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Unformatted text preview: The J ordanSchonflies Theorem
and the Classification of Surfaces Carsten Thomassen INTRODUCTION. The Jordan curve theorem says that a simple closed curve in
the Euclidean plane partitions the plane into precisely two parts: the interior and
the exterior of the curve. Although this fundamental result seems intuitively
obvious it is fascinatingly difficult to prove. There are several proofs in the
literature. For example, Tverberg [12] gave a proof involving only approximation
with polygons. Here, we give a short proof based only on a trivial part of
Kuratowski’s theorem on graph planarity (see Lemma 2.5, below), namely, that
K3,3 is not planar. Then we turn to another fundamental topological result: the classification of
(compact) surfaces. A surface is a connected compact topological space which is
locally homeomorphic to a disc (that is, the interior of a circle in the plane). The
classification of surfaces says that every surface is homeomorphic to a space
obtained from a sphere by adding handles or crosscaps. One of the first complete
proofs was given by Kerékjarto [4] and there are several short proofs based on the
assumption that every surface can be triangulated (see e.g. [1,2]). Tutte [11] gave a
proof in a purely combinatorial framework. In this paper we present a selfcon
tained proof. The proof consists of two parts: a “topological” part and a “combina
torial” part. The combinatorial part (Section 5) is very short. It differs from other
proofs in that it uses no topological results, not even the Jordan curve theorem. In
particular, it does not use Euler’s formula (which includes the Jordan curve
theorem). Thus, the combinatorial part can be read independently of the previous
results and it is of interest to those applications (for example to the Heawood
problem mentioned below) where the surfaces under consideration are already
triangulated. The topological part is a proof of the fact that every surface S can be
triangulated, i.e., S is homeomorphic to a topological space obtained by pasting
triangles together. The idea behind this is simple: First we consider, for each point
p in S, a small disc D1, around p. As S is compact, S is covered by a finite
collection of the discs Dp. If S minus the boundaries of those discs consists of a
finite number of connected components, then each of these is homeomorphic to a
disc and it is then easy to triangulate S. However, the discs D1, may overlap in
a complicated way. The previous proofs in the literature of the fact that every
surface can be triangulated are complicated and appeal to geometric intuition. In
Section 4 we present a short proof, which is perhaps not easy to follow, but which
is simple in the sense that it merely consists of repeated use of the following
extension of the Jordan curve theorem: If C1 and C2 are simple closed Jordan
curves in the plane and f is a homeomorphism between them, then f can be
extended to a homeomorphism of the whole plane. This extension, which is called
the JordanSchonﬂies theorem is a classical result, which is of interest in its own 116 CARSTEN THOMASSEN [February right. In the present paper it forms a bridge between the Jordan curve theorem
and the classification theorem. Although the JordanSchonﬂies theorem may also
seem intuitively clear, it does not generalize to sets homeomorphic to a sphere in
R3, as shown by the socalled Alexander’s Horned Sphere, see [5]. (The Jordan
curve theorem does generalize to spheres in R3.) We present a new (graphtheo
retic) proof of the JordanSchonﬂies theorem in Section 3. No previous knowledge
of graph theory and only basic topological concepts will be assumed in the paper.
In order to emphasize that the proofs are rigorous, no figures (which could be an
excuse for lack of details) are included. Instead there are, inevitably, quite a
number of technical details in the topological part (Sections 3 and 4). The difficulty
in the topological part lies precisely in the details. The classification of surfaces is not only a beautiful result of considerable
independent interest. It has turned out to be a valuable tool in combinatorial
analysis. Heawood [3] introduced the problem of determining the smallest number
h(S) such that every map on the surface S can be coloured in 11(5) colours in such
a way that no two neighbouring countries receive the same colour. Heawood
established an upper bound for h(S). He claimed that his upper bound in fact
equals h(S) (except for the sphere) and that this follows by drawing a certain
complete graph on S such that no two edges cross. While this claim, which became
known as the Heawood conjecture, turned out to be correct, it took almost 80
years before Ringel and Youngs (see [6]) completed the proof. One of the main
ideas behind the proof is the following: Instead of starting out with S and drawing
the complete graph on S, we start out with the complete graph and “paste” discs
on it such that we obtain a surface. By the classification theorem and Euler’s
formula, we know exactly which surface we get, and if we are clever enough,
we get S. The solution of the Heawood problem is an example where the classification
theorem plays a role in reducing a problem with a topological content into a purely
combinatorial one. Recently, surfaces have also played a crucial role in a purely combinatorial
result with farreaching consequences in discrete mathematics and theoretical
computer science. Let p be a graph property satisfying the following: If G is a
graph with property p, then every graph obtained from G by deleting or contract
ing edges also has property p. The RobertsonSeymour theory [7] implies an
efficient method (more precisely, a polynomially bounded algorithm) for testing if
an arbitrary graph has property p. In particular, for any fixed surface S, there is an
efficient algorithm for testing if an arbitrary graph G can be embedded into S, that
is, drawn on S such that no two edges cross. In contrast to this, the problem of
determining the smallest number of handles that must be added to the sphere in
order to get a surface on which G can be embedded is a very difficult one. More
precisely, it is NPcomplete as shown by the author [9]. 2. PLANAR GRAPHS AND THE JORDAN CURVE THEOREM. A simple arc in
a topological space X is the image of a continuous 1 — 1 map f from the real
interval [0,1] into X. We say that f(O) and f(1) are the ends of the arc and that
the arc joins f(0) and f(1). A simple closed curve is defined analogously except
that now f(0) = f(1). We say that X is connected (more precisely, arcwise
connected) if any two elements of X are joined by a simple arc. A simple polygonal
arc or closed curve in the plane is a simple are or closed curve which is the union
of a finite number of straight line segments. 1992] THE JORDANSCHO’NFLIES THEOREM 117 Lemma 2.1. If (2 is an open connected set in the plane, then any two points in Q are
joined by a simple polygonal arc in 0. Proof: Let p and q be any two points in Q and let f be a continuous map from
[0,1] to (I such that f(0) = p and f(1) = q. Let A consist of those numbers t in
[0, 1] such that 9 contains a simple polygonal are from p to, f(t). Put t0 = sup A.
