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% Section RESEARCH editor Ciesielski
% Received 11/25/96
% AMS # 03H05, 26E35, 51H05
% Accepted to RESEARCH section by Krzysztof Chris Ciesielski
% Received November 25, 1996. Final version accepted 11/7/97.
\documentclass{rae}
\usepackage{amsmath,amssymb,amsthm}
%\coverauthor{Vladimir Kanovei and Michael Reeken}
%\covertitle{A Nonstandard Proof of the Jordan Curve Theorem}
\received{November 25, 1996}
\author{
Vladimir Kanovei
\thanks{Acknowledges the support of AMS, DFG, and
University of Wuppertal.},
\ Department of Mathematics,
Moscow Transport Engineering Institute,
Moscow 101475, Russia; e-mail:
{\tt kanovei@mech.math.msu.su} \ and \
{\tt kanovei@math.uni-wuppertal.de}
\and Michael Reeken, Department of Mathematics,
University of Wuppertal, Wuppertal 42097,
Germany; e-mail: {\tt reeken@math.uni-wuppertal.de}}
\title{A NONSTANDARD PROOF OF THE JORDAN CURVE THEOREM}
\markboth{Vladimir Kanovei and Michael Reeken}{Jordan Curve
Theorem}
\keywords{Jordan curve theorem, nonstandard analysis}
\MathReviews{03H05, 54J05}
\firstpagenumber{1}
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\begin{document}
\maketitle
\begin{abstract}
We give a nonstandard variant of Jordan's proof of the Jordan
curve theorem which is free of the defects his contemporaries
criticized and avoids the epsilontic burden of the classical
proof. The proof is
self-contained, except that the Jordan theorem for polygons
is taken for granted.
\end{abstract}
%\hspep
\section*{Introduction}
\noi
The \emph{Jordan curve theorem\/} \cite{Jo1893}
(often abbreviated as JCT in the literature)
was one of the starting points in the modern development of
\emph{topology\/} (originally called \emph{Analysis Situs}).
This result is considered difficult to prove,
at least compared to its intuitive evidence.
{C.\ Jordan} \cite{Jo1893} considered the assertion to
be evident for simple polygons and reduced the case of a simple
closed continuous curve to that of a polygon by approximating the
curve by a sequence of suitable simple polygons.
Although the idea appears natural to an analyst it is not so easy
to carry through. {Jordan}'s proof did not satisfy
mathematicians of his time. On the one hand it was felt that the
case of polygons also needed a proof based on clearly stated
geometrical principles, on the other hand his proof was considered
incomplete. (See the criticisms formulated in \cite{Ve05} and in
\cite{Os12}.)
If one is willing to assume slightly more than mere continuity of
the curve, then much simpler proofs (including the case of
polygons) are available (see {Ames} \cite{Am04} and {Bliss}
\cite{Bl04} under restrictive hypotheses).
{O.\ Veblen} \cite{Ve05} is considered the first to have
given a rigorous proof which, in fact, makes no use
of metrical properties, or, in the words of {Veblen}:
\emph{We accordingly assume nothing about analytic geometry, the
parallel axiom, congruence relations, nor the existence of points
outside a plane\/.} His proof is based on the incidence and order
axioms for the plane and the natural topology defined by the basis
consisting of nondegenerate triangles. He also defines simple
curves intrinsically as specific sets without parametrizations by
intervals of the real line. He finally discusses how the
introduction of one additional axiom, existence of a point outside
the plane, allows him to reduce his result to the context {Jordan}
was working in. {Veblen} also gave a specific proof for polygons
based on the incidence and order axioms exclusively (see
\cite{Ve04}) which was later criticized as inconclusive by
{H.~Hahn} \cite{Ha08} who published his own version of a proof
based on {Veblen}'s incidence and order axioms of the plane
(which, by the way, are equivalent to the incidence and order
axioms of {Hilbert}'s system).
{Jordan}'s proof in his \emph{Cours d' analyse }of 1893 is
elementary as to the tools employed. Nevertheless the proof extends
over nine pages and, as mentioned above, cannot be considered
complete. We are interested here in this proof. It depends on some
facts for polygons and an approximation argument. It is, therefore,
a natural idea to use nonstandard arguments to eliminate the
epsilontic burden of the approximation.
