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FormalConjectures.ErdosProblems.«594»

Erdős Problem 594 #

References:

theorem Erdos594.erdos_594 :
True ∀ (V : Type) (G : SimpleGraph V), IsEmpty (G.Coloring )∃ (N : ), ∀ (k : ), N k∃ (v : V) (w : G.Walk v v), w.IsCycle w.length = 2 * k + 1

Erdős Problem 594 (Erdős–Hajnal [ErHa66], [Er69b]):

Does every graph $G$ with chromatic number $\geq \aleph_1$ contain all sufficiently large odd cycles?

The answer is Yes, proved by Erdős, Hajnal, and Shelah [EHS74].

A graph has chromatic number $\geq \aleph_1$ (i.e. uncountable chromatic number) if and only if it admits no proper colouring with countably many colours; this is encoded as IsEmpty (G.Coloring ℕ). The conclusion states that there is some $N$ such that for every $k \geq N$ the graph contains a cycle of odd length $2k + 1$.

theorem Erdos594.erdos_594.variants.erdos_hajnal (V : Type) (G : SimpleGraph V) (h : ∀ (α : Type), Cardinal.mk α Cardinal.aleph 1IsEmpty (G.Coloring α)) :
∃ (N : ), ∀ (k : ), N k∃ (v : V) (w : G.Walk v v), w.IsCycle w.length = 2 * k + 1

The earlier result of Erdős and Hajnal [ErHa66]: every graph with chromatic number $\geq \aleph_2$ contains all sufficiently large odd cycles.

Chromatic number $\geq \aleph_2$ is encoded as the nonexistence of a proper colouring with any set of at most $\aleph_1$ colours.

Sanity check for the encoding of "chromatic number $\geq \aleph_1$": a countably colourable graph, such as the empty graph on $\mathbb{N}$, does not satisfy the hypothesis IsEmpty (G.Coloring ℕ).