A complexity problem for Borel graphs

We show that there is no simple (e.g. finite or countable) basis for Borel graphs with infinite Borel chromatic number. In fact, it is proved that the closed subgraphs of the shift graph on [N]N\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$[\mathbb {N}]^\mathbb {N}$$\end{document} with finite (or, equivalently, ≤3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\le 3$$\end{document}) Borel chromatic number form a Σ21\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\Sigma }^1_2$$\end{document}-complete set. This answers a question of Kechris and Marks and strengthens several earlier results.


Introduction
by Borel sets B 0 , . . . , B n there exists an i ≤ n and an x ∈ [N] N so that [x] N ⊂ B i , in other words, all infinite subsets of x are contained in B i . This of course implies χ B (G S ) = ℵ 0 . The Galvin-Prikry theorem (in a sense that can be made precise, see [24,25]) is somewhat weaker than the Borel determinacy theorem and thus the proof of χ B (G S ) = ℵ 0 potentially can be considered as an example of a fourth kind.
Since G S is in some sense rather small (e.g. it is locally finite) but still has infinite Borel chromatic number and certain universality properties, one might wonder whether a graph G has infinite Borel chromatic number if and only if G S ≤ B G. Unfortunately, it is not hard to see that that the answer to this question is negative: the direct sum for n ∈ N of the complete finite graphs on n vertices is a counterexample. Another, much more general example to the failure of this type of basis results has been given by Conley and Miller [5].
After this, there are several natural ways to proceed. Firstly, we could restrict ourselves to a smaller class of graphs, and hope for a basis result in that class. For instance, Kechris, Solecki and Todorčević asked whether being Borel above G S characterizes the graphs with infinite Borel chromatic number of the form G f ? Or, it is also natural to consider the structure of the Borel/closed subgraphs of the shift graph: for a Borel graph G = (X, E) and B ⊂ X let us denote by G| B the graph (X, E ∩ B 2 ). A negative answer to question (1) has been given by Di Prisco and Todorčević [6] (see also [11, 4(E)] for a simple counterexample). Moreover, it has been shown recently by Pequignot [19] that (2) is false as well.
Secondly, one could hope for a different graph, or a countable basis instead of a one element basis: Question 1.2 (Kechris,Marks [11,Problem 4.23]) Is there a sequence (G n ) n∈N of Borel graphs with χ B (G n ) < ℵ 0 and χ B (G n ) unbounded such that for every Borel graph H with infinite Borel chromatic number and for every n we have that G n ≤ B H?
It follows from our results that the answer to all of these questions is negative. Roughly speaking, positive basis results typically imply that the complexity of the collection of the Borel graphs with infinite Borel chromatic number (with an appropriate coding) is low and we will show that this is not the case, even for the closed subgraphs of G S . Note that for such graphs having infinite Borel chromatic number is equivalent to having Borel chromatic number ≥ 4, see [12,Theorem 5.1]. In [2], Carroy, Miller, Schrittesser and the second author characterized the graphs with Borel chromatic number ≥ 3 similarly to the G 0 dichotomy: it is shown that there exists a Borel graph, called G odd , having the property that for every Borel graph G we have χ B (G) ≥ 3 if and only if G odd ≤ B G. Hence we obtain a complete description of the characterizability of the Borel chromatic numbers of graphs in terms of simple bases. These results reinforce the experience from the classical case of finite graphs, namely, that it is strictly more complicated to decide whether a graph has chromatic number ≥ n than to check whether a given coloring is correct for every