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20 0. Prequel Proof. Identify consecutive vertices v1,v2,v3 of P , and form the triangle ∆ determined by them. Impose coordinates on R2 with v1,v3 on the x-axis and v2 below it. Let K denote the complement in P of the two segments v1v2 and v2v3. If no point of K meets ∆, then the segment determined by v = v1 and w = v3 obviously has the desired property. Otherwise, choose w to have least y-coordinate among the finitely many vertices of K {v2} touching the 2-cell bounded by ∆ (see Figure 0.4) in this case the segment formed by v = v2 and w works.2 x v v v w y 1 2 3 K Figure 0.4. The interior of segment v2w misses P Theorem 0.11.4. Every polygonal simple closed curve P in R2 bounds a PL 2-cell. Proof. By induction on the number n of vertices in P , with the initial case n = 3 being trivial. Inductively assume the result for all polygons of fewer than n vertices, and consider a polygon P having n vertices. Apply Lemma 0.11.3 to form two polygonal simple closed curves J1,J2, each having fewer than n vertices, with J1 ∪J2 = P ∪vw and J1 ∩J2 = vw by induction, each Ji bounds a PL 2-cell Di (i = 1, 2). If D1 ∩ D2 = vw, then D1 ∪ D2 is a disk and ∂(D1 ∪ D2) = P . Otherwise, one of the Ji misses the interior of the other 2-cell Dj, for if J1 ∩ Int D2 = ∅, then J2 ∩ Int D1 = ∅, as D1 ⊂ D2 (see Figure 0.5). Let us say J1 ∩ Int D2 = ∅ for definiteness. Let C denote the closure of the bounded component of R2 P . It is left to the reader to check that there is an elementary shelling (see Rourke and Sanderson (1972), p. 40) of D1 to C across D2. Hence, C ∼ D1 is a 2-cell. Proof of Theorem 0.11.2. The theorem is an immediate corollary of The- orem 0.11.4 and the Disc Theorem of Rourke and Sanderson (1972, Theo- rem 3.34). 2 J. W. Cannon deserves credit for this elegant argument.