A Concise Proof of Discrete Jordan Curve Theorem

Suppose that $C$ is a closed curve in a simply connected surface $S$. $C$ does not reach the border of $S$, i.e. $C\cap{\partial S}=\emptyset$.

We also check $C$ if it satisfies the condition of the wide angle for each triple consecutive vertices. Also we add Veblen points and edges to make $C$ to be wide as necessary. So we can assume that:
(1) The wideness of each angle is 3 or greater, and
(2) For any two nonadjacent vertices $p$ and $q$ in $C$, any path not including edges in $C$ from $p$ to $q$ much contain three edges that belong to three different 2-cells.

Assume point $p\in C$, then suppose that $q$ and $r$ are two adjacent points of $p$ in $C$ with form of $...qpr,...$, where the direction of … $q$ to $p$ to $r$ …to $p$ is clockwise. See Fig. 2 $\{p,r\}$ is a line-cell, then there are two 2-cells containing $\{p,r\}$. Denote these 2-cells by $A$ and $B$ with clockwise orientation.

Our strategy is to prove that if there is a point $a$ in $A$ which is not in $C$, and a point $b\in B$ and $b\notin C$, then any path from $a$ to $b$ must contain a point in $C$. Then we can see that $S-C$ are not (point-) connected and we have the Jordan curve theorem.

First, we want to prove that there must exist a point in $A-C$. If each point in $A$ is in $C$, since $A$ is a simple cycle, then $C=A$. However, $C$ is not a surface-cell, so the statement can not be true. Thus, there is a point $a\in A-C$. For the same reason there is a point $b\in B-C$. We assume that $a$ is the last such point in $A$ starting with $p$, and $b$ is the first such point in $B$ starting with $p$. (see Fig. 2(a)) We always assume clockwise direction here for cell $A$ and $B$ unless we indicate otherwise.

Let us make a summery of above idea: Suppose that $C$ is a closed curve. (If $C$ is a closed simple path, we allow $C$ is a 2-cell and we allow to assign a pseudo-point in the center of the 2-cell, we still can prove this theorem.) The idea of the proof is to find two points in each sides of the curve $C$. This is because that for any 1-cell $(r,p)$ in $C$, there are two 2-cells $A$,$B$ sharing $(r,p)$ by the discrete surface definition. $A$ must contain a vertex $a$ and $B$ must contain $b$, and they are not in $C$. $a$, $b$ are adjacent to some points in $C$, respectively. We are going to prove that from $a$ to $b$, any path must cross-over $C$. That is the most important part of the Jordan curve theorem.

We assume, on the contrary, there is a simple path from $a$ to $b$ does not cross-over $C$, called $P_{(}a,b)$ in Fig. 2 (a). But we know there is $P(b,a)$ in $A\cup B$ ( i.e. $P(b,a)=b\cdots rp\cdots a$) does cross-over $C$. $P_{(}a,b)\cup P_{(}a,b)$ is a cycle in clockwise. (Fig. 2 (a) )

We know $S(r)$ is the neighborhood of $r$ in $S$. So $S(r)$ contains all 2-cells containing $r$. The boundary of $S(r)$ is a simple closed curve. (This is because we always assume that $r$ is a regular point). $a$ is on the boundary of $S(r)$. (The boundary of $S(r)$ is $S(r)-\{r\}$). $A\cup B$ is a subset of $S(r)$.

We now prove that $P_{(}a,b)$ is not a subset of $S(r)$; otherwise, it must cross-over $C$. (a 2-cell containing $r$ must have an edge on $C$, or all points of the 2-cell are on the boundary of $S(r)$ except $r$). If $P(a,b)$ does not contain $r$, must be a part of boundary of $S(r)$ which is a cycle. $r$ has two adjacent points on $C$, (If they are not pseudo points, meaning here it can be eliminated or added on an edge that does not affect to the 2-cell) so these two points are also in the boundary of $S(r)$. So there are only two paths from $a$ to $b$ on the boundary of $S(r)$. These two points are not on the same side of the cross-over path containing $r$. (The boundary of $S(r)$ was separated by the cross-over path containing $r$.) $P_{(}a,b)$ must contain such a point that is on $C$.