n, except if n ≤ 3. Now we formulate the precise statement of our results. A family A of Borel graphs is called 1 1 -parametrizable if there exist Polish spaces X, Y and a 1 1 set E ⊂ X × Y 2 so that for any G ∈ A there exists an x ∈ X with G being Borel isomorphic to (Y, {(y, z) : (x, y, z) ∈ E}) and the set {x : (Y, {(y, z) : (x, y, z) ∈ E}) is Borel isomorphic to some graph in A} is 1 1 . Recall that a subset A of a Polish space X is 1 2 -hard, if for any Polish space X and A 1 2 subset of X there exists a Borel map, called a reduction, f : X → X with x ∈ A ⇐⇒ f (x ) ∈ A for every x ∈ X . A 1 2 -hard set that is 1 2 is called 1 2 -complete. Now we are ready to state our main result. Theorem 1. 3 The collection of closed sets C ⊂ [N] N so that χ B (G S | C ) < ℵ 0 (or, equivalently, χ B (G S | C ) ≤ 3) is 1 2 -complete. Consequently, there exists no sequence of 1 1 -parametrizable collections of graphs (A n ) n∈N so that for every C ⊂ [N] N closed set χ B (G S | C ) ≥ ℵ 0 if and only if there exist an increasing sequence (n i ) i∈N and A n i ∈ A n i with A n i ≤ B G S | C . In particular, there is no one element basis, or countable basis in the sense of Question 1.2.
Let us point out that Pequignot also used complexity to answer (2) of Question 1.1. His argument is built on a result of Marcone [14], who proved that the set of quasi-orders that are better quasi-orders (bqos) is 1 2 -complete, that is, its complement is 1 2 -complete (bqos were defined by Nash-Williams [18], they form a particularly well behaving class of quasi-orders, see also [13,Chapter 9] and [20].) Pequignot's proof proceeds by showing that there is a reduction from bqos to the family of closed sets In order to prove the complexity result we isolate a general theorem about the complexity of certain families of sets. Suppose that we are given a family F of subsets of a Polish space X coming from a map that assigns to each set in F the set of the witnesses of being in F (e.g. the codes of the possible finite colorings). Suppose moreover that we put sets from F "next to each other" i. e., consider a Borel set B ⊂ N N × X and we are interested whether the sections of B are in F uniformly, that is, whether we can find witnesses of being in F in a Borel way (we will see later that in the case of graph colorings this is precisely equivalent to the existence of a finite coloring of the graph obtained by putting the graphs B s "next to each other"). How hard is it to decide the existence of such a uniform selection? Our general theorem says that if the family F is complicated enough then it is 1 2 -hard. (In our case, this will follow from the observations that non-dominating sets are complicated and that χ B (G S | B ) ≤ 3 holds if B is a non-dominating Borel set). Now we make the above idea precise. Let X, Y be uncountable Polish spaces, be a class of Borel sets and : (X ) → 1 1 (Y ) be a map. Define F ⊂ (X ) by A ∈ F ⇐⇒ (A) = ∅ and let the uniform family, U , be defined as follows: and B ∈ U ⇐⇒¯ (B) has a full Borel uniformization (that is, it contains the graph of a Borel function N N → Y ).
A family F of subsets of a Polish space X is said to be 1 1 (resp. 1 2 )-hard on , if there exists a set B ∈ (N N × X ) so that the set {s ∈ N N : B s ∈ F} is 1 1 (resp. 1 2 )-hard. One would be tempted to think that the fact that F is 1 1 -hard on is sufficient for proving the 1 2 -hard on -ness of the family U . Unfortunately, this is not the case (at least under the axiom of constructibility), see Remark 3.3. On the positive side, the typical way of proving that F is 1 1hard is to start with a given A ∈ 1 1 (N N ) and find a closed set D ⊂ N N × N N with A = proj 0 (D). Now, one constructs a set B ⊂ N N × X so that B s ∈ F iff s ∈ A and this is witnessed by every element of the set D s , i. e., D s ⊂ (B s ).
The following definition encompasses this situation.

