    # Disks in Curves of Bounded Convex Curvature

We say that a simple, closed curve γ in the plane has bounded convex curvature if for every point x on γ, there is an open unit disk U_x and ε_x>0 such that x∈∂ U_x and B_ε_x(x)∩ U_x⊂Int γ. We prove that the interior of every curve of bounded convex curvature contains an open unit disk.

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## 1 Introduction

Consider a Jordan curve , that is, a simple, closed curve in the plane. We will denote by and , respectively, the interior and exterior of . We say that has bounded convex curvature if for every point on , there is an open unit disk and such that

 x∈∂UxandBεx(x)∩Ux⊂Intγ. (1)

Here is the open disk with center and radius . Similarly, we say that has bounded concave curvature if for every point on , there is an open unit disk and such that

 x∈∂VxandBεx(x)∩Vx⊂Extγ. (2)

Finally we say that a curve has bounded curvature if it has both bounded convex and concave curvature. Curves of bounded convex curvature are the focus of this article. When we say that is a curve of bounded convex curvature it will always be understood that is a Jordan curve. Figure 1 shows an example of a curve of bounded convex curvature. Note that there may be points on a curve of bounded convex (or concave) curvature where the tangent to the curve is not defined. Our main goal is to prove the following theorem (generalizing a theorem by Pestov and Ionin  that we shall discuss later):

###### Theorem 1.

The interior of any curve of bounded convex curvature contains an open unit disk.

The theorem does not hold if we replace the word “convex” with “concave” — any circle of radius smaller than 1 provides a counterexample.

An appealing property of curves of bounded convex curvature is that they can be composed as described in the following observation (also see Figure 2). Figure 2: Illustration for Observation 2. The fat curve is the composition γ3 of γ1 (black) and γ2 (gray).
###### Observation 2.

Let and be two curves of bounded convex curvature. Consider the unbounded connected component of . If the boundary is a Jordan curve , then has bounded convex curvature.

Note that this result does not hold for curves of bounded curvature. Indeed the Jordan curves and in Figure 2 both have bounded curvature, whereas their composition only has bounded convex curvature.

In Section 2 we will explain how curves of bounded convex curvature naturally arise in problems related to computer-aided manufacturing, but first we discuss related work.

### 1.1 Related work

All previously studied notions of bounded curvature are more restrictive, and moreover defined in terms of a parameterization of the curve, contrary to our notion of bounded convex curvature. The curvature is often defined for curves that are two times continuously differentiable and parameterized by arclength. Then the (unsigned) curvature at is simply , and a curve is defined to have bounded curvature if for all . We say that such curves have strongly bounded curvature in order to avoid confusion with the curves of bounded curvature introduced in this article. Pestov and Ionin  proved that the interior of every curve of strongly bounded curvature contains an open unit disk. We denote this theorem as the Pestov–Ionin theorem.

The Pestov–Ionin theorem has often been applied to problems in robot motion planning and related fields [1, 2, 3, 4, 18]. In Section 2, we describe how curves of bounded convex curvature naturally arise in problems related to pocket machining.

Dubins  introduced the class of curves of bounded average curvature as the curves parameterized by arclength that are differentiable such that for all , we have

 ∥γ′(s1)−γ′(s2)∥≤|s1−s2|. (3)

For a curve of bounded average curvature, the second derivative is not necessarily defined everywhere, but since satisfies the Lipschitz condition (3), it follows that is defined almost everywhere. Dubins mentioned that if is a curve parameterized by arclength for which exists everywhere, then has bounded average curvature if and only if has strongly bounded curvature. Ahn et al.  proved that the Pestov–Ionin theorem holds for curves of bounded average curvature, and their proof is analogous to that of Pestov and Ionin. In particular, both proofs rely on the curve being rectifiable, i.e., having finite length. However, it is not at all clear from our more general definition that a curve of bounded convex curvature is rectifiable, so that approach cannot easily be applied in our case. Instead, our proof shows that if contains no unit disk, then there exists an such that contains infinitely many pairwise disjoint disks of radius . As is bounded, this is of course a contradiction.

Pankrashkin  gave a proof that the interior of a smooth Jordan curve of strongly bounded curvature has area at least . This of course follows from the Pestov–Ionin theorem, but Pankrashkin proved it by other means.