Therefore we proved $P_{(}a,b)$ is not a subset of $S(r)$. Then $P_{(}a,b)\cup P(b,a)$ is a simple closed curve. ($P(b,a)$ passes $r$). By the definition of the simply-connected surface, there are finite numbers of paths $P(a,b)=P_{0}(a,b)$,…, $P_{n-1}(a,b)$, such that so that $P_{i}(a,b)$ and $P_{i+1}(a,b)$ are (side-)gradually varied.

In addition, $P_{n-1}(a,b)$ is gradually varied to $P_{n}(a,b)=P^{a}(b,a)$ (a reversed $P(b,a)$ that passes $r$).

We now can assume that there is a smallest $i$ such that $P_{i}(a,b)$ cross over $C$, but $P_{i-1}(a,b)$ does not. (Fig. 2(a) ). We will prove that is impossible if $P_{i-1}(a,b)$ does not cross over $C$.

Let point $x$ in $P_{i}(a,b)\cap C$ and $x\notin P_{i-1}(a,b)$. There are two cases: (1) cross over at a single point $x$ on $C$, or (2) cross over at a sequence of points on $C$. We will prove these two cases, respectively.

Case 1: Suppose that $x=``p^{\prime\prime}=``q^{\prime\prime}$ in Definition 1 (See Fig. 1 (a)(b)). It means two curves $P_{i}(a,b)$ and $C$ share just one point $x$. and assume $P_{i}(a,b)=...uxv...$ and $C=...cxd...$, where $v\neq d$.

We know that $u,v,c,d$ are in the boundary of $S(x)$, a simple cycle $S(x)-\{x\}$ (See Lemma 1). There is a 2-cell $X$ (in between $P_{i-1}$ and $P_{i}$) contains $(u,x)$. See Fig. 2 (b).

$X$ has a sequence of points $S1$ in $P_{i-1}$ and a sequence of points $S2$ in $P_{i}$. $X$ has at most two edges $e1$, $e2$ not in $P_{i-1}\cup P_{i}$ ; $S1$, $e1$, $S2$, $e2$, are the boundary of $X$. $e1$ is the edge linking $S1$ to $S2$, and $e2$ is the edge linking $S2$ to $S1$ counterclockwise.(Again, $e1$ may or may not be directly incident to $u$, and $e1$ may be an empty edge if $P_{i-1}$ intersects $P_{i}$ at point $u$. $e2$ may also in the same situation.) We might as well assume that $x$ is the first point on $P_{i}(a,b)$ (from $a$ to $b$ in path $P_{i}$ )that is in $C$. Thus, $c,d\notin X$. (If $c$ is in $X$ $c$ must be in $P_{i-1}$. if $d$ is in $X$, $x$ is not only the cross over point. )

If $X$ contains $v$, we will have a cycle $u\cdot d\cdot v(e2)(S1)(e1)$ in the boundary of $S(x)$ $(e2)(S1)(e1)$ contains only points in $P_{i-1}$ and $u$,$v$ (that are possible end points of $e1$, $e2$). $c$ is on the boundary of $S(x)$ too. Where is $c$? It must be in the boundary curves (of $S(x)$) from $u$ to $d$ or the curve from $d$ to $v$. Then $c,d$ in $S(x)$ must in the same side of $uxv$ which is part of $P_{i}$. Therefore, $C$ and $P_{i}(a,b)$ do not cross-over each other at $x$. See Fig. 2(b) and the following extended figure.

If $X$ does not contain $v$, then there must be a 2-cell $Y$ (in between $P_{i-1}$ and $P_{i}$) containing $(x,v)$. We can see that $X$ and $Y$ are line-connected in $S(x)$. (See Fig. 2(c) and above extended figure.) This is due to the definition of regular point of $x$, all surface-cells containing $x$ are line-connected. Meaning there is a 2-cell paths they share a 1-cell in adjacent pairs.