Definition 1.5
The family F is said to be nicely 1 1 -hard on if for every and for all s ∈ N N we have We are ready to state our theorem.  Definition 2.4) and that F is nicely 1 1 -hard on . Then the family U is 1 2 -hard on . The paper is organized as follows. First we start with summarizing the most important facts and notations used in the proofs. Then in Section 3 we prove Theorem 1.6. In Section 4 we apply this result to calculate the complexity of the collection of closed subgraphs of the shift graph and also show Theorem 1.4. Finally, in the last section we discuss a counterexample to the 1 1 version of Hedetniemi's conjecture and finish with a couple of open problems.

Preliminaries and notations
For the collection of finite, (resp. infinite) sequences of elements of a set A the notations A <N , (resp. A N ) will be used, while the family of countably infinite subsets of A is denoted by [A] N . If x ∈ A N and n ∈ N then x| n will stand for the sequence (x(i)) i<n .
The standard notations 0 1 (X ), 1 1 (X ), 1 1 (X ), . . . will be used for the collection of subsets of X that are closed, Borel, analytic, etc. A coding of the Borel sets with nice properties has to be fixed, let BC(X ) be a set of Borel codes and sets A(X ) and C(X ) with the properties summarized below: • if P is a Polish space and B ∈ 1 1 (P × X ) then there exists a Borel map f : P → N N so that ran( f ) ⊂ BC(X ) and for every p ∈ P we have Similarly, there exists a so called good universal closed set for every Polish space as well: We will identify a set x ∈ [N] N with its increasing enumeration. As usually, x ≤ * y if |{n : x(n) > y(n)}| < ∞ and x ≤ y holds if for every n ∈ N we have x(n) ≤ y(n). A set S ⊂ [N] N is dominating if for any y ∈ [N] N there exists an x ∈ S with y ≤ * x. We will use the abbreviation y ≤ ∞ x for |{n : y(n) ≤ x(n)}| = ∞.
Notions and facts from effective descriptive set theory will be applied, however, the proofs can be understood using them as "black boxes".
If X is a recursively presented Polish space (in our arguments only the spaces N, N N and their finite products will appear in such a role) and p ∈ N N , 0 1 (X ; p), 1 1 (X ; p), 1 1 (X ; p) and 1 1 (X ; p) will stand for the appropriate lightface classes relative to p. In the case X = N the "N" sometimes will be omitted, and 1 1 ( p), etc. will be used. If p ∈ N N we will denote the first ordinal non-recursive in p by ω p 1 . For an ordinal α and a set A the α's level of the constructible universe relative to A will be denoted by L α [A]. If p, q ∈ N N , 0 1 (X ; p, q), ω p,q 1 etc. will abbreviate the notions 0 1 (X ; p, q ), ω p,q 1 etc., where ·, · : N N × N N → N N is a recursive bijection.
We collect the theorems of the effective theory used in the proof.

Fact 2.3
For any reals r, p ∈ N N we have For a graph G = (X, E) the -measurable chromatic number or chromatic number is defined analogously to the Borel chromatic number with requiring the coloring function to be -measurable, and denoted by χ (G). χ(G) stands for the (usual) chromatic number of the graph G. A set S ⊂ X will be called independent or E-independent if S 2 ∩ E = ∅. Note that if is closed under finite unions then for n ∈ N the existence of a -measurable n-coloring of the graph (X, E) is equivalent to the existence of a partition of X to n-many E-independent sets from .
The Effros Borel space of the closed subsets of a Polish space X will be denoted by F(X ).

Definition 2.4
Let be a class of subsets of Polish spaces. A map : Note that if for some the above condition holds for P = N N and is closed under continuous preimages then is 1 1 on : indeed, given A ∈ (P × X ) one can fix a continuous map φ : N N → P that is bijective on a closed set C ⊂ N N , and φ| C is a Borel isomorphism (see [17,Theorem 1G.2]) and pull back A with the map (φ, id X ) to obtain a set A ∈ (N N × X ). Then using the condition for