Note that the requirement on the curvature of curves of strongly bounded and bounded average curvature is completely symmetric with respect to the curve turning to the left and right when traversed in positive direction. In contrast to that, Howard and Treibergs  introduced a class of curves satisfying an asymmetric condition on the curvature, namely the curves parameterized by arclength such that is absolutely continuous and

 ⟨γ′(s+h)−γ′(s),n(s)⟩≤h

for all and , where is the dot-product and is the unit normal. They proved the Pestov–Ionin theorem for the Jordan curves in . Abrahamsen and Thorup  introduced a class of Jordan curves related to , but where the curves may have sharp concave corners without a well-defined tangent. They gave a proof of a version of the Pestov–Ionin theorem for that class of curves.

It can be shown that each of the classes of Jordan curves mentioned here are subsets of the curves of bounded convex curvature. It is therefore natural to investigate whether the Pestov–Ionin theorem holds for all curves of bounded convex curvature, which is exactly the statement of Theorem 1.

## 2 Application to Pocket Machining Figure 3: Left: A milling machine. The model is the Rabbit Mill v3.0 from SourceRabbit, who kindly provided permission to use the picture. © SourceRabbit. Right: A milling tool. Picture by Rocketmagnet, licensed under CC BY-SA 3.0. Figure 4: In each of these four situations, the thick black curve is the boundary ∂S of the pocket. The remaining material in the pocket is ensured to be between the dashed black curve and ∂S. The boundary of the tool D is the dashed circle, and the solid part of the circle between the two crosses is the maximum part that can be in engagement with the material, i.e., the largest possible portion of the tool boundary cutting away material. In the third picture, the convex corner on the path in the second picture has been rounded by an arc, thus bounding the convex curvature and reducing the maximum engagement. The two rightmost pictures show two ways of going around a concave corner of ∂S. In both cases, the maximum engagement is smaller than when the tool follows a line segment of ∂S (the case of the first picture).

In this section we explain why it is sometimes natural to restrict oneself to curves of bounded convex curvature when choosing toolpaths for pocket machining. Pocket machining is the process of cutting out a pocket of some specified shape in a piece of material, such as a block of metal or wood, using a milling machine; see Figure 3 (left).

We are given a compact region of the plane whose boundary is a Jordan curve. The task is to remove the material in using a milling machine. Suppose that we have already removed all material in except for a thin layer close to the boundary (another coarser tool has removed most of the material, but is not fine enough to do the boundary itself). In order to remove the remaining material, we are using a tool , which can be thought of as a disk of some radius , and we have to specify the toolpath. The toolpath is a curve that the center of should follow, and the material removed is the area swept over by as it does so. In practice, the tool has sharp teeth that cuts away the material as the tool spins at high speed; see Figure 3 (right). The maximum thickness of the layer of remaining material is some fraction of the tool radius , carefully chosen in order to limit the load on the tool.

It is an advantage if the tool center moves with constant speed while the tool is removing material, since that gives a higher surface quality of the resulting part. Since the tool moves at constant speed, the load on the tool is heavier in a neighborhood around a convex turn than when it follows as straight line, since it has to remove more material per time unit. In contrast to this, the load is lighter in a neighborhood around a concave turn. See Figure 4 for an illustration of this. If the load is too heavy, the accuracy and surface quality will be inferior, and the tool can even break . It has been recommended to round the convex corners of the toolpath by circular arcs of a certain radius in order to decrease the load [5, 16]. In our terminology, this is the same as requiring the toolpath to have bounded convex curvature. By restricting the convex curvature, we will inevitably leave more material that cannot be removed by the tool. This can be removed by other tools that are more expensive to use in terms of machining time.

If the toolpath consists of all points at distance to , the concave curvature will be bounded by , since the tool center will be “rolling” around any concave corner of using a circular arc of radius (as in the fourth picture in Figure 4). However, a recommended alternative way to get around is to follow the tangents to the endpoints of (as in the fifth picture in Figure 4—note that the tool will not remove any material when the center is in a neighborhood around the intersection point of the tangents). Experience shows that this results in the corner being cut much sharper and more precisely . This shows that the toolpaths arising in this context are required to have bounded convex curvature, whereas no bound can be given on the concave curvature.

