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This is to certify that the

thesis entitled

Planar Sets having Property P
presented by
Merle D. Guay
has been accepted towards fulfillment

of the requirements for

Ph.D. degree “Mathematics

%4 Q @149

Major professfi

DMC May 23, 1967

 

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IHEDI:

 

 

 

 

IHESlS

 

 

ABSTRACT
PLANAR sers HAVING PROPERTY P“

by Merle D. Guay

As a natural generalization of convexity, a subset X of a
set S in a linear Space L is said to have property Pn relative to
S if for every n distinct points of X at least two of the points are
Joined by a line segment which lies entirely in S; if X = S, then S
-is said to have property Pn. Property P2 is the usual definition

of convexity.

It is first shown that a set having property Pn may be ex-
pressed as the union of (n91) or fewer starlike sets. Several re"
sults which depend primarily upon the linearity of the containing

space are then obtained fer sets having property P“.

In an attempt to determine the number of closed convex sub-
sets which are required to express a closed, connected Pn set as
the union of convex sets several results are obtained. For n = M,
the maximum number is shown to be 5 if S bounds a bounded domain of
its complement; and to be M if S has a cut point, a oneudimensional
kernel, contains a point at which 5 is both locally convex and one«
dimensional, or has at most one point of local non-convexity which
is not in the kernel of S. If S has exactly one point of local non»
convexity Q, then S is shown to be starlike from q, without assuming

Y‘.
that S has property P“; if in addition, 8 has property P“, then it

IH SSSS

 

is shown that S may be expressed as the union of (n-l) or fewer
closed convex sets. Finally, if S has two or more points of local
non—convexity each of which is contained in the kernel of S, then
S is shown to be expressible as the union of-3 or fewer closed

convex sets, independent of’property Pn.

Finally, the higher dimensional case, the topological
preperties of Pn sets, and the problem of obtaining an upper bound
on the number of convex sets required to express a set having prop-

erty Pn as the union of convex sets are briefly considered.

 

 

PLANAR SETS HAVING PROPERTY P“

By

mrle Do Guay

A .THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirement
fbr the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1967

This thesis has been examined and approved.

 

 

 

 

 

 

 

 

Date

 

ACKNOWLEDGMENTS ‘

I am greatly indebted to Professor John G. Hocking for his

helpful guidance and encouragement during the preparation of this
thesis.

I also wish to express my gratitude to Professor B. Grunbaum
and Professor L. M. Kelly for their valuable suggestions and comments.

I am particularly grateml to the latter for suggesting a shorter

proof of Theorem 3. 3.

ii

CONTENTS

Chapter
I . DH'RODUCTION Q Q Q Q Q 0

II. SETS HAVING PROPERTY Pn . . .
. III. PLANAR SEI‘S HAVING PROPERTY P“ .
Iv. FURI‘HER CONSIDERATIONS AND EXAMPLES

BIBIJImRA-Plfl O ‘ O O O O O O

Page

17

6'4

71

CHAPTER I
INTRODUCTION'

As a natural generalization of convexity, a set S in a linear
topological space L is said to have property Pn if for every n distinct
points of S at least two of the points are Joined by a line segment
which lies entirely in S. Property P2 is equivalent to convexity. For
n a 3, valentine [17] found this concept to be usefu1 in the study of
sets each of which is the union of two convex sets. He was able to
show that a closed connected set in E2 having property P3 can always
be expressed as the union of three or fewer closed convex sets having
a mn-errpty intersection, and that the number three is best. He later
fbund this same concept userl in proving that the boundaries of two
and S

compact, convex bodies S in a Mdnkowski space Ln intersect in

1 2
a finite number of (n.- 2) - dimensional manifblds, provided that the
intersection of the interiors of S1 and 82 be contained in the interior

of the convex hull of the union of S1 and $2 [18]. The definition of
property P3 given by Valentine suggested to me the definition given
above as a natural generalization. It was later discovered that

Allen [1] and in a Joint paper, Danzer, Grunbaum and Klee [6] had
given generalizations of convexity which encompass the definition above

as a Special case. Hewever, no relevant publications have appeared to

date.

- 2'.

The results of valentine [17] suggest the possiblity that a
closed connected set in E2 having property P" should be expressible
as the union of n.or fewer closed convex sets. However, this con-
Jecture is false. For example, if the set S is closed, connected,
has property Pu, and bounds a bounded residual domain of S, then 3
may be expressed as the union of 5 or fewer closed convex sets, and
the number 5 is best. This example and the fact that a set with
property P3 is starlike suggests that the condition of starlikeness
be added to property Pn in the hypothesis of the conjecture. If, in
addition, certain restrictions are placed upon the nature of the set ‘
of'points of local non-convexity, the result is fbrthcoming. In gen-
eral, however, the result is still a conjecture for the case n = h,
while fer Values of n.greater than four starlikeness does not restrict

the number of convex sets to be n.

