COLORING LINEAR ORDERS WITH RADO’S PARTIAL ORDER
RICCARDO CAMERLO AND ALBERTO MARCONE
Abstract. Let ¹
Rbe the preorder of embeddability between countable linear orders colored with elements of Rado’s partial or- der (a standard example of a wqo which is not a bqo). We show that ¹
Rhas fairly high complexity with respect to Borel reducibil- ity (e.g. if P is a Borel preorder then P ≤
B¹
R), although its exact classification remains open.
1. Introduction
This note is a contribution to the study of the relations of embed- dability for countable colored linear orderings from the point of view of descriptive set theory. Fixing ω as the support of a countable linear or- dering, we can consider the space LO of all linear orderings on ω. This is a standard Borel space (actually a Polish space, see [Kec95]). For our purposes, to color a linear order means assigning to each element of the support an element from a fixed countable set C. A colored linear ordering on ω is thus an element of LO × C
ω, which is also a standard Borel space.
If Q is a partial order on C we have a natural relation of embed- dability ¹
Qon LO × C
ωdefined by letting (v, ϕ) ¹
Q(v
0, ϕ
0) if and only if there exists g : ω → ω such that:
(1) ∀a, b ∈ ω (a v b ⇐⇒ g(a) v
0g(b));
(2) ∀a ∈ ω ϕ(a) Q ϕ
0(g(a)).
Then ¹
Qis a Σ
11preorder which is not Borel and it can be studied in the framework of Borel reducibility.
Given binary relations P and P
0on standard Borel spaces X and X
0respectively, recall that P ≤
BP
0means that there exists a Borel function f : X → X
0such that x P y ⇐⇒ f (x) P
0f (y). A Σ
11preorder P
0is called Σ
11-complete if and only if P ≤
BP
0for any Σ
11preorder P .
For preorders of the form ¹
Qit is known that:
• if Q is a bqo then, by Laver’s celebrated theorem ([Lav71]), ¹
Qis a bqo as well; in particular it is a wqo and the relation of equality on ω is not reducible to ¹
Q; thus in this case ¹
Qis quite far from being a Σ
11-complete preorder;
2000 Mathematics Subject Classification. 03E15, 06A05.
Key words and phrases. Borel reducibility. Rado’s partial order. Colored linear orders.
1
• if Q is not a wqo, then ¹
Qis a Σ
11-complete preorder (see [Cam05], a partial result in this direction is in [MR04]).
These results leave unanswered the question of the complexity of the relation ¹
Qwhen Q is a wqo but not a bqo.
If α is a countable ordinal then a preorder Q is ω
α-wqo if ¹
Qre- stricted to well-orders of order type strictly less than ω
αis a wqo.
Therefore ω-wqo means wqo and a preorder is bqo if and only if it is ω
α-wqo for every α (details can be found e.g. in [Mar94]).
In this note we consider Rado’s partial order R defined on the set D = { (n, m) ∈ ω
2| n < m } by
(n, m) R (n
0, m
0) ⇐⇒ (n = n
0∧ m ≤ m
0) ∨ m < n
0.
R is wqo but it is not ω
2-wqo. Moreover R embeds into every wqo which is not ω
2-wqo ([Rad54]). R is the best known example of a wqo which is not bqo.
We will need witnesses for the fact that R is not ω
2-wqo. Let r
i∈ D
ωbe defined by letting r
i(n) = (i, i + 1 + n) for each n ∈ ω. Note that if i < i
0, then r
iand r
i0are incomparable under ¹
R: indeed (i
0, n) R (i, m) does not hold for any n, m, while (i, i
0) R (i
0, m) does not hold for any m. Therefore, for i 6= i
0, we can fix γ
ii0∈ ω such that r
i(γ
ii0) R r
i0(m) does not hold for any m.
In section 2 we show that for any Borel preorder P , the relation P ≤
B¹
Rholds (by the above observation the result extends to all ¹
Qwhere Q is wqo but not ω
2-wqo). This shows that ¹
Ris indeed a quite complex preorder, though the question about its Σ
11-completeness remains wide open. The proof is an application of Rosendal’s construction of a cofinal family of Borel preorders ([Ros05]). This involves comparing ¹
Rwith an ω
1-chain of Borel preorders on standard Borel spaces obtained by repeatedly applying a jump operation P 7→ P
cf.
Rosendal ([Ros05]) showed that if P is a Borel preorder satisfying a simple combinatorial condition, then P <
BP
cf. In section 3 we compare ¹
Rwith its jump (¹
R)
cfand show that ¹
R≡
B(¹
R)
cf(that is ¹
R≤
B(¹
R)
cf≤
B¹
R), giving further evidence to the fact that ¹
Rhas high descriptive complexity.