General results
In this section we prove Theorem 1.6 about the complexity of uniform families. So, let , X , Y , be as in the theorem. Before starting the proof we make two easy observations. First, without loss of generality we can assume that Y = N N : indeed, composing with a Borel bijection between Y and N N neither the families F and U , nor the fact that is 1 1 on changes. So, from now on we assume that Y = N N . Second, if one replaces N N by its homeomorphic copy in the definition of nicely 1 1 -hard on families (Definition 1.5), it yields an equivalent condition on the family F . Throughout the proof we will frequently use this, e.g. saying that "identify the space Let us roughly sketch the ideas of the proof of Theorem 1.6. Firstly, since F is 1 1 -hard for a given r ∈ N N a diagonal argument yields a set B ∈ (X ) such that (B) is nonempty, but contains no 1 1 (r ) elements. Also, one can show that the set {c : c codes the 1 1 set A and sup{ω r,c 1 : r ∈ A} < ω 1 } is 1 2 -hard, our strategy is to reduce this set to the codes of the sets in U . So, to a given A ∈ 1 1 (N N ) with code c using (a uniform version of) the diagonalization we construct a set B ∈ (N N × X ) so that for all r we have (B r ) = ∅ and (B r ) ∩ 1 1 (r, c) = ∅ ⇐⇒ r ∈ A. From this and (4) of Fact 2.3, it will easily follow that if sup{ω r,c 1 : r ∈ A} = ω 1 for some A with code c then the corresponding set B / ∈ U . Finally, the niceness of F and (2) of Fact 2.3 will yield the converse.
We start with the diagonal argument.
Proof First we construct an auxiliary set A for the sake of the diagonal argument. Take a universal set Using the fact that F is nicely 1 1 -hard (and identifying N N and D 0 (r,s) ⊂ (B 0 (r,s) ). Pick also an arbitrary pair (r * , s * ) ∈ A and let q ∈ Since is 1 1 on , the set (B 0 ) is 1 1 , and, using (4) of Fact 2.3 we get Then by the choice of A there exists an r 0 ∈ N N so that and D = D 0 r 0 . We claim that q, B and D satisfy the requirements of the Lemma. Indeed, as is closed under continuous preimages the sets {(s, t, x) : (r * ,s * ) } ∈ , while using the closedness under finite unions and intersections and 0 We check now that (1) and (2) hold. (1) and (2) hold for every such s.
In order to see (2) note that if for some t we have ∈ R, which we already have shown. Now we are ready to prove Theorem 1.6.
Proof of Theorem 1.6 Let A be a 1 2 -complete 1 2 subset of N N and find S ∈  (1) and (2) of Lemma 3.1 for every (q, r, s) triple.
Using the fact that is 1 1 on we can find a real q 0 so that (B ) ∈ 1 1 (q 0 ) and D ∈ 1 1 (q 0 ), we can also assume that q ∈ 1 1 (q 0 ).
(recall that ·, · was a recursive homeomorphism between (N N ) 2 and N N .) By (1) of Lemma 3.1 for every (q, r, s) ∈ (N N ) 3 we have that hence using the definition of q 0 and B we get that We show that for every r we have B r ∈ U if and only if r ∈ A, which is clearly sufficient to prove the theorem. Suppose that r / ∈ A and for the contradiction that B r ∈ U . By definition this implies that (B r ) has a full Borel uniformization. Consequently, by Fact 2.3 there exists a p 0 ∈ N N such that ∀s, : s / ∈ S (q 0 ,r ) }. We will show that (B r ) has a full Borel uniformization. In order to do this, by Fact 2.3 (4) it is enough to show that for every s, t ∈ N N we have that (B (r, s,t ) )∩ 1 1 ( p, q 0 , r, s, t) = ∅. If s ∈ S (q 0 ,r ) then this holds by (3.3). Now, we can assume that s / ∈ S (q 0 ,r ) and (B (r, s,t ) ) ∩ 1 1 (q 0 , r, s, t) = ∅, since if the equality is not true then we are already done. Then (B (q 0 ,r,s,t) ) ∩ 1 1 (q 0 , r, s, t) = ∅. Recall the definition of D : it has been obtained from the application of Lemma 3.1 to the set S. Hence, by (2) of Lemma 3.1 for every (q 0 , r, s) we have that for every t ∈ N N the implication 0 ,r,s,t) )) holds. Hence, in our case we have ∅ = D (q 0 ,r,s) ⊂ (B (q 0 ,r,s,t) ) = (B (r, s,t ) ). Then D (q 0 ,r,s) is a nonempty 1 1 (q 0 , r, s) set and ω  B (r, s,t ) ) which shows that (B r ) has a full Borel uniformization and finishes the proof of the theorem.
In our theorem the reason of the high complexity is the same phenomenon as in the complexity results of Adams and Kechris [1]. In fact, one of their results follows directly from our theorem.

Corollary 3.2 (Adams, Kechris) The set of trees T on N×N so that [T ] (the set of the infinite branches of T ) has a full Borel uniformization is
Proof A standard calculation shows that the set in question is 1 2 , so to show completeness, it is enough to verify that it is 1 2 -hard. One can check that   (B) is countable. The question whether 1 2 -determinacy is optimal will be investigated in an upcoming paper.

