00QB Lemma 3.6. 𝒜λ∞=∂Δλ∨∪⋃dimσ≥1𝒜λ,σ∞\mathcal{A}_{\lambda}^{\infty}=\partial\Delta_{\lambda}^{\vee}\cup\bigcup_{\dim\sigma\geq 1}\mathcal{A}_{\lambda,\sigma}^{\infty}.
00QC Proof. Let x∈𝒜λ∞x\in\mathcal{A}_{\lambda}^{\infty}. If m=0∈Δm=0\in\Delta achieves the maximum Lλ(x)L_{\lambda}(x), then x∈∂Δλ∨x\in\partial\Delta_{\lambda}^{\vee}. If not, then the maximum is achieved by at least two m∈∂Δm\in\partial\Delta, so x∈𝒜λ,σ∞x\in\mathcal{A}_{\lambda,\sigma}^{\infty} for some σ⊂∂Δ\sigma\subset\partial\Delta with dimσ≥1\dim\sigma\geq 1. ∎