Chapter 5. Infinitary Combinatorics ∗

Chapter 5. Inﬁnitary combinatorics ∗

5.1. Club and stationary sets.

Deﬁnition. Let λ be a limit ordinal. A set D ⊆ λ is unbounded in λ if ∀α < λ ∃β ∈ D α < β. A set C ⊆ λ is closed in λ if for all limit ordinals δ < λ, C ∩ δ is unbounded in δ implies that sup(C ∩ δ) ∈ C. A set that is closed and unbounded in λ is often called “club”

Examples. {γ < λ | γ is a limit ordinal} and {γ < λ | γ is a limit of limit ordinals} are closed in λ. And they are unbounded if λ is a cardinal.

You should think of club sets as being “large” in some sense.

Proposition. Let κ be an uncountable regular cardinal and F a set of ﬁnitary funtions on κ with F < κ (i.e., if f ∈ F we have that f : n(f)κ −→ κ for some n(f) < ω.) Then, C = {γ < κ | ∀f ∈ F f“ n(f)γ ⊆ γ } is closed and unbounded.

Proof. It is clear that C is closed: if λ < κ is a limit ordinal and C ∩ λ is unbounded, f ∈ F, n(f) n(f) f : κ −→ κ and (α0, . . . , αn−1) ∈ λ then there is some γ ∈ (C ∩ λ) \ max({α0, . . . , αn−1 }), and we have f(α0, . . . , αn−1) < γ < λ. S To see C is unbounded, deﬁne G0(X) = X, Gn+1(X) = Gn(X) ∪ {β | ∃α0, . . . , αm−1 ∈ Gn(X) S ∃f ∈ F f(α0, . . . , αm−1) = β }, for each, n < ω and G(X) = n∈ω Gn(X) for each subset X of κ. If ξ < κ it is clear that ξ ⊆ G(ξ) ⊆ κ and G(ξ) < κ. Because κ is regular we can choose g(ξ) = min(κ \ sup(G(ξ))). Let gn+1(ξ) = g(gn(ξ)) some each ξ < κ, and let gω(ξ) = sup{gn(ξ) | n ∈ ω }. Then we have that gω(ξ) ∈ C and ξ < gω(ξ).

Deﬁnition. Let g : κ −→ C enumerate a club set C in increasing order. Let C0 = g“{ωγ | γ < κ}. C0, ⊆ C, is the set of limit points of C.

Lemma. Let C be a club subset of a regular cardinal κ and deﬁne f : κ −→ κ by f(γ) = min(C\(γ+1)). Then C0 = {γ < κ | f“γ ⊆ γ }, and hence C0 is club.

Proof. It is clear that C0 = {γ < κ | f“γ ⊆ γ }, and thus by the previous proposition, is closed and unbounded.

Proposition. Let κ be an uncountable regular cardinal, µ < κ, and let Cα be closed and unbounded T in κ for each α < µ. Then D = {Cα | α < µ} is closed and unbounded in κ.

Proof. It is easy to see that D is closed. Now let fα(γ) = min(Cα \ γ + 1) for γ < κ and for all α < µ. D is unbounded because, by the previous proposition, {γ < κ | ∀α < µ fα“γ ⊆ γ } is unbounded (and closed), and is a subset of D by the lemma.

In fact it is easy to see that one can prove similar results for singular cardinals κ with ω < cf (κ). For T example, if µ < cf (κ) and Cα be closed and unbounded in κ for each α < µ then D = {Cα | α < µ} is closed and unbounded in κ.

First of all prove (by hand) that the intersection of any two club sets is club. Next, take a coﬁnal function f : cf (κ) −→ κ which is closed at limits, that is such that f(λ) = S{f(α) | α < λ} for

∗ version 1.0, 9/2/06

1 limit λ < cf (κ), and observe that f“ cf (κ) is a club, say C, in κ. Finally, if D ⊆ κ let D0 = D ∩ C and note that f −1“D0 is club in cf (κ). So to prove results about, for example, intersections of club subsets of κ (or the diagonal intersection lemma below) simply ‘slide’ the club sets down to cf (κ) using f −1, use the results for regular cardinals applied to the resulting sets and then use f to slide the result back up to κ

Proposition. (Diagonal intersections) Let κ be a regular cardinal and let Cα be closed and T unbounded in κ for each α < κ. Then D = {β | ∀α < β (β ∈ Cα)} (= {β | β ∈ α<β Cα)}) is closed and unbounded in κ.