Since $X$ and $Y$ are line-connected, we can assume:

a) $X$ and $Y$ share a 1-cell, i.e. $X\cap Y=(x,y)$. Then $y$ is on $P_{i-1}(a,b)$. Let $e2$ be the possible edge from $v$ to $P_{i-1}(a,b)$. ($e2$ could be empty as $e1$) and $u(e1)..y...(e2)v$ is on the boundary cycle of $S(x)$. Except $u$ and $v$, $u(e1)..y...(e2)v$ is on $P_{i-1}(a,b)$. $u...d...v$ is part of the boundary cycle of $S(x)$. In addition, $c$ (that is not in $P_{i-1}(a,b)$) must be in the boundary curves (of $S(x)$) from $u$ to $d$ or the curve from $d$ to $v$. Again, $c,d$ in $S(x)$ must in the same side of $uxv$ which is part of $P_{i}$. Therefore, $C$ and $P_{i}(a,b)$ do not cross-over each other at $x$. (See Fig. 2(c) also see the above extended figure Fig. 3(c).)

b) $X$ and $Y$ share the point $x$, and there are line-connected 2-cells as a path in between $X$ and $Y$. $X\cap Y=x$ . Let us assume that $e1$ incident to $P_{i-1}(a,b)$ at $y^{\prime}$ ($y^{\prime}$ is $u$ if $e1$ is empty ) and $e3$ incident to $P_{i-1}(a,b)$ at $y^{\prime\prime}$. We will have a set of points $y^{\prime}=y_{0},y_{1},...,y_{k}=y^{\prime\prime}$ in $P_{i-1}(a,b)$. Each $y_{t}$ is contained in a 2-cell containing $x$. All $y_{0},y_{1},...,y_{k}$ are in the boundary cycle of $S(x)$. $c$ that is not in $P_{i-1}(a,b)$. $c$ must be in the boundary curves (of $S(x)$) from $u$ to $d$ or the curve from $d$ to $v$. Thus, $c,d$ in $S(x)$ must in the same side of $uxv$ which is part of $P_{i}$. $C$ and $P_{i}(a,b)$ do not cross-over each other at $x$. (See Fig. 2 (c) and Fig. 3(c).)

Case 2: Suppose $P_{i}(a,b)$ and $C$ cross over a sequence of points on $C$: $P_{i}(a,b)=...ux_{0}x_{1}...x_{m}v...$ and $C=...cx_{0}x_{1}...x_{m}d...$, where $v\neq d$.

We still have $e1=(u,y_{0})$ and $e2=(v,y_{k})$ where $y_{0}$ and $y_{k}$ are on $P_{n-1}$ for some $k$. Each $y_{t}$ , $t=0,1,...,k$, is in a 2-cell that containing some $x_{j}$, $j=0,1,...,m$.

Note that : If $u$ does not have a direct edge linking to $P_{n-1}$, $u$ will be in a 2-cell between $P_{n}$ and $P_{n-1}$, either $u$ is a pseudo point on $P_{n}$ for the deformation from $P_{n-1}$ to $P_{n}$, or $P_{n-1}$ and $P_{n}$ intersects at $u$. That $u$ is a pseudo point means here it has a neighbor that has an edge link to $P_{n-1}$, or the neighbor’s neighbor, and so on. We can just assume here $u$ is the point that is adjacent to a point in $P_{n-1}$. In the theory, as long as $u$ is contained by a 2-cell such that all the points in the 2-cell are in $P_{n-1}$ or $P_{n}$.

The same way will apply to this case just treat $x_{0}$,…,$x_{m}$ to $x$ in Case 1. We first get the union of $S(x_{0})$,…,$S(x_{m})$. We want to prove that : The boundary of this union will be simple cycle too under the condition of Lemma 5.

Using mathematical induction we can prove it. After that, we can prove the rest of theorem using the same method presented in Case 1. See Fig. 4.

The following is the detailed proof: Let $S(x_{0},...,x_{k})=S(x_{0})\cup...\cup S(x_{k})$.

First, we will prove that the boundary of $S(x_{0})\cup S(x_{1})$ is a simple cycle (it is a simple closed curve too). We know that $(x_{0},x_{1})$ is an edge in $C\cap P_{i}(a,b)$. Also, there are two 2-cells $A,B$ in $S(x_{0})$ containing $(x_{0},x_{1})$ .

$x_{1}$ is a boundary point in $S(x_{0})$ , so no other 2-cell will contain $x_{1}$. In the same way, $S(x_{1})$ also contains $A,B$, and $x_{1}$ is only contained in two 2-cells in $S(x_{1})$. Therefore, $S(x_{0})\cap S(x_{1})=A\cup B$ and $A\cap B=(x_{0},x_{1})$.