In contrast with the latter property of ¹
R, in section 4 we prove the existence of a downward closed class of non-Borel preorders P such that P <
BP
cf.
In the sequel, we will freely use the operations of sum of colored
linear orders, and of multiplication of a colored linear order L by a
linear order L
0, indicated by L · L
0. Standard coding techniques allow
to represent the result as an element of LO × C
ωin a Borel way.
2. ¹
Ris above all Borel preorders
Definition 1. If P is a preorder on a standard Borel space X define P
cfon X
ωby setting
~x P
cf~y ⇐⇒ ∀n ∈ ω ∃m ∈ ω x
nP y
m.
Remark 2. Notice that P
0≤
BP
1implies P
0cf≤
BP
1cf. Moreover P ≤
BP
cf, as witnessed by the map x 7→ ~y where y
n= x for every n.
A (colored) linear order L is right indecomposable if it is embeddable in any of its final segments. For example each r
idefined above is right indecomposable. An ordinal is right indecomposable if and only if it is additively indecomposable: in this case we adhere to standard terminology and say that the ordinal is indecomposable.
Theorem 3. For any Borel preorder P on some standard Borel space, the relation P ≤
B¹
Rholds.
Proof. Following Rosendal ([Ros05]) define preorders P
ξwith domain X
ξ, for ξ ∈ ω
1as follows:
• X
0= ω and P
0is equality;
• given P
ξlet X
ξ+1= X
ξωand P
ξ+1= P
ξcf;
• for ξ limit let X
ξ= Q
β<ξ
X
βand ~x P
ξ~y ⇐⇒ ∀β < ξ x
βP
βy
β. Rosendal proved that this transfinite sequence is ≤
B-cofinal among Borel preorders. Therefore it suffices to show that ∀ξ < ω
1P
ξ≤
B¹
R.
We prove by induction on ξ that there exist an indecomposable countable ordinal α
ξand a Borel reduction f
ξof P
ξto ¹
Rwhose values are colored well-orders of order type α
ξwhich are right indecomposable.
Basis step. Let α
0= ω and f
0(i) = r
i.
Successor step. Let ξ ∈ ω
1and assume f
ξ, α
ξsatisfy the induction hypothesis. Set α
ξ+1= α
ξ· ω. Given ~x = (x
0, x
1, x
2, . . .) ∈ X
ξ+1, define
f
ξ+1(~x) = X
k
f
ξ(x
nk)
where (n
k)
k∈ωis an enumeration of ω where each natural number oc- curs infinitely often. Each occurrence of f
ξ(x
i) in this definition will be called a segment of f
ξ+1(~x). Notice that f
ξ+1(~x) is a right indecom- posable colored well-order of length α
ξ+1.
Suppose ~x P
ξ+1~y. For each n ∈ ω there exists m ∈ ω such that f
ξ(x
n) ¹
Rf
ξ(y
m). Since each f
ξ(y
i) occurs infinitely often in f
ξ+1(~y), it follows that f
ξ+1(~x) ¹
Rf
ξ+1(~y).
Conversely, suppose g embeds f
ξ+1(~x) into f
ξ+1(~y). Since no segment of f
ξ+1(~x) can be mapped by g cofinally into f
ξ+1(~y), a final segment of each f
ξ(x
i) must be embedded by g into some f
ξ(y
j). Since by induction hypothesis f
ξ(x
i) is right indecomposable, we have f
ξ(x
i) ¹
Rf
ξ(y
j) which implies x
iP
ξy
j. Thus ~x P
ξ+1~y.
Limit step. Let ξ be a limit ordinal. Suppose that, for each β < ξ, α
βand f
βsatisfy the induction hypothesis. Let ρ be an indecomposable
3
countable ordinal larger than each α
β. Let {β
n}
n∈ωbe an enumeration of ξ with each element occurring infinitely often. Given ~x ∈ X
ξ= Q
β<ξ
X
β, define
F
n(~x) = f
βn(x
βn) + r
n+1(γ
n+1,n) + r
n· ρ
(where r
n+1(γ
n+1,n) denotes the colored linear order with a single element colored by r
n+1(γ
n+1,n)),
F (~x) = X
n
F
n(~x),
and finally
f
ξ(~x) = F (~x) · ω.
So each f
ξ(~x) is a right indecomposable colored well-order of length α
ξ= ρ · ω
2, which is an indecomposable countable ordinal.
