0 , T 1 , . . .) be the infinite-dimensional vector tree where T i consists of all σ (i) -valuation functions of a finite level ordered by inclusion. Often σ will be implicit from the context and we will write only T for T σ . Note that if σ is constant 1 from some point on, T σ is essentially finite-dimensional (in the sense that from some point on, each T i is just the chain of constant zero valuation functions).

A key element of our construction is a correspondence between strong vector subtrees of the vector tree T and special subtrees of T 0 with shape isomorphic to initial segments of T 0 . If f is a σ (i) -valuation function of level n, g is a σ (i+1)valuation function of level n and h is a σ (i) -valuation function of level n + 1, we say that h is an extension of f by g and write h ∈ f ⌢ g if

h(x 0 , . . . , x d ) = f (x 0 , . . . , x d ) if x 0 < n, g(x 1 , . . . , x d ) if x 0 = n and d > 0.
Note that there are always σ

(i)
1 -many possible extensions of f by g which differ on their value at the singleton n, and we will use f ⌢ g to denote this set. Definition 3.1. Let σ be an arbitrary signature, consider T = T σ , let k ∈ ω+1 and let (S i : i ∈ k) be a strong vector subtree of T ↾ k of height at least k. We now define by induction on k a subset of S 0 which we call the valuation tree val(S i : i ∈ k):

S 0 S 1 val(S 0 , S 1 , S 2 ) S 2 Figure 1.
A valuation tree (right) constructed from a strong vector subtree (left).

If k = 0 then val() = ∅. For 0 < k < ω, we put S ′ = val(S 1 , . . . , S k-1 ) using the induction hypothesis and define val(S i : i ∈ k) by the following recursive rules:

(1) The root of val(S i : i ∈ k) is the root of S 0 .

(

) If f ∈ val(S i : i ∈ k), g ∈ S ′ (|f | S0 ), then ImmSucc S0 (f ) ∩ Succ T0 (f ⌢ g) ⊆ val(S i : i ∈ k) (note that this set has size σ (i) 1 2
). (3) There are no other nodes in val(S i : i ∈ k). If k = ω then we put val(S i : i ∈ k) = j∈ω val(S i : i ∈ j).

A tree T ⊆ T 0 is a valuation tree if T = val(S i : i ∈ k) for some (S i : i ∈ k) ∈ Str ≤ω (T). Note that val(S i : i ∈ k) is a subtree of S 0 and hence also a subtree of T 0 . Also, the height of val(S i : i ∈ k) equals k and the number of nodes of val(S i : i ∈ k) depends only on k and σ, see Lemma 3.1.

Note that if (S i : i ∈ k) is a strong vector subtree of T ↾ k then val(S i : i ∈ k -1) is a valuation tree and val(S i : i ∈ k -1) ⊆ val(S i : i ∈ k). Also note that T 0 (< k) = val(T i : i ∈ k) for every k ∈ ω + 1.

Example 1. Figure 1 depicts an example of a valuation tree. In this example, σ = (1, 2, 1, 1, . . .), hence T 0 consists of all functions which assign 0 or 1 to every decreasing sequence of length 2, T 1 is a binary tree and T k is just a chain for k ≥ 2. Given a strong vector subtree (S 0 , S 1 , S 2 ) of (T 0 , T 1 , T 2 ) of height 3 (depicted by thick nodes and thick successor relations) we construct the corresponding valuation tree val(S 0 , S 1 , S 2 ) ⊆ T 0 . Note that the topmost level of S 1 and the two topmost levels of S 2 are actually not used in the construction of val(S 0 , S 1 , S 2 ) ⊆ T 0 , to depict this, they have grey colour.

Let σ be an arbitrary signature, consider T = T σ , and let T and T ′ be subtrees of T 0 . A function ψ : T → T ′ is a structural embedding if it is an embedding of trees (preserving meets and relative heights of nodes), and for every x 0 , . . . ,

x d-1 ∈ T with |x 0 | > • • • > |x d-1 | it holds that ψ(x 0 )(|ψ(x 1 )| , . . . , |ψ(x d-1 )|) = x 0 (|x 1 | , . . . , |x d-1 |).
Lemma 3.1. Let σ be an arbitrary signature and consider T = T σ . For every k ∈ ω + 1 and every valuation subtree T of T 0 of height k, there exists a unique structural embedding f : T 0 (<k) → T .

Proof. For k ∈ ω we use induction on k. Cases k = 0 and k = 1 are simple. Assume now that the induction hypothesis holds for some k > 0 and let T = val(S 0 , . . . , S k ) have height k + 1. Put T ′ = val(S 1 , . . . , S k ) (it has height k). By the induction hypothesis there exists a unique structural embedding f : T 0 (<k) → T (<k) and a unique structural embedding f ′ : T 1 (<k) → T ′ .

We will now extend f to T 0 (k). Fix u ∈ T 0 (k) and denote by e : k + 1 → ω the increasing enumeration of L T0 (T ). Let v be the predecessor of u on level k -1, let w ∈ T 1 (k -1) be the unique node such that u ∈ v ⌢ w and denote x = u(k -1). Then there is a unique u

′ ∈ ImmSucc S0 (f (v))∩Succ T0 (f (v) ⌢ f ′ (w)) with u ′ (e(k -1)) = x. Put f (u) = u ′ .
It is easy to check that the extended map is a structural embedding T 0 (<k + 1) → T and that the extension was defined in the unique possible way.

If k = ω, we have proved that there are structural embeddings f i : T 0 (<i) → T (<i) for each i ∈ ω. Since these isomorphisms are unique, we get f i ⊂ f j for i < j, and f = {f i : i ∈ ω} is the desired structural embedding. On the other hand, if g : T 0 → T is a structural embedding then for each i ∈ ω the restriction g ↾ T 0 (<i) : T 0 (<i) → T (<i) is a structural embedding and thus has to be equal to f i , consequently g = f . □

L-structures on valuation trees

Most of this section will be spent proving the following proposition which will be the key ingredient in the proof of Theorem 1.1.

Proposition 4.1. Let L be a relational language which consists of no unary relations and finitely many relations of every arity, and let H be a universal countable L-hypergraph. Then H has finite big Ramsey degrees.

For this section, fix such L and H. Assume that H = ω. Let n i be the number of relations of arity i and assume that L = {R i,j : i ∈ ω, 1 ≤ j ≤ n i } such that R i,j has arity i. Let σ be the signature defined by σ i = n i-1 + 2 and let T = T σ .