Consequences on graph colorings
In this section we apply the results of the previous one and prove Theorem 1.3. We start with proving a variant of it, from which the full version will be easy to deduce.
The next lemma reduces our task to produce a Borel set B ⊂ N N × [N] N such that the set {s :

s is a bijection if B s is closed under the shift.
Note that Theorem 1.4 also follows from the above lemma: the first statement of it is obvious if χ(G S | B ) ≤ 2, while (a parametrized version of) the rest is Lemma 4.2.

Proof of Lemma 4.2
The idea of the proof is that we express B as an injective projection of a closed set. Then, by applying a homeomorphism (that serves as a coding map) to this closed set we will get another closed set so that the composition of the inverse of the projection and the homeomorphism, and also the inverse of this composition are shift-invariant.
Consider the set B = {(s, x) : (∀ j ∈ N)(S j (x) ∈ B s )}. We will build a set C and a map as in the lemma, such that is a bijection between C and B .
Let Given a pair (s, x) ∈ B there are finitely many i's with i ∈ pr ed(s, x) and so the set {ψ(s, σ i x, pr ed(s, σ i x)) : i ∈ pr ed(s, x)} is finite. Our strategy is to every (s, x) assign a natural number that encodes finite initial segments of the elements of the finite set above. The assigned number to (s, x) should be smaller than the one assigned to (s, S(x)) and the latter should encode longer initial segments of the corresponding finite set of values (this length will be determined by the first element of S(x)). We construct a map :  ∈ pr ed(s, x) .
In order to achieve the property that the natural numbers serving as codes increase as one applies the map S, we define a map 1 on the sets A n inductively. If 1 has been already defined Finally, let Note that for each s the map (s, ·) is a Borel map from B s to [N] N : by definition B s is closed under the shift, so (s, ·) is defined on B s and from the definition of 1 it follows that for any x we have Observe that an induction on n yields that if (s, x) ∈ A n then 1 (s, x) is determined by the values {(i, pr ed(s, σ i x), ψ(s, σ i x, pr ed(s, σ i x))| m ) : m ≤ x(0), i ∈ pr ed(s, x)}. In particular, for a given k ∈ N the kth coordinate of (s, x) is determined by s(k) and the set

Claim 4.3 Suppose that ((s n , x n )) n∈N is a sequence with elements in B such that the sequence ( (s n , x n )) n∈N is convergent and i, j ∈ N.
Then the sequence s n , σ i S j (x n ), pr ed(s n , σ i S j (x n )), ψ(s n , σ i S j (x n ), pr ed(s n , σ i S j (x n )) n is also convergent in the sense that either for every large enough n we have i / ∈ pr ed(s n , S j (x n )), in which case the sequence is eventually not defined, or for every large enough n we have i ∈ pr ed(s n , S j (x n )) and then it converges.
Proof Clearly, the convergence of the sequence ( (s n , x n )) n∈N implies the convergence of ( 0 (s n , S j (x n ))) n∈N and this yields that pr ed(s n , S j (x n )) must stabilize to some set I as n → ∞. If i / ∈ I then for every large enough n we have (s n , σ i S j (x n )) / ∈ B , hence the sequence is not defined. Suppose now that i ∈ I , we check the convergence of the required quadruple. The sequence (s n ) n∈N clearly converges, while the convergence of (x n ) n∈N follows from the definition of 1 and the fact that for any j ∈ N the sequence ( 1 (s n , S j (x n ))) n∈N is convergent. The convergence of the third coordinate is implied by the convergence of the sequence ( 0 (s n , S j (x n ))) n∈N .
Finally, to show the same for the fourth sequence pick an arbitrary k ∈ N. We check that the values ψ(s n , σ i S j (x n ), pr ed(s n , σ i S j (x n )))(k) stabilize. Let j ≥ max{k, j}. By the convergence of the sequence (x n ) n we can pick an i ∈ N such that σ i S j (x n ) = σ i S j (x n ) holds for every large enough n.
for every large enough n. But the sequence ( 0 (s n , S j (x n ))) n∈N converges, so the values ψ(s, σ i S j (x n ), pr ed(s, σ i S j (x n )))| S j (x n )(0) must stabilize as well, and then the fact S j (x n )(0) ≥ j ≥ k yields the convergence of the sequence ψ(s n , σ i S j (x n ), pr ed(s n , σ i S j (x n )))(k) n∈N .

