https://www.scirp.org/journal/ns Natural Science, 2019, Vol. 11, (No. 12), pp: 345-360
A Proof of the Jordan Curve Theorem
Xing Zhang
School of Mathematics and Statistics, Wuhan University, Wuhan, China Correspondence to: Xing Zhang, Keywords: Jordan Curve Theorem, JCT, Tverberg, Jordan Polygon, Sausage Domain, Jordan Curve Received: October 29, 2019 Accepted: December 17, 2019 Published: December 20, 2019 Copyright © 2019 by author(s) and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0). http://creativecommons.org/licenses/by/4.0/ Open Access
ABSTRACT The proof of the Jordan Curve Theorem (JCT) in this paper is focused on a graphic illustra- tion and analysis ways so as to make the topological proof more understandable, and is based on the Tverberg’s method, which is acknowledged as being quite esoteric with no graphic explanations. The preliminary constructs a parametrisation model for Jordan Po- lygons. It takes quite a length to introduce four lemmas since the proof by Jordan Polygon is the approach we want to concern about. Lemmas show that JCT holds for Jordan polygon and Jordan curve could be approximated uniformly by a sequence of Jordan polygons. Also, lemmas provide a certain metric description of Jordan polygons to help evaluate the limit. The final part is the proof of the theorem on the premise of introduced preliminary and lemmas.
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
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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 {( xy,|) x22+= y 1}, a Jordan curve Γ is the image of C under an injective 2 continuous mapping γ into , i.e., a simple closed curve on the plane. We have the following fundamental fact. 2 Jordan Curve Theorem [1] (JCT): \ Γ 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 M ∈ is a lower bound for a set S ⊂ if each sS∈ satisfies sM≥ . 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 glb.. ( S) or inf( S ) . Since C is compact, the mapping γ is uniformly continuous on C, its inverse γ −1 : Γ→C is also 2 uniformly continuous. Next, if A and B are nonempty disjoint compact sets in , then d( AB,) := inf{ a − b : a ∈∈ Ab , B} [9]
Note that d( AB, ) is always positive, since, otherwise we have a sequence of points (an ) , (bn ) , n∈ , so that abii−→0 as i →∞. Since A is compact, (an ) has a subsequential limit, say aA∈ , so that for any and sufficiently large N, by triangular inequality,
ab−≤−+−≤N aa N a NN b then a is a limit point of B thus aB∈ contradicts to AB = ∅. We assume that the unit circle C is consistent with the natural parametrisation t (θ) = (cos θ ,sin θθ) ,∈[ 0,2π]. A Jordan curve Γ is said to be a Jordan polygon if there is a partition Θ={θθ01,, , θn} of the interval [0,2π] (i.e., 02=<<<θθ01 θn =π ), such that
γ((cos θθ ,sin )) =++(ai θ bc ii, θ d i)
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for θθθ∈[ ii−1, ] , in∈{1, , } , abcdiii,,, i are all real constants. We call the pair (γ ,Θ) a realisation of a polygon Γ . Remark. In accordance with the choice of partition Θ , we could define edges and vertices in a natural way. It is possible that adjacent edges of a polygon Γ 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 Γ with realisation (γ ,Θ) .
Proof. Denote edges of Γ to be EE12, ,, En , and vertices to be vv12,,, vn , (so Ei= γθ(( ii−1, θ)) and vii= γθ( ) ) with
Eii E+1={ v i}( i =1, , nE ,n++1 = E 111 , v i = v) [6] 2 1) \ Γ has at most two components. 2 Let δ := min{dEE( ij, ) | E iand E j are not adjacent} and let Nii:=∈<{ q |, d( qE) δ} (Figure 2).