We define an L-hypergraph G on T 0 by putting

{x 1 , . . . , x i } ∈ R i,j G ⇐⇒ 1 ≤ j ≤ n i and |x 1 | > • • • > |x i | and x 1 (|x 2 |, . . . , |x i |) = j.
Both choices j = 0 and j = σ i + 1 represent a non-relation. We will later need both of them for technical reasons. Equip G with the enumeration defined by f ≤ g if and only if either |f | < |g|, or |f | = |g| and f (x) < g(x), where x is the lexicographically smallest tuple where f and g differ. Observe that a structural embedding T 0 → T 0 induces an (increasing) embedding G → G with this enumeration.

Our aim is to embed H into G, transfer colourings of substructures of H into colourings of vector subtrees of T and use Milliken's theorem to obtain the desired Ramsey result. However, not all embeddings H → G are created equal. In order to prove a more robust result which can later be used in characterising the exact values of big Ramsey degrees, we first need to introduce some terminology.

x = (x 0 , . . . , x m-1 ) with n > x 0 > • • • > x m-1 ≥ 0. We denote by f x the valuation function of level x m-1 defined by f x(ȳ) = f (x ⌢ ȳ) and call it the x-slice of f . An m-slice of f is an arbitrary x-slice of f where |x| = m.
In particular, a 0-slice of f is just f . Note that the elements of T i are precisely the i-slices of elements of T 0 . Definition 4.2. Let k ∈ ω, let φ : ω → T 0 be an embedding H → G and denote by R = φ[ω] its range. We say that φ is a k-enveloping embedding if there are two disjoint sets O, B ⊆ ω such that the following holds for every f ∈ R, for every m < k and for every x = (x 1 , . . . , x m ):

(1) If f x is not constant zero then x i ∈ O for every i, and (2) if g ∈ R or g is constant zero, g ′ is an m-slice of g, and f x and g ′ are incomparable then

|f x ∧ g ′ | ∈ B.
In this case we call the members of O the original levels and the members of B the branching levels. The slice f x is called original if x i ∈ O (that is, all the sliced levels are original).

The intuition is that, when embedding H → G, we "allocate" several levels before each image of a vertex of H for branching, so that we can have strong control over what happens on the branching levels. Often (e.g. when there is an upper bound on the number of relations of every arity) it is possible to get an embedding which is k-enveloping for every k. However, when the number of relations of higher arities is increasing, we were so far unable to get such a uniform embedding (and we conjecture that it does not exist).

Observation 4.2. A k-enveloping embedding is also k ′ -enveloping for every k ′ < k. □ Lemma 4.3. For every k ∈ ω, there is a k-enveloping embedding φ : H → G.
Proof. Recall that the vertex set of H is ω. Define

J = ω ∪ {branch s (x 0 , . . . , x m-1 ) : 1 ≤ s < max i∈k {σ i+m }, x 0 > • • • > x m-1 ≥ 0},
where we treat branch s (x) as a formal expression. Define a linear order ◁ on J, putting i ◁ j if and only if i < j, putting i ◁ branch s (x 0 , . . . , x m-1 ) if and only if i < x 0 and putting branch s (x)◁branch t (ȳ) if and only if either x is lexicographically smaller than ȳ, or x = ȳ and s < t.

Let ϕ : ω → ω be the function such that ϕ(i) = |{w ∈ J : w ◁ i}| (in particular, the ϕ(i)-th element of J is i). For every formal expression branch s (x) ∈ J we now define its value branch s (x) = |{w ∈ J : w ◁ branch s (x)}|. The range of ϕ will be the original levels O and the other levels (i.e. the range of branch) will be the branching levels B. Note that ϕ is an increasing function.

Let φ : H → G be the map where φ(i) ∈ T 0 is a valuation function with |φ(i)| = ϕ(i) and all entries equal to 0 except for the following two cases:

(1) If i > x 0 > • • • > x m-1 and {i, x 0 , . . . , x m-1 } ∈ R m+1,r H then we have φ(i)(ψ(x 0 ), . . . , ψ(x m-1 )) = r. (2) If n < k, i > x 0 > • • • > x n-1 > y 0 > • • • > y m-1 ,
and {i, x 0 , . . . , x n-1 , y 0 , . . . , y m-1 } ∈ R n+m+1,r H then we have φ(i)(ψ(x 0 ), . . . , ψ(x n-1 ), branch r (y 0 , . . . , y m-1 )) = σ n+m -1.

Part (1) ensures that φ is an embedding. We will prove that it is in fact kenveloping, that is, we will verify points 1 and 2 of Definition 4.2.

Point 1 follows straightforwardly from our construction of φ: The only tuples not valuated by 0 either consist of original levels only, or the very last level is branching.

To see point 2, suppose that f ′ = f x, g ′ = g ȳ , |x| = |ȳ| < k, and f ′ and g ′ are incomparable. This means that there is a level n such that

f ′ ↾ I n ω = g ′ ↾ I n ω , but f ′ ↾ I n+1 ω ̸ = g ′ ↾ I n+1 ω
. That is equivalent to n being the least integer for which there exists a decreasing sequence z such that

f (x ⌢ n ⌢ z) ̸ = g(ȳ ⌢ n ⌢ z).
If n is branching then we are done. So n is original, that is, n = ϕ(n ′ ) for some n ′ ∈ ω. Let z = (z 1 , . . . , z p ) and denote a = f (x ⌢ n ⌢ z) and b = g(ȳ ⌢ n ⌢ z). We know that a ̸ = b. We also know that z 1 , . . . , z p-1 are original (because only the last level of a tuple valuated by a non-zero integer can be non-original), so ϕ -1 (z i ) is defined for every 1 ≤ i ≤ p -1 First suppose that z p = branch c (w 1 , . . . , w q ) for some c. Then we know that exactly one of a, b is equal to zero. In this case put z′ = (ϕ -1 (z 1 ), . . . , ϕ -1 (z p-1 ), w 1 , . . . , w q ). By the construction we know that f (x ⌢ branch c (k ′⌢ z′ )) ̸ = g(x ⌢ branch c (k ′⌢ z′ )), because again exactly one of them is equal to 0. And as branch c (k ′⌢ z′ ) < ϕ(k ′ ) = k, we get a contradiction with minimality of k.

So z p is original. Put z′ = (ϕ -1 (z 1 ), . . . , ϕ -1 (z p )) and assume without loss of generality that a ̸ = 0 (at least one of a, b is non-zero). Since z p is original, from the construction it follows that a ̸ = σ |x ⌢ k ⌢ z| -1, which implies that branch a (k ′⌢ z′ ) is defined. Note that f (x ⌢ branch a (k ⌢ z′ )) ̸ = 0 while g(x ⌢ branch a (k ⌢ z′ )) = 0. Again, as branch a (k ⌢ z′ ) < k, we get a contradiction with minimality of k, which verifies that point 2 is also satisfied, hence φ is indeed k-enveloping. □ Remark 4.1. Note that if there is an absolute bound N on the number of relations of any arity then there exists an embedding which is k-enveloping for every k ∈ ω. Indeed, one can pretend in the above proof that k = ω and use the fact that we always have max i∈ω {σ i+m } ≤ N for any m ∈ ω.