We must have t0 = 1 since otherwise it is easy to find a t] in A such that t1 > to, a
contradiction. D A region of an open set in the plane is a maximal connected subset. A graph G
is the union of two finite disjoint sets V(G) and E(G) (called the vertices and
edges, respectively) such that, with every edge, there are associated two distinct
vertices x and y, called the ends of the edge. We denote such an edge by xy and
say that it joins x and y or that it is incident with x and y. If more than one edge
joins x and y we speak of a multiple edge. An isomorphism between two graphs is
defined in the obvious way. A path is a graph with distinct vertices U1, 112, . . . , on
and edges 01112, U203,...,Un_1Un. If n 2 2 and we add an edge UnUI to this path we
obtain a cycle. We denote both the above path and cycle by UIUZ . . . on. (It will
always be clear from the context if we are talking about a path or a cycle.) If G is a
graph and A g V(G) U E(G), then G —A is the graph obtained from G by
deleting all vertices of A and all those edges which are in A or are incident with a
vertex in A. We say that G is connected if every pair of vertices in G are joined by
a path, and G is 2c0nnected if it is connected and, for every vertex U, G — {U}
(which we also denote by G — v) is connected. The graph G can be embedded in
the topological space X if the vertices of G can be represented by distinct
elements in X and each edge of G can be represented by a simple arc which joins
its two ends in such a way that two edges have at most an end in common. If X is
the Euclidean plane R2, then a graph represented in X is a plane graph, and an
abstract graph which can be represented in X is a planar graph. Lemma 2.2. If G is a planar graph, then G can be drawn (embedded) in the plane
such that all edges are simple polygonal arcs. Proof: Let I‘ be a plane graph isomorphic to G. Let p be some vertex of F, and
let DP be a closed disc with p as center such that DI, intersects only those edges
that are incident with p. Furthermore, assume that DP 0 Dq = Q for every pair of
distinct vertices p, q of F. For each edge pg of F let CW be an are contained in
pq such that CM joins DP with Dq and has only its ends in common with DI, U Dq.
We can now redraw G such that all arcs C p q are in the new drawing and such that
the parts of the edges in the discs D1, are straight line segments. Using Lemma 2.1
it is now easy to replace each C M by a simple polygonal arc. D A subdivision of a graph G is a graph obtained from G by “inserting vertices on
edges.” More precisely, some (or all) edges of G are replaced by paths with the
same ends. Kuratowski’s theorem says that a graph is nonplanar if and only if it
contains a subdivision of one of the Kuratowski graph K 33 or K 5. K5 is the graph
on five vertices such that every pair of vertices are joined by exactly one edge. K3,3 is the graph with six vertices 01,112, U3, u], uz, U3 and all nine edges Uiuj, 1 < i g 3, 118 CARSTEN THOMASSEN [February 1 g j g 3. A discussion of this fundamental result (including a short proof) can be
found in [8]. We shall use here only the simple fact that K 33 is nonplanar. For this
we need the following special case of the Jordan curve theorem. Lemma 2.3. If C is a simple closed polygonal curve in the plane, then R2\ C has
precisely two regions each of which has C as boundary. Proof: We first prove that R2 \ C has at most two regions. So suppose (reductio
ad absurdum) that q1, q2, q3 belong to distinct regions of R2 \ C. Select a disc D
such that D 0 C is a straight line segment. For each i = 1, 2, 3 we can walk along
a simple polygonal arc (close to C but not intersecting C) from q, into D. Hence
some two of q1, q2, q3 are connected by a simple polygonal arc, a contradiction. Next we prove that R2 \ C is not connected. For each point q in R2 \ C we
consider a straight half line L starting at q. The intersection L O C is a finite
number of intervals some of which may be points. Consider such an interval Q. If
C enters and leaves Q on the same side of L we will say that C touches L at Q.
Otherwise C crosses L at Q. It is easy to see that the number of times that C
crosses L (reduced modulo 2) does not change when the direction of L is changed.
So that number depends only on q (and C) and is called the parity of q. Now, the
parity is the same for all points on a simple polygonal arc in R2 \ C and hence it is
the same for all points in a region of R2\C. By considering a half line that
intersects C precisely once we get points of different parity and hence in different
regions. E] The unbounded region of a closed curve C is called the exterior of C and is
denoted ext(C). The union of all other regions is the interior and is denoted
int(C). Furthermore, we write Em) = c u int(C) and EEC) = c u ext(C).
We shall extend Lemma 2.3. Lemma 2.4. Let C be a simple closed polygonal curve and P a simple polygonal arc
in H(C) such that P joins p and q on C and has no other point in common with C.
Let P1 and P2 be the two arcs on C from p to q. Then R2 \ (C U P) has precisely
three regions whose boundaries are C, P1 U P, P2 U P, respectively. Proof: Clearly, ext(C) is a region of R2 \ (C U P). As in the proof of Lemma 2.3
we conclude that the addition of P to C partitions int(C) into at most two regions.
So, we only need to prove that P partitions int(C) into (at least) two regions. Let
L1, L2 be crossing line segments such that L1 is a segment of P, and L2 has
precisely the point in L1 0 L2 in common with C U P. By the proof of Lemma 2.3,
the ends of L2 are in int(C) and in distinct regions of R2 \ (P U P1), hence also in
distinct regions of R2 \ (P U C). D Lemma 2.4 implies that, if r and s are points on P1 \{ p, q} and P2\ { p, q},
respectively, then it is not possible to join r and s by a simple polygonal arc in
H(C) without intersecting P. These remarks also hold when ext and int are
interchanged. Hence we get: Lemma 2.5. K3,3 is nonplanar. 1992] THE JORDANSCHéNFLIES THEOREM 119 Proof: K3,3 may be thought of as a cycle C: x1x2x3x4x5x6 with three chords
x1x4, xzxs, x3x6. Now if K3,3 were planar we would have a plane drawing such
that all edges are simple polygonal arcs, by Lemma 2.2. Then C would be a simple
closed polygonal curve and two of the chords x1x4, xzxs, x3x6 would either be in
E(C) or EEC). But this would contradict the remark after Lemma 2.4. El Everything so far is standard and trivial. Now we are ready for the Jordan curve
theorem. We remark again that the proof uses only the nonplanarity of K13. Proposition 2.6. If C is a simple closed curve in the plane, then R2 \ C is discon
nected. Proof: Let L1 (respectively, L2) be a vertical straight line intersecting C such that
C is entirely in the closed right (respectively, left) half plane of L1 (respectively,
L2). Let p, be the top point on L, m C for i = 1,2, and let P1 and P2 be the two
curves on C from p1 to p2. Let L3 be a vertical straight line between L1 and L2.