Note that $A$ and $B$ are adjacent 2-cells. On the other hand, $x_{1}$ is on the boundary curve (that is closed) of $S(x_{0})$. So $x_{1}$ has two adjacent points on this cycle, $y_{1}$ and $y_{2}$. (We assume that $y_{1}$ and $y_{2}$ are not pseudo points, so) $y_{1}$ and $y_{2}$ are both on the boundary of $S(x_{0})\cup S(x_{1})$. (If $y_{1}$ or $y_{1}$ is pseudo points, we can ignore $y_{1}$ or $y_{2}$ to find the a actual point that adjacent to $x_{1}$.) $(x_{1},y_{1})$ has two 2-cells containing $(x_{1},y_{1})$ in $S(x_{0})\cup S(x_{1})$. For instance, in Fig. 4 (a) , $A$ and $A_{1}$ contain $(x_{1},y_{1})$ and $B$ and $B_{1}$ contain $(x_{1},y_{2})$. Thus, the boundary of $S(x_{0})\cup S(x_{1})$ is a closed curve that is formed by the arc from $y_{1}$ to $y_{2}$ in the boundary of $S(x_{0})$, plus the arc from $y_{2}$ to $y_{1}$ in the boundary of $S(x_{1})$.

Second, we assume the boundary of $S(x_{0},...,x_{k-1})$ is a closed curve, when we consider the arc $x_{0},...,x_{k-1},x_{k}$ in $C$, we can prove the boundary of $S(x_{0},...,x_{k})$ is also a closed curve.

We know that we have two closed curves: Suppose that $Q$ is the boundary of $S(x_{0},...,x_{k-1})$ , and $R$ is the boundary of $S(x_{k})$. $(x_{k-1},x_{k})$ is in $S(x_{k})$, and $(x_{k-1},x_{k})$ is in $S(x_{0},...,x_{k-1})$ . There are two 2-cells $A$, $B$ containing $(x_{k}-1,x_{k})$ in $S(x_{k})\cap S(x_{0},...,x_{k-1})$.

$x_{k-1}$ is on the boundary cycle of $S(x_{k})$, then $x_{k-1}$ must have two adjacent points in $R$, $y_{1}$, and $y_{2}$. $(x_{k-1},y_{1})$ and $(x_{k-1},y_{2})$ are two edges in $S(x_{k})\cap S(x_{0},...,x_{k-1})$. In the same way above, we will have the cycle passing $y_{1}$ and $y_{2}$ that is the boundary curve of $S(x_{0},...,x_{k})$ . This is because there is a 2-cell in between the neighborhoods of $x_{k-2}$ and $x_{k}$ as assumed by the wider angle on each point on $C$, $y_{1}$ is not a folding point, so are $y_{2}$. In other words, the point that enters $y_{1}$ in counterclockwise in $S(x_{0},...,x_{k-1})-\{x_{0},...,x_{k-1}\}$ differs from the point follows $y_{1}$ in counterclockwise in $S(x_{k})-\{x_{k}\}$.

Thus, we have proved that the boundary curve of $S(x_{0},...,x_{k})$ is a simple closed curve. In the rest of the proof, we will treat $S(x_{0},...,x_{m})$ to be $S(x)$ in Case 1. See Fig. 5. We now use denote $X=\{x_{0},...,x_{m}\}$ and $X$ is an arc in $C$. (Please note that in Case 1, $X$ was used as a 2-cell. Now $X$ is an arc in $C$. )

In the rest of the proof, we will prove: if $P_{i-1}(a,b)$ and $C$ are not cross over each other, then, $P_{i}(a,b)$ and $C$ will not be cross over each other. Therefore, any $P(a,b)$ must cross over $C$. This completes the proof of the discrete Jordan curve theorem.

Let us first state again that $P_{i}(a,b)$ passes $x_{0}...x_{m}$ but $P_{i-1}(a,b)$ does not contain any point of $\{x_{0},...,x_{m}\}$. In addition, $P_{i-1}(a,b)$ and $P_{i}(a,b)$ is gradually varied, i.e. $P_{i}(a,b)$ was deformed from $P_{i-1}(a,b)$ directly. We also know that $S(X)=S(x_{0},...,x_{m})$ is the neighborhood of the arc in $C$, i.e. the arc $x_{0},...,x_{m}$ is a part of the closed curve $C$. The boundary of $S(X)=S(x_{0},...,x_{m})$ is a closed curve too.

