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FormalConjectures.ErdosProblems.«595»

Erdős Problem 595 #

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    Erdős Problem 595 ($250): Is there an infinite graph $G$ which contains no $K_4$ and is not the union of countably many triangle-free graphs?

    A problem of Erdős and Hajnal [Er87].

    theorem Erdos595.erdos_595.variants.folkman_finite :
    True ∀ (n : ), 1 n∃ (V : Type u_1) (x : Fintype V) (G : SimpleGraph V), G.CliqueFree 4 ∀ (H : Fin nSimpleGraph V), (∀ (i : Fin n), (H i).CliqueFree 3)G ⨆ (i : Fin n), H i

    Folkman–Nešetřil–Rödl (finite version) [Fo70, NeRo75]: For every n ≥ 1, there exists a graph G (on a finite vertex set) that contains no $K_4$ and whose edges cannot be covered by n triangle-free graphs.

    More precisely: for every n : ℕ with 1 ≤ n, there exist a finite type V and a graph G : SimpleGraph V with:

    1. G.CliqueFree 4 (no $K_4$), and
    2. For every family H : Fin n → SimpleGraph V of triangle-free graphs, G ≠ ⨆ i, H i.

    This is the finite analogue of Problem 595. The proofs of Folkman [Fo70] and Nešetřil–Rödl [NeRo75] give different explicit constructions.

    Monotonicity: If G is a countable union of triangle-free graphs and H ≤ G (i.e., H is a subgraph of G), then H is also a countable union of triangle-free graphs.

    Proof: If G = ⨆ i, G_i with each G_i triangle-free, then H = ⨆ i, H ⊓ G_i. Each H ⊓ G_i is triangle-free because it is a subgraph of G_i.

    Triangle-free graphs are trivially countable unions of triangle-free graphs: if G is already triangle-free, then G = ⨆ i : ℕ, G_i where G_0 = G and G_i = ⊥ for i ≥ 1.

    The complete graph on is a countable union of triangle-free graphs: we decompose it into the family of star graphs {H_m}_{m : ℕ}, where H_m is the graph with edges {m, n} for all n ≠ m. Each star is triangle-free (any two non-center vertices share no edge within the star), and their union covers all edges of .

    Proof sketch (star triangle-free): If {a, b, c} were a triangle in H_m, then each of the three edges {a, b}, {a, c}, {b, c} would pass through m. In particular, from {a, b} we get a = m or b = m; from {b, c} we get b = m or c = m. Case analysis shows that two vertices must equal m, contradicting the triangle having three distinct vertices.

    The complete graph on Fin 4 is not $K_4$-free: on Fin 4 equals the complete graph $K_4$, so it contains $K_4$ as a subgraph and is not $K_4$-free.

    This sanity check confirms the $K_4$-free hypothesis of Problem 595 is non-trivial.

    Reformulation via edge colourings: A graph G is a countable union of triangle-free graphs if and only if there is a colouring of the edges of G by such that no monochromatic triangle exists.

    More precisely: IsCountableUnionOfTriangleFree G is equivalent to the existence of a map c : G.edgeSet → ℕ such that for each n : ℕ, the subgraph of edges coloured n is triangle-free.