By definition no points of Ni belongs to E j , where j≠− i1, ii , + 1 (recall that EE0 = n ). so NiΓ⊂ E i−+11 EE ii and it is clear that Ni \ Γ consists of two connected components, say, Ni′ and Ni′′ . We may assume, by elementary analytic geometry, (Figure 3)
NNii′ ′++11≠∅, NN ii ′′ ′′ ≠∅, i = 1, , n ′′′′ ′′ 2 Therefore NN1 n and NN1 n are both connected. For any p in \ Ni , we could find a line segment from p with length between dp( ,Γ) and dp( ,Γ−) δ without intersecting Γ that connects p to one of Ni′ or Ni′′ , as Figure 4 shown. 2 Therefore \ Γ has at most two components. 2 2) \ Γ has at least two components. We place Γ in the coordinate system so that for vertices {vi=( xy ii,) | i = 1, , n} , all xi are different. For each vertex vi , we draw a vertical line li through vi , and notice that each such li passes through exactly one vertex, namely, vi (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.
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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
NNi i +1 into two connected components.
Figure 4. A line segment connecting p to some Ni.
We say a line li drawn as above is of type 1 if vi−1 and vi+1 are in opposite sides of li , i.e., ( xi−+11− xx ii)( −< x i) 0 , and of type 2 if vi−1 and vi+1 are in same side of li , i.e., ( xi−+11− xx ii)( −> x i) 0 . We say a vertex vi is of type 1 or 2 if the corresponding li is of type 1 or 2 respectively. Observe that every li is of either type 1 or type 2. The same is true for vi ’s. 2 We now partition \ Γ into odd and even points and show that no continuous path connects an odd point to an even point. 2 For every point p= ( xp( ), yp( )) in \ Γ we let lp be the upward vertical ray from p and mp( ) be the number of intersection between Γ and lp , with the provision that if the lp passes some
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Figure 5. Type 1 or 2 of vertical lines through each vertex, plus a point p with m(p) = 2. vertex vi of type 2 (there is only one such, by our choice of coordinate), then the intersection with vi is not counted into mp( ) . 2 None of edges Ei in Γ is vertical hence mp( )∈ for every p. We call a point p ∈Γ \ is odd 2 if mp( ) is odd, and is even if mp( ) is even. This is clearly a partition of \ Γ . 2 For every p ∈Γ \ there is an > 0 so that p and q are in the same parity whenever d( pq, ) < . Γ=Γ n ∈Γ2 To see this, we temporarily let 1 i=1li and notice that for every p \ , either 2 1) p ∈Γ \ 1 ; or 2) pl∈ i for some i, lp passes a vertex vi of type 1; or 3) pl∈ i for some i, lp passes a vertex vi of type 2; or 4) pl∈ i for some i, but lp does not pass through any vertex. For case (1) we take to be the radius of the open ball centered at p without intersecting Γ1 , For cases (2) and (4) take be the radius of the open ball centered at p intersects Γ1 on li only, For case
(3) we can find an open ball centered at p with radius intersecting Γ1 on li only, as in case (2) or (4), and by our definition of mp( ) , any q in the open ball has mq( ) = mp( ) or mp( ) + 2 (Figure 6). We have shown that every point sufficiently close to an odd (resp.even) point is odd (resp.even). If 2 Π:[ 0,1] →Γ \ is a continuous path with Π(0) is odd and Π(1) is even, let
tt0 := inf({ ∈Π[ 0,1|] ( t) iseven}) then t0 > 0 and for some > 0 we have Π(t) is odd for all t00−<< tt, while Π(t) is even for all t00<< tt + . By continuity of Π , Π(t0 ) 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 mp( ) ≠ 0 , then there is an open ball B centred at the last (or, highest) intersection between Γ and lp that counted into mp( ) , which contains only one connected component of Γ (namely, the component where our intersection lies in). B is divided into upper half and lower half by that component of Γ . Any point p0 of lp in the lower half of B will have mp( 0 ) =1 hence p0 is odd. 2 Take a sufficiently large open ball B′∈ covering Γ , any point q outside B′ would have mq( ) = 0 hence q is even (Figure 8). 2 We have shown that \ Γ 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.