Since P1 0 L3 and P2 0 L3 are compact and disjoint, L3 contains an interval L4
joining P1 with P2 and having only its ends in common with C. Let L5 be a
polygonal arc from p1 to p2 in ext(C) consisting of segments of L1, L2 and a
horizontal straight line segment above C. If L 4 is in E(C), then there is a simple
polygonal arc L6 in E(C) from L4 to L5. But then C U L4 U L5 U L6 is a plane
graph isomorphic to K13, contradicting Lemma 2.5. Hence, the midpoint of L4
does not lie in Q(C), so int(C) is nonempty. D We shall also use the nonplanarity of K3,3 to Show that int(C) has only one
region. For this we need some graph theoretic facts. First a result on abstract
graphs. Lemma 2.7. If G is a 2connected graph and H is a 2—c0nnected subgraph of G, then
G can be obtained from H by successively adding paths such that each of these paths
joins two distinct vertices in the current graph and has all other vertices outside the
current graph. Proof: The proof is by induction on the number of edges in E(G) \E(H). If that
number is zero, that is, G = H, then there is nothing to prove. So assume that
G 95 H. By the induction hypothesis, Lemma 2.7 holds when the pair G, H is
replaced by another pair G’, H’ such that E(G’)\E(H’) has fewer edges than
E(G)\E(H). Now let H’ be a maximal 2connected proper subgraph of G
containing H. If H’ ab H we apply the induction hypothesis to H’, H and then to
G, H’. So assume that H’ = H. Since G is connected, there is an edge xlx2 in
E(G)\E(H) such that x1 is in H. Since G —x1 is connected, it has a path
P: x2x3 ~ xk such that xk is in H and all xi, 2 < i < k, are not in H. Possibly
k = 2. Since H U P U {xlxz} is 2connected, we have H U P U {xlxz} = G and
the proof is complete. D If S is a set, then ISI will denote its cardinality.
Lemma 2.8. If F is a plane 2connected graph with at least three vertices, all of whose edges are simple polygonal arcs, then R2 \ F has !E(F)! — V(F)! + 2 regions
each of which has a cycle of F as boundary. 120 CARSTEN THOMASSEN [February Proof: Let C be a cycle in F. By Lemma 2.3, Lemma 2.8 holds if F = C.
Otherwise, F can be obtained from C by successively adding paths as in Lemma
2.7. Each such path is added in a region. That region is bounded by a cycle and
now we apply Lemma 2.4 to complete the proof. (Lemma 2.4 says that the number
of regions is increased by 1 when a region is subdivided). :1 For a plane graph F, the regions of R2 \ P will also be called faces of F. The
unbounded face is the outer face and, if F is 2connected, then the boundary of
the outer face is the outer cycle. The union of two abstract graphs is defined in the obvious way. For plane
graphs we shall make use of a different type of union. Lemma 2.9. If F1 and F2 are two plane graphs such that each edge is a simple
polygonal arc, then the union of F1 and F2 is a graph F3. Proof: First, let I‘; denote the plane graph such that 1“; is a subdivision of F, and
each edge of F’ is a straight line segment for i = 1,2. Secondly, let P," be the l
subdivision of 1‘; such that a point p on an edge a of I‘,’ is a vertex of 1‘,” if either I p is a vertex of F3_, or p is on an edge of F;_, that crosses a. Then the usual
union of the graphs Ff and Pg can play the role of F3. D If both F1 and F2 in Lemma 2.9 are 2connected and have at least two points in
common, then also F3 is 2connected. Lemma 2.10. Let F1, F2, . . . , Pk be plane 2c0nnected graphs all of whose edges are
simple polygonal arcs such that F, has at least two points in common with each of
E_1 and PHI and no point in common with any other (i = 2, 3,...,k — 1). Assume also that F1 0 Pk = Q. Then any point which is in the outer face of each of
F1 U 112,112 U F3 Fk_1 U Pk is also in the outerface of F1 U F2 U U Pk. Proof: Suppose p is a point in a bounded face of F1 U   U Pk. Since
F1 U U Pk is 2connected, it follows from 2.8 that there is a cycle C in
F1 U U Pk suchthat p E int(C). Choose Csuchthat C is in F, U EH U U
I} and such that j — i is minimum. We shall show that j — i < 1. So assume that
j — i 2 2. Among all cycles in F, U U having p in the interior we assume
that C is chosen such that the number of edges in C and not in I‘j_1 is minimum.
Since C intersects both and Fj_2, C has at least two disjoint maximal segments
in Fj_1; let P be one of these; let P’ be a shortest path in Fj_1 from P to
C — V(P); the ends of P’ divide C into arcs P1 and P2, each of which contains
segments not in Fj_1. One of the cycles P’ U P1 and P’ U P2 contains p in its
interior; it has fewer edges not in Fj_1 than C has. This contradicts the minimality
of C, so a minimal C does not lie in a minimal union F, U T, +1 U ~ ' U with
i < j — 2. D Proposition 2.11. If P is a simple arc in the plane, then R2 \P is connected.
