= (Σ, E), where E ⊆ [Σ]2. • A proper coloring of G is a function c with domain Σ such that c(σ) = c(τ) for all {σ, τ} ∈ E. • The chromatic number of G, denoted χ(G), is the least cardinal χ such that there is a proper coloring c : Σ → χ.
= (Σ, E), where E ⊆ [Σ]2. • A proper coloring of G is a function c with domain Σ such that c(σ) = c(τ) for all {σ, τ} ∈ E. • The chromatic number of G, denoted χ(G), is the least cardinal χ such that there is a proper coloring c : Σ → χ. • Given a natural number ≥ 3, a cycle of length in G is an injective sequence σi | i < from Σ such that • {σi , σi+1} ∈ E for all i < − 1; • {σ −1 , σ0} ∈ E.
= (Σ, E), where E ⊆ [Σ]2. • A proper coloring of G is a function c with domain Σ such that c(σ) = c(τ) for all {σ, τ} ∈ E. • The chromatic number of G, denoted χ(G), is the least cardinal χ such that there is a proper coloring c : Σ → χ. • Given a natural number ≥ 3, a cycle of length in G is an injective sequence σi | i < from Σ such that • {σi , σi+1} ∈ E for all i < − 1; • {σ −1 , σ0} ∈ E. Note that the following are equivalent for a graph G:
= (Σ, E), where E ⊆ [Σ]2. • A proper coloring of G is a function c with domain Σ such that c(σ) = c(τ) for all {σ, τ} ∈ E. • The chromatic number of G, denoted χ(G), is the least cardinal χ such that there is a proper coloring c : Σ → χ. • Given a natural number ≥ 3, a cycle of length in G is an injective sequence σi | i < from Σ such that • {σi , σi+1} ∈ E for all i < − 1; • {σ −1 , σ0} ∈ E. Note that the following are equivalent for a graph G: • G has no cycles of odd length;
= (Σ, E), where E ⊆ [Σ]2. • A proper coloring of G is a function c with domain Σ such that c(σ) = c(τ) for all {σ, τ} ∈ E. • The chromatic number of G, denoted χ(G), is the least cardinal χ such that there is a proper coloring c : Σ → χ. • Given a natural number ≥ 3, a cycle of length in G is an injective sequence σi | i < from Σ such that • {σi , σi+1} ∈ E for all i < − 1; • {σ −1 , σ0} ∈ E. Note that the following are equivalent for a graph G: • G has no cycles of odd length; • G is bipartite;
= (Σ, E), where E ⊆ [Σ]2. • A proper coloring of G is a function c with domain Σ such that c(σ) = c(τ) for all {σ, τ} ∈ E. • The chromatic number of G, denoted χ(G), is the least cardinal χ such that there is a proper coloring c : Σ → χ. • Given a natural number ≥ 3, a cycle of length in G is an injective sequence σi | i < from Σ such that • {σi , σi+1} ∈ E for all i < − 1; • {σ −1 , σ0} ∈ E. Note that the following are equivalent for a graph G: • G has no cycles of odd length; • G is bipartite; • χ(G) ≤ 2.
G is a graph, k is a natural number, and χ(H) ≤ k for every finite subgraph H of G. Then χ(G) ≤ k. As a result, if a graph G has infinite chromatic number, there is a function fG : N → N defined by letting fG (k) be the least number of vertices in a subgraph of G with chromatic number k.
G is a graph, k is a natural number, and χ(H) ≤ k for every finite subgraph H of G. Then χ(G) ≤ k. As a result, if a graph G has infinite chromatic number, there is a function fG : N → N defined by letting fG (k) be the least number of vertices in a subgraph of G with chromatic number k. fG is clearly an increasing function.
G is a graph, k is a natural number, and χ(H) ≤ k for every finite subgraph H of G. Then χ(G) ≤ k. As a result, if a graph G has infinite chromatic number, there is a function fG : N → N defined by letting fG (k) be the least number of vertices in a subgraph of G with chromatic number k. fG is clearly an increasing function. Question How fast can fG grow for graphs G with large chromatic number?
