## 7 Proofs of lemmas in Section 2
Since no edge of \(V_{p}\) intersects \(\sigma\) and some edge of \(V_{q}\) intersects \(\sigma\), there exists two consecutive vertices \(p_{i}\) and \(p_{i+1}\) in this sequence so that no edge of \(V_{p_{i}}\) intersects \(\sigma\) whereas some edge of \(V_{p_{i+1}}\) does intersect \(\sigma\). By Lemma 2 we can claim that \(\sigma_{p_{i+1}}\) is a disk. A boundary cycle of \(\sigma_{p_{i}}\) overlaps with the boundary of \(\sigma_{p_{i+1}}\). This is impossible as the curves on the boundary of \(\sigma_{p_{i+1}}\) intersect Voronoi edges whereas those on the boundary of \(\sigma_{p_{i}}\) do not. Now we are ready to prove the main claim of Lemma 2. Figure 7 explains more about its implications. **Lemma 14**: _There is a \(\lambda>0\) so that if H1 and H2 of Lemma 2 are satisfied, then \(\sigma_{p}\) is a topological disk. Furthermore, each Voronoi edge and facet that intersects \(\sigma_{p}\) does so in a single point and single open curve respectively._  
Recall \(B=B(p,2\lambda)\). Consider an edge of \(V_{p}\) that intersects \(B\cap\sigma\). By Lemma 10(ii), \(\angle\mathbf{n}_{p},e\leqslant 5\sqrt{2(\lambda/\rho_{\sigma})^{2}}\). Then, by Lemma 10(i.a), \(B\cap\sigma\) is monotonic in the direction of \(e\) when \(\lambda\) is sufficiently small. Therefore, \(e\) intersects \(B\cap\sigma\) and hence \(\sigma_{p}\) at most once. Each facet of \(V_{p}\) that intersects \(B\cap\sigma\) intersects it in a single open curve by Lemma 12. This completes the proof of the second part of the lemma. We show that the argument of Lemma 4.12 in [4] can be used here to establish that \(\sigma_{p}\) is a topological disk. Following the notation in the proof of this lemma, let us call a cycle in \(\sigma_{p}\) that intersects a Voronoi edge a type 1 cycle. Let \(X\) denote the set of type 1 cycles. Each Voronoi facet containing a point in \(X\) has a Voronoi edge intersecting \(\sigma_{p}\). By H1, this intersection point is within \(\lambda\) distance of \(p\). Therefore, \(B\) intersects all Voronoi facets that contain a point in \(X\). These Voronoi facets intersect \(B\cap\sigma\) in a single open curve which is in \(B\). Also, if \(V_{p}\) intersects \(\operatorname{bd}\sigma\), the entire curve segment \(V_{p}\cap\operatorname{bd}\sigma\) is contained in the protecting ball of \(p\). This means each point of \(X\) is within \(\lambda\) distance of \(p\). Since each cycle in \(X\) is a piecewise smooth closed curve where each piece is either an open curve contained in a Voronoi facet or a segment of \(V_{p}\cap\operatorname{bd}\sigma\), it is contained completely in \(B\). Therefore, each boundary of \(\sigma_{p}\) which intersects a Voronoi edge lies in \(B\). Consider the arrangement of closed curves in \(X\) on \(B\cap\sigma\) which is a topological disk according to the Feature Ball Lemma [21]. Let \(C\in X\) be a closed curve which does not contain any other curve of \(X\) inside. At least one such curve exists since \(X\) is not empty by Lemma 13. The closed curve \(C\) must bound a disk, say \(D\) in \(\sigma\). At this point we can invoke the proof of Lemma 4.12 in [4] to prove that \(D=\sigma_{p}\) establishing our first claim. The boundary \(\operatorname{bd}\sigma_{p}\) lies in \(X\) which has at most a single interval in any Voronoi facet as we argued above. This establishes our second claim. [Proof of Lemma 2.] Proof of R1: Follows from Lemma 14. Figure 7: (left) A 2-face \(\sigma\) has intersected \(V_{p}\) where \(\sigma_{p}\) is not topological disk, this is not possible by R1. (middle) \(\sigma_{p}\) is a disk but another face \(\tau\) where \(p\
ot\in\tau\) intersects some edge of \(V_{p}\). It does not violate R1 but violates Lemma 8. (right) Within \(B\), \(\sigma\) intersects \(V_{p}\) in a topological disk. There is a different component (\(\tau\)) which does not intersect any Voronoi edge and hence does not contribute any dual restricted triangle incident to \(p\). R1 does not prevent it though Lemma 8 does. Proof of R2: Since conditions C1-C3 hold, Lemma 1 can be applied.