Proof. It is easy to see that D is closed: if λ is a limit ordinal and D ∩ λ is unbounded in λ we have that ∀α < λ ∃β (α ≤ β < λ & ∀γ < β β ∈ Cγ ). Hence ∀α < λ ∃β (α ≤ β < λ & β ∈ Cα). T Thus Cα ∩ λ is unbounded in λ for all α < λ. Thus λ ∈ {Cα | α < λ}. T In order to see that D is unbounded, deﬁne g(γ) = min( α<γ Cα \ γ) for all γ < κ. (Note that this intersection is unbounded by the proposition that the intersection of fewer than κ club sets is club, thus g(γ) is well-deﬁned.) Now let β < κ, β0 = β, βn+1 = g(βn) for n < ω, and βω = sup{βn | n < ω }. Then βω ∈ D.

Deﬁnition. Let κ be a cardinal with ω < cf (κ). A set S ⊆ κ is stationary if S ∩ C 6= ∅ for all C which are closed and unbounded in κ. A set X ⊆ κ is non-stationary if it is not stationary.

Stationary sets are “medium-sized” as subsets of κ and non-stationary sets are “small” in this sense – even though they can have cardinality κ. The proposition that any intersection of fewer than cf (κ) many club sets is club in κ says that any union of fewer than cf (κ) non-stationary sets is non-stationary.

Examples. Let µ < cf (κ) be a regular cardinal. Then {α < κ | cf (α) = µ} is stationary, because if C is closed and unbounded in κ the µth element in the increasing enumeration of C has coﬁnality µ. For example, both {α < ω2 | cf (α) = ω } and {α < ω2 | cf (α) = ω1 } are both stationary in ω2.

Proposition. (Fodor’s Lemma) Let κ be a regular cardinal, ω < κ, let S be a subset stationary of κ and let f : S −→ κ be such that f(α) ∈ α for each α ∈ S. Then there is some γ < κ such that {α ∈ S | f(α) = γ } is stationary in κ.

Proof. If the conclusion fails, then choose for each γ ∈ κ a set Cγ which is closed and unbounded −1 in κ and such that Cγ ∩ f “{γ } = ∅. Let D = {β | ∀γ < β (β ∈ Cγ )}. The diagonal intersections proposition says that D is closed and unbounded in κ. But D ∩ S = ∅, because if β ∈ D we have f(β) 6= α for each α < β. This is a contradition to the supposition that S is stationary.

An illustration that stationary sets are not as “fat” as club sets are is given by the following proposition.

+ Proposition. Let κ = µ and let S be stationary in κ. Then there is some Sα for α < κ with S S = {Sα | α < κ} and Sα ∩ Sβ = ∅ for each pair α, β < κ with α 6= β.

ξ Proof. Let fγ : γ −→ µ be an injection for each γ < κ. Define Sα = {γ | α < γ < κ & ξ ξ fγ (α) = ξ } ⊆ κ. Then Sα ∩ Sβ = ∅ if α 6= β for each ξ < µ because each fγ is a injection. Also S ξ we have that ξ<µ Sα = {γ | α < γ < κ} is stationary in κ. Because the union of fewer than κ non-stationary subsets of κ is non-stationary, it must be that for each α there is some ξα < µ ξα such that Sα is stationary. Since µ < κ there is some ξ < κ such that {α < κ | ξ = ξα } = κ. ξ Let A = {α < κ | ξ = ξα } and α0 = min(A). Define Sα = Sα for α ∈ A \{α0 }, and define S = S \ S{Sξ | α ∈ A \{α }} (since Sξ ⊆ S ). We have that S ⊆ S is stationary for each α0 α 0 α0 α0 α α ∈ A and Sα 6= Sβ where α, β ∈ A and α 6= β.