: N → N, there is a graph G such that |G| = χ(G) = ℵ0 and fG grows faster than f . Theorem (Erd˝ os-Hajnal-Szemer´ edi, 1982) For every n ∈ N and every cardinal κ, there is a graph G such that χ(G) ≥ κ and fG grows faster than expn .
: N → N, there is a graph G such that |G| = χ(G) = ℵ0 and fG grows faster than f . Theorem (Erd˝ os-Hajnal-Szemer´ edi, 1982) For every n ∈ N and every cardinal κ, there is a graph G such that χ(G) ≥ κ and fG grows faster than expn . Question (EHS, 1982) Is it true that, for every f : N → N, there is a graph G such that χ(G) > ℵ0 and fG grows faster than f ?
• For every ∈ N, there is a graph G such that χ(G) = ℵ0 and G has no cycles of length less than . • If χ(G) > ℵ0, then G contains every finite bipartite graph as a subgraph. In particular, it has cycles of all even lengths.
• For every ∈ N, there is a graph G such that χ(G) = ℵ0 and G has no cycles of length less than . • If χ(G) > ℵ0, then G contains every finite bipartite graph as a subgraph. In particular, it has cycles of all even lengths. Theorem (R¨ odl; Komj´ ath-Shelah) • If χ(G) = ℵ0, then there is a triangle-free subgraph H of G such that χ(H) = ℵ0.
• For every ∈ N, there is a graph G such that χ(G) = ℵ0 and G has no cycles of length less than . • If χ(G) > ℵ0, then G contains every finite bipartite graph as a subgraph. In particular, it has cycles of all even lengths. Theorem (R¨ odl; Komj´ ath-Shelah) • If χ(G) = ℵ0, then there is a triangle-free subgraph H of G such that χ(H) = ℵ0. • There is consistently a graph G such that χ(G) = ℵ1 but χ(H) ≤ ℵ0 for every triangle-free subgraph H of G.
: N → N, there is a graph G such that |G| = χ(G) = ℵ1 and fG grows faster than f . They force over a model of ♦ with a length-ω1 finite-support iteration of c.c.c. posets, dealing one at a time with each function f : N → N. CH holds in the resulting model.
: N → N, there is a graph G such that |G| = χ(G) = ℵ1 and fG grows faster than f . They force over a model of ♦ with a length-ω1 finite-support iteration of c.c.c. posets, dealing one at a time with each function f : N → N. CH holds in the resulting model. Main Theorem (CLH, 2019) For every f : N → N, there is a graph G such that |G| = 2ℵ0 , χ(G) = ℵ1, and fG (k) ≥ f (k) for all k ≥ 3.
type of width n is a function t : 2n → 2 such that |t−1(0)| = |t−1(1)| = n. • If a, b ∈ [On]n are disjoint, then we say tp(a, b) = t if, letting a ∪ b be enumerated in increasing order as αi | i < 2n , we have a = {αi | t(i) = 0} and b = {αi | t(i) = 1}.
type of width n is a function t : 2n → 2 such that |t−1(0)| = |t−1(1)| = n. • If a, b ∈ [On]n are disjoint, then we say tp(a, b) = t if, letting a ∪ b be enumerated in increasing order as αi | i < 2n , we have a = {αi | t(i) = 0} and b = {αi | t(i) = 1}.
is a disjoint type of width n, and α is an ordinal. Then G(α, t) is the graph with vertex set [α]n and edge set E(α, t), where {a, b} ∈ E(α, t) if tp(a, b) = t or tp(b, a) = t.
is a disjoint type of width n, and α is an ordinal. Then G(α, t) is the graph with vertex set [α]n and edge set E(α, t), where {a, b} ∈ E(α, t) if tp(a, b) = t or tp(b, a) = t. Theorem (Erd˝ os-Hajnal) Suppose that 1 ≤ s < n < ω and α is an infinite ordinal.
is a disjoint type of width n, and α is an ordinal. Then G(α, t) is the graph with vertex set [α]n and edge set E(α, t), where {a, b} ∈ E(α, t) if tp(a, b) = t or tp(b, a) = t. Theorem (Erd˝ os-Hajnal) Suppose that 1 ≤ s < n < ω and α is an infinite ordinal. 1 χ(G(α, tn s )) = |α|.