Proof: Let p, q be two points in R2 \P and let d be a positive number such that
each of p, q has distance > 3d from P. We shall join p, q by a simple polygonal arc in R2\P. Since P is the image of a continuous (and hence uniformly
continuous) map we can partition P into segments P1,P2,...,Pk such that P, 1992] THE JORDANSCHO'NFLIES THEOREM 121 joins pi and pi+1 for i = 1,2, . . . , k and such that each point on P, has distance
less than d from p, (i = 1,2, . . . , k — 1). Let d’ be the minimum distance between
P, and Pj,1<i<j— 2 <k — 2. Note that d’ <d. For each i= 1,2,...,k, we
partition P, into segments PM, Pm, . ..,P,’ki such that PM joins pm with p,,j+1
for j = 1,2, . .., k, — 1 and such that each point on PM has distance less than
d’ / 4 to pi, j, and let P, be the graph which is the union of the boundaries of the
squares that consist of horizontal and vertical line segments of length d’ / 2 and
have a point pi, , as midpoint. Then the graphs F1, F2, . . . , Pk satisfy the assump
tion of Lemma 2.10. Hence both of p and q are in the outer face of F1 U    U Pk
(because they are outside the disc of radius 3d and with center pi while F, U T, +1
is inside that disc) and P does not intersect that face. Therefore, p and q can be joined by a simple polygonal arc disjoint from P. D If C is a closed subset of the plane and Q is a region of R2 \ C, then a point p
in C is accessible from (I if for some (and hence each) point q in (I, there is a
simple polygonal arc from q to p having only p in common with C. If C is a
simple closed curve, then p need not be accessible from 0. However, if P is any
are of C containing p, then Proposition 2.11 implies that R2\(C \P) is con
nected and therefore contains a simple polygonal arc P’ from q to a region of
R2\ C distinct from (I. Then P’ intersects C in a point on P. Since P can be
chosen to be arbitrarily small we conclude that the points on C accessible from Q
are dense on C. We also get Theorem 2.12 (The Jordan Curve Theorem). If C is a simple closed curve in the
plane, then R2 \ C has precisely two regions, each of which has C as boundary. Proof: Assume (reductio ad absurdum) that q1, q2, q3 are points in distinct regions
01,02, 03 of R2 \C. Let Q1, Q2, Q3 be pairwise disjoint segments of C. By the
remark following Proposition 2.11, 0, has a simple polygonal arc P131 from qt. to
Q for i,j = 1,2,3. We can assume that PM. 0 PM, = {q} for j aéj’. (If we walk
along Pl.)2 from Q2 towards q, and we hit Pi,1 in q; 95 qt, then we can modify Pi,2
such that its last segment is close to the segment of Pi,1 from q} to q, and such
that the new Pi,2 has only q, in common with Pi, 1. Pi,3 can be modified similarly, if
necessary.) Clearly, PM 0 P,” j, = @ when i ah i’. We can now extend (by adding a
segment in each of Q1, Q2, Q3) the union of the arcs PM. (i,j = 1, 2, 3) to a plane
graph isomorphic to K13. This contradicts Lemma 2.5. Thus R2 \ C has precisely
two regions ext(C) and int(C). As above, Proposition 2.11 implies that every point
of C is a boundary point of ext(C) and int(C). El The Jordan Curve Theorem is a special case of the JordanSchonﬂies theorem
which we prove in the next section. For this we shall generalize some of the
previous results. First, Lemma 2.4 generalizes as follows. Lemma 2.13. Let C be a simple closed curve and P a simple polygonal arc in int(C)
such that P joins p and q on C and has no other point in common with C. Let P1 and
P2 be the two arcs on C from p to q. Then R2 \ (C U P) has precisely three regions
whose boundaries are C, P1 U P, and P2 U P, respectively. Proof: As in the proof of Lemma 2.4 the only nontrivial part is to prove that E(C)
is partitioned into (at least) two regions. If the ends of L2 (defined as in the proof
of Lemma 2.4) are in the same region of R2 \ (P U C), then that region contains 122 CARSTEN THOMASSEN [February a polygonal arc P3 such that P3 U L2 is a simple closed polygonal curve. By the
proof of Lemma 2.3, the ends of L) are in distinct regions of R2 \ (P3 U L2). But
they are also in the same region of R2\(P3 U L2) since they are joined by a simple arc (in P U C) not intersecting P3 U L2. This contradiction proves Lemma
2.13. D We also generalize Lemma 2.8. Lemma 2.14. If F is a plane 2connected graph containing a cycle C (whirl is a
simple closed curve) such that all edges in F \ C are simple polygonal arcs in int(C),
then R2\I‘ has IE(F) — IV(F) + 2 regions each of which has a cycle of F as
boundary. Proof: The proof is as that of Lemma 2.8 except that we now use Lemma 2.13
instead of Lemma 2.4. D Finally, we shall use the fact that Lemma 2.9 remains valid if F1 and F2 are
plane graphs whose intersection containﬂ cycle C such that all edges in F1 or F2
(not in C) are simple polygonal arcs in int(C). 3. THE JORDANSCHONFLIES THEOREM. If C and C’ are simple closed
curves and F and I" are 2connected graphs consisting of C (respectively, C’) and
simple polygonal arcs in ﬁ(C) (respectively, R(C’)), then I‘ and I" are said to be
planeisomorphic if there is an isomorphism of F to I" such that a cycle in I‘ is a
face boundary of F iff the image of the cycle is a face boundary of 1" and such
that the outer cycle of F is mapped onto the outer cycle of F’. Theorem 3.1. If f is a homeomorphism of a simple closed curve C onto a simple
closed curve C’, then f can be extended into a homeomorphism of the whole plane. Proof: Without loss of generality we can assume that C’ is a convex polygon. We
shall first extend f to a homeomorphism of in—t(C) to E(C’). Let B denote a
countable dense set in int(C) (for example the points with rational coordinates).
Since the points on C accessible from int(C) are dense on C, there exists a
countable dense set A in C consisting of points accessible from int(C). Let
p,, p2, . . . be a sequence of points in A U B such that each point in A U B occurs
infinitely often in that sequence. Let 1‘0 denote any 2—connected graph consisting
of C and some simple polygonal arcs in E(C). Let Ff, be a graph consisting of C’
and simple polygonal arcs in ﬁ(C’) such that To and F6 are planeisomorphic
(with isomorphism g0) such that go and f coincide on C O V(1“(,). We now extend
f to C U V(F(,) such that g0 and f coincide on V(F(,). We shall define a sequence
of 2connected graphs F0, I], F2 . .. and Fl), F; . .. such that, for each n 2 1, F" is
an extension of a subdivision of F,,_ 1, I); is an extension of a subdivision of BL],
there is a plane isomorphism g" of F" onto 1“; coinciding with g,,_I on V(I‘n_l),
and F" (respectively 17,) consists of C (respectively C’) and simple polygonal arcs
in H(C) (respectively E(C’)). Also, we shall assume that I); \ C’ is connected for
each n. We then extend f to C U VG”) such that f and g" coincide on V(1“,,).