Now we check that (B )
is a closed set. In order to see this suppose that ((s n , x n )) n∈N is a sequence in B so that ( (s n , x n )) n∈N is convergent. By Claim 4.3 the sequence (s n , x n , pr ed(s n , x n ), ψ(s n , x n , pr ed(s n , x n ))) n is convergent, and by the fact that the graph of ψ is closed, it converges to some (s, x, pr ed(s, x), ψ(s, x, pr ed(s, x))). To see that (s n , x n ) → (s, x) holds, pick an arbitrary k ∈ N. We show that (s n , x n )(k) → (s, x)(k). Using the convergence of (s n , S j (x n ), pr ed(s n , S j (x n ))) n∈N for every j ∈ N (which follows from Claim 4.3) we can assume that for each n and j ≤ k we have S j (x n )(0) = S j (x)(0) and pr ed(s n , S j (x n )) = pr ed(s, S j (x)). By Claim 4.3 for any j ≤ S k (x)(0) = S k (x n )(0) and i ∈ pr ed(s, S j (x)) = pr ed(s, S j (x n )) the sequence s n , σ i S j (x n ), pr ed(s n , σ i S j (x n )), ψ(s n , σ i S j (x n ), pr ed(s n , σ i S j (x n )) n is convergent and again by the closedness of the graph of ψ its limit is necessarily s, σ i S j (x), pr ed(s, σ i S j (x)), ψ(s, σ i S j (x), pr ed(s, σ i S j (x))) . Hence, using the observation made before Claim 4.3 for a large enough n all the values determining (s n , x n )(k) and (s, x)(k) will be the same. Thus C = (B ) is indeed closed. Note that for any (s, y) = (s, x) and j ∈ N the exponent of 2 in y( j) is x( j). Hence, is invertible and −1 (s, S(y)) = (s, S(x)). Thus, we obtain that := −1 is a continuous bijection, so that for each s ∈ N N the map s back-and-forth shift-invariant. This implies that It is not hard to check that c is a Borel max{2, n}-coloring of G S | B s , hence Thus, in order to show Theorem 4.1 it is enough to construct the required Borel set. This will be done in two steps.
First we will notice that G S contains an isomorphic copy of H and then using Theorem 1.6 we will show that the finitely chromatic Borel subsets of the graph H are already 1 2 -hard. We will use an observation of Di Prisco and the first author that says that the restrictions of the shift graph to non-dominating subsets of [N] N have finite Borel chromatic number. In the latter part of the paper a uniform version of this statement is needed, so for the sake of completeness we include a proof of the uniform version. We will use the sets BC, A, C from Fact 2.1. Fix also a homeomorphism ·, ·, · : (N N ) 3 → N N .
We will construct a Borel set U ⊂ ([N] N ) 2 such that whenever (x, y) ∈ D, then there exists a i ∈ N with (x, S i (y)) ∈ U and for each x the set {y : (x, y) ∈ D ∩ U } is G S -independent. This is enough, as the map is an appropriate coloring. Now, let x(−1) = 0 and define (x, y) ∈ U ⇐⇒ for the minimal n ∈ N with |[x(2n −1), x(2n +1))∩ y| = 0 we have that |[x(2n −1), x(2n +1))∩ y| is even. Observe that if y ≤ ∞ x then there exist infinitely many n's such that From this, one easily checks that U satisfies the requirements.
which set is 1 1 . Now, we show that F is nicely 1 1 -hard on 0 2 . Let A ⊂ N N be analytic and take a closed set We show that the complement of B is 0 2 , hence B ∈ 0 2 . For every σ ∈ N N that is eventually zero define B σ = {(s, y) : Let  ((s m , y m )) m∈N ⊂ B σ and suppose that (s m , y m ) → (s, y). Then for each m there exists an x m ≤ y m + σ so that (s m , x m ) ∈ F. For every fixed n we have that the set {m : (∃k ≤ n)(x m (k) > y(k) + σ (k))} is finite. Thus, applying König's Lemma to the tree formed by {x m | k : k, m ∈ N, x m | k ≤ (y + σ )| k } we get that (x m ) m∈N contains a convergent subsequence, and its limit witnesses (s, y) ∈ B σ . Let where f dom is the function from Lemma 4.5. We will show that B and D witness that F is nicely 1 1 -hard. We have already seen that B ∈ 0 2 and by definition D is analytic. Suppose that s ∈ A. Then for each x ∈ F s we have B s (= {y : Borel isomorphic to a subgraph of G S , so if it has finite Borel chromatic number then it has one ≤ 3 by [12]. Let S 0 , S 1 , S 2 witness this fact. Using Fact 2.1 there are Borel maps f 0 , f 1 , f 2 so that for any t ∈ N N we have that In order to see that this set is 1 2 , similarly to the proof of Lemma 4.6 just notice that {s :