When there is no such bound, we have been unable to produce such an embedding and we conjecture that actually there is no embedding where all envelopes would be bounded: Conjecture 4.4. If for every n ∈ ω there is a ∈ ω such that the number of relations of arity a is at least n then there is no embedding φ : H → G with the property that for every k ∈ ω there exists R(k) ∈ ω such that every set S ⊆ φ[H] with |S| = k has an envelope of height at most R(k).

4.1. Envelopes. As was noted earlier, our goal is to transfer colourings of substructures of a nice copy of H in G to colourings of valuation subtrees of T 0 . Definition 4.3. Given S ⊆ T 0 and a valuation subtree T ⊆ T 0 of height k ∈ ω + 1, we say that T is an envelope of S if S ⊆ T .

Having an enveloping embedding allows us to envelope finite subsets of its range in bounded-height valuation trees: Lemma 4.5. For every k ∈ ω there exists R(k) ∈ ω such that for every kenveloping embedding φ : H → G and every set S ⊂ H of size k it holds that φ[S] has an envelope of height at most R(k).

Proof. Fix a set S of k elements from H. Put E 1 0 = φ[S] and E 2 0 = {f ∧ T0 g : f, g ∈ φ[S]} ⊂ T 0 . Define by induction sets E 0 i , E 1 i and E 2 i for 1 ≤ i < R(k)
, where R(k) will be defined later, as follows:

(1)

E 0 i = {f (|g|) : f, g ∈ E 2 i-1 , |g| < |f |} ⊆ T i , (2) E 1 i = E 0 i ∪ {z}, where z is the constant zero valuation function of level max{|f | : f ∈ E 0 i }, (3) E 2 i = {f ∧ Ti g : f, g ∈ E 1 i }. Claim 4.6.
The following properties hold for every 0 ≤ i < R(k):

(1)

E 2 i ⊇ E 1 i ⊇ E 0 i and E 2 i is a subtree of T i , (2) each element of E 1
i if either constant zero or an original i-slice of a member of φ[S] and each element of E 2 i is a restriction of an element of E 1 i , (3) there are at most max(0, k -i) levels with non-zero members of E 1 i , (4) for i > 0, every level in

L Ti (E 2 i ) \ L Ti-1 (E 2 i-1 ) is branching, (5) if ℓ ∈ L Ti (E 2 i ) then either ℓ is branching or there is f ∈ φ[S] with |f | = ℓ. (6) for i > 0 we have that L Ti-1 (E 2 i-1 ) \ L Ti (E 2 i ) = {max(L Ti-1 (E 2 i-1 ))}, (7) for i > 0, every level in L Ti (E 2 i ) is strictly smaller than max(L Ti-1 (E 2 i-1 )
). We will proceed by induction on i. For i = 0 this is immediate. So i > 0 and we know that all properties hold for i -1. Property (1) is straightforward from the definition. We know that (2) holds for i -1, so E 0 i consists of slices of members of φ[S] and constant zero functions, so this is true also for E 1 i , and in constructing E 2 i we are only adding restrictions of members of E 1 i . Note that if i > k then, by (3), we know that we only had constant zero functions in E 1 i-1 and consequently also in E 2 i-1 . So it suffices to prove (3) for i ≤ k. By (5) for i -1 we have that the only original levels we can slice with correspond to members of φ[S] and by (2) we know that the only non-zero members of E 1 i are original i-slices of members of φ[S]. To get an i-slice, we need i + 1 members of φ[S] (one to slice and the other i for a decreasing sequence of original levels to slice with), hence there are only k -i possible last elements of the decreasing sequence. This verifies (3).

To see (4), note that we only add new levels compared to E 2 i-1 when taking meets in the construction of E 2 i . This means that we can assume that i < k as otherwise all elements of E 1 i are comparable. By (2) we know that each member of E 1 i is either constant zero or an original i-slice of a member of φ[S], and since φ is k-enveloping and i < k it follows that their meets happen on branching levels.

Property (5) is immediate from (4) and the induction hypothesis. Property (6) is also easy, in the construction of E 0 i we only lose the highest level. Having (6), ( 7) is again easy, because we only add new levels as meets of functions from existing levels. This finishes the proof of Claim 4.6.

□

Property (3) implies that each member of E 1 k is constant zero, hence E 2 k = E 1 k . It follows that for every i > k we have E 1 i = E 2 i and all its elements are constant zeros. Consequently L Ti (E 2 i ) ⊊ L Ti-1 (E 2 i-1
) and these two level sets differ precisely by max(L Ti-1 (E 2 i-1 )) by ( 6).

Put L = k i=0 L Ti (E 2 i ). Clearly, for every i ≥ 1 it holds that |L Ti (E 1 i )| ≤ |L Ti-1 (E 2
i-1 )| and for every i ≥ 0 we have

|L Ti (E 2 i )| ≤ 2|L Ti (E 1 i )| -1.
Together with |L T0 (E 1 0 )| ≤ k this implies that |L| is bounded from above by some R(k) which is a function of k. In particular, |L Tj (E 2 j )| ≤ R(k) -j, and so E 2 R(k)-1 is a singleton set containing a constant zero function.

For every i ∈ R(k) define

E 3 i = {f ↾ ℓ : f ∈ E 2 i , ℓ ∈ L, |f | ≥ ℓ}. Note that L Ti (E 3 i ) = L and that E 3 i is a subtree of T i for every i ∈ R(k).
For every i ∈ R(k) let S i be some strong subtree of T i containing E 3 i such that L Ti (S i ) = L. Such trees exist by Observation 2.1.

It remains to prove that φ[S] ⊆ val(S i : i ∈ R(k)). We will prove the following stronger result:

Claim 4.7. For every i ∈ R(k) and ℓ ∈ L it holds that if f ∈ E 3 R(k)-i-1 and ℓ ≤ |f | then f ↾ ℓ ∈ val(S R(k)-i-1 , . . . , S R(k)-1 ).
We will prove this statement by double induction on i (outer induction) and ℓ ∈ L (inner induction). For i = 0 this is easy because E 3 R(k)-1 consists only of the root of S R(k)-1 which is also the constant zero function of level min(L) and is the root of val(S R(k)-1 ).

