must lie on ∂U, because f is a homeomorphism between D and U and thus a sequence converging in U has to be the image under f of a sequence converging in D. By assumption there exist a homeomorphism β between the circle ∂D and ∂U. Since β−1 is uniformly continuous, the distance between the two points ξn and ηn corresponding to an and bn in ∂U must tend to 0. So eventually the smallest circular arc in ∂D joining ξn and ηn is defined. Denote τn image of this arc under β. By uniform continuity of β, diameter of τn in ∂U tends to 0. Together τn and f ∘ γrn form a simple Jordan curve. Its interior Un is contained in U by the Jordan curve theorem for ∂U and ∂Un: to see this, notice that U is the interior of ∂U, as it is bounded, connected and it is both open and closed in the complement of ∂U; so the exterior region of ∂U is unbounded, connected and does not intersect ∂Un, hence its closure is contained in the closure of the exterior of ∂Un; taking complements, we get the desired inclusion. The diameter of ∂Un tends to 0 because the diameters of τn and f ∘ γrn tend to 0. Hence the diameter of Un tend to 0. (For is compact set, hence contains two points u and v such that distance between them is maximal. It is easy to see that u and v must lie in ∂U and diameters of both U and ∂U equal .) Now if Vn denotes the intersection of D with the disk |z − ζ| < rn, then for all sufficiently large n f(Vn) = Un. Indeed, the arc γrn divides D into Vn and complementary region , so under the conformal homeomorphism f the curve f ∘ γrn divides U into and a complementary region ; Un is a connected component of U \ f ∘ γrn, as it is connected and is both open and closed in this set, hence equals either or . Diameter of does not decrease with increasing n, for implies . Since diameter of Un tends to 0 as n goes to infinity, it is eventually less than the diameter of and then necessarily f(Vn) = Un. So the diameter of f(Vn) tends to 0. On the other hand, passing to subsequences of