1. Introduction
Though the definition of the Jordan Curve Theorem is not hermetic at all, the proof of the theorem is quite formidable and has experienced ups and downs throughout history.
Bernard Bolzano was the first person who formulated a precise conjecture: it was not self-evident but required a hard proof. However, the Jordan Curve Theorem was named after Camille Jordan, a mathematician who came up with the first proof in his lectures on real analysis and published his findings in his book [1], yet critics doubted the completeness of his proof. After that, using very complicated methods in 1905, Oswald Veblen was generally recognized as the first person who rigorously proved the Jordan Curve Theorem [2]. Veblen also commented that “His (Jordon’s) proof, however, is unsatisfactory to many mathematicians. It assumes the theorem without proof in the important special case of a simple polygon, and of the argument from that point on, one must admit at least that all details are not given” [2]. However, in 2007, Thomas C. Hales demurred Veblen’s comments and wrote that “Nearly every modern citation that I have found agrees that the first correct proof is due to Veblen ... In view of the heavy criticism of Jordan’s proof, I was surprised when I sat down to read his proof to find nothing objectionable about it. Since then, I have contacted a number of the authors who have criticized Jordan, and each case the author has admitted to having no direct knowledge of an error in Jordan’s proof” [3]. Hales also explained that the special case of simple polygons is not an easy exercise, but is not really in the way of Jordan’s proof. He quoted Michael Reeken as saying “Jordan’s proof is essentially correct ... Jordan’s proof does not present the details in a satisfactory way. But the idea is right, and with some polishing, the proof would be impeccable” [3]. Just like Hales’s judgement that vindicates the significance of Jordan’s work, Schoenflies critically analyzed and completed Jordan’s proof in 1924 [4]. However, the settlement of most controversies brought by Jordan’s prove didn’t diminish the enthusiasm of proving the Jordan Curve Theorem. Elementary proofs were presented by Filippov [5] in 1950 and Tverberg [6] in 1980. Maehara in 1984 used the Brouwer fixed point theorem to prove it [7]. A proof using non-planarity of the complete bipartite graph K3,3 was given by Thomassen in 1992 [8]. Although the Jordan Curve Theorem had been successfully proved in the 20th century, mathematicians still sought more formal ways to prove it in the 21st century. The formal proof was provided by an international team of mathematicians using the Mizar system in 2005 and Hales in 2007 [3], respectively, and both of which rely on libraries of previously proved theorems.
This paper is intended to strengthen Tverberg’s claim. Tverberg once suggested in his paper that “Although the JCT is one of the best known topological theorems, there are many, even among professional mathematicians, who have never read a proof of it. The present paper is intended to provide a reasonably short and self-contained proof or at least, failing that, to point at the need for one” [6]. Tverberg’s paper is hard to understand generally reflected by those who have read it. So the following parts will support more details and graphs, so as to make the way of the proof more scrutable and credible.
It is comparatively easy to prove that the Jordan curve theorem holds for every Jordan polygon in Lemma 1, and every Jordan curve can be approximated arbitrarily well by a Jordan polygon in Lemma 2. Then Lemma 3 and Lemma 4 deal with the situation in limiting processes to prevent the cases from the polygons that may thin to zero somewhere.
2. Preliminaries
Let C be the unit circle
, a Jordan curve
is the image of C under an injective continuous mapping
into
, i.e., a simple closed curve on the plane. We have the following fundamental fact.
Jordan Curve Theorem [1] (JCT):
has exactly two connected components.
We introduce here some terms from analysis that will be used later.
First, we may recall some basic knowledge in real analysis.
We say that
is a lower bound for a set
if each
satisfies
. And a lower bound for S that is greater than all other lower bounds for S is a greatest lower bound for S. The greatest lower bound for S is denoted
or
.
Since C is compact, the mapping
is uniformly continuous on C, its inverse
:
is also uniformly continuous. Next, if A and B are nonempty disjoint compact sets in
, then
[9]
Note that
is always positive, since, otherwise we have a sequence of points
,
,
, so that
as
. Since A is compact,
has a subsequential limit, say
, so that for any
and sufficiently large N, by triangular inequality,
then a is a limit point of B thus
contradicts to
We assume that the unit circle C is consistent with the natural parametrisation
. A Jordan curve
is said to be a Jordan polygon if there is a partition
of the interval
(i.e.,
), such that
for
,
,
are all real constants.
We call the pair
a realisation of a polygon
.
Remark. In accordance with the choice of partition
, we could define edges and vertices in a natural way. It is possible that adjacent edges of a polygon
lie on the same line (Figure 1).