Assume now that the statement is true for every j < i, every ℓ ∈ L and also for i and every ℓ ′ ∈ L such that ℓ ′ < ℓ, we will prove it for i and ℓ. Pick an arbitrary

f ∈ E 3 R(k)-i-1 . If ℓ = min(L) then f ↾ ℓ is the root of S R(k)-i-1
and hence the root of val(S R(k)-i-1 , . . . , S R(k)-1 ).

So ℓ > min(L). Let ℓ ′ be the largest member of L smaller than ℓ. By the construction, f ↾ ℓ ′ ∈ E 3 R(k)-i-1 , and so f ↾ ℓ ′ ∈ val(S R(k)-i-1 , . . . , S R(k)-1 ) by the induction hypothesis for i and ℓ ′ . If ℓ ′ is branching then f (ℓ ′ ) is the constant zero valuation function of level ℓ ′ and as such is in E 3 R(k)-i and consequently in val(S R(k)-i , . . . , S R(k)-1 ) by the induction hypothesis. Otherwise ℓ ′ is original, but then

f (ℓ ′ ) ∈ E 0 R(k)-i ⊆ E 3 R(k)-i
and hence also in val(S R(k)-i , . . . , S R(k)-1 ) by the induction hypothesis.

So we know that

f ↾ ℓ ′ ∈ val(S R(k)-i-1 , . . . , S R(k)-1 )
and

f (ℓ ′ ) ∈ val(S R(k)-i , . . . , S R(k)-1 ).
The definition of valuation tree then gives that f ↾ ℓ ∈ val(S R(k)-i-1 , . . . , S R(k)-1 ) which concludes the proof of Claim 4.7. □

As a special case of Claim 4.7 we get that φ

[S] ⊆ E 1 0 ⊆ E 3 0 ⊆ val(S i : i ∈ R(k)
). This means that there indeed is a valuation tree of height at most R(k) which contains φ[S].

□

We can now proceed with the proof of Proposition 4.1.

Proof of Proposition 4.1. Fix a finite L-hypergraph A with |A| = k, a k-enveloping embedding φ : H → G (for example one given by Lemma 4.3), and a colouring χ 0 : H A → p. Since H is universal, it follows that there is an embedding θ : G → H. Consider the colouring χ : G A → p obtained by setting χ A = χ 0 θ A for every A ∈ G A . Let h = R(k) be given by Lemma 4.5 and let G h be the induced sub-Lhypergraph of G on T 0 (<h). We enumerate the copies of A in G h A as { A i : i ∈ ℓ} for some ℓ ∈ ω.

By Lemma 3.1, for every valuation tree T of height h, there is a structural embedding f T : G h → T that is also an isomorphism of the corresponding substructures of G. Let S = (S i : i ∈ h) be a strong subtree of T ↾ h of height h and consider the structural embedding f = f val(Si:i∈h) :

G h → val(S i : i ∈ h). Put χ(S) = χ f A i : i ∈ ℓ ,
which is a finite colouring of Str h (T ↾ h ). By Theorem 2.2, there is an infinite strong subtree of T ↾ h monochromatic with respect to χ. Extend it arbitrarily to an infinite strong subtree of T with the same level set and let U be its corresponding valuation subtree. Note that the extension does not influence in any way the valuation subtrees of height h. The structural embedding ψ : T 0 → U given by Lemma 3.1 is a hypergraph embedding ψ : G → G.

We claim that θ [ψ [φ[H]]] is the desired copy g[H]

of H, in which copies of A have at most ℓ different colours in χ 0 . This is true, because by Lemma 4.5 every copy of A in φ[H] is contained in a valuation subtree of height h. All of these subtrees have the same colour (with respect to χ) in ψ[φ[H]], and so we know that χ takes at most ℓ different values on ψ

[φ[H]] A . Consequently, χ 0 attains at most ℓ different values on θ[ψ[φ[H]]] A .
□

The main results

In this section we do three simple constructions on top of Proposition 4.1 in order to prove Theorem 1.1. 5.1. Unary relations. First, we introduce a general construction for adding unary relations in order to prove the following theorem.

Theorem 5.1. Let L be a relational language with finitely many relations of every arity greater than one and with finitely or countably many unary relations. Let H be a countable universal L-hypergraph. Then H has finite big Ramsey degrees.

Proof. Assume that the unary relations of L are {U i+1 : i ∈ u} for some u ∈ ω + 1. Let L -be the language which one gets from L by removing all unary relations and let M be a universal L --hypergraph (it exists for example by the Fraïssé theorem [START_REF] Fraïssé | Sur certaines relations qui généralisent l'ordre des nombres rationnels[END_REF]). Without loss of generality we assume that the vertex set of both M and H is ω. We now define an L-structure G as follows:

(1) The vertex set of G is G = {(v, i) : v ∈ ω, i ∈ min(v + 1, u)} with enumeration given by the lexicographic order, (2) vertex (v, i) is in unary relation U j if and only if i = j (in particular, (v, 0) is in no unary relations), and (3) ((x 0 , i 0 ), . . . , (x n , i n )) ∈ R G if and only if (x 0 , . . . , x n ) ∈ R M . Fix some finite L-hypergraph A and let A -be its L --reduct. By Proposition 4.1 there is ℓ ∈ ω such that M -→ (M) A - k,ℓ for every k ∈ ω. We will now prove that G -→ (H) A k,ℓ for an arbitrary k ∈ ω. Let π : G → M be the map sending (v, i) → v. We say that a map f :

X → G is transversal if π • f is injective. Note that if f : A → G is a transversal embedding then π • f is an embedding A -→ M. A colouring χ 0 : G A → k then induces a partial colouring χ : M
A -→ k by ignoring non-transversal copies and composing with π. There may be copies of A -which do not get any colour, we assign them a colour arbitrarily.

Since M -→ (M) A - k,ℓ , there is an embedding ψ : M → M with χ attaining at most ℓ colours on ψ [M] A -. Let ψ 0 : G → G be the embedding mapping (v, i) → (ψ(v), i). We have that χ 0 attains at most ℓ colours on transversal copies from

ψ0[G]

A

. Let H -be the L --reduct of H forgetting unary relations. Since H -is a countable L --hypergraph, there is an embedding φ 0 : H -→ M with the property that φ 0 (x) ≥ i x , where i x = 0 if x is in no unary relation of H and i x = j if x ∈ U j H . It is straightforward to check that the map φ : H → G, defined by φ(x) = (φ 0 (x), i x ), is a transversal embedding H → G, hence χ 0 attains at most ℓ colours on ψ0

[φ[H]] A . Consequently, G -→ (H) A k,ℓ . Knowing that G -→ (H) A k,ℓ , proving that H -→ (H) A k,ℓ is straightforward: G is a countable L-hypergraph
and so there is an embedding θ : G → H, which means that a colouring of H A restricts to a colouring of G A exactly as in the proof of Proposition 4.1. □ 5.2. Non-L-hypergraphs. For the constructions above it was convenient to work with L-hypergraphs. In this section we state a folkloristic result showing that one does not lose any generality working with L-hypergraphs only. Its proof is a straightforward verification of the construction.