Suppose we have already defined F0, Fl, . . . , an1, Fl), Ff, . . . , 17,; l, and
g0, g1, . . . , gngl. We shall define F", F}, and g" as follows. We consider the point
p". If p" E A, then we let P be a simple polygonal are from p” to a point an of
1‘";l \ C such that Pn_l H P = {pn, an}. We let I), denote the graph I‘n_l U P. P 1992] THE JORDANSCHONFLIES THEOREM 123 is drawn in a face of Fn_1 bounded by a cycle S, say. We add to 17H1 a simple
polygonal arc P’ in the face bounded by gn_1(S) such that P’ joins f(pn) with
gn_1(qn) (if q" is a vertex of Fn_1) or a point on gnal(a) (if a is an edge of Fnzl
containing the point q”). Then we put I); = I);1 U P’ and we define the planeiso
morphism gn from F” to 11’, in the obvious way. We extend f such that f(qn) =
g,,(q,.). If p" E B we consider the largest square which has vertical and horizontal
sides, which has 1),, as midpoint and which is in ﬁ(C). In this square (whose sides
we are not going to add to I‘n_1 as they may contain infinitely many points of C)
we draw a new square with vertical and horizontal sides each of which has distance
< l/n from the sides of the first square. Inside the new square we draw vertical
and horizontal lines such that p" is on both a vertical line and a horizontal line
and such that all regions in the square have diameter < l/n. We let H,1 be the
union of Fnzl and the new horizontal and vertical straight line segments possibly
together with an additional polygonal arc in int(C) in order to make H" 2con
nected and H” \ C connected. By Lemma 2.7, H” can be obtained from I‘n_1 by
successively adding paths in faces. We add the corresponding paths to I‘é_1 and
obtain a graph H; which is planeisomorphic to H". Then we add vertical and
horizontal lines in E(C’) to H}, such that the resulting graph has no (bounded)
region of diameter 2 1/2n. If necessary, we displace some of the lines a little such
that they intersect C’ only in f(A) and such that all bounded regions have
diameter < 1/11 and such that each of the new lines has only finite intersection
with H,’. This extends Hf, into a graph we denote by 171. We add to H” polygonal
arcs such that we obtain a graph 1‘" planeisomorphic to 17;. Then we extend f
such that it is defined on C U V(Fn) and coincides with the planeisomorphism g”
on V(I‘n). When we extend H; into I‘,’ and H" into T" we are adding many edges
and it is perhaps difficult to visualize what is going on. However, Lemma 2.7 tells
us that we can look at the extension of H}, into 11’, as the result of a sequence of
simple extensions each consisting of the addition of a path (which in this case is a
straight line segment in a face). We then just perform successively the correspond
ing additions in H". Note that we have plenty of freedom for that since the current
f is only defined on the current vertex set. The images of’the points on the current
edges have not been specified yet. In this way we extend f to a 1 — 1 map defined
on F = C U V(I‘0) U V(F1) U ~_and with image C’ U V(F(’,) U V(F{) U .
These sets are dense in E(C) and int(C’), respectively. If p is a point in int(C) on
which f is not yet defined, then we consider a sequence q1, q2, . . . converging to p
and consisting of points in V(F0) U V(F1) U  ' ' . We shall show that
f(ql), f(qz), . . . converges and we let f( p) be the limit. Let d be the distance from
p to C and let p" be a point of B of distance < 03/3 from p. Then p is inside the
largest square in iFdC) having p" as midpoint (and also inside what we called the
new square if n is sufficiently large). By the construction of 1‘" and 11" it follows
that F" has a cycle S such that p E int(S) and such that both S and gn(S) ar_e in
discs of radius < l/n. Since f maps F n int(S) into int(gn(S)) and F n ext(S)
into §(gn(S)), it follows in particular, that the sequence f(qm), f(qm +1), . . . is in
int(gn(S)) for some m. Since 11 can be chosen arbitrarily large, f(ql), f(qz), . .. is a
Cauchy sequence and hence convergent. It follows that f is welldefined. More
over, using the above notation, f maps int(S) into int(gn(S)). Hence f is continu
ous in int(C). Since V(I‘(’,) U V(F{) is dense in int(C’) the same argument shows
that f maps int(C) onto int(C’) that f is 1 — 1 and that f‘1 is continuous on
int(C’). It only remains to be shown that f is continuous on C. (Then also f‘1 is
continuous since in_t(C) is compact). In order to prove this it is sufficient to 124 CARSTEN THOMASSEN [February consider a sequence q], (12,. .. of points in int(C) converging to q on C and then
show that f(q_1), f(qz), . . . converges to f(q). Suppose therefore that this is not the
case. Since int(C’) is compact we can assume (by considering an appropriate
subsequence, if necessary) that lim,1 _m f(qn) = 61’ 9E f(q). Since 1“1 is continuous
on int(C’), q’ is on C’. Since A is dense in C, f(A) is dense in C’ and hence each
of the two arcs on C’ from (1’ to f(q) contain a point ﬂat) and f(qz), respectively,
in f(A). For some n, F” has a path P from q1 to q2 having only q1 and q2 in
common with C. By Lemma 2.13, P separates int(C) into two regions. These two
regions are mapped on the two distinct regions of int(C’) \gn(P). One of these
contains almost all the f(qn) while the other has f(q) on its boundary, but not the
boundary common to both regions. Hence we cannot have limn gm f(qn) = (1’. This
contradiction shows that f has the appropriate extension to int(C). By similar arguments, f can be extended to ext(C). We consider a coordinate
system in the plane. Without loss of generality we can assume that int(C) contains
the origin and that both C and C’ are in the interior of the quadrangle T with
corners (i1, i 1). Let L1,L2,L3 be the line segments (on lines through the
origin) from (1, 1), (— 1, — 1) and (1, — 1), respectively, to C. Let p, be the end of
L, on C for i = 1,2,3. Let L’1 and L’2 be polygonal arcs from f(pl) to (1,1) and
from f(pz) to (— 1, — 1), respectively, such that L’l ﬂ L’2 = Q and U, has only its
ends in common with C’ and T for i = 1,2. It is easy to see that we can find a
polygonal arc L’3 from f(p3) to either (1, — 1) or (— 1, 1) such that L’3 is disjoint
from L’l U L’2 and has only its ends in common with C’ and T. After a reﬂection
of C’ in the line through (1, 1) and (— 1, — 1), if necessary, we can assume that L’3
goes to (1, — 1). Now we can use the method of the first part of the proof to extend
f to a homeomorphism of EU) such that f is the identity on T. Then f extends
to a homeomorphism of the whole plane such that f is the identity on ext(T). El If F is a closed set in the plane, then we say that point p in F is carveaccessi
ble if, for each point q not in F, there is a simple arc from q to p having only p in
common with F. The JordanSchonﬂies theorem implies that every point on a
simple closed curve is curveaccessible. Hence we have the following extension of
part of Theorem 2.12. Theorem 3.2. If F is a closed set in the plane with at least three curveaccessible
points, then R2 \ F has at most two regions. Proof: If p1, p2, p3 are curveaccessible on F and ql,qz,q3 belong to distinct
regions of R2\F, then we get, as in the proof of Theorem 2.12, a plane graph isomorphic to K3,3 with vertices p1,p2,p3,q1, q2, (13, a contradiction to Lemma
2.5. CI In Theorem 3.2, “three” cannot be replaced by “two.” To see this, we let F be
a collection of internally disjoint simple arcs between two fixed points. Theorem 3.3. Let F and 1" be 2connected plane graphs such that g is a homeomor—