The results contained herein were obtained in an attempt to de-
termine the properties of sets having property Pn (n 3.3), and to de-
termine how such sets may be expressed as the union of their convex

subsets.

CHAPTER II
SEI‘S HAVING PROPERTY Pn

The results of this chapter are of an intrinsic nature, de-
pending primarily upon the properties attributed to the set itself.
The linearity of the containing space is indispensable, of course.
While it is assumed that the sets being considered are embedded in
Euclidean m—dimensional Space, Em, many of the results could have
been stated with a general linear topological space as the contain-
ing space. Since the results are also of a heterogeneous nature,
and are not required, fer the most part, in the proofs of later re-
sults, they are numbered as prepositions rather than being called
lemmas.

With rare exception familiarity with the common terminology
of convexity and topology is assumed. Notation used is explained
as its introduction becomes necessary. The following less familiar
definitions are essential to the understanding of’most of that which

fellows. Each is a natural generalization of convexity.

Definition 2.1 A set 8 contained in a linear Space L, is said to be
starlike if there exists a point x in S such that for each y in 8,

it is true that the line segment xy lies entirely in 8.
Remark: A non-empty convex set is starlike from each of its points.
Definition 2.2 A subset X of a set S in a linear space L is said to

have preperty Pn, (n :_2), relative to S if fbr every n distinct points

of X at least two of the points are joined by a line segment which lies

-ll-

entirely in S. If X = S, the set X is said to have property P”,
Remark: Property P2 is the usual definition of convexity.

The next two results help to explain the intimate relationship

between the two concepts.

Proposition 2.1 Let S C Lmhave property P“. Then S may be expressed

as the union of (n - l) or fewer starlike sets.

M. For n = 2, the set is convex. Assume then that the
result is true for n = k - l, and consider the case n = k. There
must exist (k - 1) points of S no two of which are Joined by a line
segment lying entirely in S, since otherwise S has property P1 for
j < k and the induction Wpotheses applies. Hence, let p1,... ’pk-l’
be the (k - 1) points no two of which are Joined by a line segment
lying entirely in S, and let x be an element of S different from pi,
i as l,...,k-l. Then the line segment xp1 is contained in S for some
1, since otherwise x,pl,... ’pk-l would violate property Pk. Thus S

is a union of sets X1 starlike from p1 and the result follows.

As trivial examples of sets in E2 having property Pn one might
consider the boundary of a regular n - sided polygon as a set having
property Pml for n _>_ 3. A set consisting of n distinct line segments
which intersect in the origin is an example of a starlike set having

property PM1 .

Remarks: It is clear that any set which is the union of exactly

(n - l) convex sets has property Pu,

-5-

It is also clear that property Pn implies property Pm for

m>n.

The well-known definition of local convexity proves to be

extremely useful and so is included.

Definition 2.3 A set S is said to be locally convex at a point p in S
if there exists an open Spherical neighborhood N of p such that S n N A
is convex. If a set is locally convex at each of its points, it is
' said to be locally convex. A point p of S is a point of local non-

convexity if S is not locally convex at p.

Proposition 2.2 If S C Em is a closed connected set having property P”,
n > 3, then S is the union of a starlike set and a set having property
Pn-2 relative to S.

M. Tietze [114] has shown that a closed connected set in
Em which is locally convex is in fact convex. Hence, if S has no points
of local non-convexity we are done. Let t be a point of local non-
convexity and T = {x c S l xtc: S} let xl,...,xn_2 be points of S - T
which are not Joined in S, and N be a Spherical neighborhood of t of

1

radius in By the closure of S and the definition of t, for i suffi-

ciently large, there exist yi and z in N1 n S such that y:‘.,zi,xl,...,lttn__2

i
are not Joined in S, contradicting the fact that S has property Pn.
Hence, S - T is contained in a subset of S which has property Pm"2

relative to S, and the result follows.

Remark: Instinctively one considers the above result as an invitation
to attempt an induction on n when seeking to prove a given result. While

this is sometimes effective, the set T will, in general, have the same

- 5 _

property Pn as did 3, and S - T is only contained in a subset of S

having property Pn"2 relative to S. For n = A, consider the following
example which illustrates the difficulty: the shaded area corresponds
to S - T.

 

 

 

 

 

/’ .z

/ /

As the union of three convex sets S quite obviously has property Pu,
and also quite obviously S - T is contained in a set having property
P2, a convex set, while T again has property Pu.

‘While almost all examples given are polygonal, no proof given
depends upon this property. It is Simply easier to construct such
examples, and, having constructed them, to determine whether or not
they do indeed have property Pn fer some predetermined value of n.

