... =k +1 kt − 1 kt ∈ [ 1 k +1 , 1] 0 t ∈ [0, 1 k +1 ].g0(·, t) t ∈ T bk→ b0. p0 := (f0, g0, b0), pk := (fk, gk, bk), fk := f0, gk := g0k ≥ 1. pk→ p0. ˆx := ( 1 4, 1 4) ... t) = 1 2t − 1 4< t = b0(t) ∀t ∈ T.C(p0).C p0.C(p0) = Ω, C(pk) ={(x 1 , x2) |0 ≤ x 1 ≤ 1 k + 1 , 0 ≤ x2≤ kx 1 }∪{(x 1 , x2) |x 1 ≥ 1 k + 1 , 0 ≤ x2≤ 1 − x 1 } ∀k ≥ 1, S(p0) ... ≤x2≤ 1} , S(pk) = {(0, 0)} ∀k ≥ 1. x0 := (0, 1 2) ∈ S(p0) V (x0) := B 1 4(x0) ∩ Ω B 1 4(x0) := {x ∈ R2|||x −x0||2< 1 4}S p0.g(x, t) := (0, , 0) ∈ Rmb(t) := (1, , 1) ...