Shift and Better quasi-order
Yann Pequignot June 5, 2014
Abstract
Viewing infinite subsets ofωas their increasing and injective enumer- ation, the shift map is defined asf7→f◦s, wheresdenotes the successor function on ω. We consider variants of the formf 7→f◦g for some in- creasing and injective g : ω → ω different from the identity. We show that as far as the theory of better quasi-order is concerned they are all equivalent.
We denote by[ω]∞ the set of infinite subsets of natural numbers and we let S : [ω]∞→[ω]∞ denoteshift map, S(X) =X\ {minX}. We let[Z]∞ denote the infinite subsets ofZ∈[ω]∞.
The pair ([ω]∞, S)is notably studied in [2, 3,8]. The associated graph [ω]∞,
{X, S(X)} |X ∈[ω]∞
also plays a role in the theory of Borel chromatic numbers. Notably it has chro- matic number2 and Borel chromatic numberℵ0 (see [4, 1]). Most importantly ([ω]∞, S)plays a central role in the definition of abetter quasi-order which we now recall.
LetA, Bbe topological spaces,Ra binary relation onA,S a binary relation on B. A continuous morphism from (A, R) to (B, S) is a continuous map ϕ:A→Bsuch that for everya, a0 ∈A,aRa0impliesϕ(a)Sϕ(a0). An important particular case is when the binary relation is a continuous function. Letf :A→ Aand g :B →B be continuous functions. A continuous function ϕ: A→B is a morphism from(A, f)to(B, g)exactly in caseϕ◦f =g◦ϕ. For a binary relationR onAwe denote byR{ the binary relation(A×A)\R.
In case R is a binary relation on a discrete spaceA, it follows from Nash- Williams theorem [7] that the following are equivalent (see for example [9, Proposition 4.8]):
1. for every continuous mapϕ: [ω]∞→Athere existsX ∈[ω]∞ such that ϕ(X), ϕ(S(X))
∈R,
2. there exists no continuous morphismψ: ([ω]∞,S)→(A, R{),
3. for every continuous mapϕ: [ω]∞ →A there existsZ ∈[ω]∞ such that the restrictionϕ: ([Z]∞,S)→(A, R)is a continuous morphism.
If one of these conditions holds, we say thatR is abetter relation onA.
For a reflexive and transitive relation RonA, i.e. a quasi-order onA, this is the notion of abetter quasi-order onA, the famous notion invented by Nash- Williams in [6], here in a modern reformulation. For an arbitrary binary relation R, this notion was first defined by Shelah in [11] and it plays an important role in [5].
As in [12, 10], we opt for the language of increasing injections rather than that of sets. We denote byII the monoid of embeddings ofω into itself under composition,
II={f :ω→ω|f is injective and increasing}.
For everyX ∈[ω]∞, we letfX ∈IIdenote the unique increasing and injective enumeration of X. Conversely we associate to every f ∈IIthe infinite subset Imf ofω consisting in the range off. Observe that for allX, Y ∈[ω]∞we have
X⊆Y ←→ ∃g∈IIfX =fY ◦g.
The monoidIIis endowed with the topology induced by the Baire spaceωω of all functions from ω to ω. In particular, the composition ◦ : II×II →II, (f, g)7→f◦g is continuous for this topology.
In the terminology of increasing injections, the shift map S : II → II is simply the composition on the right with the successor functions∈II, s(n) = n+ 1. Indeed for everyX
fS(X)=fX◦s.
This suggests to consider arbitrary injective increasing functiong, g6= idω, in place of the successor function. For anyg∈II, we write~g:II→II,f 7→f◦g for the composition on the right byg. In particular,~s=S is the shift.
In this note we show that these generalised shifts~g are all equivalent as far as the theory of better relations is concerned.
Definition. Let g ∈ II, R a binary relation on a discrete space A. We say (A, R) is a g-better relation if for every continuous ϕ : II → A there exists f ∈IIsuch thatϕ(f)≤ϕ(f◦g). In case≤is a quasi-order on a discrete space Q, we say that (Q,≤)isg-bqo instead of(Q,≤)is a g-better relation.