There is an article by {L.\ Narens} \cite{Na71} in which
this point of view is adopted. Unfortunately, some part of this
proof has been criticized recently as inconclusive and, in any
case, the reasoning is not essentially
shorter than, or as elementary as, {Jordan}'s proof.
It is certainly true that not all classical arguments can be replaced
in some useful or reasonable way by simpler nonstandard arguments.
But as we shall show it is possible to simplify the approximation
argument specific to {Jordan}'s proof. We shall follow the
proof quite closely but take a
somewhat different approach when proving path-connectedness.
That nonstandard analysis can even give some additional insight
into the geometric problem is manifest from the proof by
{N.\ Bertoglio} and {R.\ Chuaqui} \cite{BeCh94} which
avoids polygons and approximations
entirely by looking at a nonstandard discretization of the plane and
reducing the problem to a combinatorial version of the JCT
proved by {L.\ N.\ Stout} \cite{St88}. This reduction of the
problem to a (formally) discrete one is interesting and leads to a
proof which establishes a link to a context totally different from
{Jordan}'s.
As a curiosity we note in passing that {Jordan} speaks of
\emph{infinitesimals\/} in his proof but it is only a figure of
speech for a number which may be chosen as small as one wishes or
for a function which tends to zero.
For reference we state the JCT.
\vbox{
\begin{jcth}
A simple closed continuous curve\/ $\cK$ in the plane separates
its complement into two open sets of which it is the common
boundary$;$ one of them is called the\/
\emph{outer (}or\/ \emph{exterior) region}
$\cK\exte$ which is an open,
unbounded, path-connected set and another set called the\/
\emph{inner (}or\/ \emph{interior) region} $\cK\inte$ which
is an open, simply path-connected, bounded set.
\end{jcth}
}
\section*{Notation}
By {\it simple\/} (polygon, curve) we shall always mean one
having no self-intersec\-tions. A {\it broken line\/} will
be a curve consisting of finitely (or hyperfinitely -- in a
nonstandard domain) many nonzero segments.
The reader is assumed to have a basic knowledge of nonstandard
analysis.~\footnote
{\ We refer to {Lindstr\o m} \cite{li} and
{Nelson}~\cite{ne77}.}
In what follows we shall always identify reals and
points from the standard domain with their ``asterisk'' images
in the nonstandard domain (although the standard curve $\cK$
will be distinguished from $\aK$). We shall understand the words
{\it point, polygon, real, curve\/}
\etc\ as meaning internal objects in the nonstandard domain
unless otherwise specified, for instance by the
adjective ``standard''. Hopefully this way of exposition will
be equally understandable by both $\IST$ followers and those who
prefer the model--theoretic version of nonstandard analysis
(although the latter should understand as hyperreals, hyperpoints
\etc\ what we will call reals, points \etc).
\section*{Plan of the Proof}
Starting the proof of the Jordan theorem, we consider a standard
simple closed curve $\cK =\ans{K(t):0\leq t<1}$ where
$K:\dR\lra\dR^2$ is a (standard) continuous \dd1 periodic
function which is injective modulo $1$ (\ie\ $K(t)=K(t')$ implies
$t-t'\equiv 0\bmod 1$). From $K(t)\approx K(t')$ it follows then
that $t\approx t'\;\bmod1$.\vom
{\it Section \ref{appr}.}
Working in a fixed nonstandard domain,
we infinitesimally approximate $\cK$ by a simple
(nonstandard) polygon $\Pi,$ using a construction,
essentially due to Jordan, of consecutively cutting off
loops from an originally self-intersecting approximation.\vom
{\it Section \ref{domain}.} We define the interior region
$\cK\inte$ as the open set of all standard points which belong to
$\Pi\inte$ but do not belong to the monad of $\Pi.$
(The Jordan theorem for polygons is taken for granted; this
attaches definite meaning to $\Pi\inte$ and $\Pi\exte$ in the
nonstandard domain.) $\cK\exte$ is defined accordingly.\vom
{\it Section \ref{cbor}.}
We prove that any point of $\cK$
is a limit point for both $\cK\inte$ and $\cK\exte\,;$ this
also implies the non-emptiness of the regions.