is a disjoint type of width n, and α is an ordinal. Then G(α, t) is the graph with vertex set [α]n and edge set E(α, t), where {a, b} ∈ E(α, t) if tp(a, b) = t or tp(b, a) = t. Theorem (Erd˝ os-Hajnal) Suppose that 1 ≤ s < n < ω and α is an infinite ordinal. 1 χ(G(α, tn s )) = |α|. 2 If n ≥ 2s2 + 1, then G(α, tn s ) contains no odd cycles of length 2s + 1 or shorter.
is a graph and E = i∈I Ei . For each i ∈ I, let Gi = (Σ, Ei ). Then χ(G) ≤ i∈I χ(Gi ). Fact Suppose that G and H are graphs and there is a graph homomorphism from G to H. If is a natural number and H contains no odd cycles of length or shorter, then G also contains no odd cycles of length or shorter.
→ N, there is a graph G such that χ(G) = ℵ1 and fG (k) ≥ f (k) for all k ≥ 3. Proof: Fix f : N → N. We will actually construct G such that, for every k ∈ N and every subgraph H with at most f (k) vertices, we have χ(H) ≤ 2k+1.
→ N, there is a graph G such that χ(G) = ℵ1 and fG (k) ≥ f (k) for all k ≥ 3. Proof: Fix f : N → N. We will actually construct G such that, for every k ∈ N and every subgraph H with at most f (k) vertices, we have χ(H) ≤ 2k+1. This suffices to prove the theorem.
→ N, there is a graph G such that χ(G) = ℵ1 and fG (k) ≥ f (k) for all k ≥ 3. Proof: Fix f : N → N. We will actually construct G such that, for every k ∈ N and every subgraph H with at most f (k) vertices, we have χ(H) ≤ 2k+1. This suffices to prove the theorem. We construct a graph G = (Σ, E). Let us first describe the set of vertices.
= Mσ, fσ, Cσ such that 1 Mσ is a transitive element of H(ℵ1) such that δσ := Mσ ∩ ω1 is a limit ordinal; 2 fσ : Mσ ∩ Σ → ω; 3 Cσ is an ω-sequence of ordinals cofinal in δσ.
= Mσ, fσ, Cσ such that 1 Mσ is a transitive element of H(ℵ1) such that δσ := Mσ ∩ ω1 is a limit ordinal; 2 fσ : Mσ ∩ Σ → ω; 3 Cσ is an ω-sequence of ordinals cofinal in δσ. • The M’s provide a partial ordering of Σ: σ ≺ τ iff σ ∈ Mτ .
= Mσ, fσ, Cσ such that 1 Mσ is a transitive element of H(ℵ1) such that δσ := Mσ ∩ ω1 is a limit ordinal; 2 fσ : Mσ ∩ Σ → ω; 3 Cσ is an ω-sequence of ordinals cofinal in δσ. • The M’s provide a partial ordering of Σ: σ ≺ τ iff σ ∈ Mτ . • The f ’s will be used to anticipate potential proper colorings of G using only countably many colors in order to sabotage them.
= Mσ, fσ, Cσ such that 1 Mσ is a transitive element of H(ℵ1) such that δσ := Mσ ∩ ω1 is a limit ordinal; 2 fσ : Mσ ∩ Σ → ω; 3 Cσ is an ω-sequence of ordinals cofinal in δσ. • The M’s provide a partial ordering of Σ: σ ≺ τ iff σ ∈ Mτ . • The f ’s will be used to anticipate potential proper colorings of G using only countably many colors in order to sabotage them. • The C’s will place constraints on whether an edge can be drawn between two vertices, ensuring that the chromatic numbers of finite subgraphs do not grow too quickly.
the least natural number s ≥ 1 such that 2s + 1 ≥ f (k), and let nk = 2s2 k + 1. (Point: G(ω1, tnk sk ) has no odd cycles of length f (k) or shorter.) Partition N into adjacent intervals I0, I1, I2, . . . such that |Ik | = nk
the least natural number s ≥ 1 such that 2s + 1 ≥ f (k), and let nk = 2s2 k + 1. (Point: G(ω1, tnk sk ) has no odd cycles of length f (k) or shorter.) Partition N into adjacent intervals I0, I1, I2, . . . such that |Ik | = nk , i.e., I0 = {0, . . . , n0 − 1}, I1 = {n0, . . . , n0 + n1 − 1}, . . ..
