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
Definition 2.9.1 A grating G is a finite set of horizontal and vertical lines, each of which separates the plane. The grating divides the plane into two dimensional domains the closures of which are called 2 cells of G. The 1 cells of G are the edges of the 2 cells and the 0 cells of G are the end points of the 1 cells.
For k = 0,1,2, one speaks of k chains. For

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,

so
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.
Definition 2.9.2 Next, the boundary operator is defined. This is denoted by ∂. ∂ takes k cells to k − 1 chains. If a is a 2 cell, then ∂a consists of the edges of a. If a is a 1 cell, then ∂a consists of the ends of the 1 cell. If a is a 0 cell, then ∂a ≡ 0. This extends in a natural way to k chains. For


A k chain C is called a cycle if ∂C = 0.
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? C_{1} 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. C_{2} 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 C_{2}? 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.
Lemma 2.9.3 If C is a bounded 1 cycle (∂C = 0), then there are exactly two disjoint 2 chains D_{1},D_{2} such that

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 D_{1},D_{2} 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 D_{1} and D_{2} are uniquely determined assuming there are only two such 2 chains having boundary C. If you had a third, D_{3}, then let e be an edge of d ∈ ∂D_{3}. It would then need to be an edge of two 2 cells, one in D_{1} and one in D_{2} 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.
Lemma 2.9.4 Let C be a bounded 1 chain such that ∂C = x + y. Then both x,y are contained in a continuous curve which is a subset of
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
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 F_{1} and you can miss F_{2} in going from x to y, then you can miss both F_{1} and F_{2} by climbing around F_{1}.
Lemma 2.9.5 Let F_{1} be compact and F_{2} closed. Suppose C_{1},C_{2} are two bounded 1 chains in a grating. Suppose also that this grating has no unbounded 2 cells intersecting F_{1}. Suppose ∂C_{i} = x+y where x,y
Proof: Let a_{1},a_{2},

This is a 1 chain and ∂C = x + y because ∂∂a_{k} = 0. Then
Does it also avoid F_{1}? Suppose to the contrary that l is a 1 cell of C which does intersect F_{1}.
If
Therefore, if b is the 2 cell adjacent to a_{k} which shares l as an edge, it must follow b
Lemma 2.9.6 Let C be a bounded 1 cycle such that
Proof: If p is a limit point of
Definition 2.9.7 A Jordan arc is a set of points of the form Γ ≡ r
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
Then C_{1} + C_{2} is a 1 cycle and so by Lemma 2.9.3 there are exactly two 2 chains whose boundary is C_{1} + C_{2}. Since z

Thus, continuing this way, there is a sequence of Jordan arcs p_{k}q_{k} where 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
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,

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}
Definition 2.9.10 Let r map
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 ∈
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.
Lemma 2.9.11 Let K be a compact set in ℝ^{n} and let f : K → ℝ^{m} be continuous and one to one. Then f^{−1} : f
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
Lemma 2.9.12 J is a simple closed curve if and only if there exists a mapping θ : S^{1} → J where S^{1} is the unit circle

such that θ is one to one and continuous.
Proof: Suppose J = r

The last equality holds because r

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
Conversely suppose J is the image of the unit circle as just explained. Then let R :
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
Corollary 2.9.13 The following are equivalent.
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 S^{1} 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.
Lemma 2.9.14 Let U be a connected open set and let x,y be points of U. Then there is a square curve which joins x and y.
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
Theorem 2.9.15 Let J be a Jordan curve in the plane. Then J^{C} consists of exactly two components, a bounded component, and an unbounded component, and J is the frontier of both of these components. Furthermore, J has empty interior.
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