%To prove that $\cK\inte$ is nonempty,
%we separate $\Pi$ into two complementary non-infinitesimal arcs
%and locate a segment in $\Pi\inte$ between the arcs which contains
%a point not in the monad of $\Pi.$
\vom
{\it Section \ref{pconn}.} To prove that $\cK\inte$ is
path-connected we define a simple
nonstandard polygon $\Pi'$ which
lies entirely within $\Pi\inte,$ does not intersect $\cK,$ and
contains all (standard) points of $\cK\inte.$
This easily implies the path-connectedness.\vom
%{\it Section \ref{bord}.} We prove that $\cK$ is the common border
%for both $\cK\inte$ and $\cK\exte$.
\section{Approximation by a Simple Polygon}
\label{appr}
We say that a polygon $\Pi=P_1P_2\ldots P_nP_1$
($n$ may be infinitely large)
\emph{approximates\/} $\cK$ if there is an internal sequence
of (perhaps nonstandard) reals $0\leq t_10$ such that\/ $\aK$
is in the\/ \dd\vep neighborhood of\/ $\Pi$ and\/
$\Pi$ is in the\/ \dd\vep neighborhood of\/ $\aK\,$ and
\item\label{n}
if $P\approx Q$ are on $\Pi$, then precisely one of the two arcs
$\Pi$ is decomposed into by these points must be included in the
monad of $P$.
\een
\ele
\pf
\ref{i}\
The requirement $(\ddag)$ does not allow the hyperreals $t_k$ to
collapse into a sort of infinitesimal ``cluster'' or into a pair
of them around $0$ and $1,$ which are compatible with
$\Da(\Pi)\approx 0$ alone. (Note that the injectivity modulo $1$ of
$\aK$ is used in the proof that $t_1\approx 0$ and
$t_n\approx 1$.) \vom
\ref{ii}\
$\da_i=\max_{t_i\leq t\leq t_{i+1}}\left| K(t)-K(t_i)\right|$ is
infinitesimal for each $1\leq i\leq n$ and therefore
$\vep=2\max_{1\leq i\leq n}\da_i$ is infinitesimal and proves
the assertion.\vom
\ref{n}
Since all edges of $\Pi$ are infinitesimal by \ref{i}, we may
assume that $P$ and $Q$ are vertices, say $P=P_i$ and $Q=P_j.$
Then either
$t_i\approx t_j$ or $t_i\approx 0$ while $t_j\approx 1.$ (Indeed
otherwise $\aK$ would have a self-intersection.) Consider the first
case. The arc determined by $t_i\le t\le t_j$ is clearly within the
monad of $P.$ To see that the other arc is not included in the monad
consider any $t_k$ which is $\not\approx$ any of $t_i,\,0,\,1.$ Then
$P_k\not\approx P$ as otherwise $\cK$ would have a self-intersection.
\qed
\ble
\label{apol}
There is a simple polygon which infinitesimally approximates
$\cK$.
\ele
\pf
Taking $t_i=\frac in$ for some infinitely large $n$ results
in a polygon which infinitesimally approximates $\cK$. But
it may have self-intersections.
%First of all, given two adjacent sides, $P_kP_{k+1}$ and
%$P_{k+1}P_{k+2}$ one may be included in the other. It suffices
%then to reduce the polygon $P_1\dots P_kP_{k+1}P_{k+2} P_nP_1$
%to $P_1\dots P_kP_{k+2}\dots P_nP_1$
%which obviously satisfies $(\ddag)$ by Lemma~\ref{app}\ref{i}.
Assume two non-adjacent sides intersect, \ie\ $P_iP_{i+1}$
intersects $P_jP_{j+1}$ for some $1\leq i\frac 12$
we replace the complementary arc $P_j\ldots P_nP_1\ldots P_i$
by a new side $P_jP_i$ such that again $(\ddag)$ is satisfied.