Abrahamsen and Thorup  studied the computational problem of computing the maximum region with a boundary of bounded convex curvature inside a given region in the plane, which defines the maximum region that can be cleared by the tool using a toolpath of bounded convex curvature. A version of the Pestov–Ionin theorem (mentioned in the introduction) was used to establish the maximality of the region returned by the algorithm described in the article.

## 3 Proving Theorem 1

The proof of Theorem 1 is by contradiction. We assume that is a curve of bounded convex curvature with an interior containing no open unit disk. We then show that there exists an such that contains infinitely many pairwise disjoint disks of radius . As is bounded, this is a contradiction.

To construct these disks we need to prove a special property of curves of bounded convex curvature, namely that the radii of disks having are lower bounded by some constant depending only on .

Our first step is to set up an alternative condition that guarantees that does not have bounded convex curvature, as stated in Lemma 5 below. We start with the following lemma; see Figure 5. Figure 5: The situation described in Lemma 3. The curve γ does not have bounded convex curvature.
###### Lemma 3.

Let be a Jordan curve and consider a point on . If there exists an open unit disk where such that

1. there exists such that , and

2. for all we have ,

then does not have bounded convex curvature.

###### Proof.

Assume for contradiction that has bounded convex curvature, and choose and such that condition (1) in the definition of bounded convex curvature is satisfied for . We must show that for any unit disk with , either condition 1 or 2 of Lemma 3 fails. Let be such a unit disk and suppose is such that condition 1 of the lemma is satisfied. Let . Then,

 Bη(x)∩Ux⊂Bη(x)∩Intγ⊂D.

This implies that : Indeed, and are two unit disks with on the boundary, so if , then would contain points arbitrarily close to . Now , so . As , condition 2 is not satisfied. This completes the proof. ∎

For any Jordan curve and two distinct points and on , we denote by the closed interval on from to in the positive direction. We may for example apply this notation to the boundary curve for a disk . By , we denote the open interval . While it might be intuitively clear what it means to traverse in the positive or negative direction, we give a precise definition in Appendix A.

We require the following lemma which phrased informally states that if is traversed positively, the interior of is “to the left” of the curve.

###### Lemma 4.

Let be a point on a Jordan curve , and let be an open disk with center , sufficiently small so that is not contained in . The intersection of and is a collection of intervals of of which one, say , contains . Consider the Jordan curves

 α+=γ[a,b]∪∂U[b,a]andα−=γ[a,b]∪∂U[a,b]

Then and coincide near , that is, there exists a small disk centered at such that . Similarly and coincide near .

We believe the result to be standard but we were unable to find an equivalent one in the literature, phrased for completely arbitrary Jordan curves, e.g., with no assumptions on the differentiability of the curve. We will provide a proof in Appendix A.

Suppose is a Jordan curve, are distinct points on , and is an open disk satisfying and . If is the open region bounded by the Jordan curve , then either or . In the former case we say that winds negatively around from to and in the latter that winds positively around from to . Figure 6: The two cases in the proof of Lemma 5.
###### Lemma 5.

Let be a Jordan curve and consider an interval of such that is contained in an open unit disk . Suppose there is an open disk of radius such that and winds positively around from to . Then does not have bounded convex curvature.

###### Proof.

The general outline of the proof is as follows: We first make a translation of into a disk such that still contains and such that meets in at least one point. We then argue that we may choose a point for which Lemma 3 applies to show that does not have bounded convex curvature.

By translating and rotating we may assume about the coordinates that is centered at the origin and that and for some with and (note that may be negative). Suppose that is centered at and that (the case is dealt with in a symmetric way). Also define the Jordan curve .

We start the proof by showing the following two claims.

###### Claim 1.

Let be a point on the arc . Let be the midpoint of segment and . Consider the ray . Then intersects .

###### Proof of Claim 1.

Let be an open disk centered at , so small that . Further let be such that . Then is the disjoint union of two open connected sets and satisfying and . Moreover, and are both subsets of and, being connected, they are each fully contained in either or . Now as and by the Jordan curve theorem, it follows that either or . But by the assumption on the winding direction of from to , we have , and so . It follows that . Furthermore, we trivially have that and so must intersect . This cannot happen at a point of so must intersect as claimed. ∎

###### Claim 2.

Let be an open disk satisfying that . Then .