the least natural number s ≥ 1 such that 2s + 1 ≥ f (k), and let nk = 2s2 k + 1. (Point: G(ω1, tnk sk ) has no odd cycles of length f (k) or shorter.) Partition N into adjacent intervals I0, I1, I2, . . . such that |Ik | = nk , i.e., I0 = {0, . . . , n0 − 1}, I1 = {n0, . . . , n0 + n1 − 1}, . . .. Given an ω-sequence of ordinals C and a k ∈ N, we let C[Ik ] denote {α ∈ C | |C ∩ α| ∈ Ik }.
constraints. • If {σ, τ} ∈ E, then σ ≺ τ or τ ≺ σ. It thus suffices to define N≺ G (τ) := {σ ∈ Mτ | {σ, τ} ∈ E} for each τ ∈ Σ. • For all τ ∈ Σ and k ∈ N, there is at most one σ ∈ N≺ G (τ) such that fτ (σ) = k.
constraints. • If {σ, τ} ∈ E, then σ ≺ τ or τ ≺ σ. It thus suffices to define N≺ G (τ) := {σ ∈ Mτ | {σ, τ} ∈ E} for each τ ∈ Σ. • For all τ ∈ Σ and k ∈ N, there is at most one σ ∈ N≺ G (τ) such that fτ (σ) = k. • For all τ ∈ Σ and σ ∈ N≺ G (τ), if fτ (σ) = k, then, for all j ≤ k, we have tp(Cσ[Ij ], Cτ [Ij ]) = tnj sj .
constraints. • If {σ, τ} ∈ E, then σ ≺ τ or τ ≺ σ. It thus suffices to define N≺ G (τ) := {σ ∈ Mτ | {σ, τ} ∈ E} for each τ ∈ Σ. • For all τ ∈ Σ and k ∈ N, there is at most one σ ∈ N≺ G (τ) such that fτ (σ) = k. • For all τ ∈ Σ and σ ∈ N≺ G (τ), if fτ (σ) = k, then, for all j ≤ k, we have tp(Cσ[Ij ], Cτ [Ij ]) = tnj sj . Constructing E: For each τ ∈ Σ and each k ∈ N, ask whether there is σ ∈ Mτ ∩ Σ such that fτ (σ) = k and tp(Cσ[Ij ], Cτ [Ij ]) = tnj sj for all j ≤ k.
constraints. • If {σ, τ} ∈ E, then σ ≺ τ or τ ≺ σ. It thus suffices to define N≺ G (τ) := {σ ∈ Mτ | {σ, τ} ∈ E} for each τ ∈ Σ. • For all τ ∈ Σ and k ∈ N, there is at most one σ ∈ N≺ G (τ) such that fτ (σ) = k. • For all τ ∈ Σ and σ ∈ N≺ G (τ), if fτ (σ) = k, then, for all j ≤ k, we have tp(Cσ[Ij ], Cτ [Ij ]) = tnj sj . Constructing E: For each τ ∈ Σ and each k ∈ N, ask whether there is σ ∈ Mτ ∩ Σ such that fτ (σ) = k and tp(Cσ[Ij ], Cτ [Ij ]) = tnj sj for all j ≤ k. If the answer is “yes”, then choose one such σ and put {σ, τ} in E.
constraints. • If {σ, τ} ∈ E, then σ ≺ τ or τ ≺ σ. It thus suffices to define N≺ G (τ) := {σ ∈ Mτ | {σ, τ} ∈ E} for each τ ∈ Σ. • For all τ ∈ Σ and k ∈ N, there is at most one σ ∈ N≺ G (τ) such that fτ (σ) = k. • For all τ ∈ Σ and σ ∈ N≺ G (τ), if fτ (σ) = k, then, for all j ≤ k, we have tp(Cσ[Ij ], Cτ [Ij ]) = tnj sj . Constructing E: For each τ ∈ Σ and each k ∈ N, ask whether there is σ ∈ Mτ ∩ Σ such that fτ (σ) = k and tp(Cσ[Ij ], Cτ [Ij ]) = tnj sj for all j ≤ k. If the answer is “yes”, then choose one such σ and put {σ, τ} in E. If the answer is “no”, then do nothing for this pair (τ, k).