2 Fact. With a little more work one can show the proposition for all regular cardinals and not only for successor cardinals as is done above.

5.2 Silver’s Theorem.

Now we are going to use the concepts of closed and unbounded and of stationary sets to prove Silver theorem about exponentiation of singular cardinals of uncountable coﬁnality that was mentioned at the end of §4.4. It is useful to have one more deﬁnition.

ω1 Deﬁnition. F ⊆ ωω1 is called (an) eventually diﬀerent (family) if ∀f, g ∈ F (f 6= g −→ {α | f(α) = g(α)} ≤ ω.

ℵα ℵω1 Theorem. (Silver, 1975) If 2 = ℵα+1 for all α < ω1, then 2 = ℵω1+1.

α Proof. By the hypotheses of the theorem, supposethatP(ωα) = {Xβ | β < ωα+1 } for all α < ω1. α Then for every X ⊆ ωω1 and α < ω1 there is some β < ωα+1 such that X ∩ ωα = Xβ . So for each α X ⊆ ωω1 we can deﬁne a function fX : ω1 −→ ωω1 by fX (α) = β(< ωα+1) if X ∩ ωα = Xβ . Note that F = {fX | X ∈ P(ωω1 )} is eventually diﬀerent because X 6= Y implies that there is some α < ω1 such that fX (α) 6= fY (α), and fX (α) 6= fY (α) implies fX (γ) 6= fY (γ) for every γ ≥ α.

Now partially order F by f < g if and only if {α < ω1 | g(α) ≤ f(α)} is a non-stationary subset. of ω1. By Zorn’s Lemma let ≺ be any total order which extends <. (See Exercise 3.2.a.) So if f ≺ g we have that {α < ω1 | f(α) < g(α)} is stationary in ω1. Let Fg = {f ∈ F | f ≺ g }. We shall show that Fg ≤ ωω1 for each g ∈ F, which will give us the theorem.

In order to do this we use the following two lemmas.

Lemma. If G is eventually different and ∀g ∈ G {α | g(α) < ωα } is stationary, then G ≤ ωω1 . S Proof. Suppose f ∈ G. Let g : {α < ω1 | α = α} −→ ω1 be defined by g(α) = the minimum γ < α such that f(α) < κγ . By Fodor’s Lemma there is a stationary subset Sf of S {α < ω1 | α = α} and some γf < ω1 such that g(α) = γf for all α ∈ Sf . Hence, if ωω1 < G 0 0 there is a subset G ⊆ G, a γ < ω1, and a stationary set S ⊆ ω1 such that ωω1 < G and 0 0 ∀f ∈ G (γf ,Sf ) = (γ, S). As G is eventually different, ∀f, g ∈ G (f 6= g −→ f S 6= g S). But we have

S ω1 ωγ ω1 ωγ ω1 + ωγ = ωγ ≤ max({ωγ , ω1 }) = max({2 , 2 }) = max({ωγ , ω2 }) ≤ κ.

A contradition! Thus G ≤ κ, as required.

The other lemma crudely reduces what remains of the proof to a situation in which this last lemma can be applied.

Lemma. If f : ω1 −→ ωω1 is such that f(α) < ωα+1 for each α ∈ ω1, G is eventually diﬀerent and

∀g ∈ G {α | g(α) < f(α)} is stationary, so G ≤ ωω1 .

0 Proof. For each α < ω let hα : f(α) −→ ωα be an injection. For each g ∈ G deﬁne g : ω1 −→ ωω1 0 0 0 0 by g (α) = hα(g(α)) and let G = {g | g ∈ G }. Then G satisﬁes the hypothesis of the previous 0 0 0 0 lemma. But because all the functions hα are injections, g = k −→ g = k. Thus G = G ≤ ωω1 .

ωω1 Hence Fg ≤ ωω1 for each g ∈ F, and, so, F ≤ ωω1+1. Thus F = 2 = ωω1+1.