The case
$\left| P_iP_j\right| > \left| P_{i+1}P_{j+1}\right|$ is treated
in the same way. (The dots $\dots$ indicate that all indices in
between are involved.)
In all the cases the resulting polygon $\Pi\new$ still
infinitesimally
approximates $\cK$ because $(\dag)$ and $(\ddag)$ are satisfied
(for the accordingly reduced system of parameter values $t_i$)
%by Lemma~\ref{app}\ref{i}
and $\Da(\Pi\new)\leq\Da(\Pi)$.
This (internal) procedure does not necessarily reduce the number of
self-intersections because for the one which is removed there may
be others appearing on the newly introduced side of the reduced
polygon $\Pi\new.$ But the number of vertices of
$\Pi\new$ is strictly less than that of $\Pi.$
%
Therefore the internal sequence of polygons arising from
$\Pi$ by iterated applications of this reduction procedure
eventually ends with a simple polygon $\Pi'$ which approximates
$\cK$ infinitesimally.
\qed
\section{Definition of the Interior and Exterior Region}
\label{domain}
Let us fix for the remainder a simple (nonstandard) polygon
$\Pi =P_1P_2\ldots P_nP_1$ which approximates $\cK$ infinitesimally.
Let $\cK\inte$ be the open standard set of all standard points
$A\in\Pi\inte$ which have a non-infinitesimal distance from $\Pi.$
We
%put $\cK\inte=I$ and
call this the
\emph{interior region\/}
of the curve $\cK.$ In the same way we define the open standard
set $\cK\exte$ of all standard points from $\Pi\exte$ which have
non-infinitesimal distance from $\Pi
%,$ put $\cK \exte=E
$ and call
this the \emph{exterior region\/} of the curve $\cK$.
Omitting rather elementary proofs that $\cK\inte$ is bounded,
$\cK\exte$ is unbounded, and the complement of the union of both
sets equals the curve $\cK,$
%Let $P$ be a (standard) point of that complement.
%Then any standard circle around $P$ intersects $\Pi.$ By overspill
%there is a point $P'\in \Pi$ with $P\approx P'.$ This implies
%by Lemma \ref{app} that there is a point $P''\approx P$ on $\cK .$
%Then $P$ equals $P''$ as both are standard.
%
let us prove that for $A\in \cK\inte$ and $B\in \cK\exte$ any
standard continuous arc $\al$ from $A$ to $B$ intersects $\cK.$
Indeed $\ups\al$ must intersect $\Pi$ in some point $P$
because it starts in $\Pi\inte$ and ends in $\Pi\exte.$
(The JCT, transferred to the nonstandard domain, is
applied.) By Lemma \ref{app}, and the fact that $\cK$ is
compact, there is a (standard) point
$P'\in\cK$ infinitesimally close to $P\in\Pi.$ As $\cK$ and the
arc are standard and closed, $P'$ is in $\cK\cap\al$.
\section{The Curve is the common Boundary}
\label{cbor}
We prove that each (standard) point of $\cK$ is a limit point
for both the interior region $\cK\inte$ and the exterior region
$\cK\exte\,;$ this clearly implies that the interior region is
not empty (that the exterior region is not empty is trivial).
By the choice of $\Pi$ and the definition of $\cK\inte$ and
$\cK\exte,$ it suffices to prove the following:
{\it given a vertex\/ $A$ on\/ $\Pi,$, then for any square\/ $S$
with center in\/ $A$ and non-infinitesimal\/} ({\it possibly
nonstandard\/}) {\it size the domain $S\inte$ contains
points in both\/ $\Pi\inte$ and\/ $\Pi\exte$ which have
non-infinitesimal distance from\/ $\Pi.$} We prove this
assertion for $\Pi\inte$ only; the proof for the exterior
region is similar.
Let $B$ be another vertex of $\Pi$ chosen such that the distance
$|AB|$ is non-infinitesimal. We can assume that $B$ lies in
$S\exte$ and has non-infinitesimal distance from $S,$ and
in addition $S$ itself does not contain any vertex of $\Pi.$
%
Let $\al$ and $\ba$ be the simple broken lines -- connecting $A$
with $B$ -- into which $\Pi$ is partitioned by the vertices $A$
and $B$.