Lemma 5.2. Let L be a relational language and let C be the class of all L-structures where every relation is injective and every vertex is in exactly one unary relation.

For every i ∈ ω, put

M i = {(R, π) : R ∈ L, i = a(R), π ∈ Sym(i)}.
Given A ∈ C and vertices

x 1 < • • • < x n ∈ A, put M A (x 1 , . . . , x n ) = {(R, π) ∈ M n : (x π(1) , . . . , x π(n)) ) ∈ R A }.
Define a language L ′ containing an i-ary relation R S for every nonempty finite S ⊆ M i , for every i ∈ ω, and let C ′ be the class of all L ′ -hypergraphs. If L has only finitely many relations of every arity greater than one and finitely or countably many unary relations then so does L ′ . Define T to be the map assigning to every A ∈ C a structure T (A) ∈ C ′ on the same vertex set such that

{x 1 < • • • < x n } ∈ R M A (x1,...,xn) T (A) if M A (x 1 , . . . , x n ) is nonempty.
There are no other relations in T (A). Define U to be the map assigning to every A ∈ C ′ a structure U (A) ∈ C on the same vertex set such that whenever we have

x 1 < • • • < x n ∈ A with {x 1 , . . . , x n } ∈ R S A , we put (x π(1) , . . . , x π(n)) ) ∈ R U (A)
for every (R, π) ∈ S. There are no other relations in U (A).

Then T and U are mutually inverse and define a bijection between C and C ′ . Moreover, given A, B ∈ C and a function f : A → B, it holds that f is an embedding A → B if and only if it is an embedding T (A) → T (B). □ Corollary 5.3. Let L be a relational language with finitely many relations of every arity greater than one and with finitely or countably many unary relations and let H be a countable universal L-structure where all relations are injective and every vertex is in exactly one unary relation. Then H has finite big Ramsey degrees.

Note that this statement is the same as the statement of Theorem 1.1 for F = ∅.

Forbidding structures.

Proof of Theorem 1.1. Since H is unrestricted, it is countable and hence there are only countably many types of vertices. Hence, without loss of generality, we can assume that each vertex of H is in exactly one unary relation (by changing the language similarly as in Lemma 5.2) and that F forbids no single unary relation (otherwise we can remove it from the language). Let M be a countable universal L-structure where all relations are injective and every vertex is in one unary relation (it exists for example by the Fraïssé theorem [START_REF] Fraïssé | Sur certaines relations qui généralisent l'ordre des nombres rationnels[END_REF]). We say that a subset S ⊆ M is bad if M induces a structure from F on S and we say that a tuple x of elements of M is bad if it contains a bad subset. Let G be the L-structure on the same vertex set as M such that x ∈ R G if and only if x ∈ R M and x is not bad (i.e. we remove bad tuples from all relations). Clearly, G is F-free, and note that G and M have the same unary relations. We will prove that G has finite big Ramsey degrees. Since G embeds into H and any embedding H → M is also an embedding H → G, this would imply that H has finite big Ramsey degrees, thereby proving the theorem.

Fix a finite F-free L-structure A where all relations are injective. Let ι be the identity map understood as a function G → M and let A 0 , . . . , A m be some enumeration of all isomorphism types of structures from {ι•f [A] : f is an embedding A → G}, that is, it is an enumeration of all possible isomorphism types which, after removing bad tuples from relations, are isomorphic to A. There are only finitely many of them because they have the same unary relations as A and there are only finitely many L-structures on a given number of vertices with given unary relations, up to isomorphism.

For every 0 ≤ i ≤ m, let ℓ i be the big Ramsey degree of A i in M (ℓ i is finite by Corollary 5.3) and put ℓ = m i=0 ℓ i . We now prove that G -→ (G) A k,ℓ for every k ∈ ω.

Fix a colouring χ : G A → k and let χ i : M Ai → k, 0 ≤ i ≤ m, be the colourings obtained from χ by composing with ι. By inductive usage of Corollary 5.3 we get an embedding f : M → M such that, for every 0 ≤ i ≤ m, χ i attains at most ℓ i colours on f [M] Ai . By the construction it follows that f is also an embedding G → G, and since every copy in f [G] A corresponds to a copy of A i in f [M] for some i, it follows that χ attains at most ℓ colours on f [G] A , hence the big Ramsey degree of A in G is indeed finite. □

Infinite big Ramsey degrees?

As soon as one has infinite branching, the Milliken theorem stops being true even for colouring vertices (see Proposition 6.3). This means that one cannot generalise our methods directly for languages with infinitely many relations of some arity at least 2. In fact, no known methods generalise because all of them find very specific tree-like copies and one can construct infinite colourings which are persistent on these copies. Doing this in full generality requires developing the theory of weak types and it will appear elsewhere. Here we only show a special case which is technically much simpler but only works for binary relations. Definition 6.1. Let H be a countable relational structure with vertex set ω. Given X ⊆ ω, we define the type of X, denoted by tp H (X), to be the isomorphism type of the substructure of H induced on X. (Note that this corresponds to the quantifierfree type over the empty set from model theory if we consider the enumeration to be part of the language.) Let f : H → H be an embedding. We say that f is 1-tree-like (or, in this paper, simply tree-like) if for every finite

X = {x 0 < • • • < x m } ⊂ ω, every 0 ≤ i ≤ m and every x ∈ ω such that x > x m there exists y > x m such that tp H (f [X] ∪ {f (y)}) = tp H (X ∪ {x}) and moreover tp H ({0, . . . , f (x 0 ) -1, f (y)}) = tp H ({0, . . . , f (x 0 ) -1, f (x i )}).
For every embedding f we have that tp H (f [X] ∪ {f (x)}) = tp H (X ∪ {x}). The second condition says that for tree-like embeddings we have some control even over types with respect to the ambient structure H within which our copy lies. Note that every structural embedding T σ → T σ is a tree-like embedding of the corresponding hypergraphs.

Example 2. Let L be a language consisting of infinitely many binary relations and let R be the countable homogeneous L-hypergraph (that is, an infinite-edgecoloured countable random graph where the colour classes are generic). One can repeat our constructions for R and get the everywhere infinitely branching tree T = [ω] <ω as T 0 . If the Milliken theorem was true for infinitely branching trees, it would produce tree-like copies (which would arise simply as strong subtrees isomorphic to [ω] <ω ).