Suppose ~x P
ξ~y. Then F
n(~x) ¹
RF
n(~y) for every n, so F (~x) ¹
RF (~y) and finally f
ξ(~x) ¹
Rf
ξ(~y).
Conversely, let g witness f
ξ(~x) ¹
Rf
ξ(~y). Then a final segment of the first occurrence Φ of F (~x) in the definition of f
ξ(~x) is embedded by g into some occurrence Ψ of F (~y) in f
ξ(~y). Let i ∈ ω be least such that the occurrence of F
i(~x) in Φ is embedded by g into Ψ. So for each j ≥ i a final segment of the occurrence of r
j· ρ in Φ must be sent by g cofinally into the corresponding occurrence in Ψ. As, for j > i, the occurrence of r
j+1(γ
j+1,j) in F
j(~x) just preceding the occurrence of r
j·ρ cannot be sent by g to an element of the occurrence of r
j· ρ in F
j(~y), it follows that g embeds f
βj(x
βj) into f
βj(y
βj), witnessing x
βjP
βjy
βj. Since each β ∈ ξ occurs as β
jfor some j > i, it follows that ~x P
ξ~y. ¤ Corollary 4. If Q is a countable wqo but not an ω
2-wqo then, for all Borel preorders P on standard Borel spaces, P ≤
B¹
Q.
Proof. By Rado’s Theorem R embeds into Q and from this it follows
easily that ¹
R≤
B¹
Q. ¤
3. A closure property for ¹
RThe goal of this section is proving the following result:
Theorem 5. (¹
R)
cf≡
B¹
R.
The proof of Theorem 5 uses the following definition and a couple of lemmas.
Definition 6. If Q is a partial order on a countable set C of colors, define ¹
∗Qon LO × C
ωby L ¹
∗QL
0if and only if L · ω ¹
QL
0· ω.
Remark 7. For any Q the map L 7→ L · ω witnesses that ¹
∗Q≤
B¹
Q.
Moreover it is easy to check that L ¹
∗QL
0is equivalent to the exis-
tence of k ∈ ω such that L ¹
QL
0· k.
Lemma 8. If Q is a partial order on a countable set then we have (¹
∗Q)
cf≤
B¹
Q.
Proof. Fix a sequence (n
k)
k∈ωenumerating each natural number infin- itely many times and use the map ~ L 7→ P
k∈ω
(L
nk· ω). ¤ Lemma 9. Let P be any preorder on LO×D
ωsuch that ¹
R⊆ P ⊆ ¹
∗R. Then ¹
R≤
BP . In particular ¹
R≤
B¹
∗Rand thus (by Remark 7)
¹
R≡
B¹
∗R.
Proof. Given L ∈ LO × D
ωlet L
(i)be the colored linear order obtained from L by replacing each color (n, m) with (2(i + 1)n + 1, 2(i + 1)m + 1) (the underlying linear order is unchanged). Notice that L
(i)¹
RM
(i)if and only if L ¹
RM.
To show that ¹
R≤
BP we use the map L 7→ F (L) where F (L) = X
i∈ω
(L
(i)+ r
2i· Q).
It is immediate that if L ¹
RM then F (L) ¹
RF (M) and hence F (L) P F (M).
Now assume that F (L) P F (M), which implies F (L) ¹
∗RF (M), so that for some k ∈ ω we have F (L) ¹
RF (M) · k via some g. We need to show that L ¹
RM. There are two cases:
• First suppose that for some i, j we have r
2i· Q ¹
RM
(j)via a function p = p(n, q) (the domain of r
2i· Q is ω × Q, and (n, q) has color (2i, 2i + n + 1)). In this case we claim that N ¹
RM
(j)for any D-colored countable linear ordering N.
To prove the claim fix N and an order preserving map h : N → Q (here we are looking at N as a linear order). For any a ∈ N let (ϕ(a), ψ(a)) the label assigned by N to a. Define f : N → M
(j)order preserving by f (a) = p(ψ(a), h(a)). Since M
(j)does not use any label of the form (2i, k), the first component of the label of f (a) is greater than 2i + ψ(a) + 1 > ψ(a). Therefore f witnesses N ¹
RM
(j)and the claim is proved.
Then for any N ∈ LO ×D
ωwe have N
(j)¹
RM
(j), and hence N ¹
RM. In particular L ¹
RM.
• Now assume that r
2i· Q
RM
(j)for all i and j. As g maps F (L) into F (M) · k, for some `, g maps
X
i≥`