Corollary to the Proof. Let κ be a limit cardinal and singular with cf (κ) = θ < κ and let c : θ −→ κ be coﬁnal. If there is a stationary subset T ⊆ θ such that 2c(µ) = c(µ)+ for each µ < θ,

3 then 2κ = κ+.

5.3. Ramsey’s Theorem.

In the next section we shall see another example of the usefulness of the notions of club and stationary sets. This application will be in graph theory.

Definition. A graph is a pair (X,A) where X is a set and A ⊆ X × X \{(x, x) | x ∈ X }. X is the set of vertices or nodes and A is the set of edges). Another way of saying the same thing is to define a graph to be a set X with a binary relation A such that A is symmetric and anti-reflexive.1 We are interested in graphs only up to isomorphism where (X,A) =∼ (Y,B) if there is a bijection f : X −→ Y such that (x, y) ∈ A ⇐⇒ (f(x), f(y)) ∈ B.

There are many examples of graphs, both ﬁnite and inﬁnite. Among them are the complete graphs: graphs of the form (X,B) where for each pair {x, y } of members of X with x 6= y one has (x, y) ∈ B. We write Kκ for the complete graph of size κ. (Clear Kκ is unique up to isomorphism.) For example, we have that K3 is a triangle and K4 is a square with its diagonals.

Deﬁnition. H = (Y,B) is a subgraph of G = (X,A) if Y ⊆ X and B = A ∩ (Y × Y ). Thus H has all of the edges that G does between the vertices in Y .

Deﬁnition. A colouring of a graph G = (X,A) with γ colours is a function f : A −→ γ.

Ramsey’s theorem in its simplest form says that for any finite colouring f of the complete countably ∼ infinte graph Kω there is a subgraph G = (X,A) of Kω such that G = Kω and all of the edges of G have the same colour: {f(e) | e ∈ A} = 1. We say that G is monochromatic. There is also a finite version of the theorem which is deducible from this infinite version: for every n < ω and r < ω there is m < ω such that for any colouring f of Km there is a subgraph G of Km such that ∼ G = Kn. Actually it is worthwhile proving a slight extension to Ramsey’s theorem that has more or less the same proof.

Deﬁnition. For any set x and any cardinal κ let [x]κ = {y ⊆ x | y = κ}.

Ramsey’s Theorem. (1930) Let n, r < ω. If f :[ω]n −→ r there is some X ∈ [ω]ω (that is, an inﬁnite subset of ω) and there is some p < r such that f“[X]n = {p}.(I.e., for any colouring of the set of subsets of size n of ω with r colours there is an inﬁnite subset X of ω such that all the subsets of X of size r have the same colour.)

Firstly, observe the following lemma, a pigeonhole principle, which we will use several times during the proof.

Lemma. Let n, r < ω and f :[ω]n+1 −→ r. Suppose that Z ∈ [ω]ω, x ∈ [Z]n, and for each p < r let xp = {s ∈ Z | s∈ / x & f(x ∪ {s}) = p}. Then at least one of the sets xp is infinite. S Proof of the Lemma. If each xp was finite we would have that Z \ x = {xp | p < r} would be a finite union of finite sets, hence finite. A contradiction!

Proof of Ramsey’s theorem. The proof is by induction on n. The ﬁrst case is n = 1, and in 1 this case we have some f :[ω] −→ r. Let yp = {m ∈ ω | f({m}) = p} for each p < r. So by the lemma at least one of the sets yp is inﬁnite, as required.

1 For more information about graph theory look at, for example, Bela Bollobas, Combinatorics, Cambridge University Press, 1986? or Bela Bollobas’s Graph Theory: An introductory course, Springer, 1979.

4 Now suppose we have shown the theorem for n and any r < ω. Let r < ω and f :[ω]n+1 −→ r. We deﬁne a colouring g :[ω]n −→ r using f: if x ∈ [ω]n, using the Lemma, let g(x) = p if p is minimal such that there is an inﬁnite set of s < ω with f(x ∪ {s}) = p.

