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10. χ(G_D) countable?

problem

Let D⊆(0,∞)D⊆(0,∞) and define a graph GDG_D with vertex set Rℝ and edge set {{x,y}:∣x−y∣∈D}\{\{x,y\}:\lvert x-y\rvert∈D\}. Show that χB(GD)≤ℵ0χ_B(G_D)≤ℵ₀ if and only if 00 is not in the closure of DD.

  • [⟸] Suppose 0∉Dˉ0∉\bar D. We wish to show that χB(GD)≤ℵ0χ_B(G_D)≤ℵ₀.

    proof

    0∉Dˉ0∉\bar D means that ϵ≔inf⁡D>0ϵ≔\inf D>0. Then define the coloring c ⁣:R→Zc\colon ℝ→ℤ by c(x)=⌊x/ϵ⌋c(x)=⌊x/ϵ⌋.

    Suppose x,yx,y are joined by an edge in GDG_D; that is, ∣x−y∣∈D\lvert x-y\rvert∈D. Then ∣x−y∣>ϵ\lvert x-y\rvert>ϵ, so x/ϵx/ϵ and y/ϵy/ϵ are at least distance 11 apart, so c(x)≠c(y)c(x)≠c(y).

  • [⟸] Suppose χB(GD)≤ℵ0χ_B(G_D)≤ℵ₀. We wish to show that 0∉Dˉ0∉\bar D.

    lemma

    If 0∈Dˉ0∈\bar D, then in any Borel proper coloring of GDG_D the set of points assigned any given color is meager.

    proof

    Fix SS any set of points with the same coloring. Suppose SS is not meager. Since it is Borel and therefore trivially Baire-measurable, so by the Baire alternative there exists an open interval U=(x−ϵ,x+ϵ)U=(x-ϵ,x+ϵ) so that U∖SU∖S is meager. By the assumption that 0∈Dˉ0∈\bar D, there exists δ∈Dδ∈D satisfying 0<δ<ϵ0<δ<ϵ.

    Consider S′=S∩(S+δ)S'=S∩(S+δ), and observe that

    (x,x+ϵ)∖S′=[(x,x+ϵ)∖S]⏟⊆U∖S∪[(x,x+ϵ)∖(S+δ)]⏟⊆(U+δ)∖(S+δ). (x,x+ϵ) ∖ S' = \underbrace{[(x,x+ϵ) ∖ S]}_{⊆U∖S} ∪ \underbrace{[(x,x+ϵ) ∖ (S+δ)]}_{⊆(U+δ)∖(S+δ)}.

    Both sets in the union above are contained in translates of U∖SU∖S, so (x,x+ϵ)∖S′(x,x+ϵ)∖S' is meager. Then S′S' is comeager in (x,x+ϵ)(x,x+ϵ) and therefore nonempty. That is, there exists x∈S′=S∩(S+δ)x∈S'=S∩(S+δ). Then x,x−δ∈Sx,x-δ∈S, so they have the same color.

    At the same time δ∈Dδ∈D implies {x,x−δ}\{x,x-δ\} is an edge in GDG_D, so xx and x−dx-d must have different colors, a contradiction.

    claim

    0∉Dˉ0∉\bar D.

    proof

    Fix a countable Borel proper coloring c ⁣:R→Nc\colon ℝ→ℕ. Suppose towards a contradiction that 0∈Dˉ0∈\bar D. Then by the lemma, for each n∈Nn∈ℕ the set Sn≔c−1({n})S_n≔c⁻¹(\{n\}) of points with color nn is meager.

    Then

    R=⋃n∈Nc−1({n}) ℝ = ⋃_{n∈ℕ} c⁻¹(\{n\})

    is a countable union of meager sets and therefore also meager, a contradiction.