phism and planeisomorphism of F onto F’. Then g can be extended to a homeomor
phism of the whole plane. Proof: The proof is by induction on the number of edges of F. If F is a cycle, then Theorem 3.3 reduces to Theorem 3.1. Otherwise it follows from Lemma 2.7 that F
has a path P and a 2connected subgraph F1 containing the outer cycle of F 1992] THE JORDANSCHéNFLIEs THEOREM 125 such that F is obtained from I‘1 by adding P in E(C) where C bounds a face of
F1. Now we apply the induction hypothesis first to I‘1 and then to the two cycles of
C U P containing P. 4. TRIANGULATING A SURFACE. Consider a finite collection of pairwise dis
joint convex polygons (together with their interiors) in the plane such that all side
lengths are 1. Form a topological space S as follows: Every side in a polygon is
identified with precisely one side in another (or in the same) polygon. This also
defines a graph G whose vertices are the corners and the edges the sides. Clearly
S is compact. Now S is a surface iff S is connected (i.e., G is connected) and S is
locally homeomorphic to a disc at every vertex U of G. If this is the case then we
say that G is a 2—cell embedding in S. If all polygons are triangles, then we say that
G is a triangulation of S and that S is a triangulated surface. In case of a
triangulation we shall assume that there are at least four triangles and that there
are no multiple edges. Theorem 4.1. Every surface S is homeomorphic to a triangulated surface. Proof: Since the interior of a convex polygon can be triangulated it is sufficient to
prove that S is homeomorphic to a surface with a 2—cell embedding. For each point
p on S, let D( p) be a disc in the plane which is homeomorphic to a neighbour
hood of p on S. (Instead of specifying a homeomorphism we shall use the same
notation for a point in D( p) and the corresponding point on S.) In D( p) we draw
two quadrangles Q1( p) and Q2( p) such that p E int(Q1(p)) C int(Q2(p)). Since S
is compact, it has a finite number of points p1, p2, . . . , 1),, such that S =
U111 int(Q1(pi)). Viewed as subsets in the plane, D(p1),...,D(pn) can be as
sumed to be pairwise disjoint. In what follows we are going to keep
D(p1), D(p2), . . . , D(p,,) fixed in the plane (keeping in mind, though, that they
also correspond to subsets of S). However, we shall modify the homeomorphism
between D(p,) and the corresponding set on S and consider new quadrangles
Q1(p,). More precisely, we shall show that Q1(p1), . . . , Q1(pn) can be chosen such
that they form a 2cell embedding of S. Suppose, by induction on k, that they have been chosen such that any two of
Q1( p1), Q1( p2), . . . , Q1(pk_1) have only a finite number of points in common on S.
We now focus on Q2(pk). We define a bad segment as a segment P of some
Q1(pj) (1 <j < k — 1) which joins two points of Q2(pk) and which has all other
points in int(Q2(pk)). Let Q3(pk) be a square between Q1( pk) and Q2(pk). We say
that a bad segment inside Q2(pk) is very bad if it intersects Q3(pk). There may be
infinitely many bad segments but only finitely many very bad ones. The very bad
ones together with Q2(pk) form a 2—connected graph I‘. We redraw F inside
Q2(pk) such that we get a graph F’ which is planeisomorphic to I‘ and such that
all edges of F’ are simple polygonal arcs. This can be done using Lemma 2.7. Now
we apply Theorem 3.3 to extend the planeisomorphism from I‘ to l" to a
homeomorphism of EQ2(pk) keeping Q2(pk) fixed. This transforms Q1(pk) and
Q3(pk) into simple closed curves Q1 and Q3 such that pk E int Q’1 g int Q’3. We
consider a simple closed polygonal curve ’3’ in int Q2(pk) such that Q’1 g int Q’3’
and such that Q; intersects no bad segments except the very bad ones (which are
now simple polygonal arcs). (The existence of Q’3’ can be established as follows: For
every point p on Q’3 we let R( p) be a square with p as midpoint such that R( p)
does not intersect either Q’1 nor any bad segment which is not very bad. We
consider a (minimal) finite covering of Q’3 by such squares. The union of those 126 CARSTEN THOMASSEN [February squares is a 2connected plane graph whose outer cycle can play the role of Q’3’).
By redrawing 1" U Q’g (which is a 2connected graph) and using Theorem 3.3 once
more we can assume that '3' is in fact a quadrangle having Q’1 in its interior. If we
let Q; be the new choice of Q1(pk), then any two of Q1(p1), . . . , Q1(pk) have only
finite intersection. The inductive hypothesis is proved for all k. Thus we can assume that there are only finitely many very bad segments inside
each Q2(pk) and that those segments are simple polygonal arcs forming a 2con
nected plane graph. The union U 7:1Q1( pl.) may be thought of as a graph F drawn
on S. Each region of S \ F is bounded by a cycle C in I‘. (We may think of C as a
simple closed polygonal curve inside some Q2(p,)). Now we draw a convex polygon
C’ of side length 1 such that the corners of C’ correspond to the vertices of C. The
union of the polygons C’ forms a surface S’ with a 2cell embedding I" which is
isomorphic to T. An isomorphism of F to T’ may be extended to a homeomor
phism f of the point set of F on S onto the point set of I" on S’. In particular, the
restriction of f to the above cycle C is a homeomorphism onto C’. By Theorem
3.1, f can be extended to a homeomorphism of in—t(C) to H(C’). This defines a
homeomorphism of S onto 5’. El 5. THE CLASSIFICATION OF SURFACES. Consider now two disjoint triangles
T1, T2 (such that all six sides have the same length) in a face F of a surface S with
a 2cell embedding G. We form a new surface S’ by deleting from F the interior
of T1 and T2 and identifying T1 with T2 such that the clockwise orientations
around T1 and T2 disagree. (We recall that S consists of polygons and their
interiors in the plane. So when we speak of clockwise orientation we are simply
referring to the plane. We are not discussing orientability of surfaces.) If the
orientations agree we obtain instead a surface S”. Finally, we let S’” denote the
surface obtained by deleting the interior of T1 and identifying “diametrically
opposite” points on T]. We say that S’, S”, S’” are obtained from S by adding a
handle, a twisted handle, and a crosscap, respectively. It is easy to extend G to a
2cell embedding of S’, S” and S’”, respectively. Also, it is an easy exercise to show
that S’, S” and S’” are independent (up to homeomorphism) of where T1 and T2
are located since it is easy to continuously deform a pair of triangles into another
pair of triangles inside a given triangle. In fact, they may belong to distinct faces,
also, except that then (at this stage) we cannot distinguish between a handle and a
twisted handle. When adding a crosscap it is sufficient that T1 is a simple closed
polygonal curve, which can be continuously deformed into a point (and hence to a
triangle in a face). We shall now consider all surfaces obtained from the sphere S0 (which we here
think of as a tetrahedron) by adding handles, twisted handles and crosscaps. If we
add to S0 h handles we obtain the surface Sh, and if we add to S0 k crosscaps we
obtain Nk. SI,N1,N2 are the torus, the projective plane and the Klein bottle,
respectively. N2 is also S0 plus a twisted handle. One way to see this is as follows:
Let T1 and T2 be two disjoint tetrahedra (which are homeomorphic to SO). Select
a triangle in T1 and T2 and add in that triangle a twisted handle or two crosscaps.
