THE JORDAN CURVE THEOREM FOR PIECEWISE SMOOTH CURVES AKSHAA VATWANI Abstract. The Jordan curve theorem is one of the classical theorems of mathematics whose statement is deceptively simple. It states that a simple, closed curve C in the plane separates it into precisely two components, of which C is the common boundary. A companion theorem called the Jordan arc theorem states that a simple arc does not separate the plane. There exist many proofs of these theorems; ranging from those using the theory of fundamental groups and covering spaces [1], to those using the Brouwer Fixed Point Theorem [2]. We present here a proof of the Jordan arc theorem followed by the Jordan curve theorem, both for the special case of piecewise smooth curves. The proof uses induction and elementary tools of analysis.

Contents 1. Introduction 2. The Proof For Polygons 2.1. The Jordan arc theorem for a simple polygon 2.2. The Jordan curve theorem for a simple closed polygon 3. Parallel arcs 4. Sides of an arc 5. Some More Results 6. The Jordan Arc Theorem 7. The Jordan Curve Theorem References

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1. Introduction The statement of the Jordan curve theorem seems obvious, but it was a very difficult theorem to prove. The result was easy to establish for simple curves such as polygonal lines but the problem came in generalising it for even “complicated” curves like nowhere differentiable curves. It took a while for it to be realised that even this self-evident statement needed a proof and the first to formulate this need was Bernard Bolanzo. After that there have been many proofs, including those by Jordan, Veblen, Maehara[2], etc. We present here an attempt to generalise the proof used for polygons to the case of piecewise smooth curves. In order to understand the material, the notions of arc, closed curve and related terms need to be made precise. Definition 1.1. An arc is a continuous mapping C given by z = φ(t) : [a, b] ∈ R → C. It is customary to identify the arc with the image of the interval [a, b] under the map φ. An arc is said to be simple or a Jordan arc, if it does not intersect itself i.e φ(t1 ) = φ(t2 ) ⇐⇒ t1 = t2 . In other words, φ is a one-one map. Definition 1.2. A closed curve is an arc whose endpoints coincide i.e φ(a) = φ(b). A closed curve is called simple or a Jordan curve if it does not intersect itself except at the endpoints. Definition 1.3. An arc is said to be smooth if it has C 1 parametrisation. This means that φ′ (t) exists and is continuous. A piecewise smooth arc is an arc which is smooth except at finitely many points. Thus, it can be constructed by joining finitely many smooth arcs. We can now state formally the two relevant theorems: Theorem 1.4. (Jordan Arc Theorem) A simple arc C does not separate the plane. The complement of C in the plane is connected and its boundary is C. Theorem 1.5. (Jordan Curve Theorem) The complement of a simple closed curve C in the plane is composed of two (connected) components, each having C as its boundary. Also, we note that since R2 is locally path connected and any arc is closed as a set in R2 , the components and the path components of the open set R2 − C are the same. Thus connected ⇐⇒ path connected for our purpose. 2. The Proof For Polygons By a polygon we mean an arc which is piecewise linear i.e. it can be constructed by joining together finitely many line segments. A closed polygon is one whose endpoints coincide.