The interior region $\Pi\inte$ is decomposed by $S$ into a number
of polygonal domains. Let $\Pi'$ be the polygon which bounds that
domain among them the boundary of which contains $A.$ Then $\Pi'$
consists of parts of
the broken lines $\al$ and $\ba$ and connected parts of $S.$
Since $A$ is the only common point of $\al$ and $\ba$ except
for $B$ (which is far away from $S$), going around $\Pi'$ we
find a connected ``interval'' $C_1C_2$ of $S$ (which may
occasionally contain one or more of the four vertices of $S$)
such that the points $C_1$ and $C_2$ belong to different curves
among $\al,\,\ba.$ Since $C_1C_2$ is also a part of $\Pi',$
any inner point $E$ of $C_1C_2$ belongs to $\Pi\inte$.
Consider a point $E$ in $C_1C_2$ which has
equal distance $d=\dist(E,\al)-\dist(E,\ba)$ from both $\al$ and
$\ba.$ Note that $d$ is not infinitesimal. Indeed
otherwise there are points $A'\in \al$ and $B'\in \ba$ such that
$A'\approx E\approx B',$
which is impossible by Lemma~\ref{app}\ref{n} as $S$ has
non-infinitesimal distance from both $A$ and $B$.
Thus $E\in\Pi\inte$ has a non-infinitesimal distance from $\Pi,$
as required.
\section{Path-Connectedness}
\label{pconn}
Let $A$ and $B$ be two (standard) points in $\cK\inte$. We have
to prove that there is a (standard) broken line joining $A$ with
$B$ and not intersecting $\cK.$ This is based on the following lemma.
\ble
\label{rl}
There exists a simple polygon\/ $\Pi'$ lying entirely
within\/ $\Pi\inte,$ containing no point of\/ $\aK$ in\/
$\Pi'\inte,$ and containing every standard point of\/ $\cK\inte$
%both\/ $A$ and\/ $B$
in\/ $\Pi'\inte$.
\ele
%
The lemma clearly implies the result:
%path-connectedness of $\cK\inte:$
indeed, by the JCT for polygons, $A$
can be connected to $B$ by a broken line which lies within
$\Pi'\inte$ therefore does not intersect $\aK.$ By Transfer
we get a standard broken line which connects $A$ and $B$ and
does not intersect $\cK,$ as required.
Moreover, the lemma implies the {\em simple\/} path-connectedness
of $\cK\inte.$ Indeed we have to prove that every standard simple
closed curve $\cK_1$ lying entirely within $\cK \inte$ can be
appropriately contracted into a point. To see this note that $\aK_1$
is evidently situated within $\Pi'\inte,$ which is the interior of
a simple polygon, so that $\aK_1$ has the required property in the
nonstandard domain by the JCT for polygons. It remains to apply
Transfer.
As for the path-connectedness of the {\it exterior}\/ region
$\cK\exte,$ we choose a point in $\cK \inte$ and apply an
inversion with center in this point. The interior region becomes
a neighborhood of $\infty$ and the exterior region becomes the
interior region of the image of the curve. To this we apply the
result above.
\medskip
\noindent{\sc Proof of Lemma \ref{rl}.} Let an infinitesimal
$\vep>0$ be defined as in Lemma~\ref{app}\ref{ii}, so that $\cK$
and $\aK$ are included in the \dd\vep neighborhood of $\Pi$.
Note that each side of $\Pi$ is infinitesimal by definition.
For any side $PQ$ of $\Pi$
we draw a rectangle of the size $(|\wed PQ|+4\vep)\ti(4\vep)$ so
that the side $PQ$ lies within the rectangle at equal distance
$2\vep$ from each of the four sides of the rectangle.
Let us say that a point $E$ is the {\it inner intersection\/} of
two straight segments $\sg$ and $\sg'$ iff $E$ is an inner point of
both $\sg$ and $\sg',$ and $\sg\cap\sg'=\ans{E}.$
%
For any point $C\in\Pi\inte$ which is either a vertex of some
of the rectangles above, or an inner intersection of sides of two
different rectangles in this family let $\wed C{C'}$ be a
shortest straight segment which connects $C$ with a point $C'$
on $\Pi\,;$ obviously each $CC'$ is infinitesimal.