This transforms T1 into T1’ and T2 into Té. Now choose your favourite representa
tion of the Klein bottle and your favourite triangulation G of it. Then for each
i = 1,2, draw G on T,’ such that the face boundaries are the same triangles in G
in all three triangulations. Then the graph isomorphism of G on T; to G on T;
can be extended to a homeomorphism of TI’ onto Té. Moreover, if we have already
added a crosscap, then adding a handle amounts to the same, up to homeomor phism, as adding a twisted handle. (First observe that when we add a crosscap, it 1992] THE JORDANSCHONFLIES THEOREM 127 does not matter where we add it; we get always the same surface up to homeomor
phism. So we only need to verify the statement when we add a crosscap and then a
handle or twisted handle inside the same triangle of the surface. This can be done
by triangulating the two surfaces by the same graph G as above). So, the surfaces
obtained from S0 by adding handles, twisted handles and crosscaps are precisely
the surfaces Sh (h 2 0) and Nk (k 2 1). Theorem 5.1. Let S be a surface and G a 2cell embedding of S with n vertices, e
edges and f faces. Then S is homeomorphic to either S h or Nk where h and k are
deﬁned by the equations n—e+f=2—2h=2—k. Proof: We first show that n — e + f g 2. For this we successively delete edges
from G until we get a minimal connected subgraph of G, that is, a spanning tree
H of G. For each edge deletion the number of faces (which are now not
necessarily 2cells) is unchanged or decreased by 1. Since H has n vertices, n — 1
edges and only one face it follows that n — e + f g 2. We next extend G to a triangulation of S as follows: For each face F of G
which is a convex polygon with corners vl,vz,...,vq, where q 2 4 and their
indices are expressed modulo q, we add new vertices a, 141,. . ., a in F and we 4
add the edges uv uiui+1,uiui+1,aiu for i=1,2,...,q. Let n’, e’,f’ be the l 1’
number of vertices edges and faces, respectively, of G’. Clearly, n’ — e’ + f’ = n —
e + f. Thus it is sufficient to prove the Theorem in the case where G is a
triangulation which we now assume. Suppose (redactio ad absurdum) that S, G are
a counterexample to Theorem 5.1 such that G is a triangulation with at least four
vertices and (1)2 — n + e — f is minimum. (2) n is minimum subject to (1), and (3) the minimum valency m of G is minimum subject to (1), (2). (The valency of
a vertex is the number of edges incident with it.) Let u be a vertex of minimum valency. Let 0,, U2, . . . , um be the neighbours of U
such that UUIUZ, uuzu3, . . . , uumul are the faces incident with U and the indices are
expressed modulo m. Since 01 and um are joined only by one edge, m 2 3. If
m = 3, then G — u is a triangulation of S unless n = 4 in which case S is the
tetrahedron. This contradicts (2) or the assumption that S, G are a counterexam
ple to the Theorem. So m 2 4. If for some i = 1, 2, . . . , m, U, is not joined to 0H2 by an edge, then we let G’
be obtained from G by deleting the edge UUi+l and adding the edge up,” instead.
Clearly, G’ triangulates S, contradicting (3). So we can assume that G contains all
edges vial”, i = 1, 2, . . . , m, when u is a vertex of minimum valency. Intuitively, we complete the proof by cutting the surface (using a pair of scissors,
say) along the triangle T: 00103. This transforms T into either two triangles T1 and
T2 or into a hexagon H (in case S has a Mobius strip that contains T). We get a
new surface S’ by adding two new triangles (and their interior) or a hexagon (and
its interior which we triangulate) and identify their sides with T1 and T2 or with H,
respectively. Then S’ is a triangulated surface with smaller 2 — n + e — f than S.
By the minimality of this parameter, S’ is of the form Sg, or Nk. Then S is of that
form, too. Formally, we argue as follows. Recall that S is a triangulated surface, i.e., S is
obtained by identifying sides of pairwise disjoint triangles in the plane. Let M
denote the topological space which is formed by using the same triangles and the 128 CARSTEN THOMASSEN [February same side identifications, except that those six sides that correspond to the edges
uul,vlv3,u3u are not identified with any other side. Let us call those six sides
boundary sides of M. Let G’ be the graph whose vertices are the corners of the
triangles of M and whose edges are the sides of the triangles. It is easy to see that
G’ has precisely six vertices which are incident with boundary sides and that each
of these six vertices is incident with precisely two boundary sides. Thus the
boundary sides are a subgraph C of G’ with vertices each Of which has valency 2.
There are only two such graphs (up to isomorphism): C is either a hexagon or two
disjoint triangles. If C is two disjoint triangles, then we add to M two disjoint
triangles (and their interior) in the plane and identify their sides with the edges of
C such that we obtain a new surface S’ which is triangulated by G’. If C is a
hexagon, then we add to M a hexagon in the plane together with its interior (which
we triangulate) and then we identify the sides of this hexagon with the edges of C.
In this way M is extended to a surface S” and G’ is extended to a graph G” which
triangulates S”. Thus we have transformed G and S into a triangulation G’ with n’
vertices e’ edges and f’ faces of a surface S’, or a triangulation G” with n”
vertices e” edges and f” faces of a surface S”. In the former case we have e’—n’+f’=e—n+f+2.
In the latter case we have
e”—n”+f”=e—n+f+1. By (1), S’ or S” is homeomorphic to a surface of the form S h, or Nk/. (Note that G’
is obtained from G by “cutting” the triangle vulu3. Then G’ is connected because
of the edge 020m. Hence also the spaces M, S’, S” are connected.) If C consists of
two triangles, then clearly S is obtained from S’ by adding a handle or a twisted
handle. If C is a hexagon, then in S”, C can be continuously deformed into a
point, and hence S is obtained from S” by adding a crosscap (see the discussion
preceding Theorem 5.1). In the latter case (where C is a hexagon) S is homeomor
phic to Nk'+1 or NM+1 (by the discussion preceding Theorem 5.1). This contra
dicts the assumption that S and G are a counterexample to Theorem 5.1.
Similarly, if C is two triangles, then S is homeomorphic to either Nk, +2 or S h, +1 or
NM +2, and again we obtain a contradiction which finally proves the theorem. [:1 We have now completed the proof of the classification theorem without refer
ring to orientability of surfaces or using Euler’s formula (which consists of the
equations of Theorem 5.1 and which is therefore now a corollary of Theorem 5.1).