Remark 4.7
In the proof of Lemma 4.6 we actually show that the collection of non-dominating 0 2 sets is 1 1 -hard in the codes. The proof presented here is an alternate non-effective version of an unpublished result of Hjorth [9]. A similar argument has been also used by Solovay [26]. We would like to mention here that more is true: even the collection of non-dominating closed sets is 1 1 -hard in the codes.
We conclude this section with proving our main result, Theorem 1.3. In order to formulate the precise statement we use the set U   In order to show the statement that talks about the closed sets with the Effros Borel structure we state a general lemma which essentially follows from the work of Sabok [21].

Lemma 4.9
Suppose that P is a property of closed subsets of N N and there exists a closed set C ⊂ N N × N N so that {x ∈ N N : C x has P} is 1 2 -hard. Then {F ⊂ N N : F has P} is also 1 2 -hard as a subset of the Effros Borel space.
Proof Consider the map f : N N → F(N N ) given by f (x) = C x . As usual, we can identify F(N N ) with the collection of pruned trees on N <N , hence it becomes a Borel subset of 2 N <N , let us endow F(N N ) with the inherited topology. By [21,Theorem 2] it is enough to show that f is 1 Proof of Theorem 1.3 In (1) and (2) the fact that the sets are 1 2 can be easily seen directly, similarly to (*). Moreover, Lemma 4.9 and the fact that N N is homeomorphic to [N] N shows that (1) implies (2).
Take the set C from Theorem 4.
shows that the latter set is 1 2 -complete. For the last statement, suppose that such collection of A i 's exists with the appropriate parametrizations Note that (c) and (d) express that c does not code a total function and that the function coded is not a homomorphism from (Y n i , (E n i ) x ) to G S | C s . Clearly, the formula (a) is 1 1 and the formulas (b) and (d) are 1 1 . It follows from the (uniform version of) Luzin's unicity theorem [10,Theorem 18.11] that (c) is 1 1 (an alternative proof can be given using (4) of Fact 2.3). Consequently, the existence of the families (A n ) n∈N would imply that the set {s ∈ N N : χ B (G S | C s ) < ℵ 0 } is 1 2 , contradicting Theorem 4.1.

Relation to Hedetniemi's conjecture and open problems
In this section we collect several open problems and discuss the relation of our results to Hedetniemi's conjecture. Let G = (X, E) and G = (X , E ) be Borel graphs. The product of the graphs G and G , G × G , is the graph (X × X , E G×G ), where (x, x )E G×G (y, y ) ⇐⇒ (x Ey and x E y ). It is clear that G × G is a Borel graph and note also that The classical Hedetniemi's conjecture is the above statement for finite graphs (and thus with usual chromatic numbers). Clearly, the Borel version of the conjecture for graphs with finite Borel chromatic numbers implies the classical one. However, there are substantial differences between the Borel and classical cases for infinite chromatic numbers.
On the one hand, note that if for some graph G we have G ≤ B G, G then G ≤ B G × G , thus, in such a situation χ B (G ) gives a lower bound for the value χ B (G × G ). For instance, the G 0 dichotomy implies Hedetniemi's conjecture for analytic graphs of Borel chromatic number > ℵ 0 .
On the other hand, it has been proved by Hajnal [8] that there exist graphs G and G so that χ(G) = χ(G ) = ℵ 1 , but χ(G × G ) < ℵ 1 . Moreover, it has been shown in [12] that it is consistent that there exist graphs G, G with coanalytic edge relation such that Note also that a compactness argument implies that if χ(G) = ℵ 0 and χ(G ) = n then the conjecture holds.
Concerning the conjecture for finite graphs 1 it is known that for any n > 2 there are graphs with chromatic number n and arbitrarily high odd girth, thus, there is no finite graph H with χ(H ) = n that would admit a homomorphism to each finite G with χ(G) ≥ n. So, Hedetniemi's conjecture cannot be solved by a basis result in the collection of finite graphs (see e. g. [22]). However, we would like to remark that the finite conjecture is in fact equivalent to a basis result if we are allowed to consider infinite graphs and the right notion of chromatic number: i∈N be an enumeration of all the finite graphs with chromatic number n. Let G ∞ be their infinite product, that is, is a Borel graph with a closed edge relation and it is not hard to see that Hedetniemi's conjecture for n implies that G ∞ has clopen chromatic number n. Conversely, since G ∞ admits a continuous homomorphism into each G i , if the clopen chromatic number of G ∞ is n, then Hedetniemi's conjecture holds for n.
Unfortunately, it is not clear whether it is possible to turn antibasis results to counterexamples to the Borel version of Hedetniemi's conjecture. But, if one considers 1 1 -measurable colorings instead of Borel ones our construction yields an example, which works for a rather simple reason: there exist 1 1 sets B and C so that G S | B has a finite Borel chromatic number but has no finite 1 1 coloring and C contains reals which code finite Borel colorings of B, hence the 1 1 chromatic number of the product graph will be finite.

