Calculus of Real and Complex Variables

2.9 The Jordan Curve Theorem^∗

This short chapter is devoted to giving an elementary proof of the Jordan curve theorem. This is normally done in courses on algebraic topology or it can be done using degree theory. I am following lecture notes from a topology course given by Fernley at BYU in the 1970s.

The reason for this theorem is to enable a complete proof of Green’s theorem for a simple closed curve having finite length. This will be used to give a treatment of the Cauchy integral theorem which is like what was originally proposed by Cauchy. Rather than simply citing the result, I am trying to give a proof for those who need to see why the theorem is true. The theorem itself is pretty easy to believe. A unit circle divides the plane into two connected components, an inside and an unbounded outside with the circle itself being the complete boundary of both of these connected open sets. This is obvious. The Jordan curve says that if h is a one to one continuous function defined on such a circle C, then the wriggly circle h

also divides the plane into two components, an inside and an outside just like the circle does. Proving this turns out to be surprisingly difficult. It was first done by Veblun in 1917.

For k = 0,1,2, one speaks of k chains. For

_j=1ⁿ a set of k cells, the k chain is denoted as a formal sum

C = a + a + ⋅⋅⋅+ a 1 2 n

where the sum is taken modulo 2. The sums are just formal expressions like the above. Thus for a a k cell, a + a = 0,0 + a = a, the summation sign is commutative. In other words, if a k cell is repeated an even number of times in the formal sum, it disappears resulting in 0 defined by 0 + a = a + 0 = a. The symbol 0 indicates there is no k cell. For a a k cell,

denotes the points of the plane which are contained in a. For a k chain C as above,

|C | ≡ {x : x ∈ |aj| for some aj}

so

is the union of the k cells in the sum remembering that when a k cell occurs twice, it is gone and does not contribute to .

The following picture illustrates the above definition. The following is a picture of the 2 cells in a 2 chain. The dotted lines indicate the lines in the grating.

Now the following is a picture of the 1 chain consisting of the sum of the 1 cells which are the edges of the above 2 cells. Remember when a 1 cell is added to itself, it disappears from the chain. Thus if you add up the 1 cells which are the edges of the above 2 cells, lots of them cancel off. In fact all the edges which are shared between two 2 cells disappear. The following is what results.

In the second of the above pictures, you have a 1 cycle. Here is a picture of another one in which the boundary of another 2 cell has been included over on the right.

This 1 cycle shown above is the boundary of exactly two 2 chains. What are they? C₁ consists of the 2 cells in the first picture above along with the 2 cell whose boundary is the 1 cycle over on the right. C₂ is all the other 2 cells of the grating. You see this clearly works. Could you make that 2 cell on the right be in C₂? No, you couldn’t do it. This is because the 1 cells which are shown would disappear, being listed twice.

This illustrates the fundamental lemma of the plane which comes next.

Proof: First consider the unbounded 2 cells. These must all be in a single 2 chain. To see this, imagine moving around a very large rectangle which contains the bounded 1 cycle in its interior. (The Jordan curve theorem is obvious for a rectangle.) If these 2 cells are assigned to more than one 2 chain, then an unbounded 1 cell is in the boundary of two adjacent 2 cells and would appear in C contrary to C being bounded. Therefore, one is forced to assign all unbounded 2 cells to a single 2 chain if the lemma is to be true. Label each of these with an A to indicate they are all in the same 2 chain.

Now starting from the left of the bounded 1 cycle and moving toward the right, toggle between A and B every time you hit a vertical 1 cell of C. This will label every 2 cell with either A or B. Next, starting above the bounded 1 cycle, toggle between A and B every time you encounter a horizontal 1 cell of C.  This also labels every 2 cell as either A or B. Then let one of the 2 chains be those labeled with A and the other be the ones labeled with B. If the lemma is to be true, then these are the two 2 chains D₁,D₂ which satisfy ∂D_i = C since this accounts for all edges of D exhibiting each as part of the boundary of D_i.

The difficulty is whether this is well defined. Indeed, we have not yet used the fact that this is a 1 cycle having zero boundary. Consider the following picture.

Suppose you have a 2 cell which is labeled A from left to right and B from top to bottom. Pick the contradictory 2 cell which is farthest to the left and as high as possible out of all contradictory 2 cells. Beginning with the label B move over the circle counterclockwise changing the label whenever you encounter an edge of C. Since you end up with A, you must have encountered an odd number of 1 cells of C and so C was not a 1 cycle after all. The point at the center in the above picture must be in ∂C.