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We give an informal ouline of the proof of the two theorems for the case of polygons. 2.1. The Jordan arc theorem for a simple polygon. Proof. The proof works by induction. Induction Hypothesis: Assume that the complement in R2 is connected for all polygons having atmost n segments. If Pn+1 is a simple polygon having n + 1 segments, then it can be obtained by adjoining a single segment σ to a simple polygon Pn having n segments. Let P ′ denote the complement of P. Then Pn′ is connected by the induction hypothesis. ′ Now, any two points z and ζ in Pn+1 also lie in Pn′ and can hence be joined by a path γ ′ and we are done. If it does in Pn′ . If γ does not intersect σ, then clearly it lies in Pn+1 intersect σ, then we can draw parallel lines l1 and l2 on either side of σ, they will intersect the path. Then we can construct a new path as follows: start from z, follow γ till it touches one of the parallel lines, say l1 , go along l1 , eventually make a “turn” to reach l2 and then resume γ from wherever it intersects l2 . Thus γ is replaced by a new path γ ′ which does ′ ′ is connected. . This shows that Pn+1 not intersect σ i.e. γ ′ ∈ Pn+1 To complete this induction proof, we note that for the case n = 1, the complement of a single segment σ is indeed connected.  2.2. The Jordan curve theorem for a simple closed polygon. Proof. Let P be a simple closed polygon. We want to prove that P ′ consists of two connected components. For this proof, we will use the previous theorem. Step 1: By removing an appropriate segment σ from P , we can obtain a simple polygon Γ. Since P is bounded as a set, there exists a disk containing P in its interior. Choose some point outside this disk and call it ζ. Now we can can precisely define two “components” of P ′ as follows: First note that, using P ′ ⊂ Γ′ and the previous theorem, every point in P ′ can be joined to ζ by a path which is contained in Γ′ . Define E to be the set of all points in P ′ which can be joined to ζ using a path in Γ′ which does not cross σ. Define I to be the set of all other points in P ′ . Step 2: E is connected: Any two points in E can be joined to ζ and hence to each other, using a path in Γ′ which does not cross σ. This path lies entirely in E since any point on this path can be joined (using the same path) to ζ without crossing σ. Thus any two points in E can be joined by a path lying entirely in in E, which shows that E is connected. Step 3: I is connected: To join any point z ∈ I to ζ in Γ′ , the path has to cross σ. Hence z and ζ lie on opposite “sides” of σ. Similarly, any w ∈ I and ζ lie on opposite “sides” of σ. This means

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that z and w lie on the same “side” of σ and it can be proved that they can be joined by a path γ which does not cross σ. We must show that this path lies entirely in I. Suppose some point r of the path lies in E. Then we can join z to r using γ without crossing σ and then r to ζ without crossing γ (since r ∈ E). This means that z ∈ E, which is a contradiction. Hence, I is connected.  Note that the notion of “turn” as well as “sides” of the segment σ used in theses proofs need to be made precise, which is why this proof is only a sketch of the steps involved. We will try to rigorously imitate the above arguments for piecewise smooth curves.

3. Parallel arcs In the preceding argument, the only property of straight line segments that was used is the idea of parallelism i.e. two parallel lines do not intersect. Hence we try to define an analogous system of “parallel” arcs for a given simple smooth arc. This is our motivation for the first Lemma: Lemma 3.1. Let C : z = φ(t), 0 ≤ t ≤ L, be a simple smooth arc parametrized by arc length. Define, for each real ǫ, Cǫ to be the arc parametrized by z = φǫ(t) ≡ φ(t)+iǫφ′ (t), 0 ≤ t ≤ L. Then, there exists a d > 0 such that Cǫ ∩ C = ∅ when 0 < |ǫ| < d. Proof. For our purpose, arc length parametrization simply means that φ′ (t) is a unit vector i.e |φ′ (t)| = 1. We first show that portions of C and Cǫ corresponding to sufficiently small neighborhoods of the interval [0, L] are disjoint. Let t, τ ∈ [0, L]. Then, φǫ (t) − φ(τ ) = φ(t) + iǫφ′ (t) − φ(τ ) = (t − τ + iǫ)φ′ (t) + [φ(t) − φ(τ )] − [(t − τ )φ′ (t)] = (t − τ + iǫ)φ′ (t) + Iτt [φ′ (s) − φ′ (t)]ds By continuity of φ′ , there exists a δ > 0 such that |φ′ (s) − φ′ (t)| < 1/2 if |s − t| < δ. Since s ranges from τ to t, it is sufficient to ensure that |t − τ | < δ. Then we have: |Iτt [φ′ (s) − φ′ (t)]ds| ≤ Iτt |φ′ (s) − φ′ (t)|ds t−τ < Iτt (1/2)ds = 2 |t − τ | ≤ 2 From this we get:

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|φǫ (t) − φ(τ )| ≥ |(t − τ + iǫ)| |φ′ (t)| − |Iτt [φ′ (s) − φ′ (t)]ds| |t − τ | > |(t − τ + iǫ)| − (using |φ′ (t)| = 1) 2 |t − τ | ≥ |t − τ | + |iǫ| − 2 For ǫ 6= 0, the R.H.S is always positive, which means that: (3.1)

φǫ (t) 6= φ(τ ) if |t − τ | < δ and ǫ 6= 0

Now since [0, L] is compact, we can choose points 0 = t0 < t1 < . . . < tn = L such that |tk − tk−1 | < δ/4 ∀k ∈ {0, 1, . . . , n}. For an any such k, we denote B k,η = {t ∈ [0, L] : |t − tk | ≤ η}. For any t ∈ B k, δ and 2

τ ∈ B k, δ , 4

|t − τ | ≤ |t − tk | + |τ − tk | δ δ 3δ ≤ + = 2 4 4 < δ Hence, using 3.1, we find that the images of the intervals B k, δ and B k, δ under the arcs Cǫ 2 4 and C respectively, are disjoint i.e (3.2)

Cǫ {B k, δ } ∩ C{B k, δ } = ∅ 2

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Let Sk = {t ∈ [0, L] : |t − tk | ≥ δ/2}. Then Sk and B k, δ cover the whole interval [0, L]. 2

We now want to prove that Cǫ {Sk } ∩ C{B k, δ } = ∅, for small enough ǫ. 4

Since the map C is simple, continuous and the sets B k, δ and Sk are disjoint and compact, 4

this means that the image sets C{B k, δ } and C{Sk } are disjoint and compact. Hence these 4 image sets have a positive distance dk between them i.e. dist(C{B k, δ }, C{Sk }) = dk . 4

Assume that Cǫ {Sk } intersects C{B k, δ } at some point R = φǫ (r). 4 Then, φǫ (r) ∈ Cǫ {Sk } =⇒ φ(r) ∈ C{Sk }. Now considering R = φǫ (r) and φ(r) as points of Cǫ {Sk } and C{Sk } respectively, we get dist(R, φ(r)) = |ǫ|. But considering R and φ(r) as points of C{B k, δ } and C{Sk } respectively, and we get 4 dist(R, φ(r)) ≥ dk . Hence, |ǫ| < dk gives us a contradiction. This means that (3.3)

Cǫ {Sk } ∩ C{B k, δ } = ∅ if |ǫ| < dk 4

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Clubbing together equations 3.2 and 3.3, we get: Cǫ ∩ C{B k, δ } = ∅ if |ǫ| < dk . This is 4

true for all k ∈ {0, 1, . . . , n} and C can be covered using the sets C{B k, δ }. Hence, we have: 4