Proposition 5.2
There exist sets B, C ∈ 1 1 ([N] N ) so that χ B (G S | C ) = ℵ 0 , G S | B has no 1 1 finite coloring, but the product G S | B × G S | C has a 1 1 3coloring.
Proof sketch Instead of constructing a set C ⊂ [N] N we construct a set C ⊂ N N × [N] N and prove that H| C and G S | B has the required properties, from this it is easy to deduce the proposition using Lemma 4.4.
It is not hard to check (similarly to the proof of Lemma 4.6) that B ∈ 1 1 (N × [N] N ), the set {n : G S | B n has a finite 1 1 coloring} is 1 1 and A contains this set. Consequently, for some n ∈ A we have that B n has no 1 1 finite coloring, let C = {(x, r ) : x ∈ C n , r ∈ [N] N } and B = B n . Then clearly C ∈ 1 1 (N N × [N] N ) and as C n is nonempty, χ B (H| C ) = ℵ 0 . Note now that for every (x, r ) ∈ C clearly B ⊂ {y : y ≤ ∞ x}. Thus, by (the lightface version of) Lemma 4.5 for each (x, r ) ∈ C the graph G| B has a 1 1 (x, r ) 3-coloring. Using 4 of Fact 2.3, the graph H| C × G S | B has a 1 1 3-coloring: we can construct a coloring from the 1 1 (x, r )-colorings uniformly.
As we have seen, Theorem 1.3 excludes the possibility of a simple Borel/analytic basis. However, the following is still possible: Question 5.3 Does there exist a graph G = (X, E) where X is a Polish space and E is a 1 1 edge relation so that for any Borel graph G we have χ B (G ) ≥ ℵ 0 if and only if G ≤ B G ?
Note that the above question makes sense even with some finite number instead of ℵ 0 . A possibility of a positive answer is even more intriguing in the light of Remark 5.1: it would be very interesting if in both cases the large chromatic number of a certain class of graphs was witnessed by a graph outside of this class.
On the other hand we don't know whether the idea of Proposition 5.2 can be turned to a counterexample to the Borel version Hedetniemi's conjecture. A fundamental tool for the investigation of Hedetniemi's conjecture is the n-coloring graph C n (G) of a graph G defined by El-Zahar and Sauer [7]. It is not clear, however, whether there exist analogous well-behaving objects for Borel graphs. Problem 5.5 Let G be a Borel graph. Define a graph C n (G) of n-colorings of G for which the results of El-Zahar and Sauer [7] can be generalized.
One could hope for a positive result after excluding the sort of examples constructed in this paper. More precisely, our example can be viewed as follows: a smooth equivalence relation E has been constructed so that there are no E S edges between the classes (in other words E is a smooth super-equivalence relation of a restriction of E 0 to some Borel set) and each E class has finite Borel chromatic number, but the union of E classes has infinite Borel chromatic number. Note also that such a graph still has a 1 2 -measurable finite coloring. Hence, the following questions are natural: Question 5. 6