Let us fix a standard point $A$ in $\cK\inte$.
The parts of the rectangles lying within $\Pi$ and the
segments $\wed C{C'}$ decompose the interior region $\Pi\inte$
into a (possibly hyperfinite) number of polygonal domains.
Let the polygon $\Pi'$ be the boundary of the domain
containing $A.$ (Note that all the lines involved lie in
the monad of $\Pi,$ hence none of them contains $A$.) It
remains to prove that ${\Pi'}\inte$ also contains any other
standard point $B$ of $\cK\inte$.
%the other point.~\footnote
%{\ It is clear then that ${\Pi'}\inte$ contains all standard points
%in $\cK\inte.$ -- {\it Pointed out by the referee\/}. }
Note that each side of $\Pi'$ is a part of either a side of
one of the rectangles covering $\Pi$ or of a segment
of the form $\wed{C}{C'}$ --- therefore it is infinitesimal.
Let $\Pi'=C_1C_2\dots C_n.$ We observe that by construction,
for any $k=1,...,n,$
there is a shortest segment $\sg_k=\wed{C_k}{C'_k},$ connecting
$C_k$ with
a point $C'_k$ in $\Pi$ which does not intersect ${\Pi'}\inte$.
Moreover, by the triangle equality, the segments
$\sg_k$ have no inner intersections. Therefore
any two of them intersect each other only in such a manner that
either the only intersection point is the common endpoint
$C'_k=C'_l$ or one of them is an end-part the other one.
Then the segments $\sg_k$ decompose the ring-like polygonal region
$\cR$ between $\Pi$ and $\Pi'$ into
$n$ open domains $\cD_k\msur$ $(k=1,...,n)$ defined as follows.
If $\sg_k$ and $\sg_{k+1}$ are disjoint (${\sg}_{n+1}$ equals
${\sg}_1$), then the border of $\cD_k$ consists of $\sg_k,\msur$
$\sg_{k+1},$ the side $\wed{C_k}{C_{k+1}}$ of $\Pi',$ and that
arc $\curve{C'_k}{C'_{k+1}}$ of $\Pi$ which does not contain any
of the points $C'_l$ as an inner point.
If $\sg_k$ and $\sg_{k+1}$ have the common endpoint $C_k=C_{k+1}$
and no more common points, then the border shrinks to $\sg_k,\msur$
$\sg_{k+1},$ and $\wed{C_k}{C_{k+1}}.$ If, finally, one of
the segments is included in the other, then $\cD_k$ is
empty.
If now $B\in {\Pi'}\exte$, then $B$ belongs to one of the domains
$\cD_k.$ If this is a domain of the first type, then the
infinitesimal simple arc $C'_kC_kC_{k+1}C'_{k+1}$ separates $A$
from $B$ within $\Pi,$ which easily implies, by
Lemma~\ref{app}\ref{n}, that either $A$ or $B$ belongs to the
monad of $\Pi,$ which contradicts the choice of the points. If
$\cD_k$ is a domain of second type, then the
barrier accordingly shrinks, leading to the same contradiction.
\qed
\noindent{\bfit Acknowledgments} \ The authors acknowledge with
pleasure useful discussions with C.\ W.\ Henson, H.\ J.\ Keisler,
S.\ Leth, P.\ A.\ Loeb in matters of the Jordan curve theorem in
the course of the Edinburgh meeting on nonstandard analysis
(August 1996).
The authors are thankful to the organizers of the meeting for
the opportunity to give a preliminary talk. The authors are thankful
to the referee for substantial comments and suggestions.
\begin{thebibliography}{99}
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Situs relating to the division of a plane or of space by a closed
curve or surface, \emph{%
Bull. Amer. Math. Soc.} (2), \textbf{10} (1904), 301.
\bibitem{BeCh94} {N.\ Bertoglio, R.\ Chuaqui},
An elementary geometric nonstandard proof of the Jordan curve
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