Next we define a sequence of infinite sets Xi for i < ω with X0 = ω and Xi+1 ⊆ Xi and write ai for min(Xi). For i < n let Xi+1 = Xi \{ai }. Otherwise suppose that we have hai | i ≤ j i and n hXi | i ≤ j i for some n − 1 ≤ j < ω. Let hyk | k < li be an enumeration of [{ai | i < j }] for some l < ω. We shall define infinite sets Yk for k ≤ l with Y0 = Xj \ (aj + 1) and Yk+1 ⊆ Yk and we will set Xj+1 = Yk. By the Lemma again, let qk < r be minimal such that there is an infinite subset Y of Yk for which for all b ∈ Y we have f(yk ∪ {b}) = qk, and let Yk+1 = {b ∈ Yk | f(yk ∪ {b}) = qk }.

ω Let X = {ai | i < ω }. By the theorem for n we have that there is some Z ∈ [X] and a colour n n+1 n+1 q < r such that g“[Z] = {q }. We show that f“[Z] = 1 as well. Let x0, x1 ∈ [Z] with b0 = max(x0) and b1 = max(x1), and let y0 = x0 \{b0 } and y1 = x1 \{b1 }. Then f(x0) = g(y0) n and f(x1) = g(y1). But g(y0) = g(y1), because y0, y1 ∈ [Z] , hence f(x0) = f(x1).

5.4. Todorcevic’s Theorem.

In contrast to Ramsey’s theorem, however, Sierpinski already showed in the 1930s that the anal- ogous theorem for successor cardinals is false, at least assuming GCH. First we prove a useful lemma.

Deﬁnition. Let κ be a cardinal. The lexicographic order on P(κ) is given by x < y if when z is minimal such that z ∈ (x \ y) ∪ (y \ x) we have that z ∈ y.

Note. It is clear that < is a linear order. And note that this really is the lexicographic order on κ κ 2 where we have the isomorphism χ : P(κ) −→ 2 given by χx(n) = 1 if and only if n ∈ x for x ∈ P(κ).

Lemma. Let κ be an inﬁnite cardinal and let < be the lexicographic order on P(κ). There is no subset of P(κ) which is well-ordered by < and which has cardinality greater than κ.

Proof. Suppose that Y ⊆ P(κ) is well-ordered by < and κ < Y . Then there is some ordinal ρ ∼ + such that (Y, <) = (ρ, ∈) by an o.p. isomophism g and κ ≤ ρ. Let Y0 be the initial segment of Y ∼ + such that (Y0, <) = (κ , ∈) by g. Let γα be the least γ < κ such that ∀β < α γβ < γ and there is T x ∈ Yα such that γ ∈ x, let Yα+1 = {x ∈ Yα | γα ∈ x}, and let Yλ = {Yα | α < λ} for limit λ < κ. Further, let aα ∈ Yα if Yα 6= ∅ and β = {α | Yα 6= ∅}.

Then the function f : β −→ Y0 deﬁned by f(α) = aα for α < β and is coﬁnal into Y0 and we have that g · f shows that cf (κ+) ≤ cf (β) ≤ κ. A contradiction!

Proposition. (Sierpinski) There is a colouring f of [P(κ)]2 with 2 colour such that there is no subset X of P(κ) of cardinality greater than κ with f“[X]2 = 1.

Proof. Let < be the lexicographic order on P(κ). Let ≺ be any well-ordering of P(κ). Deﬁne f :[P(κ)]2 −→ 2 by setting f({x, y }) = 0 if x < y and x ≺ y or y < x and y ≺ x, and f({x, y }) = 1 if x < y and y ≺ x or y < x and x ≺ y.

If H ⊆ P(κ) is such that f“[H]2 = {0} then < is a well-ordering of H. And if H ⊆ P(κ) is such that f“[H]2 = {1} then {κ \ x | x ∈ H } is well-ordered by <. So, in either case H ≤ κ by the Lemma.