###### Proof of Claim 2.

It clearly suffices to show that contains . Take any point and consider the line from Claim 1 that intersects in some point where . Now by assumption contains , hence also . Since , and is convex, contains the midpoint of segment . Finally is on the line segment between and so by convexity contains . Since was arbitrary, this establishes the claim. ∎

We now let be the disk . We split the proof into two cases depicted in Figure 6.

Case 1: . In this case, we let be minimal such that the closure of the unit disk contains .

Consider the set of intersection points , which is nonempty by construction. We claim that contains neither nor . To see this, note that the ray intersects by Claim 1. Now if contained (and thus by symmetry ) then, as , we would have and hence that , a contradiction. We conclude that contains neither nor .

The set is nonempty as , and consists of pairwise disjoint open arcs. Let be an endpoint of such an arc. We will now show that and satisfy the conditions of Lemma 3, from which it follows that does not have bounded convex curvature.

That for all is immediate as is an endpoint of one of the open arcs in . It thus suffices to check condition 1, as follows. First note that either or . However, if , we could apply Lemma 4 to with to conclude that . But we assumed that and so it follows that . Since , we conclude that . Another application of Lemma 4, this time with , gives that and coincide locally near , that is, there exists an such that . Now contains and hence by Claim 2. Thus, and it follows that , as desired.

Case 2: . In this case, let be minimal such that the closure of contains . As , contains neither nor . Letting , the same argument as in Case 1 finishes the proof. ∎

We are slowly setting up the stage for the proof of Theorem 1. Intuitively, the following lemma is unsurprising. The lemma will be helpful for checking one of the conditions of Lemma 5, hence making it easier to apply.

###### Lemma 6.

Let be a Jordan curve and an open disk contained in . Suppose that are distinct points on such that . Then winds positively around from to , that is, ). Similarly, .

###### Proof.

We only prove the first statement in the theorem as the proof of the second part is similar. Letting we must show that . As it must hold that . Now either or . Suppose first that and let be any point on . Applying Lemma 4 we find that and coincide near . This is a contradiction as . It follows that . Now choose any point . Again applying Lemma 4 we find that and coincide near . As , it immediately follows that , as desired. ∎

Now we can prove that if has bounded convex curvature, then certain maximal disks contained in cannot be too small.

###### Lemma 7.

Let be a curve of bounded convex curvature. There exists a constant with the following property: If is an open disk of radius , and meets in at least two points, then .

###### Proof.

We show the contrapositive. Suppose that no such exists and take a sequence of balls satisfying for all and . Further suppose that for all .

For each , let be two distinct points in . Since is compact, we may assume that for some by passing to an appropriate subsequence. As we must have that .

Let be an open ball centered at of radius . Then is a collection of open intervals one of which, say , contains . Let be an open ball centered at and so small that .

As we must have that for sufficiently large. But then , so either or . In particular, either or is contained in an open unit disk.

We now wish to apply Lemma 5 to show that this implies that does not have bounded convex curvature. Assume without loss of generality that is such that is contained in an open unit disk. Now is a collection of open intervals of . Moreover, the collection is nonempty as otherwise , and as and , this would violate the bounded convex curvature condition. We may thus choose distinct such that is such an interval.

Since winds positively around from to by Lemma 6, we are in a position to apply Lemma 5 and we conclude that does not have bounded convex curvature.

For the proof of Theorem 1 we will also need the following easy lemma.

###### Lemma 8.

Let be a Jordan curve. Let be the supremum over all such that contains an open disk of radius . Then contains an open disk of radius .

###### Proof.

The proof is a standard compactness argument, using only that is a bounded open set. To be precise, let be defined by

 f(x)=sup{r≥0:Br(x)⊂Intγ}.

If we put , then clearly , so we may in fact write . Now, for any and thus is continuous. Furthermore, , and since is compact, attains this maximum at some point . But then . ∎

We are now ready to prove Theorem 1.

###### Proof of Theorem 1.

Let be a curve of bounded convex curvature and assume for contradiction that contains no open unit disk.

By Lemma 7 and Lemma 8, we may choose and with , such that any disk with satisfies .

Let be any point of and let the disk be tangent to in and of maximal radius. We note that , apart from , contains at least one other point. Otherwise, and then we can enlarge , contradicting the maximality of . Thus . The set consists of some (at least two) open intervals of . Let be distinct points on such that is such an open interval.

In general for , we will recursively define distinct points , and an open disk such that . Letting be the open region bounded by the Jordan curve , the construction satisfies, for all , that

1. [label=()]

2. , and

3. contains an open disk of radius at least .