On the other hand, assume that the vertex set of R is ω and consider any embedding f : R → R such that f (i) is connected to vertex 0 by the i-th relation (such an embedding exists by the extension property). This embedding is somehow as far as possible from being tree-like, because everything branches at level 0 and one cannot say anything about the behaviour of this copy with respect to the external vertices.

All known big Ramsey methods produce tree-like copies, and whenever the exact big Ramsey degrees are known, the proof can be adapted to show that every copy contains a subcopy which is "weakly tree-like". Example 2 shows that such a property fails for the infinite-edge-coloured random graph. We believe that working in the category of tree-like embeddings (or some other variant of nice embeddings) would be a completely reasonable thing to do as it still captures (a lot of) the combinatorial complexity of the problems, in addition to allowing one to prove negative results.

In this section we will see an instance of this. Given a relational structure A, its Gaifman graph is the graph on the same vertex set where vertices x ̸ = y ∈ A are connected by an edge if and only if there exists a tuple z of vertices of A containing both x and y which belongs to a relation of A. A relational structure A is irreducible if its Gaifman graph is a complete graph. Theorem 6.1. Let L be a relational language, let F be a set of finite irreducible Lstructures and let H be a universal F-free structure. Assume that there are F-free structures B and U consisting of one vertex, and infinitely many pairwise nonisomorphic 1-vertex F-free extensions C 0 , C 1 , . . . of B such that the added vertices come last in the enumeration and each of them is isomorphic to U. Then there exists an F-free structure A on three vertices and a colouring c :

H A → ω such that if f : H → H is a tree-like

embedding then c attains all values on f [H]

A . In other words, the big Ramsey degree of A is infinite for tree-like copies. Note that if F either contains no enumeration or all enumerations of every finite A, the class of all finite F-free structures is a free amalgamation class when one ignores the enumerations. Corollary 6.2. Let L be a relational language containing infinitely many binary relations and let H be a countable universal L-hypergraph. Then there exists an Lhypergraph A on three vertices and a colouring c :

H A → ω such that if f : H → H is a tree-like

embedding then c attains all values on f [H]

A . Proof. Note that being an L-hypergraph can be described by a set F of forbidden irreducible structures (namely those where a tuple is in a relation, but not all possible permutations of the tuple are). Let B be the empty hypergraph on 1 vertex which is in no unary relations and let C 0 , C 1 , . . . be all possible one-vertex extensions of B by a vertex in no unary relation. There are infinitely many of them as they correspond to binary relations in L. Therefore the conditions of Theorem 6.1 are satisfied and the conclusion follows. □

A particular example of a structure satisfying this corollary is the infinite-edgecoloured random graph from Example 2.

Our proof of Theorem 6.1 is derived from a proof that the Halpern-Läuchli theorem does not hold for the tree [ω] <ω , which we now present as a nice warmup. We believe that this proof is folkloristic, but we were unable to find it in the literature. Proposition 6.3. Let T = [ω] <ω be the tree of all finite sequences of natural numbers and let ⊑ be the usual tree order by end-extension. There is a colouring c : T → ω such that whenever T ′ is a strong subtree of T of infinite height then c[T ′ ] = ω.

Proof. Given t ∈ T , we denote by |t| the length (level) of t, we put w(t) = |t| + i<|t| t(i) and we define ℓ(t) to be the least ℓ such that w(t ↾ ℓ ) ≥ |t|. Note that ℓ(t) always exists as w(t) ≥ |t|.

We define a colouring c : T → ω putting c(t) = w(t ↾ ℓ(t) ) -|t|. Let T ′ be an arbitrary strong subtree of T of infinite height, let r be the root of t ′ and let n ∈ ω be such that n > w(r) and T ′ ∩ T (n) ̸ = ∅. Put k = n -w(r) -1. Now we will prove that c[T ′ ] = ω. For that, fix a colour x ∈ ω and find t ∈ T ′ such that |t| = n and r ⌢ (k + x) ⊑ t (such t exists as T ′ is a strong subtree of T ). Now, w(r) < n, so ℓ(t) > |r|. On the other hand, w(r ⌢ (k + x)) = k + x + w(r) + 1 = n + x ≥ n, hence ℓ(t) = |r| + 1 and c(t) = x. This means that for every x ∈ ω we can find t ∈ T ′ such that c(t) = x, hence indeed c[T ′ ] = ω. □

Note that in the construction of the colouring c we essentially only needed to be able to address a particular level ℓ(t) on which we knew that the passing numbers attain all possible values. We will now show how this idea can be adapted to prove infinite big Ramsey degrees.

Proof of Theorem 6.1. Let B 0 , B 1 , . . . be an enumeration of all copies of B in H such that the vertex of B i comes in the enumeration of H before the vertex of B j whenever i < j. There are infinitely many of them because the infinite disjoint union of copies of B is F-free and thus embeds into H.

Given a vertex v ∈ H isomorphic to U (i.e. it has the same unary relations), we let b(v) be the least integer such that there is w ∈ B b(v) with w ≥ v, and let s(v) : b(v) → ω be the sequence satisfying s(v) i = x if and only if the structure induced by H on B i ∪ {v} is isomorphic to C x .

Given a vertex v ∈ H isomorphic to B, we let j(v) be such that {v} = B j(v) . Given s ∈ [ω] <ω and n ≤ |s|, we put w(s) = |s| + i∈|s| s(i) and define ℓ(s, n) to be the least ℓ such that w(s ↾ ℓ ) ≥ n. It exists since n ≤ |s|.

Let A be an enumerated structure on three vertices such that the second vertex in the enumeration is isomorphic to B, the third one is isomorphic to U and there are no non-unary relations in A. Clearly, A is F-free.

For a copy A ⊆ H, we denote by n( A) the level of the first vertex and we put s( A) = s(v) ↾ j(w) where w is the second and v is the third vertex of A. We define a colouring c : Let f : H → H be an arbitrary tree-like embedding of H to H. We will prove that c[ f [H] A ] = ω. Given p ∈ ω, we will construct a copy

A ∈ f [H]
A with c( A) = p. Assume that the vertex set of H is ω and let r 0 < r 1 be arbitrary vertices of H such that r 0 is isomorphic to B and r 1 is isomorphic to U. Let r 2 ∈ ω be chosen such that:

(1) r 2 > r 1 , (2) r 2 has the same unary relation as the first vertex of A, and

(3) f (r 2 ) is the n-th vertex of H for some n with n > w(s(f (r 1 )) ↾ j(f (r0)) ). Such a vertex exists, because the infinite disjoint union of vertices with the unary is F-free, and so H contains infinitely many such vertices. Let r 3 ∈ ω be an arbitrary vertex such that r 3 > r 2 , j(f (r 3 )) > n, r 3 is isomorphic to B and there are no relations on {r 2 , r 3 }.