Cǫ ∩ C = ∅ if ǫ < min{dk }  Hence, the class of arcs Cǫ , with ǫ small enough can be used to play the role of parallel arcs for a given simple smooth arc C. Note that these parallel arcs need not be simple themselves. 4. Sides of an arc For the arcs Cǫ defined above to be useful in constructing paths which do not intersect C, it is needed that any point sufficiently close to an interior point of C lies on one of these arcs. We give a standard variational argument to show this. Lemma 4.1. If z ∈ / C is closer to C than it is to either end point of C, then there exists a t0 ∈ (0, L) and an ǫ0 = 6 0 such that z = φ(t0 ) + iǫ0 φ′ (t0 ). In other words, z ∈ Cǫ0 . Proof. Since z is closer to C than it is to φ(0) or φ(L) and C is a closed set, there exists a point t0 ∈ (0, L) such that |z − φ(t0 )| = dist(z, C). Using the Taylor expansion for φ(t) in a neighborhood of φ(t0 ), we can write: z − φ(t) = [z − φ(t0 ) + (−1)φ′ (t0 )(t − t0 )] + o(t − t0 ) where o(t − t0 ) denotes terms of higher order. Since the minimum distance of z is from φ(t0 ), |z − φ(t0 )|2 ≤ |z − φ(t)|2 ≤ |z − φ(t0 )|2 + 2Re[(z − φ(t0 ))φ′ (t0 )](t − t0 ) + o(t − t0 ) Since t − t0 can be either positive or negative, we must have 2Re[(z − φ(t0 ))φ′ (t0 )] = 0. This means that (z − φ(t0 ))φ′ (t0 ) = iǫ0 f or some ǫ 6= 0, Hence z = φ(t0 ) + iǫ0 φ′ (t0 ), i.e. z ∈ Cǫ0 . Note we have not imposed any condition here that Cǫ0 should be disjoint from C.  As a consequence of this Lemma, we can now define what we mean by “sides” of a smooth simple arc C. We say that two points z ∈ Cǫ and ζ ∈ Cη are on same or opposite sides of C according to whether ǫ and η have the same or opposite signs. According to this definition, an arc has only two sides and the notion of side of an arc only makes sense for points which are close enough to the arc for the above Lemma can be applied.

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5. Some More Results Lemma 5.1. C be a simple smooth arc as defined before. There exists a d > 0 such that the “half neighborhood” z = φ(L) + ǫφ′ (L)eiθ , 0 < ǫ < d, −π/2 ≤ θ ≤ π/2, is disjoint from C. Proof. We have, φ(L) + ǫφ′ (L)eiθ − φ(t) = ItL [φ′ (s) − φ′ (L)]ds + φ′ (L)[(L − t) + ieiθ ] By continuity of φ′ , there exists a δ > 0 such that |φ′ (s) − φ′ (t)| < 1/2 if |s − t| < δ. Then, as was done in Lemma 3.1, if L − t < δ, we can write: |φ(L) + ǫφ′ (L)eiθ − φ(t)| > |L − t + ieiθ | − ≥

|L − t| 2

|L − t| + |ieiθ | 2

The above expression is seen to be positive for ǫ > 0 and all θ ranging from − π2 to π2 . Thus the half neighborhood at L is seen to be disjoint from the image set C{L − δ ≤ t ≤ L}. Since C is simple, φ(L) is disjoint from the image set C{0 ≤ t ≤ L − δ} and hence has a positive distance d from it. By arguing similarly as in Lemma 3.1, we see that the half neighborhood at L is disjoint from the image set C{0 ≤ t ≤ L − δ} if 0 < ǫ < d. Thus, the half neighborhood is disjoint from C for 0 < ǫ < d and π/2 ≤ θ ≤ π/2.  This half neighborhood will be useful to make a turn of radius ǫ about the endpoint of the arc without intersecting the arc. Lemma 5.2. Let C be a simple smooth arc as defined before. Let A be a compact set and z and ζ be two points of (C ∪ A)′ , each of which is closer to an interior point of C than it is to A or to an endpoint of C. Then z can be joined to ζ by a path in (C ∪ A)′ if any of the following two conditions hold: (i)C ∩ A = φ(0) or (ii)C ∩ A =φ(0) ∪ φ(L) and z, ζ are on the same side of C. Proof. Case (i): Since the minimum distances of z and ζ from the set C ∪ A are attained, let z1 = φ(t1 ) and ζ1 = φ(τ1 ) be points on C having minimum distances from z and ζ respectively. From Lemma 4.1, z = φ(t1 ) + iǫ0 φ′ (t1 ) and z = φ(τ1 ) + iη0 φ′ (τ1 ). The segments [z, z1 ] and [ζ, ζ1 ] intersect C only at the points z1 and ζ1 respectively. The segment [z, z1 ] is a continuous mapping and hence we can go along this segment from z to a point z2 arbitrarily close to z1 . Since z2 satisfies the criteria of Lemma 4.1, it lies on Cǫ for some ǫ and we can make this ǫ as small as we want. Similarly for ζ, there is a point ζ2 ∈ Cη and this η can be made arbitrarily small. It follows that ǫ and η have the same signs as ǫ0 and η0 respectively. We choose ǫ and η such that |ǫ| = |η|.