$u,v,c,d$ are on the boundary of $S(x_{0},...,x_{m})$ (Assume $u,v,c,d$ are not pseudo points, otherwise, we can find corresponding none-pseudo on the boundary of $S(x_{0},...,x_{m})$.) $u,(x_{0},...,x_{m}),v$ is a part of $P_{i}$ We also know that $c$ and $(x_{0},...,x_{m})$ are not in $P_{i-1}$. There will be two 2-cells, $U$ and $V$, are in between $P_{i}(a,b)$ and $P_{i-1}(a,b)$ (all points of $U$ and $V$ are in $P_{i}(a,b)\cup P_{i-1}(a,b)$) such that $(u,x_{0})\in U$ and $(x_{m},v)\in V$.

Let $P_{i-1}\cap U=S1$ and $P_{i}\cap U=S2$. Let $e1$ be the edge in $U$ linking $S1$ to $S2$ (in most cases, $e1$ incident to $u$, but not necessarily ), and let $e2$ be the edge in $U$ linking $S2$ to $S1$ (possibly starting at $x_{0}$). So, $(e2)(S1)(e1)(S2)$ are the boundary of $U$, counterclockwise.

Subcase (i): If $U$ contains $v$ ($U=V$), all points in $U$’s boundary are contained in $S(\{x_{0},...,x_{m}\})$ by the definition of $S(x_{0})$. we will have a cycle $u\cdot d\cdot v(e2)(S1)(e1)$ in the boundary of $S(X=\{x_{0},...,x_{m}\})$. $c$ is on the boundary of $S(X)$ too. But $c\notin P_{i-1}$. It must be in the boundary curves (of $S(X)$) from $u$ to $d$ or the curve from $d$ to $v$. Then $c,d$ in $S(X)$ must in the same side of $uXv$ which is part of $P_{i}$. Therefore, $C$ and $P_{i}(a,b)$ do not cross-over each other at $X$. (See Fig. 5.)

Subcase (ii): If $U$ does not contain $v$, then there must be a 2-cell $V$ (in between $P_{i-1}$ and $P_{i}$) containing $(x_{m},v)$.

Let $e1=(p1,p2)$ be the edge in $U$ incident to a point in $P_{i-1}$ and a point in $P_{i}$, respectively. (In most cases, $e1$ incident to $u$, i.e. $u=p2$, but not necessarily ). And let $e2=(r2,r1)$ be the edge in $V$ incident to a point in $P_{i}$ and a point in $P_{i-1}$, respectively. $r2$ is usually $v$.

Note that: $c$ must not be in $U$. Gradual variation (direct deformation) means that each point in each 2-cell of $U$ and $V$ in between $P_{i}$ and $P_{i-1}$ must be in $P_{i}\cup P_{i-1}$. Formally, $P_{i}$ $XoRSum$ $P_{i-1}$ is a set of 2-cells; every point in these 2-cells is in $P_{i}\cup P_{i-1}$ .

We can see that $U$ and $V$ are line-connected in $S(X)$ by the definition of line-connected paths, meaning there is a path of 2-cells where each adjacent pair shares a 1-cell. (See Fig. 4(d) and Fig. 5.)

From $r1$ to $p1$, there is an arc in $P_{i-1}$ . To prove that all points in this arc are in the boundary of $S(X)$ we need to prove each point on the arc must be in a 2-cell that contains a point in $\{x_{0},...,x_{m}\}$, and this 2-cell is other than (except this 2-cell is) $U$ or $V$. It gives us some difficult to prove it. The above discussion seems not very productive.

We found a more elegant way to prove this case by finding another simple path (or pseudo curve) that cross-over $C$. The method is the following: If $U\neq V$, there must be a $x_{k}$ in $\{x_{0},...,x_{m}\}$, $x_{k}$ has an edge linking to $P_{i-1}$. (Otherwise, $u,x_{0},...,x_{m},v$ are in a 2-cell that contains some points in $P_{i-1}$. Therefore, $U=V$.) We can also assume that $k$ is not $m$, otherwise, $v$ is in $P_{i-1}$, so $U=V$. See Fig. 6.

