Theorem 4.4.2504CCB65ED650

Proof. An interior point of a segment is regular for a disk slit along the segment by the explicit exterior-segment barrier. The boundary nonthinness theorem therefore makes the segment nonthin at that point. If it were thin at an endpoint, reflect an amplified barrier across the endpoint and add the two barriers. The doubled segment would then be thin at its interior point, a contradiction. The radial projection $y\mapsto a+|y-a|e_1$ is a contraction preserving distances from $a$. If every sufficiently small circle met a thin complement, its projection would contain a nondegenerate segment ending at $a$, contradicting contraction invariance of thinness. Hence the required circles exist. An exterior continuum containing $a$ and another point $b$ meets every circle of radius less than $|a-b|$, by connectedness of its distance image. This contradicts the circle conclusion if the complement were thin. Nonthinness gives regularity. For the final assertion use the standard planar topological characterization of a simply connected domain: its complement in the sphere is connected. For a bounded domain this complement is a nondegenerate compact continuum containing infinity; its radial distance image at each finite boundary point meets every sufficiently small circle. The same argument applies directly on the sphere, without assuming a finite exterior arc. This planar topological characterization is an explicit topological prerequisite, not a new potential-theoretic premise. $\square$