Erdős Problem 595 #
References:
- erdosproblems.com/595
- [Er87] Erdős, Paul, Problems and results on set systems and hypergraphs. Extremal problems for finite sets (Visegrád, 1991), Bolyai Soc. Math. Stud. (1994), 217-227.
- [Fo70] Folkman, Jon, Graphs with monochromatic complete subgraphs in every edge coloring. SIAM J. Appl. Math. (1970), 19:340-345.
- [NeRo75] Nešetřil, Jaroslav and Rödl, Vojtěch, Type theory of partition problems of graphs. Recent advances in graph theory (Proc. Second Czechoslovak Sympos., Prague, 1974), Academia, Prague (1975), 405-412.
Equations
- Erdos595.IsCountableUnionOfTriangleFree G = ∃ (H : ℕ → SimpleGraph V), (∀ (i : ℕ), (H i).CliqueFree 3) ∧ G = ⨆ (i : ℕ), H i
Instances For
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].
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:
G.CliqueFree 4(no $K_4$), and- For every family
H : Fin n → SimpleGraph Vof 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 ℕ 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.
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 ℕ 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.