Put q = n -w(s(f (r 1 )) ↾ j(f (r0)) ) -1 + p, X = {r 0 , r 1 , r 2 , r 3 } and i = 1. Pick any x ∈ ω such that x > r 3 , {r 0 , x} is isomorphic to C q and there are no relations on {r 2 , x} and {r 3 , x}. Since f is tree-like, we get y ∈ ω such that y > r 3 , tp H ({f (r 0 ), f (r 2 ), f (r 3 ), f (y)}}) = tp H ({r 0 , r 2 , r 3 , x}) and tp H ({0, . . . , f (r 0 ) -1, f (y)}) = tp H ({0, . . . , f (r 0 ) -1, f (r 1 )}). Put v = f (y) and observe that v satisfies the following:

(1) {f (r 0 ), v} is isomorphic to C p , so in particular v is isomorphic to U, (2) there are no relations on {f (r 2 ), v} and {f (r 3 ), v}, (3) v > f (r 3 ) and thus b(v) > j(f (r 3 )) > n, (4) s(f (r 1 )) ↾ j(f (r0)) ⊑ s(v), (5) s(v) j(f (r0)) = q, and (6) H induces a copy of A on {f (r 2 ), f (r 3 ), v}.

Let A be the copy of A induced on {f (r 2 ), f (r 3 ), v}. Put

s = s( A) = s(v) ↾ j(f (r3)) = s(f (r 1 )) ↾ j(f (r0)) ⌢ q ⌢ s ′
for some sequence s ′ and note that n( A) = n (which was defined when we chose r 2 ). By the choice of r 3 we know that |s| > n and so ℓ(s, n) is defined. By the choice of n we know that

w(s(f (r 1 )) ↾ j(f (r0)) ) < n
and from the choice of q we have that

w(s(f (r 1 )) ↾ j(f (r0)) ) + q + 1 = n + p, hence ℓ(s, n) = j(f (r 0 )) + 1. Thus indeed c( A) = w(s ↾ j(f (r0))+1 ) -n = p. □ 7.

Conclusion

Being the first positive big Ramsey result for random structures in infinite languages, this paper helps locate the boundaries of finiteness of big Ramsey degrees. However, there still remain a lot of open problems even regarding unrestricted structures.

While we have proved finiteness of big Ramsey degrees, we do not know their exact values or descriptions. We believe that this is an important problem (which is open even for 3-uniform hypergraphs), but at the same time it is tractable. We think it makes sense to start with L-hypergraphs with no unary relations, where we expect the main combinatorial difficulties will already present themselves, while it will hopefully avoid some technical complications.

We believe that for finite L the problem might be easier, or at least the description of the exact big Ramsey degrees might be simpler. For this reason we state it as a separate problem.

Problem 7.1. Characterise the exact big Ramsey degrees of countable universal L-hypergraphs for finite L with no unary relations.

Note that big Ramsey degrees are a property of the bi-embeddability type, hence all countable universal L-hypergraphs have the same big Ramsey degrees.

Problem 7.2. Characterise the exact big Ramsey degrees of countable universal L-hypergraphs for L which has no unary relations and has finitely many relations of every arity.

The next natural step seems to be adding unary relations.

Problem 7.3. Characterise the exact big Ramsey degrees of countable universal L-hypergraphs for L which has at most countably many unary relations and has finitely many relations of every arity greater than one. Of course, our result is just a small step towards the grand goal of characterising big Ramsey degrees of all structures (or perhaps, more realistically, all structures with known small Ramsey degrees). At this point, however, it seems that even free amalgamation classes behave in unexpectedly complex ways. 7.1. Infinite big Ramsey degrees. In Section 6 we proved that one cannot hope to strengthen our methods to prove big Ramsey results with infinitely many binary relations. We are confident that this argument generalises to higher arities and the following problem has a solution (we believe that we have such a strengthening of tree-likeness, it will appear elsewhere).

Problem 7.5. Define a concept of tree-likeness such that all structural embedding are tree-like and prove an analogue of Corollary 6.2 for the cases when there are infinitely many relations of some higher arity.

However, even for binary relations the situation is interesting. Corollary 6.2 shows that with respect to tree-like embeddings, the infinite-edge-coloured random graph has infinite big Ramsey degree for a particular triple. On the other hand, it is indivisible by the standard argument (try to embed it into one colour class, if it fails then all realisations of some type are in the other colour class which hence contains a monochromatic copy). We do not know what the big Ramsey degree of an edge of some particular colour is: Question 7.6. Given the infinite-edge-coloured random graph (see Example 2) and its substructure A on two vertices, is the big Ramsey degree of A finite? Is it finite with respect to tree-like embeddings?

Of course, big Ramsey degrees in general for this structure remain open as well: Question 7.7. Does the infinite-edge-coloured random graph have finite big Ramsey degrees?

A positive answer to this question would likely require developing new non-treelike methods for finding oligochromatic copies, which would be a major event for the area. Note that this shows that big Ramsey degrees are much more subtle than small Ramsey degrees, because when we only want to find a finite oligochromatic structure, the whole problem only touches finitely many colours and hence reduces to the finite-language problem.

In fact, it is even more subtle than this: Consider a countable infinite-edgecoloured graph G such that edges containing the i-th vertex only use the first i colours. This is a countable infinite-edge-coloured graph which is universal for all finite infinite-edge-coloured graphs as well as all countable finitely-edge-coloured graphs, but it does not embed the infinite-edge-coloured random graph. At the same time, an application of the Milliken theorem on the tree which branches n + 1 times on level n shows that this graph has finite big Ramsey degrees, and the Laflamme-Sauer-Vuksanovic arguments [START_REF] Laflamme | Canonical partitions of universal structures[END_REF] give the exact big Ramsey degrees for this structure (which in turn recovers the exact big Ramsey degrees of all finite-edge-coloured random graphs).

We conjecture that the answers to these questions are negative. In fact, we believe that Theorem 1.1 is tight if there are finitely many unary relations (see the following paragraphs why one has to assume this):

Conjecture 7.8. Let L be a relational language with finitely many unary relations and infinitely many relations of some arity a ≥ 2. Let H be an unrestricted Lstructure realising all relations from L. Then there is a finite L-structure A whose big Ramsey degree in H is infinite. Moreover, the number of vertices of A only depends on a.