Thus D₁ and D₂ are uniquely determined assuming there are only two such 2 chains having boundary C. If you had a third, D₃, then let e be an edge of d ∈ ∂D₃. It would then need to be an edge of two 2 cells, one in D₁ and one in D₂ and so some 2 cell is in two different D_i and they are not disjoint after all. ■

The next lemma is interesting because it gives the existence of a continuous curve joining two points.

Proof: There are an odd number of 1 cells of C which have x at one end. Otherwise ∂C≠x + y. Begin at x and move along an edge leading away from x. Continue till there is no new edge to travel along. You must be at y since otherwise, you would have found another boundary point of C. This point would be in either one or three one cells of C. It can’t be x because x is contained in either one or three one cells of C. Thus, there is always a way to leave x if the process returns to it. It follows that there is a continuous curve in

joining x to y. ■

The next lemma gives conditions under which you can go around a couple of closed sets. It is called Alexander’s lemma. The following picture is a rough illustration of the situation. Roughly, it says that if you can miss F₁ and you can miss F₂ in going from x to y, then you can miss both F₁ and F₂ by climbing around F₁.

Proof: Let a₁,a₂,

,a_m be the 2 cells of D which intersect the compact set F₁. Consider

∑ C ≡ C2 + ∂ak. k

This is a 1 chain and ∂C = x + y because ∂∂a_k = 0. Then

∩F₂ = ∅. This is because ∩F₁≠∅ and none of the 2 cells of D intersect both F₁ and F₂ by assumption. Therefore, C is a bounded 1 chain which avoids intersecting F₂.

Does it also avoid F₁? Suppose to the contrary that l is a 1 cell of C which does intersect F₁.

If

⊆, then it would be an edge of some 2 cell of D and would have to be a 1 cell of C₂ since it intersects F₁ so it would have been added twice, once from C₂ and once from ∑ _k∂a_k and therefore could not be a summand in C. Therefore, is not in . It follows l must be an edge of some a_k ∈ D and it is not a 1 cell of C₁ + C₂.

Therefore, if b is the 2 cell adjacent to a_k which shares l as an edge, it must follow b

D since otherwise l would not appear in the sum because it would be added out. But now, it follows that l ∈ ∂D = C₁ + C₂ which was ruled out above. Therefore C misses F₁ also. ■

Proof: If p is a limit point of

and p ∈, then p must be contained in an edge of some 2 cell of D since otherwise it could not be a limit point, being contained in an open set whose intersection with is empty. If p is a point of an even number of edges of 2 cells of D, then it is likewise an interior point of which cannot be a limit point of . Therefore, if p is a limit point of , it must be the case that p ∈. A similar observation holds for the case where p ∈ and is a limit point of . Thus if H ∩ and H ∩ are both nonempty, then they separate the connected set H and so H must be a subset of one of or . If you have a limit point of either one, then it is in and so not in the other since H has no points of . ■

Proof: Suppose this is not so. Then there exists x,y points in Γ^C which are in different components of Γ^C. Let G be a grating having x,y as points of intersection of a horizontal line and a vertical line of G and let p,q be the points at the ends of the Jordan arc. Also let G be such that no unbounded two cell has nonempty intersection with Γ. Let p = r

and q = r. Now let z = r and consider the two arcs pz and zq. If ∂C = x + y then it is required ∩ Γ≠∅ since otherwise these two points would not be in different components. Suppose there exists C₁,∂C₁ = x + y and ∩ zq = ∅ and C₂,∂C₂ = x + y but ∩ pz = ∅.

Then C₁ + C₂ is a 1 cycle and so by Lemma 2.9.3 there are exactly two 2 chains whose boundary is C₁ + C₂. Since z

, it follows z = pz ∩ zq can only be in one of these 2 chains because it is a single point. Then by Lemma 2.9.5, Alexander’s lemma, there exists C a 1 chain with ∂C = x + y and ∩ = ∅ so by Lemma 2.9.4 x,y are not in different components of Γ^C contrary to the assumption they are in different components. Hence one of pz,zq has the property that every 1 chain C with ∂C = x + y intersects it. Say every such 1 chain goes through zq. Then let zq play the role of pq and conclude every 1 chain C such that ∂C = x + y goes through either zw or wq there

(( ) ) w = r a+-b-+ b 1 2 2

Thus, continuing this way, there is a sequence of Jordan arcs p_kq_k where r

= q_k and r = p_k with < , ⊆ such that every C with ∂C = x + y has nonempty intersection with p_kq_k. The intersection of these arcs is r where s = ∩_k=1^∞. Then all such C must go through r because such C with ∂C = x + y must intersect p_kq_k for each k and their intersection is r. But now there is an obvious contradiction to having every 1 chain whose boundary is x + y intersecting r.