We select the smallest $k$ having an edge linking $x_{k}$ to $P_{i-1}$, $0\leq k\leq m-1$. $x_{k}$ is in both $P_{i}$ and $C$. We might as well let $(x_{k},v^{\prime})$ is such an edge, and $v^{\prime}$ is a point in $P_{i-1}$. Therefore, we will have the new path (simple path), $P^{\prime}_{i}$. This new path has two parts: The first part is the same as $P_{i}$ before and including the point $x_{k}$, and the second part is the partial path (curve) of $P_{i-1}$ after point $v^{\prime}$ . This path $P^{\prime}_{i}=...,u,x_{0},...,x_{k},v^{\prime},...$ does cross-over $C=...,c,x_{0},...,x_{k},x_{k+1},...,x_{m},d,...$ . It is obvious that $P_{i-1},P^{\prime}_{i},P_{i}$ are gradually varied. This is because we just inserted a path in between of $P_{i-1}$ and $P_{i}$.

This new path $P^{\prime}_{i}$ has such a good property that is $\{x_{0},...,x_{k-1}\}$ do not incident (link) an edge that has an end vertex in $P_{i-1}$ . Since $v^{\prime}$ is in $P^{\prime}_{i}$, the 2-cell $V^{\prime}$ (in $P_{i-1}$ $XoRSum$ $P^{\prime}_{i}$ ²²2Here $XoRSum$ is exclusive “OR” operation defined in [6], just like $sum(modulo2)$ in [12] ) contains $v^{\prime}$ also contains $(x_{k-1},x_{k})$ and $(x_{k},v^{\prime})$ (in $S(x_{k})$). Since no edge from $x_{0},...,x_{k-1}$ to $P_{i-1}$ , $U$ containing $u$ is just $V^{\prime}$. We will have just Subcase (i) using $P^{\prime}_{i}$ to replace $P_{i}$.

The entire theorem is proven. ∎

In this proof, we can put the central pseudo points (the Veblen points) for each 1-cells and 2-cells to assist our proof. The reason is that if we embed 1-cells and 2-cells into Euclidean plane or higher dimensional space. We can always find the central points for each cell. The idea of the central pseudo points is at least valid in Euclidean space. In fact, the central pseudo points also have two normal directions for a 2-cell. It also has two directions for a 1-cell. For instance, $\{a,b\}$ is an edge, $a\to b$ and $b\to a$ are two directions.

In the proof of Theorem 1, we know that we have two 2-cells $A$ and $B$ at the different side of the cycle $C$. We proved that point $a\in A$ and $b\in B$ are not connected in $S-C$.

$C$ has the orientation of clockwise (or counterclockwise as we first made). $(p,r)$ is clockwise in $A$, but $(p,r)$ is counterclockwise in $B$. So we call $A$ is clockwise, and $B$ is counterclockwise. For each edge $e_{i}$ (e.g. $(p,r)$) in $C$, we will have two 2-cells containing $e_{i}$, denoted by $A_{i}$ and $B_{i}$. There must be one in clockwise and another is in counterclockwise.

We always assume that $A_{i}$ is clockwise and $B_{i}$ is counterclockwise. We now add all the central pseudo points to all 1-cells and 2-cells in $S$. And immediately remove all central pseudo points from 1-cells in $C$. (This operation is to stop a path will go through the central pseudo points on $C$.)

We also know in our assumption: each 2-cell must have at least three edges (1-cells) in its boundary. This is because $S$ is a simple graph. (We can always add a point to make it in Euclidean space.) We also assume that $C$ does not reach the border of $S$ meaning that $C\cap{\partial S}=\emptyset$.

Case 1: A special case $C$ is the boundary of a single 2-cell, denoted as $A$. A simple path could be just the boundary of a 2-cell. In this case, we have a central pseudo point in the cell $A$. So this theorem is virtually true if we can prove that $S-C$ is point-connected. That is to say that except the central pseudo point in $A$, $S-C$ is a point-connected component. In other word, $S-A$ is one point-connected component.

We can prove this is because of the following facts: For each cell $B$ in $S-A$, if $B$ has an edge in $C$. The boundary of $B$ is a simple closed path. If the boundary of $B$ is $C$, then $S=A\cup B$ since every edge has shared by two 2-cells ($A$ and $B$) already. We have two components in $S-C$: the central pseudo point of $A$ and the central pseudo point of $B$.