E≥k ) denote the set of edges {σ, τ} ∈ E such that σ ≺ τ and fτ (σ) = k (resp. fτ (σ) ≥ k). Let Gk = (Σ, Ek ) and G≥k = (Σ, E≥k ) (and similarly define Hk and H≥k for subgraphs H of G).
E≥k ) denote the set of edges {σ, τ} ∈ E such that σ ≺ τ and fτ (σ) = k (resp. fτ (σ) ≥ k). Let Gk = (Σ, Ek ) and G≥k = (Σ, E≥k ) (and similarly define Hk and H≥k for subgraphs H of G). Observe: • For each τ ∈ Σ, there is at most one σ ≺ τ such that {σ, τ} ∈ Ek .
E≥k ) denote the set of edges {σ, τ} ∈ E such that σ ≺ τ and fτ (σ) = k (resp. fτ (σ) ≥ k). Let Gk = (Σ, Ek ) and G≥k = (Σ, E≥k ) (and similarly define Hk and H≥k for subgraphs H of G). Observe: • For each τ ∈ Σ, there is at most one σ ≺ τ such that {σ, τ} ∈ Ek . As a result, Gk has no cycles.
E≥k ) denote the set of edges {σ, τ} ∈ E such that σ ≺ τ and fτ (σ) = k (resp. fτ (σ) ≥ k). Let Gk = (Σ, Ek ) and G≥k = (Σ, E≥k ) (and similarly define Hk and H≥k for subgraphs H of G). Observe: • For each τ ∈ Σ, there is at most one σ ≺ τ such that {σ, τ} ∈ Ek . As a result, Gk has no cycles. • The map σ → Cσ[Ik ] induces a graph homomorphism from G≥k to G(ω1, tnk sk ).
E≥k ) denote the set of edges {σ, τ} ∈ E such that σ ≺ τ and fτ (σ) = k (resp. fτ (σ) ≥ k). Let Gk = (Σ, Ek ) and G≥k = (Σ, E≥k ) (and similarly define Hk and H≥k for subgraphs H of G). Observe: • For each τ ∈ Σ, there is at most one σ ≺ τ such that {σ, τ} ∈ Ek . As a result, Gk has no cycles. • The map σ → Cσ[Ik ] induces a graph homomorphism from G≥k to G(ω1, tnk sk ). As a result, G≥k has no odd cycles of length f (k) or shorter.
of G with at most f (k) vertices. We get an edge-decomposition of H as H = H≥k ∪ j<k Hj . • For each j < k, Hj has no cycles, so χ(Hj ) ≤ 2. • H≥k has no odd cycles of length f (k) or shorter.
of G with at most f (k) vertices. We get an edge-decomposition of H as H = H≥k ∪ j<k Hj . • For each j < k, Hj has no cycles, so χ(Hj ) ≤ 2. • H≥k has no odd cycles of length f (k) or shorter. But H≥k has at most f (k) vertices, so H≥k has no odd cycles.
of G with at most f (k) vertices. We get an edge-decomposition of H as H = H≥k ∪ j<k Hj . • For each j < k, Hj has no cycles, so χ(Hj ) ≤ 2. • H≥k has no odd cycles of length f (k) or shorter. But H≥k has at most f (k) vertices, so H≥k has no odd cycles. Thus, χ(H≥k ) ≤ 2.
of G with at most f (k) vertices. We get an edge-decomposition of H as H = H≥k ∪ j<k Hj . • For each j < k, Hj has no cycles, so χ(Hj ) ≤ 2. • H≥k has no odd cycles of length f (k) or shorter. But H≥k has at most f (k) vertices, so H≥k has no odd cycles. Thus, χ(H≥k ) ≤ 2. But then χ(H) ≤ χ(H≥k ) · j<k χ(Hj ) ≤ 2k+1, as desired.
χ(G) ≤ ℵ1 is easy. Thus, suppose that f : Σ → N is given. We show that f is not a proper coloring. Let Ni | i ≤ ω be a continuous ∈-increasing chain of countable elementary submodels of (H(c+), ∈, , G, f ).