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Suppose ǫ = −η : Now we choose |ǫ| smaller than both the d′ s of Lemmas 3.1 and 5.1, so that Cǫ and Cη are disjoint from C, hence parallel to it and the half neighborhood at φ(L) defined by φ(L) + ǫφ′ (L)eiθ , −π/2 ≤ θ ≤ π/2, is disjoint from C. Now we are ready to construct our path in (C ∪ A)′ from z to ζ as follows: (1) Go from z to z2 using the line segment [z, z2 ]. (2) Use the parallel arc Cǫ on which z2 lies to go till the point φ(L) + iǫφ′ (L). (3) Use the half neighborhood to go from φ(L) + iǫφ′ (L) to the point φ(L) − iǫφ′ (L) which lies on C−ǫ . (4) Use the parallel arc C−ǫ to go till the point ζ2 , which can then be joined to ζ using the segment [ζ2 , ζ]. This path lies in C ′ and will work provided it does not intersect A at any step. To ensure this, a further condition needs to be imposed on |ǫ| : We can assume t1 ≤ τ1 without loss of generality. The image set C{[t1 , L]} and the set A are disjoint compact sets and hence have a positive distance δ between them. The distance of the path used above from the image set C{[t1 , L]} is atmost |ǫ|. Therefore, if |ǫ| < δ, then the path used to join z and ζ will be disjoint from A and will hence lie in (C ∪ A)′ . The argument for ǫ = η is similar to the argument given above. Case (ii) : The proof works on similar lines and the complete details can be worked out. We only mention that in this case we can choose ǫ = η and there will be no half neighborhood used since z and ζ now lie on the same side of C. Thus, there are only two restrictions on |ǫ| : one that it should be smaller than the d of Lemma 3.1 and the second that |ǫ| < δ where δ is as defined above. The first condition will ensure that Cǫ is parallel to C and the second will ensure that the path used to join z and ζ does not intersect A.  We can now proceed towards the main results of these notes. 6. The Jordan Arc Theorem Theorem 6.1. The complement of a simple piecewise smooth arc C is a connected set having C as its boundary. Proof. C ′ is open and hence its boundary is contained in C. Now take any neighborhood Nδ around a smooth point of C. Then Cǫ ∩ Nδ 6= ∅ for |ǫ| < min{δ, d} where d is as used in Lemma 3.1. Since Cǫ ∈ C ′ , this means that C ′ ∩ Nδ 6= ∅ ∀ δ > 0. Hence every smooth point is a boundary point and since set of boundary points is closed, the “corner” points at which C is not smooth must also be boundary points. Therefore, boundary of C ′ is precisely C. To prove that C ′ is connected, we proceed by induction on the number of smooth segments of C: Let C be a simple smooth arc. If z, ζ ∈ C ′ , then they can be joined by a path Γ which does not intersect either end point of C, since removing two points from an open connected set retains the connectedness. If Γ intersects C, then due to the continuity of Γ, we can join