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2 Figure 6. Points in \ Γ belong to four different cases.
Figure 7. Behaviour of Π around t0.
Figure 8. An odd point p0 and an even point q.
Next we show that any Jordan curve Γ could be approximated uniformly by a sequence of Jordan polygons ( Γn ). Lemma 2. Every Jordan Curve Γ with parametrisation γ could be approximated arbitrarily well by a Jordan polygon Γ′ with parametrisation γ ′ . Moreover, it should be noted, the vertices on Γ′ all lie on Γ .
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Proof. We want to show that given any > 0 , there is some Jordan polygon Γ′ so that γ(cos θθγ ,sin) −<′( cos θθ ,sin ) , ∀∈(cosθθ ,sin ) C 2
By uniform continuity we can choose 1 > 0 such that pq−≤⇒ γγ( p) −( q) < [6] 1 2 and 2 > 0 such that
γγ( p) −( q) ≤⇒21 pq − −1 diam(γ ( S )) < min(1 , 3) [6] − 2π Each γ 1 (S ) is contained in a unique minimal arc A on C, with radian shorter than (Figure 9). 3 We now construct Γ′ stated in the lemma in a finite process. Observe that Γ meets only finitely many of the squares, we label them SS12,,, Sn . We first change Γ=Γ0 into another Jordan curve Γ1 . Let A1 be the unique minimal circular arc −−11 on C containing γγ01(SS)(= ( 1)) . Define γγ10= (= γ ) outside A1 , and for (cosθθ ,sin )∈ A1 , γ1(cos θθ ,sin ) =++(a1 θ bc 11, θ d 1) , where abcd111,,, 1 are chosen so that γ1 is continuous on C. Note −−11 that when i ≥ 2 , γγ10(SSii) ⊆ ( ) , and thus has diameter less than 3 ; it could even be empty (Figures 10-12). The procedure would end in at most n steps as Γ intersects only n squares. We claim that Γn is our desired Γ′ (Figure 13). −1 By above argument each circular arc Aj on C containing γ jj−1 (S ) , if nonempty, has radian less 2π than . Consider now any aC∈ for which γγ(aa) ≠ ( ) . There is i so that γ(aa) = γγ( ) ≠ − ( a) . 3 n nii1 By construction a belongs to the arc Ai , with endpoints b, c, where γγ(bc), ( ) in the same square Si , say. By above procedure again we have γγi (bb) = ( ) , γγi (cc) = ( ) . Then γγ(a) −n( a) = γγγ( ab) −+( ) ii( b) − γ( a) ≤−+γγδ(ab) ( ) ≤−+γγ(ab) ( ) 2 Because γγ(b) −( c) ≤≤ δ21 ⇒−≤ bc and ab−≤− bc, we have ab−≤1 . Therefore γγ(ab) −<( ) , showing that γγ(aa) −<( ) 2 n https://doi.org/10.4236/ns.2019.1112037 351 Natural Science Figure 9. The inscribed equilateral triangle of a circle. The triangle has side length equals to 3 . By restricting the diameter of inverse image of each square to be less than 3 , the Jordan curve meets at least three different squares so the approximation would not degenerate to a single line segment. (replace 3 by the side length of any inscribed regular polygon in a circle should also work.) Figure 10. Having produced Γi−1 , let Ai be the unique −1 −1 minimal arc on C containing γ ii−1 ( S ) . If γ ii−1 ( S ) = ∅ then let Ai := ∅ and Γ=Γii: −1 . Otherwise define γγii:= −1 outside Ai , and for (cosθθ, sin )∈ Ai , γi(cos θθ,, sin) :=++(ai θ bc ii θ d i) , where abcdiii,,, i are chosen so that γ i is continuous on C. https://doi.org/10.4236/ns.2019.1112037 352 Natural Science Figure 11. Each square corresponds to an arc on the parametrising circle C. Figure 12. The action on square S168. Figure 13. Polygonal approximation for curve in Figure 11. Remark. Notice the two points on the unit circle C of distance 3 apart correspond to endpoints of 2π an arc of radian . This radian produced an inscribed equilateral triangle which is the “smallest” regular 