χ(G) ≤ ℵ1 is easy. Thus, suppose that f : Σ → N is given. We show that f is not a proper coloring. Let Ni | i ≤ ω be a continuous ∈-increasing chain of countable elementary submodels of (H(c+), ∈, , G, f ). For i ≤ ω, let Mi = Ni ∩ H(ℵ1) and δi = Mi ∩ ω1.
χ(G) ≤ ℵ1 is easy. Thus, suppose that f : Σ → N is given. We show that f is not a proper coloring. Let Ni | i ≤ ω be a continuous ∈-increasing chain of countable elementary submodels of (H(c+), ∈, , G, f ). For i ≤ ω, let Mi = Ni ∩ H(ℵ1) and δi = Mi ∩ ω1. Let C = {δi | i < ω} and τ = Mω, f (Mω ∩ Σ), C .
χ(G) ≤ ℵ1 is easy. Thus, suppose that f : Σ → N is given. We show that f is not a proper coloring. Let Ni | i ≤ ω be a continuous ∈-increasing chain of countable elementary submodels of (H(c+), ∈, , G, f ). For i ≤ ω, let Mi = Ni ∩ H(ℵ1) and δi = Mi ∩ ω1. Let C = {δi | i < ω} and τ = Mω, f (Mω ∩ Σ), C . Then τ ∈ Σ. Let k = f (τ).
disjoint type t of width n, there is σ ∈ Mω such that f (σ) = k and tp(Cσ[n], C[n]) = t. Proof sketch: For concreteness, we prove the claim for t = t3 1 = 001011.
disjoint type t of width n, there is σ ∈ Mω such that f (σ) = k and tp(Cσ[n], C[n]) = t. Proof sketch: For concreteness, we prove the claim for t = t3 1 = 001011. Let ∃∞γ . . . abbreviate “there are unboundedly many γ < ω1 . . .”.
disjoint type t of width n, there is σ ∈ Mω such that f (σ) = k and tp(Cσ[n], C[n]) = t. Proof sketch: For concreteness, we prove the claim for t = t3 1 = 001011. Let ∃∞γ . . . abbreviate “there are unboundedly many γ < ω1 . . .”. For every β < δ2, the following sentence holds in H(c+): ∃γ2 > β ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, δ1, γ2}), as witnessed by δ2 and τ.
disjoint type t of width n, there is σ ∈ Mω such that f (σ) = k and tp(Cσ[n], C[n]) = t. Proof sketch: For concreteness, we prove the claim for t = t3 1 = 001011. Let ∃∞γ . . . abbreviate “there are unboundedly many γ < ω1 . . .”. For every β < δ2, the following sentence holds in H(c+): ∃γ2 > β ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, δ1, γ2}), as witnessed by δ2 and τ. By elementarity, the following holds in N2: ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, δ1, γ2}),
disjoint type t of width n, there is σ ∈ Mω such that f (σ) = k and tp(Cσ[n], C[n]) = t. Proof sketch: For concreteness, we prove the claim for t = t3 1 = 001011. Let ∃∞γ . . . abbreviate “there are unboundedly many γ < ω1 . . .”. For every β < δ2, the following sentence holds in H(c+): ∃γ2 > β ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, δ1, γ2}), as witnessed by δ2 and τ. By elementarity, the following holds in N2: ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, δ1, γ2}), and hence also in H(c+).
δ1, the following sentence holds in H(c+): ∃γ1 > β ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, γ1, γ2}), as witnessed by δ1. By elementarity, the following holds in N1, and hence in H(c+): ∃∞γ1 ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, γ1, γ2}).
δ1, the following sentence holds in H(c+): ∃γ1 > β ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, γ1, γ2}), as witnessed by δ1. By elementarity, the following holds in N1, and hence in H(c+): ∃∞γ1 ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {δ0, γ1, γ2}). Similarly, in N0 and H(c+) we get: ∃∞γ0 ∃∞γ1 ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {γ0, γ1, γ2}).
(σ) = k ∧ Cσ[3] = {γ0, γ1, γ2}). and t = t3 1 = 001011. Working inside N0, choose γ∗ 0 < γ∗ 1 < δ0 such that ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {γ∗ 0 , γ∗ 1 , γ2}) holds (in N0 and hence also in N1).