Pick a 1 chain C whose boundary is x + y. Let D be the two chain of at most four 2 cells consisting of those two cells which have r

on some edge. Then ∂ = ∂C = x + y but r. Therefore, this contradiction shows Γ^C must be connected after all. ■

The other important observation about a Jordan arc is that it has no interior points. This will also follow later from a harder result but it is also easy to prove.

Proof: Suppose to the contrary that Γ has an interior point p. Then for some r > 0,

B (p,r) ⊆ Γ .

Consider the circles of radius δ < r centered at p. Denoting as C_δ one of these, it follows the C_δ are disjoint. Therefore, since r is one to one, the sets r⁻¹

are also disjoint. Now r is continuous and one to one mapping to a compact set. Therefore, by Theorem 2.2.29, r⁻¹ is also  continuous. It follows r⁻¹ is connected and compact. Thus by Theorem 2.7.8 each of these sets is a closed interval of positive length since r is one to one. It follows there exist disjoint open nonempty intervals consisting of the interiors of r⁻¹,_δ<r. This is a contradiction to the density of ℚ and the fact that ℚ is at most countable. ■

Note that if J is a Jordan curve, then it is the union of two Jordan arcs whose intersection is two distinct points of J. You could pick z ∈

and consider r and r as the two Jordan arcs.

The next lemma gives a probably more convenient way of thinking about a Jordan curve. It says essentially that a Jordan curve is a wriggly circle. Recall Theorem 2.2.29. This is stated as the following lemma.

Next is the lemma which says that a Jordan curve is like a wiggly circle. Thus a Jordan arc is the one to one continuous image of

and a Jordan curve is the one to one continuous image of the unit circle.

Proof: Suppose J = r

where r is continuous on one to one on [a,b),r = r. Is J also the continuous one to one image of S¹? From Lemma 2.9.11, r⁻¹ is continuous on r for each small δ. Therefore, r⁻¹ is continuous on

( [ 1]) ∪∞k=1r a,b− -- = r ([a,b)) = J. k

The last equality holds because r

= r. Now define R : J → S¹ by

( ( ) ( ) ) R (x) ≡ cos -2π--(r− 1(x)− a) ,sin -2π--(r−1(x)− a) b− a b − a

It was just shown that this is continuous. It is the composition of a continuous function with one which is continuous. R is also one to one. Indeed, if

then = and so x = since r is one to one on [a,b). By Lemma 2.9.11 again, R⁻¹ : S¹ → J is one to one and onto. Let θ = R⁻¹.

Conversely suppose J is the image of the unit circle as just explained. Then let R :

→ S¹ be defined as R ≡. Then consider r ≡ θ. r is one to one on [0,2π) and (0,2π] with r = r and is continuous, being the composition of continuous functions. ■

A similar result could be proved if J were in ℝ^p for p > 2, but we are primarily concerned with simple closed curves in the plane.

Note that we could have assumed r is continuous on

and one to one on (a,b] with r = r and proved that this is also equivalent to J being the continuous one to one image of S¹. The proof would be essentially the same as what was given above.

Before the proof of the Jordan curve theorem, recall Theorem 2.7.14 which says that the connected components of an open sets are open and that an open connected set is arcwise connected. If J is a Jordan curve then it is the continuous image of the compact set S¹ and so J is also compact. Therefore, its complement is open and the connected components of J^C are connected. The following lemma is a fairly obvious conclusion of this. A square curve is a continuous curve which consists entirely of line segments which are either horizontal or vertical. Thus, given a square curve, you can always consider it as part of a grating.

Proof: Let V denote those points of U which can be joined to x by a square curve. Then if z ∈ V, there exists B

⊆ U. It is clear that every point of B can be joined to z by a square curve. Also V ^C must be open since if z ∈ V ^C,B ⊆ U for some r. Then if any w ∈ B is in V, one could join w to z by a square curve and conclude that z ∈ V after all. The fact that both V,V ^C are both open would result in a contradiction unless both x,y ∈ V since otherwise, U is separated by V,V ^C. ■

Proof: To begin with consider the claim there are no more than two components. Suppose this is not so. Then there exist x,y,z each of which is in a different component of J^C. Let J = H ∪ K where H and K are two Jordan arcs joined at the points a and b. If the Jordan curve is r

where r = r as described above, you could take H = r and K = r. Thus the points on the Jordan curve illustrated in the following picture could be

(c+ d ) a = r (c),b = r----- 2