Reduction of Erdős Problem 111 to countable colorings of linear graphs #
- Countable deletion. If removing a countable set leaves a countably colorable graph, the graph is countably colorable: give the removed vertices fresh colors.
- Packing. Uncountably many pairwise disjoint finite witnesses with cost above
2 Kper vertex include infinitely many with the same sizevand costb, since these are natural numbers. Then⌊n / v⌋of them, padded tonvertices, cost at leastb ⌊n / v⌋ > K nforn ≥ 2 v. - The limit. If
Gis not countably colorable, fixKand take a maximal family of pairwise disjoint witnesses. If it were countable, the remainder would be2 K-linear by maximality, hence countably colorable, and so wouldG. So it is uncountable, and packing givesh G n > K nfor all largen.
theorem
Erdos111.nonempty_coloring_of_countable_compl
{V : Type u}
{G : SimpleGraph V}
{S : Set V}
(hS : S.Countable)
(h : Nonempty ((SimpleGraph.induce Sᶜ G).Coloring ℕ))
:
Countable deletion.
theorem
Erdos111.eventually_lt_h_of_uncountable
{V : Type u}
{G : SimpleGraph V}
{ι : Type u_1}
[Uncountable ι]
(K : ℝ)
(s : ι → Finset V)
(E : ι → Finset (Sym2 V))
(hdisj : Pairwise fun (i j : ι) => Disjoint (s i) (s j))
(hE : ∀ (i : ι), ↑(E i) ⊆ G.edgeSet)
(hEs : ∀ (i : ι), ∀ e ∈ E i, ∀ v ∈ e, v ∈ s i)
(hcost : ∀ (i : ι), 2 * K * ↑(s i).card < ↑(edgeCost (E i)))
:
Packing.
theorem
Erdos111.tendsto_of_isEmpty_coloring
{V : Type u}
(hmain : ∀ (W : Type u) (H : SimpleGraph W) (C : NNReal), Linear H C → Nonempty (H.Coloring ℕ))
(G : SimpleGraph V)
(hG : IsEmpty (G.Coloring ℕ))
:
Filter.Tendsto (fun (n : ℕ) => ↑(h G n) / ↑n) Filter.atTop Filter.atTop
If every linear graph is countably colorable, then h G n / n → ∞ for every graph G
that is not countably colorable.
theorem
Erdos111.isEmpty_coloring_of_chromaticCardinal
{V : Type u}
{G : SimpleGraph V}
(hG : chromaticCardinal G = Cardinal.aleph 1)
:
A graph of chromatic number ℵ₁ is not countably colorable.