(σ) = k ∧ Cσ[3] = {γ0, γ1, γ2}). and t = t3 1 = 001011. Working inside N0, choose γ∗ 0 < γ∗ 1 < δ0 such that ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {γ∗ 0 , γ∗ 1 , γ2}) holds (in N0 and hence also in N1). Next, working in N1, choose γ∗ 2 and σ such that δ0 < γ∗ 2 < δ1, f (σ) = k, and Cσ[3] = {γ∗ 0 , γ∗ 1 , γ∗ 2 }.
(σ) = k ∧ Cσ[3] = {γ0, γ1, γ2}). and t = t3 1 = 001011. Working inside N0, choose γ∗ 0 < γ∗ 1 < δ0 such that ∃∞γ2 ∃σ (f (σ) = k ∧ Cσ[3] = {γ∗ 0 , γ∗ 1 , γ2}) holds (in N0 and hence also in N1). Next, working in N1, choose γ∗ 2 and σ such that δ0 < γ∗ 2 < δ1, f (σ) = k, and Cσ[3] = {γ∗ 0 , γ∗ 1 , γ∗ 2 }. Then σ ∈ Mω, f (σ) = k, and tp(Cσ[n], C[n]) = t, as required. Claim
= tn0 s0 tn1 s1 . . . tnk sk , there is σ ∈ Mω such that f (σ) = k and tp(Cσ[Ij ], C[Ij ]) = tnj sj for all j ≤ k. Thus, when we were building G, we chose such a σ and added {σ, τ} to E. But then we have f (σ) = k = f (τ), so f is not a proper coloring. Theorem
our main theorem. Question Is it true that for every f : N → N, there is a graph G such that |G| = χ(G) = ℵ1 and fG grows faster than f ? Question Is it true that for ever f : N → N and every cardinal κ, there is a graph G such that χ(G) ≥ κ and fG grows faster than f ?
our main theorem. Question Is it true that for every f : N → N, there is a graph G such that |G| = χ(G) = ℵ1 and fG grows faster than f ? Question Is it true that for ever f : N → N and every cardinal κ, there is a graph G such that χ(G) ≥ κ and fG grows faster than f ? What about just κ = ℵ2?
os and Hajnal about finite subgraphs of uncountably chromatic graphs. Question Suppose that G and H are uncountably chromatic graphs. Must they have a common 4-chromatic subgraph?
os and Hajnal about finite subgraphs of uncountably chromatic graphs. Question Suppose that G and H are uncountably chromatic graphs. Must they have a common 4-chromatic subgraph? Question Is there an uncountably chromatic graph G such that every triangle-free subgraph of G has countable chromatic number?
os and Hajnal about finite subgraphs of uncountably chromatic graphs. Question Suppose that G and H are uncountably chromatic graphs. Must they have a common 4-chromatic subgraph? Question Is there an uncountably chromatic graph G such that every triangle-free subgraph of G has countable chromatic number? Question Is there a K4-free graph that cannot be written as the union of countably many triangle-free graphs?
A tribute to Paul Erd˝ os, 467–478, Cambridge Univ. Press, Cambridge, 1990. Lambie-Hanson, Chris. On the growth rate of chromatic numbers of finite subgraphs. Adv. Math. To appear, 2020. https://arxiv.org/pdf/1902.08177.pdf
A tribute to Paul Erd˝ os, 467–478, Cambridge Univ. Press, Cambridge, 1990. Lambie-Hanson, Chris. On the growth rate of chromatic numbers of finite subgraphs. Adv. Math. To appear, 2020. https://arxiv.org/pdf/1902.08177.pdf All artwork on these slides by Joan Mir´ o.
clambiehanson
fufufukakaka
ontheslope
.png){kind=avatar}
satai
mela_dayo
skoyamalab
analokmaus
trafficbrain
mickey_kubo
sansantech
1manken
nanoshimarobot
bkeepers
kimpetersen
reverentgeek
cassininazir
ipullrank
marketingsoph
axbom
pwiseman
honnibal
bryan
frankvandijk
kastner