The disks are pairwise disjoint and all contained in , and moreover they have radius at least . As is bounded, this gives the desired contradiction, thus completing the proof of the theorem. Figure 7: The construction in the proof of Theorem 1, where γ is the black Jordan curve. The region An+1 is bounded by the fat Jordan curve. The small disk En+1 is contained in An (not excplicitly shown), but disjoint from An+1.

We have already constructed and . We now describe the construction of , and given , , and , and then argue that with this construction, 1 and 2 above are satisfied. Figure 7 illustrates the construction. First of all, winds positively around from to by Lemma 6, so we may apply Lemma 5 and conclude that no open unit disk contains . In particular, this applies to the open unit disk having the same center as , and as the radius of is at most , there exists a point with .

Consider now the Jordan curve

 γ1:=γ[xn,yn]∪∂Dn[yn,xn]

which, by Lemma 6, contains . We let be the open disk of maximal radius contained in and tangent to in .

By the same reasoning that we used to argue about above, we must have that meets in at least two points. None of these points can be in since this would imply that and hence that , a contradiction. It follows that . The set is a collection of open intervals of , and since , at least one of them, call it , is contained in . This completes the construction of , and .

It remains to argue that with this construction, the conditions 1 and 2 are satisfied.

1. [label=()]

2. We make use of the following claim.

###### Claim 3.

We have that .

###### Proof of Claim 3.

Let the Jordan curve be defined by

 γ2:=γ[xn+1,yn+1]∪∂Dn+1[yn+1,xn+1].

By Lemma 6, , from which it follows that . Suppose for contradiction that . Since , must then intersect at least twice. But

 ∂Dn[xn,yn]∩γ(xn+1,yn+1)⊂∂Dn[xn,yn]∩γ(xn,yn)=∅,

so in fact must intersect at least twice. It follows that intersects at least thrice, which is a contradiction as . We conclude that , as desired. ∎

The arc separates into two regions, namely and . The claim thus gives that either or . Now observe that or : Indeed, but contains no point of . In particular either or and it is therefore the case that . Since now , we get that .

3. We define to be the disk of radius , tangent to in , and contained in . The radius of is at most , and since has distance at least to , it follows that . Moreover, , and we conclude that , as desired.

Having argued that the conditions 1 and 2 are satisfied, the proof is complete. ∎

## 4 Open problems

We mention here two open problems that we find interesting.

### 4.1 Are curves of bounded convex curvature rectifiable?

As mentioned in the introduction, some earlier proofs of the Pestov–Ionin theorem have used that the length of is finite. In contrast, our proof relies on having finite area which is an immediate property of Jordan domains. It is, however, easy to verify that if curves of bounded convex curvature are rectifiable, i.e., has finite length, then the proof given by Pestov and Ionin  would carry through almost unchanged. We believe this to actually be the case. Is there a (simple) proof that curves of bounded convex curvature are rectifiable?

### 4.2 What is the picture in higher dimensions?

The Jordan-Brouwer separation theorem states that if is an -dimensional topological sphere in , i.e., is obtained as the image of an injective continuous map , then the complement of in consists of exactly two connected components, one being bounded (the interior) and one being unbounded (the exterior).

It is easy to generalize our notion of bounded convex curvature to this setting. We say that has bounded convex curvature if for every point on , there is an open -dimensional unit ball and such that

 x∈∂UxandBεx(x)∩Ux⊂Intγ. (4)

The natural question is: If has bounded convex curvature, does contain an open -dimensional unit ball? This turns out to be false. Indeed, Lagunov and Fet [12, 13] studied connected -dimensional -hypersurfaces in having all principal curvatures . They showed, for instance, that topological -spheres with these properties all contain an -ball in their interior of radius at least , and that this is sharp when . Other relevant work was made by Lagunov [9, 10, 11], who showed that all compact, connected, , -dimensional hypersurfaces embedded in , for which all principal curvatures satisfy , contain a ball of radius and that this is sharp.

As our class of hypersurfaces of bounded convex curvature is less restricted (there is no assumption on differentiability and we make no requirement that the concave curvature be bounded) it is natural to ask whether it still holds that topological -spheres of bounded convex curvature contain a ball of radius (or in the case of general compact, connected, -dimensional hypersurfaces embedded in ) in their interior. Even for we find this an interesting question.