3. Lemmas
We divide the proof of JCT into several steps. Lemma 1 below shows that JCT indeed holds for Jordan polygons. Lemma 2 shows every Jordan curve could be approximated uniformly by a sequence of Jordan polygons. Lemmas 3 and 4 provide certain metric description of Jordan polygons, which helps to evaluate the limit.
Lemma 1. The Jordan curve theorem holds for every Jordan polygon
with realisation
.
Proof. Denote edges of
to be
, and vertices to be
, (so
and
) with
[6]
1)
has at most two components.
Let
and let
(Figure 2).
By definition no points of
belongs to
, where
(recall that
). so
and it is clear that
consists of two connected components, say,
and
. We may assume, by elementary analytic geometry, (Figure 3)
Therefore
and
are both connected. For any p in
, we could find a line segment from p with length between
and
without intersecting
that connects p to one of
or
, as Figure 4 shown.
Therefore
has at most two components.
2)
has at least two components.
We place
in the coordinate system so that for vertices
, all
are different. For each vertex
, we draw a vertical line
through
, and notice that each such
passes through exactly one vertex, namely,
(Figure 5).
![]()
Figure 1. An example of Jordan polygon. Letters are vertices. Particularly, there exist letters, as the point M suggests, that might be on the edge, which shows the case that the neighbouring edges of the very point are on the same line.
![]()
Figure 2. Ni consists of a rectangle and two semicircles on opposite edges. It sometimes refers to a sausage domain.
![]()
Figure 3. Edges of Jordan polygon partition
into two connected components.
![]()
Figure 4. A line segment connecting p to some Ni.
We say a line
drawn as above is of type 1 if
and
are in opposite sides of
, i.e.,
, and of type 2 if
and
are in same side of
, i.e.,
. We say a vertex
is of type 1 or 2 if the corresponding
is of type 1 or 2 respectively. Observe that every
is of either type 1 or type 2. The same is true for
’s.
We now partition
into odd and even points and show that no continuous path connects an odd point to an even point.
For every point
in
we let
be the upward vertical ray from p and
be the number of intersection between
and
, with the provision that if the
passes some
![]()
Figure 5. Type 1 or 2 of vertical lines through each vertex, plus a point p with m(p) = 2.
vertex
of type 2 (there is only one such, by our choice of coordinate), then the intersection with
is not counted into
.
None of edges
in
is vertical hence
for every p. We call a point
is odd if
is odd, and is even if
is even. This is clearly a partition of
.
For every
there is an
so that p and q are in the same parity whenever
. To see this, we temporarily let
and notice that for every
, either
1)
; or
2)
for some i,
passes a vertex
of type 1; or
3)
for some i,
passes a vertex
of type 2; or
4)
for some i, but
does not pass through any vertex.
For case (1) we take
to be the radius of the open ball centered at p without intersecting
, For cases (2) and (4) take
be the radius of the open ball centered at p intersects
on
only, For case (3) we can find an open ball centered at p with radius
intersecting
on
only, as in case (2) or (4), and by our definition of
, any q in the open ball has
or
(Figure 6).
We have shown that every point sufficiently close to an odd (resp.even) point is odd (resp.even). If
is a continuous path with
is odd and
is even, let
![]()
then
and for some
we have
is odd for all
, while
is even for all
. By continuity of
,
would be neither odd or even, which is absurd (Figure 7).
It remains to show that there do exist even points and odd points.
Take any p so that
, then there is an open ball B centred at the last (or, highest) intersection between
and
that counted into
, which contains only one connected component of
(namely, the component where our intersection lies in). B is divided into upper half and lower half by that component of
. Any point
of
in the lower half of B will have
hence
is odd.
Take a sufficiently large open ball
covering
, any point q outside
would have
hence q is even (Figure 8).
We have shown that
has indeed two connected components. Therefore the proof of Lemma 1 is complete.
Remark. The above proof has shown that a Jordan polygon divides the plane into a bounded connected component O and an unbounded connected component V. Moreover, O and V are nonempty and open.
![]()
Figure 6. Points in
belong to four different cases.
![]()
Figure 8. An odd point p0 and an even point q.
Next we show that any Jordan curve
could be approximated uniformly by a sequence of Jordan polygons (
).
Lemma 2. Every Jordan Curve
with parametrisation
could be approximated arbitrarily well by a Jordan polygon
with parametrisation
. Moreover, it should be noted, the vertices on
all lie on
.
Proof. We want to show that given any
, there is some Jordan polygon
so that
, ![]()
By uniform continuity we can choose
such that
[6]
and
such that
[6]
Put
.