To complete the discussion we indicate a proof of the fact that all the surfaces
S0, S1, . . . , N1, N2 . . . are pairwise nonhomeomorphic. In this discussion, however,
many details will be left for the reader. First we observe that Euler’s formula holds for all 2cell embeddings since any
such embedding can be extended to a triangulation. Now let us consider any
connected graph G with n vertices and e edges drawn on Sh. Using Lemma 2.2 we
assume that all edges are simple polygonal arcs. Let f be the number of faces for
this drawing. If G’ is a 2cell embedding of S h, then G L! G’ is a 2cell embedding
satisfying Euler’s formula and containing a subdivision of G. By successively
deleting edges (and isolated vertices) from G L! G’ until we get a subdivision of G
we conclude that n—e+f>2—2h.
Since
3f<26 1992] THE JORDANSCHéNFLIES THEOREM 129 we conclude that e<3n—6+6h with equality if and only if G is a triangulation of S h. Thus a triangulation of S h
has too many edges in order to be drawn on S ht when h’ < h, and hence S h and S h,
are nonhomeomorphic for h’ < h. More generally, this argument shows that
S0, S1, . . . , N1, N2, . . . are pairwise nonhomeomorphic except that S h and N2},
might be homeomorphic. We sketch an argument which shows that they are not.
It is easy to describe a simple closed polygonal curve C in N2h such that, when
we traverse C, left and right interchange. Also it is easy (though a little tedious) to
show that S h has no such simple closed polygonal curve C’. (It is convenient to
consider a 2cell embedding G such that G contains no such C’ and then extend
the argument to an arbitrary C’ in S.) So it suffices to show the following: If there
exists a homeomorphism f: NM —> Sh, then there exists a homeomorphism f’:
NM —> S h such that f ’(C ) is a simple closed polygonal curve. To see this we let G
be a 2cell embedding of N”. Then also G U C may be regarded as a 2cell
embedding, and G L! C can be extended to a triangulation H of NM. We
construct H such that it has no other triangles than the face boundaries. Then
@(H) is a graph drawn on Sh and we apply Lemma 2.2 to redraw gD(H) (resulting
in a graph H’) such that all edges are simple polygonal arcs. Since H’ and H are
isomorphic and H is a triangulation of NM, it follows from Euler’s formula that
H’ is a triangulation of S h. Hence the face boundaries of H’ are the same as the
face boundaries of H. So, any isomorphism H —> H’ can be extended into a
homeomorphism go’: NM —> Sh taking C into a simple closed polygonal curve. ACKNOWLEDGMENT. Thanks are due to the referee for numerous comments on the paper. REFERENCES P. Andrews, The classification of surfaces, Amer. Math. Monthly, to appear. J. L. Gross and T. W. Tucker, Topological Graph Theory, Wiley and Sons, New York, 1987. P. J. Heawood, Mapcolor theorem, Quart. J. Math. Oxford Ser., 24 (1890) 332—338. B. Kerékjarté, Vorlesungen iiber Topologie, Springer, Berlin, 1923. E. E. Moise, Geometric Topology in Dimensions 2 and 3, Graduate Texts in Mathematics, Springer,
New York 1977. G. Ringel, Map Color Theorem, SpringerVerlag, Berlin, 1974. N. Robertson and P. D. Seymour, Graph minors XIII. The disjoint paths problem, to appear. C. Thomassen, Kuratowski’s theorem, J. Graph Theory, 5 (1981) 225—241. C. Thomassen, The graph genus problem is NPcomplete, J. Algorithms, 10 (1989) 568—576. C. Thomassen, Embeddings and minors, in: Handbook of Combinatorics (eds., M. Grotschel,
L. Lovasz and R. L. Graham), NorthHolland, to appear. 11. W. T. Tutte, Graph Theory, AddisonWesley, Reading, Mass, 1984. 12. H. Tverberg, A proof of the Jordan Curve Theorem, Bull. London Math. Soc. 12 (1980) 34—38. .U‘PS’JNE" PPWS? 1 Mathematical Institute The Technical University of Denmark
Building 303 DK2800 Lyngby, DENMARK 130 CARSTEN THOMASSEN [February Are Mathematics and Poetry Fundamentally Similar?
JoAnne S. Growney If you doubt their intrinsic similarity, consider the following quotations. In each of the
following, the key word (“mathematics” or “poetry” or “mathematician” or “poet” or a
variation of one of these terms) has been left out, although the name of the author may provide
a give—away clue. Can you guess which art form is being described in each case? The missing
words are supplied at the end of the quotations. (1) is the art of uniting pleasure with truth. —Samuel Johnson
(2) To think is thinkable—that is the ’s aim. —Cassius J. Keyser
(3) All [is] putting the infinite within the finite. —Robert Browning (4) The moving power of invention is not reasoning but imagination.
—A. DeMorgan (5) When you read and understand , comprehending its reach and formal meanings,
then you master chaos a little. —Stephen Spender (6) practice absolute freedom. —Henry Adams (7) I think that one possible definition of our modern culture is that it is one in which
ninetenths of our intellectuals can’t read any . —Randall Jarrell (8) Do not imagine that is hard and crabbed, and repulsive to common sense. It is
merely the etherealization of common sense. —L0rd Kelvin (9) The merit of , in its wildest forms, still consists in its truth; truth conveyed to the
understanding, not directly by words, but circuitously by means of imaginative associations,
which serve as conductors. —T. B. Macaulay (10) It is a safe rule to apply that, when a or philosophical author writes with a misty
profundity, he is talking nonsense. —A. N. Whitehead (11) is a habit. —C. DayLewis (12) . .. in you don’t understand things, you just get used to them.
—John von Neumann (13) are all who love—who feel great truths
And tell them. —P. J. Bailey
Festus (14) The is perfect only in so far as he is a perfect being, in so far as he perceives the
beauty of truth; only then will his work be thorough, transparent, comprehensive, pure,
clear, attractive, and even elegant. —Goethe (15) . . . [In these days] the function of as a game . . . [looms] larger than its function
as a search for truth .. . . —C. DayLewis (16) A thorough advocate in a just cause, a penetrating facing the starry heavens,
both alike bear the semblance of divinity. —Goethe (17) is getting something right in language. —Howard Nemerov See pg. 133 for answers. These quotations are taken from an article by Professor Growney entitled “Mathematics and
Poetry: Isolated or Integrated” which appeared in the Humanistic Mathematics Network Newslet
ter #6 (May 1991), 60—69. To subscribe contact Alvin White, Harvey Mudd College. 131 ...
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