Deﬁnition. A cardinal κ is weakly compact if for any colouring f of [κ]2 with 2 colours there is

5 some x ∈ [κ]κ such that f“[x]κ = 1.

Example. ω is weakly compact by Ramsey’s theorem.

Proposition. If κ is weakly compact, then κ is strongly inaccessible.

Proof. If κ is not strongly inaccessible then either there is some µ < κ such that κ ≤ 2µ or κ is singular. If µ < κ and κ ≤ 2µ we have that there is a colouring of [2µ]2 with 2 colours without a monochromatic subset of 2µ of size µ+ by Sierpinski’s theorem. But µ+ ≤ κ, so the restriction of this colouring to [κ]2 has no monochromatic subset of cardinality κ.

If κ is singular, let g : cf (κ) −→ κ be coﬁnal and strictly increasing. Deﬁne f :[κ]2 −→ 2 by f(γ, δ) = 0 if and only if there is some α < cf (κ) such that g(α) ≤ γ, δ < g(α + 1). Because for each α < cf (κ) we have that {γ < κ | g(α) ≤ γ < g(α + 1)} < κ, there is no x ∈ [κ]κ such that + f“[x]2 = {0}. But also there is no x ∈ [κ]cf (κ) such that f“[x]2 = {1}.

Much later Stevo Todorˇcevi´cproved a beautiful theorem which greatly strengthens Sierpinski result. This theorem serves as another example of how useful the notions of ‘closed and unbounded’ and of ‘stationary’ are. (The proof that we give here is a variation by Dan Velleman on Todorˇcevi´c’s original proof.)

2 2 2 Theorem. (Todorˇcevi´c ) There is a colouring f of [ω1] with ω1 colours such that f“[x] = ω1 for ω1 each x ∈ [ω1] . In other words, there is a colouring of Kω1 with ω1 colours such that any subgraph

isomorphic to Kω1 must contain edges coloured with each of the colours.

ω Proof. Let {rα | α < ω1 } ⊆ 2. Also let eα : α −→ ω be an injection for each α < ω1. Deﬁne σ(α, β) = the least n < ω such that rα(n) 6= rβ(n) for α < β < ω, and let f(α, β) = the least δ < ω1 such that α ≤ δ < β and eβ(δ) ≤ σ(α, β) if there is some such δ, and = 0 otherwise.

ω1 To prove the theorem it is enough to show that if Z ∈ [ω1] , then there is a club set C such that C ⊆ f“[Z]2. In order to see this suﬃces remember that by the proposition immediately following the deﬁnition of stationarity there are stationary sets Si ⊆ ω1 for each i < ω1, with S ω1 = {Si | i < ω1 } and Si ∩ Sj = ∅ if i, j < ω1 and i 6= j. If we set g(α) = i if and only if α ∈ Si 2 ω1 we have that g · f“[Z] = g“CZ where CZ is closed and unbounded for each Z ∈ [ω1] . But for each i < ω1 we have that there is some αi ∈ Si ∩ CZ because Si is stationary. So if i < ω1 there is some {x, y } ∈ [Z]2 such that f({x, y }) = α and g(α) = i. Hence g · f instantiates the theorem.

ω1 S n <ω So, let Z ∈ [ω1] and for each g ∈ { 2 | n < ω } = 2 let Bg = {α ∈ Z | g ⊆ rα }. Also set <ω S C = {δ < ω1 | ∀g ∈ 2, either Bg ⊆ δ or (Bg = ω1 & δ = (Bg ∩ δ))}. I claim that C is club in ω1.

In order to see that C is closed suppose that λ is a limit ordinal and S C ∩ λ = λ. If g ∈ <ω2 and Bg 6⊆ λ, we have Bg 6⊆ δ for each δ < λ, and so we have Bg = ω1, because C ∩ λ 6= ∅, and S S S S λ = C ∩ λ = ( {Bg ∩ δ | δ ∈ C ∩ λ}) = Bg ∩ λ. Thus C is closed.