3 polygon one could produce on the unit circle. it helps us, in the above process, really produce a polygon for each Γi . We can now approximate Γ by taking (uniform) limit of a sequence of Jordan polygons ( Γn ). To show JCT holds for general Γ we need to eliminate cases that https://doi.org/10.4236/ns.2019.1112037 353 Natural Science 22 2 a) \Γ= \ lim Γn has no bounded connected component while each \ Γn has; 2 ( n→∞ ) b) \ Γ 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 pO∈ 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 Γ 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 Γ . If they meet in more than one point we are done. Otherwise suppose γ (a) is the only intersection, we consider a variable circle through γ (a) , with center moves from p in the direction γ (ap) until it meets Γ in another point γ (b) . a and b are on the unit circle C thus ab− is bounded. By Bolzano-Weierestrass there exists a disc D as described with ab− maximal (Figure 14). 2 Lemma 3. Let Γ be a Jordan polygon. Then the bounded component of \ Γ contains a disc, on the boundary circle of which are two points γ (a) and γ (b) , with ab−≥3 . Proof. The existence of the disc D with ab− maximal is clear from above discussion. Assume 4π ab−<3 . Then a and b are the endpoints of an arc A of radian > . The boundary circle cannot meet 3 γ( A) \,{ γγ( ab) ( )} as max{ acbc− , −} >− ab for every c in A\,{ ab} (Figure 15). Let γγ(vv1 ),, ( n ) be the vertices of Γ on γ ( A) as met when passing from γ (a) to γ (b) . By Lemma 1, the chord γγ(ab) ( ) divides the interior of Jordan polygon into two parts, say P and Q. We have γγ(vv1 ),, ( n ) are in the same component, say P . We claim that γ (v1 ) and γ (vn ) are in the same side of the γγ(ab) ( ) (Figure 16). Suppose γ (vn ) and γ (v1 ) are in opposite sides of γγ(ab) ( ) . Notice that there is no vertex between γ (a) and γ (v1 ) , neither between γ (b) and γ (vn ) . From the proof in Lemma 1, we can find δ > 0 such that any open ball centred at a point in Γ γγ(ab) ( ) intersects Γ γγ(ab) ( ) on a single connected component of Γ . Take αβ, ∈ P such that dv(γ( 1 ), αδ) < and that dv(γ( n ), βδ) < . Let π1 be a path in P from α to β , π 2 the path from α to α′ a point with distance to the midpoint of γγ(ab) ( ) less than δ , similarly, π 3 the path from β to β ′ a point with distance to the midpoint of γγ(ab) ( ) less than δ , notice that α′ , β′ are in different sides of γγ(ab) ( ) . But this leads to a contradiction: γγ(ab) ( ) doesn’t separate the interior of Γ . Now a and b could either belong to those vertices or not. We shall divide the situation into different cases (Figure 17). 1) av≠ 1 and bv≠ n . Figure 14. There is always a circle containing p meets Γ at two points. https://doi.org/10.4236/ns.2019.1112037 354 Natural Science Figure 15. Positions on A of vertices as met when moving from γ (a) to γ (b) . Figure 16. The path πππ213 in P connecting α′ and β ′ in both sides of γγ(ab) ( ) . Figure 17. Case 1. A circle touching Γ at points very close to γ (a) and γ (b) . https://doi.org/10.4236/ns.2019.1112037 355 Natural Science Figure 18. Case 2. A circle passing γ (a) and touching Γ at a point very close to γ (b) . In this case the boundary of D is tangent to both γγ(av) ( 1 ) and γγ(bv) ( n ) . Consider a circle touching segment γγ(av) ( 1 ) at a point γ (a′) very near γ (a) and touching segment γγ(bv) ( n ) at a point γ (b′) very near γ (b) , where ab′′, ∈ A. Provided they are close enough, the now circle would meet Γ in those two points only. As a′′− b >− ab contradiction arises (Figure 18). 