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z and ζ to points z2 and ζ2 which are arbitrarily close to interior points of C. Then, as was done in the proof of Lemma 5.2, we can construct a path between z2 and ζ2 and hence between z and ζ such that it lies in C ′ . Thus, the theorem is true for a simple smooth arc. Now suppose that the theorem is true for all simple arcs having n smooth arc-segments. If Cn+1 has n + 1 smooth arc-segments, let Cn denote the first n smooth segments and σ ′ . Then by the induction hypothesis, they can be joined denote the last. Consider z, ζ ∈ Cn+1 ′ by a path Γ in Cn . Since removing a point from an open set does not disconnect it, we may assume that Γ does not pass through the endpoint of σ. If Γ does not pass through σ, then ′ ′ to points z ′ and ζ ′ which and we are done. Else, we can join z and ζ in Cn+1 Γ ∈ Cn+1 are arbitrarily close to interior points of σ. Using Lemma 5.2 for the points z ′ and ζ ′ with ′ A = Cn , it follows that z ′ can be joined to ζ ′ by a path in Cn+1 . This gives us a path in ′ Cn+1 between z and ζ. This proves the theorem.  7. The Jordan Curve Theorem Theorem 7.1. The complement of a simple closed piecewise smooth curve C consists of two connected components, E and I, each having C as its boundary. Proof. We first construct the components of C. There exists a disk containing the compact set C in its interior. Choose a point ζ outside this disk. Define E to be the set of all points which can be joined to ζ by a path in C ′ . Then E is clearly connected. Let I be the set of all the other points in C ′ . We want to show that I is connected: Remove a smooth arc σ from C, what we get is a simple piecewise smooth arc Γ. By Theorem 10, any point z1 ∈ I can be joined to ζ in Γ′ . This path necessarily crosses σ since z1 ∈ I. As in the proof of Theorem 10, we can choose points z1′ and ζ ′ arbitrarily close to interior points of σ such that z1 and ζ can be joined to z1′ and ζ ′ respectively in C ′ . We claim that z1′ and ζ ′ are on opposite sides of σ, since otherwise we can join z1′ and hence z1 to ζ without crossing σ. If z2 ∈ I, with z2′ playing the role analogous to z1′ , then the point z2′ must lie on the same side of σ as z1′ since otherwise, z2 could be connected to ζ without crossing σ. Since z1′ and z2′ are on the same side of σ, we can use Lemma 5.2 to construct a path in Γ′ between z1 and z2 , such that it does not cross σ. Then this path must lie in I, which shows that I is connected. In order to show that I is nonempty, one can consider the difference between the winding number of C with respect to two points on opposite sides of a smooth portion of C. If z0= φ(t0 ) is such a point, then we find ∆ = n(C, z0 + iǫφ′ (t0 )) − n(C, z0 − iǫφ′ (t0 )). Using continuity of φ′ at t0 and tools of analysis, ∆ is found to have an integral value, which means that there are atleast two components i.e I 6= ∅. Thus C ′ = E ∪ I, where E and I are connected and nonempty. To prove that both components of C ′ have C as boundary, we begin by showing that each component U of R2 − C is open. Let x ∈ U . As C is closed in R2 , R2 − C is open and hence

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there is a small neighborhood of x which is disjoint from C. Then U ∪ D must be connected and hence, since U is a component of R2 − C, we must have D ⊂ U . Thus any component U is open and does not contain its boundary. This means that boundary of U ⊆ C. Any smooth point of C can be shown to lie in the boundary and since the boundary is a closed set, the corner points at which C is not smooth must also lie in the boundary. Hence boundary of any component of C ′ is C. This proves the theorem.  Unfortunately, these proofs rely on the piecewise smooth nature of the arcs and cannot be extended to the most general case of an arbitrary arc in R2 . For the general case, we need to use more advanced tools. References [1] J.R.Munkres, Topology, Prentice-Hall, Inc., 2000. [2] Ryuji Maehara, The Jordan Curve Theorem via the Brouwer Fixed Point Theorem, The American Mathematical Monthly, 641-643, December 1984. [3] R.N. Pederson, The Jordan Curve Theorem For Piecewise Smooth Curves, The American Mathematical Monthly, December 1984, 641-643. [4] L.V.Ahlfors, Complex Analysis, McGraw-Hill, Inc., 1979 IISER, Central tower, Sai Trinity building, Pashan circle, Pune 411021 INDIA E-mail address: [email protected]