### Acknowledgments

We thank Anders Thorup for his very careful reading of the manuscript and numerous suggestions for improving the presentation, in particular by pointing out steps in our proofs that seemed intuitively clear, but in fact required detailed arguments.

A big obstacle in our work has been that many of the relevant papers are written in Russian. We thank Richard Bishop for providing us copies of his English translations of  and . We furthermore wish to thank the teams behind and , the first of which we used to convert the cyrillic script in  to machine-encoded cyrillic text and the second of which to translate the resulting text to English, together making it possible for us to understand the proof given by Pestov and Ionin.

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## Appendix A Appendix

In this appendix we will discuss orientations of Jordan curves and eventually provide a proof of Lemma 4. For this purpose it will be necessary to view curves as continuous maps where is a closed and bounded interval. With this notation is closed if , and simple if is injective, where in the closed case we allow . A Jordan curve is the image of a simple closed curve.

The starting point will be the classic theorem by Jordan.

###### Theorem 9 (Jordan curve theorem).

Let be a Jordan curve. Then the complement consists of two connected components. Moreover, is the boundary of each of these components.

We now recall some simple facts concerning argument variation. For a given point and an argument for with respect to

is an argument for the vector

, that is, an angle such that for some . If , is a curve, and , then a continuous argument function for with respect to is a continuous map such that is an argument for for all . The argument variation around is defined as and this does not depend on the choice of , nor is it changed if we use an orientation preserving reparametrization of . Importantly, the function is continuous on .

From the above it follows that for a Jordan curve , the argument variation around any is a multiple of , constant on each of the two connected components of the complement of , and on the unbounded component. In fact, the argument variation is when , as we will see shortly. We say that a parametrization of is positively oriented if the argument variation of around any is . Otherwise we say that is negatively oriented. If and are distinct points on , we write for the interval of obtained by traversing from to along the positive orientation. Slightly abusing notation we will sometimes write for a parametrization of this interval traversed from to . We also define .

Finally, if and are curves satisfying , we let be the continuous curve obtained by first traversing and then . Also, if is a curve, we write for the curve obtained by traversing in the opposite direction. We finally write when the addition is well-defined. We now restate Lemma 4 in a more general form. Figure 8: The setting of Lemma 10. The dashed lines represent γ potentially reentering U.
###### Lemma 10.

Let be a point on a Jordan curve , and let be an open disk with center , sufficiently small so that is not contained in . The intersection of and is a collection of open intervals of of which one, say , contains . Consider the two Jordan curves

 α+=γ[a,b]+∂U[b,a]andα−=γ[a,b]−∂U[a,b].

Then is the disjoint union

 U=γ(a,b)∪Intα+∪Intα−.

Moreover, and coincide near , that is, there exists a small disk centered at such that . (See Figure 8.)

#### Remark.

We prove the lemma by first proving the statement about the decomposition of . Strictly speaking, is not defined at this point and and may a priori be chosen in two different ways. Likewise, and are not defined, but it is obvious what it means to traverse a circle in the positive and negative direction. The statement about the decomposition of is correct regardless of how and are chosen, so the ambiguity does not matter for that part of the lemma. The second part of the proof starts by showing that the argument variation around any is , which by the introductory comments lets us define intervals such as unambiguously. This is important for the statement that and coincide near , which is finally proven.

###### Proof of Lemma 10.

Define . For any point let be the angle from to in the positive direction. Clearly, and . Hence, if is in and not on we have the two equations

 Argxα+ =Argxγ[a,b]+2π−θx, (5) Argxα− =Argxγ[a,b]−θx. (6)

If then the left side of (5) vanishes, and if then the left side of (6) vanishes. Consequently, by (5) and (6),

 Argxγ[a,b] =θx−2π for x∈Extα+∩U, (7) Argxγ[a,b] =θx for x∈Extα−∩U. (8)

In turn, when the latter two equations are inserted into (5) we obtain the equations

 Argxα+ =2π for x∈Extα−∩U, (9) Argxα− =−2π for x∈Extα+∩U. (10)

We now observe that is nonempty, as follows. Let be a point on and an open disk centered at so small that . Since a part of is outside (and thus in ), it follows that . Hence and since is nonempty, the claim follows. Let .

We next prove that . Consider a point . It follows from (9) that