2) av= 1 and bv≠ n . The boundary of D passes γ (a) and is tangent to γγ(bv) ( n ) . We choose a circle passing γ (a) and touching γγ(bv) ( n ) at a point γ (b′) very close to γ (b) where bA′∈ . A contradiction similar to case (1) arises (Figure 19). 3) av= 1 and bv= n . We consider a variable circle through γ (a) and γ (b) and move its center from center of D to the domain bounded by the radii of D to γ (a) and γ (b) together with γ ( A) . The circle will eventually either meet Γ at some point γ(c)∈ γ( A) \,{ γγ( ab) ( )} contradict to maximality of ab− as max{ acbc− , −} >− ab for every c in A\,{ ab}; or become tangent to segment γγ(av) ( 2 ) or γγ(bv) ( n−1 ) reduced to cases (1) or (2). 2 Consider Γ 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 Γ . By Lemma 1, XS\ consists of two connected open sets. Fix two points a and b, suppose d(Γ≥,,{ ab}) 1, and assume that for every S of length less than 2, a and b are in the same component of XS\ . Then we have: Lemma 4. Under the assumptions stated there is a continuous path Π from a to b such that d (ΠΓ,1) ≥ . 2 Proof. We first note that if a′ is any point in , connected to a by a continuous path Π′ , where d (ΠΓ≥′,1) . If S is any chord of length less than 2, Π′ would not pass S so a and a′ are in the same component of XS\ . By assumption a and b are in the same component, so are a′ and b. Hence we may assume now that da( ,Γ=) db( ,1 Γ=) . Choose γ (ua ) and γ (ub ) on Γ (so uuab, ∈ C). Let D be a mobile circle initially placed with its center c in a. We roll the circle along Γ starting at γ (ua ) until c falls in b. The desired path Π will be obtained as the curve followed by c. D arrives at γ (ub ) . Suppose not. At some position, D and Γ has some common chord S= γγ( uu12) ( ) of length less than 2 (see Figure 20). XS\ consists of two components, say Y and Z. By assumption a and b are in the https://doi.org/10.4236/ns.2019.1112037 356 Natural Science Figure 19. Case 3. As the variable circle passing two fixed points γ (a) and γ (b) while its center moves away, it will either meet the third point of Γ , which contradicts our construction, or it will eventually tangent to γγ(av) ( 1 ) or γγ(bv) ( n ) or both. Figure 20. The radius of E towards γ (ub ) would never meet γ (ub ) . Note that E meets γγ(uu12) ( ) at two points. same component, say Y, replacing a′ by c in the first paragraph c is also in Y. For D doesn’t reach γ (ub ) , γ (ub ) must lie in the boundary of Z, which is between γ (u1 ) and γ (u2 ) on Γ . Now draw a unit circle E centred at b. The circle meets Γ at γ (ub ) . Draw the radius of E to γ (ub ) . Since b and γ (ub ) are in the opposite sides of S, the radius crosses S. As E lies in X, and ED≠ , E encompasses neither γ (u1 ) nor γ (u2 ) while D does. So E intersects S at two points. Now b and c lie to the same side of S, with d( bS,,) > d( cS) and thus d( bZ,,) > ( cZ) . The radius of E from b to γ (ub ) must fall inside or on D so can never reach γ (ub ) , which is absurd. https://doi.org/10.4236/ns.2019.1112037 357 Natural Science c arrives at b We have verified that D reaches γ (ub ) , a point on Γ of distance 1 away from b. We have to show that at this position, c is in b. The statement is clear when γ (ub ) is not a vertex. The problem happens when γ (ub ) is a vertex, and there is a common chord S= γ ( uyb ) of length less than 2 between c and b, where y is some point on Γ (see Figure 21). 