If we fill the plane by squares with vertices
, then the side length of each square is
. And the distance between every pair of points in the same square, say S, is smaller than
, therefore either
, or
[6]
Each
is contained in a unique minimal arc A on C, with radian shorter than
(Figure 9).
We now construct
stated in the lemma in a finite process. Observe that
meets only finitely many of the squares, we label them
.
We first change
into another Jordan curve
. Let
be the unique minimal circular arc
on C containing
. Define
outside
, and for
,
, where
are chosen so that
is continuous on C. Note
that when
,
, and thus has diameter less than
; it could even be empty (Figures 10-12).
The procedure would end in at most n steps as
intersects only n squares. We claim that
is our desired
(Figure 13).
By above argument each circular arc
on C containing
, if nonempty, has radian less than
. Consider now any
for which
. There is i so that
. By construction a belongs to the arc
, with endpoints b, c, where
in the same square
, say. By above procedure again we have
,
. Then
![]()
Because
and
, we have
. Therefore
, showing that ![]()
![]()
Figure 11. Each square corresponds to an arc on the parametrising circle C.
![]()
Figure 13. Polygonal approximation for curve in Figure 11.
Remark. Notice the two points on the unit circle C of distance
apart correspond to endpoints of
an arc of radian
. This radian produced an inscribed equilateral triangle which is the “smallest” regular
polygon one could produce on the unit circle. it helps us, in the above process, really produce a polygon for each
.
We can now approximate
by taking (uniform) limit of a sequence of Jordan polygons (
). To show JCT holds for general
we need to eliminate cases that
a)
has no bounded connected component while each
has;
b)
has more than two bounded components.
The following lemmas 3 and 4 ensure that (a) and (b) mentioned above would not happen in the limiting process.
For every point
where O is the open bounded component enclosed by a Jordan polygon, we claim that there exists a circle in O containing p that meets
in more than one point. There is an open disc D centred at p inside O. We enlarge the radius of D until boundary of D meets
. If they meet in more than one point we are done. Otherwise suppose
is the only intersection, we consider a variable circle through
, with center moves from p in the direction
until it meets
in another point
. a and b are on the unit circle C thus
is bounded. By Bolzano-Weierestrass there exists a disc D as described with
maximal (Figure 14).
Lemma 3. Let
be a Jordan polygon. Then the bounded component of
contains a disc, on the boundary circle of which are two points
and
, with
.
Proof. The existence of the disc D with
maximal is clear from above discussion. Assume
. Then a and b are the endpoints of an arc A of radian
. The boundary circle cannot meet
as
for every c in
(Figure 15).
Let
be the vertices of
on
as met when passing from
to
. By Lemma 1, the chord
divides the interior of Jordan polygon into two parts, say P and Q. We have
are in the same component, say
. We claim that
and
are in the same side of the
(Figure 16).
Suppose
and
are in opposite sides of
. Notice that there is no vertex between
and
, neither between
and
. From the proof in Lemma 1, we can find
such that any open ball centred at a point in
intersects
on a single connected component of
. Take
such that
and that
. Let
be a path in
from
to
,
the path from
to
a point with distance to the midpoint of
less than
, similarly,
the path from
to
a point with distance to the midpoint of
less than
, notice that
,
are in different sides of
. But this leads to a contradiction:
doesn’t separate the interior of
.
Now a and b could either belong to those vertices or not. We shall divide the situation into different cases (Figure 17).
1)
and
.
![]()
Figure 14. There is always a circle containing p meets
at two points.
![]()
Figure 15. Positions on A of vertices as met when moving from
to
.
In this case the boundary of D is tangent to both
and
. Consider a circle touching segment
at a point
very near
and touching segment
at a point
very near
, where
. Provided they are close enough, the now circle would meet
in those two points only. As
contradiction arises (Figure 18).
2)
and
.
The boundary of D passes
and is tangent to
. We choose a circle passing
and touching
at a point
very close to
where
. A contradiction similar to case (1) arises (Figure 19).
3)
and
.
We consider a variable circle through
and
and move its center from center of D to the domain bounded by the radii of D to
and
together with
. The circle will eventually either
meet
at some point
contradict to maximality of
as
for every c in
; or
become tangent to segment
or
reduced to cases (1) or (2).
Consider
a Jordan polygon in
which divides the plane into two components. Let X be one of those components. By a chord S in X we mean a line segment in X (except for its endpoints) between two distinct points on
. By Lemma 1,
consists of two connected open sets. Fix two points a and b, suppose
, and assume that for every S of length less than 2, a and b are in the same component of
. Then we have:
Lemma 4. Under the assumptions stated there is a continuous path
from a to b such that
.