<ω It is also not diﬃcult to see that C is also unbounded in ω1. 2 = ω, so there is some δ < ω1 S S such that ( {Bg | Bg < ω1 }) < δ. Let {Bn | n < ω } = {Bg | Bg = ω1 }. So given α < ω1 one can recursively choose βω.n+m to be the least member of Bm such that α, δ, and βω.n0+m0 < βω.n+m if 0 0 0 S n < n or n = n and m < m. It is clear that α < γ = {βω.n+m | n, m ∈ ω } ∈ C.

2 Stevo Todorˇcevi´c, Partitioning pairs of countable ordinals, Acta Mathematica, 159 (1987), pp. 261-294. This great paper has many other extremely interesting facets. Highly recommended reading!

6 We shall show that C ⊆ f“[Z]2. Suppose that δ ∈ C and β ∈ Z with δ < β. We shall ﬁnd some α ≤ δ such that f(α, β) = δ.

Let n = eβ(δ) and g = rβ n. Then β ∈ Bg, and so Bg 6⊆ δ and we have that Bg = ω1.

For each γ ∈ Bg such that β < γ let mγ = σ(β, γ) and hγ = rγ (mγ + 1). Note that ω1 hγ mγ = g mγ , but hγ (mγ ) 6= g(mγ ). Because Bg = ω1 there is some subset B ∈ [Bg] and some m < ω and h : m −→ ω such that m = mγ and h = hγ for each γ ∈ B. Thus Bh = ω1 S and by the deﬁnition of C we have that δ = δ ∩ Bh.

Let F = { < δ | eβ() ≤ m}. F is ﬁnite because eβ is an injection. So there is some α ∈ Bh ∩ δ such that F ⊆ α. Now we have that α ∈ Bh, so h ⊆ rα, and we have that h m = rβ m and h(m) 6= rβ(m). Hence σ(α, β) = m. If α ≤ < δ we have that m < eβ(), but eβ(δ) = m. So δ is least such that α ≤ δ < β and eβ(δ) ≤ m. Thus δ = f(α, β).

It would be very interesting to have similar results for graphs other than Kω1 . For example András Hajnal and Péter Komjáth’s, Some remarks on the simultaneous chromatic number, gives some information on this problem, but there is still lots of room for improvement on their results.

To take one example, let us say that G = (X,A) has chromatic number ω1 if ω1 is the least cardinal

κ such that there is some f : κ −→ X such that f(x) 6= f(y) if {x, y } ∈ A. It is clear that Kω1 has chromatic number ω1.

+ The principle ♦ says that there is some collection {Bα,n | α < ω1 & n < ω } such that Bα,n ⊆ α for each α < ω1 and n < ω, and if B ⊆ ω1, then there is some club set, C, such that

∀γ ∈ C ∃n < ω (B ∩ γ = Bγ,n).

+ Thus ♦ says that there is a system {Bα,n | α < ω1 & n < ω } which predicts any subset B of ω1 correctly on at large (closed and unbounded) number of places. ♦+ is a strengthening of (CH), and so is not provable in ZF (although it is true if V = L, for example).

[♦+ implies (CH) because if x ⊆ ω there is some C such that ∀γ ∈ C \ ω there is some n < ω with ω x = x ∩ γ = Bγ,n. Thus 2 = P(ω) ≤ {Bα,n | α < ω1 & n < ω } = ω1.]

+ Hajnal and Komj´ath show that ♦ implies that if G = (ω1,A) has chromatic number ω1, then ω1 there is some f : A −→ ω1 such that if X ∈ [ω1] and (X,A∩(X ×X)) also has chromatic number ω1 one has that f“(A ∩ (X × X)) = ω1. But, for example, it is still an open question as to whether one can prove in ZFC (alone) the following natural slightly weaker statement.

Whenever G = (ω1,A) has chromatic number ω1, there is some f : A −→ ω1 (with f“A = ω1) such −1 that if g : ω1 −→ ω there is some i ∈ ω such that f“A∩g “{i} = ω1. (In words, if f is a colouring of G with ω1 colours and g is a partition of the vertices of G into countably many disjoint sets then f takes all of the colours on one of the sets.)