Of course this notion trivialises for g = idω, since an idω-better relation is simply a reflexive relation. Moreover better relation corresponds tos-better relation. The following is the main result of this note.
Theorem 1. Let g, h ∈ II\ {idω}, R a binary relation on a discrete space A. Then R is a g-better relation if and only if R is an h-better relation. In particular, a quasi-order(Q,≤)is g-bqo if and only if (Q,≤)is bqo.
Therefore, a qoQis bqo if and only if for every continuous ϕ:II→Q and everyh∈IIthere existsg∈II such that
ϕ(g)≤ϕ(g◦h).
The proof of Theorem1directly follows from the following proposition.
Proposition 2. Letg, h∈II\ {idω}. Then there exists a continuous morphism T : (II, ~g) → (II, ~h), i.e. a continuous map T : II → II such that for every f ∈II
T(f ◦g) =T(f)◦h.
First the proof of Theorem1:
Proof of Theorem1. Ifϕ: (II, ~h)→(A, R{)is a continuous morphism, then for T given by Proposition2we obtain thatϕ◦ T : (II, ~g)→(A, R{)is continuous morphism since
ϕ◦ T(f)6≤ϕ(T(f)◦h) =ϕ(T(f ◦g)).
We break the proof of Proposition2 into two lemmas.
Lemma 3. There exists a continuous morphism R : (II, ~g) → (II,~s), i.e. a continuous mapR:II→IIsuch that for everyf ∈II
R(f◦g) =R(f)◦s.
Proof. Since g 6= idω, there existskg = min{k∈ω |k < g(k)}. DefineG∈II byG(n) =gn(kg), whereg0= idω andgn+1=g◦gn. We letR(f) =f ◦Gfor everyf ∈II. ClearlyR:II→IIis continuous and for everyf ∈II and every nwe have
R(f ◦g)(n) =f◦g◦gn(kg) =f ◦G(n+ 1) = (R(f)◦s)(n).
Lemma 4. There exists a continuous morphism E : (II,~s) → (II, ~h), i.e. a continuous mapE:II→IIsuch that for everyf ∈II
E(f ◦s) =E(f)◦h.
Proof. Letkh= min{k|k < h(k)}. Again we defineH ∈IIbyH(n) =hn(kh).
For everyf ∈IIand everyl∈ω, we let E(f)(l) =
(l ifl < H(0),
hf(n)−n(l) ifH(n)≤l < H(n+ 1), forn∈ω.
Let us check thatE(f)is indeed an increasing injection fromωtoωfor every f ∈II. Since E(f) is increasing and injective on each piece of its definition, it is enough to observe the following: Ifl < H(0), then
E(f)(l) =l < H(0)≤H◦f(0) =hf(0)(H(0)) =E(f)(H(0)),
and ifH(n)≤l < H(n+ 1)then
E(f)(l) =hf(n)−n(l)< hf(n)−n(H(n+ 1))
=hf(n)+1(kh)≤hf(n+1)(kh) =H(f(n+ 1)),
but we have
H(f(n+ 1)) =hf(n+1)−(n+1)(H(n+ 1)) =E(f)(H(n+ 1)),
One easily checks thatE:II→IIis continuous. Now on the one hand E(H◦f ◦s)(l) =
(l ifl < H(0),
hf(n+1)−n(l) ifH(n)≤l < H(n+ 1), forn∈ω.
and on the other hand E(H◦f)(h(l)) =
(h(l) ifh(l)< H(0),
hf(n)−n(h(l)) ifH(n)≤h(l)< H(n+ 1).
By definition ofH,h(l)< H(0)if and only ifl=h(l). Moreover if H(n)≤l <
H(n+ 1) thenH(n+ 1)< h(l)< H(n+ 1)and so
E(H◦f◦s)(l) =hf(n+1)−n(l) =hf(n+1)−(n+1)(h(l)) =E(H◦f)(h(l)),
which proves the lemma.
The mapT =E ◦ Rclearly satisfies the requirements of Proposition2.
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