Proof. We first note that if
is any point in
, connected to a by a continuous path
, where
. If S is any chord of length less than 2,
would not pass S so a and
are in the same component of
. By assumption a and b are in the same component, so are
and b.
Hence we may assume now that
.
Choose
and
on
(so
). Let D be a mobile circle initially placed with its center c in a. We roll the circle along
starting at
until c falls in b. The desired path
will be obtained as the curve followed by c.
D arrives at
.
Suppose not. At some position, D and
has some common chord
of length less than 2 (see Figure 20).
consists of two components, say Y and Z. By assumption a and b are in the
same component, say Y, replacing
by c in the first paragraph c is also in Y. For D doesn’t reach
,
must lie in the boundary of Z, which is between
and
on
.
Now draw a unit circle E centred at b. The circle meets
at
. Draw the radius of E to
. Since b and
are in the opposite sides of S, the radius crosses S.
As E lies in X, and
, E encompasses neither
nor
while D does. So E intersects S at two points. Now b and c lie to the same side of S, with
and thus
. The radius of E from b to
must fall inside or on D so can never reach
, which is absurd.
c arrives at b
We have verified that D reaches
, a point on
of distance 1 away from b. We have to show that at this position, c is in b.
The statement is clear when
is not a vertex. The problem happens when
is a vertex, and there is a common chord
of length less than 2 between c and b, where y is some point on
(see Figure 21).
4. Proof of Jordan’s Theorem
has at least two components:
The existence of unbounded component is clear. For the bounded one, draw a large closed disc D with boundary circle
around
, let
be a sequence of Jordan polygons converging uniformly to
. We may assume that all
lie inside
. By Lemma 3, there is a circle
centred
in the bounded component on
so that there are points
with
. By Bolzano-Weierstrass theorem again there is a limit point, say z, in D. Passing to a subsequence of the original one we may assume that
as
.
By uniform continuity there exists
such that
By our choice of
we have
. since
, There exists
such that
Therefore
, so
.Since
we also have ![]()
Taking
we have for
![]()
Figure 21. But this is not possible: again replacing
in the first paragraph by c we have a, b, and c are in the same component of
, so b and c should be in the same side.
![]()
so z and
are in the same component of
. The same is true for
.
If z is in the unbounded component of
, we could find a continuous path
from z to a point outside
. Set
. Again by our choice of (
), there exists
such that for
,
![]()
so for such n,
![]()
and some
such that for
,
![]()
Taking
, we have
so
and z are both in the unbounded component of
, which contradicts to the definition of
.
has at most two components: Suppose not. Let
be points from three distinct components of
, with
. Let
uniformly as we did. Then
.
For any
, two of the three points should be in the same component
of
. By passing to a subsequence we may assume that p and q are in
for all n.
Assume first that there is some
and infinitely many n so that p and q are connected by a
continuous path
with
, as
, for large n,
, i.e., p and q could be
connected by a path not intersecting
, which is not possible. Hence there is no such
.
By above argument we conclude that for any
any path
connecting p and q would have
for all but finitely many
. It yields, together with Lemma 4, an increasing sequence
, and a sequence of chords
, so that one has:
1) p and q are in different components of
;
2)
as
, where
and
are endpoints of
.
Let T be the component of
bounded by
and
where
is the smaller arc on C with endpoints
and
. Suppose, without loss of generality,
. Now since
and
is homeomorphic, as
,
, thus
and therefore
. For large i,
![]()
In particular, we have
, contradicts to
.
5. Conclusion
So far the proof of Jordan Curve Theorem is done. The four lemmas are of paramount importance in the whole proof. Especially, the first lemma, which Jordan was in an inappropriate way of demonstration, seems to be intuitive, however, requires rigorous proof. This theorem is discussed in the
and it is a useful theorem to bolster the basis of the edifice in Topology.
AcknowledgementS
I’m very grateful that the Professor Charles C. Pugh from the Department of Mathematics, University of California of Berkeley helped me a lot with my study of Topology. And I appreciate my groupmate Haiqi Wu from Oxford helped me with my paper. And thanks to the platform provided by CIS organization, I have the chance to study topology with the emeritus professor and hence accomplish the paper. Also if it were not for the teachers from my alma mater Wuhan University, I wouldn’t have been able to walk on my avenue of Mathematics with determination. Last but not least, I would like to thank my parents, and it’s they who are always supporting me behind.