4. PROOF OF JORDAN’S THEOREM 2 \ Γ 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 C0 around Γ , let ΓΓ12,, be a sequence of Jordan polygons converging uniformly to Γ . We may assume that all Γn lie inside C0 . By Lemma 3, there is a circle Cn centred 2 zn in the bounded component on \ Γn so that there are points γγnn(ab), nn( )∈Γ n C n with abnn−≥3 . 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 zzn → as n →∞. By uniform continuity there exists > 0 such that ab−≥3 ⇒γγ( a) −( b) ≥ By our choice of (abnn, ) we have γγ(abnn) −≥( ) . since γγn → , There exists N1 ∈ such that γγ(a) −>( b) if nN > nn2 1 Therefore diam( C ) > , so dz( ,Γ>) . n 2 nn 4 Since zzn → we also have N2 ∈ d( zz, ) =−< z z nn8 Taking N:= max{ NN12 , } we have for nN> Figure 21. But this is not possible: again replacing a′ in the first paragraph by c we have a, b, and c are in the same component of XS\ , so b and c should be in the same side. https://doi.org/10.4236/ns.2019.1112037 358 Natural Science dz( ,Γ≥) d( Γ ,, z) − dz( z) ≥ n nn n 8 2 2 so z and zn are in the same component of \ Γn . The same is true for \ Γ . 2 If z is in the unbounded component of \ Γ , we could find a continuous path Π from z to a point outside C0 . Set δ =d ( ΠΓ, ) . Again by our choice of ( Γn ), there exists N1 ∈ such that for nN> 1 , δ d (γγ, ) = γ −< γ nn2 so for such n, δ dz( ,,Γ) ≥ d( ΠΓ) > nn2 and some N2 ∈ such that for nN> 2 , δ dz( , z) =−< z z nn2 Taking N> max{ NN12 , } , we have dzz( ,,nn) ≤Γ dz( ) so zn and z are both in the unbounded 2 component of \ Γn , which contradicts to the definition of zn . 2 2 \ Γ has at most two components: Suppose not. Let pqr,,∈Γ \ be points from three distinct 2 components of \ Γ , with d(Γ=,{ pqr ,,}) . Let Γ12,, Γ →Γ uniformly as we did. Then d(Γ≥,{ pqr ,,}) . n 2 2 For any n∈ , two of the three points should be in the same component X n of \ Γ . By passing to a subsequence we may assume that p and q are in X n for all n. Assume first that there is some δ ∈(0,) and infinitely many n so that p and q are connected by a δ continuous path Π with d (ΠΓ, ) ≥δ , as Γ →Γ, for large n, d (ΓΠ, ) > , i.e., p and q could be n nn n n 2 connected by a path not intersecting Γ , which is not possible. Hence there is no such δ . By above argument we conclude that for any δ > 0 any path Π connecting p and q would have d (Γn , Π<) δ for all but finitely many δ . It yields, together with Lemma 4, an increasing sequence nn,, S⊂⊂ XS,, X 12 , and a sequence of chords 12nn12, so that one has: XS\ 1) p and q are in different components of nii ; γγab−→0 i →∞ γ a γ b S 2) niii( ) ni( ) as , where nii ( ) and nii ( ) are endpoints of i . XS\ S γ A A Let T be the component of nii bounded by i and nii ( ) where i is the smaller arc on C with endpoints ai and bi . Suppose, without loss of generality, pT∈ . Now since γγn → and γ is i →∞ γγab−→0 ab−→0 diamγ A → 0 homeomorphic, as , ( ii) ( ) , thus ii and therefore ( nii ( )) . For large i, diam T≤ diamγ A +< diam S ( ) ( nii ( )) ( i) In particular, we have pa−<γ ( i ) , contradicts to d(Γ>,{ pqr ,,}) . 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, 2 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 https://doi.org/10.4236/ns.2019.1112037 359 Natural Science 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. 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