If one allows infinitely many unary relations, there are new related structures with yet different behaviour. Consider, for example, a language with infinitely many unary relations and the universal structure where each vertex is in exactly one unary. One can define binary relations on this structure by the pair of unaries on the respective vertices. This structure has infinitely many binary relations but it has finite big Ramsey degrees by Theorem 1.1 (in fact, to prove it it suffices to repeat the proof of Theorem 5.1 on ω which has all big Ramsey degrees equal to one by the Ramsey theorem).

Or consider language L with unary relations U 1 , . . . and binary relations R 1 , . . ., let every vertex have exactly one unary relation, only allow any binary relations between vertices from the same unary relation, and only allow binary relations R 1 , . . . , R i between vertices from unary U i . In other words, look at the disjoint union of infinitely many edge-coloured random graphs such that the i-th of them has all vertices in the unary U i and has i colours. This structure is unrestricted, but Theorem 1.1 does not capture it because of the infinitely many binary relations. However, for a fixed finite A, we can restrict ourselves to the substructure induced on the unaries which appear in A for which Theorem 1.1 can be applied. Since the substructures on different unaries are disjoint, one can then simply add the remaining unaries back.

However, if one allows binary relations even between vertices with different unary relations, but vertices in U i can only participate in R 1 , . . . , R i then the argument from the previous paragraph no longer applies, because we have no guarantee that the oligochromatic copy in finitely many unaries will be generic with respect to the rest, that is, that one will be able to extend it to a full copy. (For example, what might happen is that U i is present in A, U j is not and the monochromatic copy of the restriction will be such that there are no edges between its vertices from U i and vertices from U j .)

We believe that such structures still have finite big Ramsey degrees and that in order to prove it, one only needs to use a stronger statement of Theorem 1.1 which promises that the oligochromatic copy comes from a product tree which will make it possible to extend this copy to a copy using all unaries. It seems that a proper formulation of Conjecture 7.8 should speak about the tree of types having a dense set of vertices on which it branches uniformly. 7.2. Small Ramsey degrees and the partite lemma. Note that the proof of Proposition 4.1 actually gives something stronger than just a finite number of colours: It proves that the colour of every copy only depends on how it embeds into its envelope (this is usually called the embedding type). Part of the job when characterising the exact big Ramsey degrees is constructing embeddings which realise as few embedding types as possible.

It was discovered by Hubička [Hub20] that even without knowing the exact big Ramsey degrees, one can often use the big Ramsey upper bound to give exact small Ramsey degrees. For unrestricted structures, the argument is as follows: Given a finite (enumerated) structure A, pick an arbitrary relation R ∈ L of arity a ≥ 2 and extend A to A ′ adding a vertex b v for every v ∈ A, and putting b v < w for every v, w ∈ A and b v < b w if and only if v > w. Finally, add vertices c 1 , . . . , c a-2 which come very first in the enumeration. A will be a substructure of A ′ and we will only add relations (c 1 , . . . , c a-2 , b v , v) ∈ R A ′ for every v ∈ A.

Note that the lexicographic order on A within A ′ is the same as the enumeration of A and that if B is a substructure of A then B ′ is a substructure of A ′ . Moreover, A ′ describes one particular embedding type of A (namely the one where first all vertices branch and only then they are coded, and they branch so that the lex-order coincides with the enumeration).

If we only colour this particular embedding type of A ′ , we get a monochromatic subtree. Given any finite B which contains A as a substructure, this subtree will contain a copy of (the embedding type described by) B ′ in which all copies of (embedding types described by) A ′ will be monochromatic. However, as we noted in the previous paragraphs, every copy of A inside this B has embedding type described by A ′ , hence in fact all copies of A inside this B will be monochromatic. By compactness we did not need to find the whole monochromatic subtree, just a finite initial segment of it, which contains a unique copy of C ′ for some C which hence satisfies C -→ (B) A k,1 . Thus we get, in particular, a new proof of the Abramson-Harrington theorem [START_REF] Abramson | Models without indiscernibles[END_REF], or in other words, the Nešetřil-Rödl theorem without any forbidden substructures [START_REF] Nešetřil | Partitions of finite relational and set systems[END_REF]. (This will appear in full detail elsewhere.)

Using the Abramson-Harrington theorem one can give a simple proof of the partite lemma of Nešetřil and Rödl [START_REF] Nešetřil | The partite construction and Ramsey set systems[END_REF]. This goes as follows: We can consider A-partite structures (say, induced, but the non-induced variant can also be done in this way), as structures with unary marks on vertices which describe the projection to A. The constraints of A-partiteness are saying that some particular combination of unaries is forbidden to be in a relation together (i.e. they have a projection to A where the relation is not present).

Given an A-partite structure B let B -be its reduct forgetting the unaries. By the Abramson-Harrington theorem there is C -such that C --→ (B -) A 2 . Let C be an A-partite structure with vertex set (C -× A) where the unaries are given by the projection to A and a tuple ((u i , x i )) i∈n is in a relation R C if and only if (u i ) i∈n ∈ R C -and (x i ) i∈n ∈ R A .

A substructure of C is transversal if it does not contain any two vertices of the form (u, x), (u, y) for some u ∈ C -and x ̸ = y ∈ A. Note that a substructure of a transversal structure is again transversal. By a similar argument as in the proof of Theorem 5.1 we can show that a colouring of transversal copies of A in C gives a transversal copy of B in which all transversal (and thus actually all) copies of A have the same colour, thereby proving the partite lemma. Note that conditions of begin A-partite can be described by a set of forbidden substructures which are covered by a relation. From this point of view, Theorem 1.1 can be considered as a big Ramsey generalisation of the partite lemma.

Computer Science Institute of Charles University (I ÚUK), Charles University, Malostranské náměstí 25, Praha 1, Czech Republic Email address: sbraunfeld@iuuk.mff.cuni.cz Department of Applied Mathematics (KAM), Charles University, Malostranské náměstí 25, Praha 1, Czech Republic Email address: chodounsky@math.cas.cz

Definition 4. 1 .

 1 Let f be a valuation function of dimension ω with |f | = n and let

  H A → ω putting c( A) = w(s ↾ ℓ(s,n) ) -n if n ≤ |s|, where s = s( A) and n = n( A), 0 otherwise.

Finally

  , one can ask for the exact variant of Theorem 1.1. Problem 7.4. Characterise the exact big Ramsey degrees for structures considered in Theorem 1.1.

Acknowledgments

We would like to thank Stevo Todorčević for discussions about the infinite-edgecoloured random graph, and Andy Zucker for helpful comments regarding the presentation of our results.

This paper is part of a project that has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement No 810115). S. Braunfeld, D. Chodounský, J. Hubička, J. Kawach, and M. Konečný were further supported by project 21-10775S of the Czech Science Foundation (GA ČR).