Since a set having property Pn is the union of a finite number
of starlike sets, it is not surprising that the following concept and

result are of interest when considering such sets.

Definition 2.h Let S be contained in a linear topological space L.
The kernel K of S is the set of all points of S with respect to which

S is starlike. That is, K = {z e S I zx (:8 for all x e S}.

N’—

Wr—

 

-7-

Brunn showed that K is a closed convex set, provided that S is
a closed subset of E2. The following result generalizes the Brunn

theorem [’4] and provides a useful characterization of the kernel of 8.

Proposition 2.3 Let S be a set in a linear topological Space L. Then

the kernel K of S is the intersection of all maximal convex subsets of

S.

£1293. First of all, every point x of S is contained in a maximal
convex subset, Mx of S. Let x be in S. Then {x} is a convex set. Par-
tially order by inclusion the collection C: of all convex subsets of 8
containing x. Using the maximal principal extract a maximal simply -

ordered subcollection {C2,}, and let Mx - L3,! {02.}.

Now, let v be in K. Then v is contained in every maximal convex
set in S for otherwise vile, the Join of v with Mx would be a convex set
containing Mx properly. Next, assume that V is in the intersection of
all maximal convex sets of S. Then, since every point x of X is in some

naximal convex set, x? is in S. It follows that V s x.

Corollggy. The kernel K of S c: L is a convex set which is closed if S

15 Closed.

Proof. The set X is convex Since the intersection of any number
of convex sets is known to be convex. If S is closed, Mx is closed for
x in S since the closure of a convex set is a convex set. Finally, the

intersection of an arbitrary number of closed sets is a closed set.

Remark: Helly [8] proved the following interesting result: If F is a

family of compact convex sets in an N-dimensional Minkowski Space LN,

.. 8 ..
then a necessary and sufficient condition that all members of F have a
point in comnon is that every N + 1 members of F have a point in common.

Using the third result and the fact that K, as a closed subset of
a compact set, is compact, the theorem could be stated: If F is a family
of compact starlike sets in an N-dimensional Minkowski space LN, then a
necessary and sufficient condition that all members of F be starlike from
a point common to all of their kernels, is that the intersection of each
N + 1 members of the family contain a point common to the kernels of the
N + 1 members.

The Helly number'of a family F of sets is defined to be the small-
est cardinal k such that whenever G is. a finite subfamily of F and n G #

E for all 6: G with card G < k + l, thenn G # fl. Helly's theorem
asserts that the Helly number of the family F of compact convex sets in
Em is m + 1. An intriguing but extremely difficult question is: Does the
family of compact (connected) sets in Em having property Pn have a finite
Helly number?

It is of possible interest to mention, in passing, that the analogue
'of the separation theorem for convex sets and the Krein-r-iillman theorem for
convex sets are obtainable for starlike sets using the concepts of a homeo-
morphism [2] and relative extreme points [12], reSpectively.

Although there are a number of elementary results which one may
prove for starlike sets which are the natural analogues of those usually
encountered for convex sets, our interest here is in sets having property
P”, and so only a‘ re. elementary results which do not hold for sets having
property P“ will be included.

-9-

'me following results are an indication of the fact that property
Pn is preserved under many of the usual operations which are in some

{sense "linear" operations .

Proposition 2.18 If S C Em has property P" and L is a linear transforma-'
tion of S, then L(S) has property P“.

Proof. Let yl,...,yn be n distinct points of MS). Then there
exist distinct points xl""’xn in S such that y1 = L(x1), i = l,2,...,n.

Since S has property Pn, x1 C S for some i and J. But L(dx1 + (1 - a)x )=

"J J
”1 + (l - “”3 for o i“ _<_ 1. Hence yiyJ C L(S) for some i and J, as was

Owned.

Proposition 2.5 let S C Em have property Pn and A be any real number.
Then A S =- {is I s e S} has property Pn.

Proof. A S = {is | s c S} defines a linear transformation.

Proposition 2.6 let SC Em be a set having property Pr1 which is con-
tained in a linear variety T of dimension m - 1. Let v be some point of
if" - T. Then vS, the cone over S with vertex v, has property 1’“.

Proof. Let x1, x be points of vS - v such that xlxzcz VS (which

2
clearly do not lie along the same generator of the cone), and let u be

the proJection map which carries x1 and x2

of the cone through x1 and x3, respectively. Then the segment m(xl)m(x2) C S,

into S along the generators

Since «(x1), «(x2) and v determine a plane containing x1 and x2: by the

:th2 would then be in vS. Thus, if x1,x2,'...,xn were
distinct points of vS which were not Joined in v3, then m(xl), «(x2),...,W(Xn)

definition of vS,

would be distinct points of S which would violate property Pn.

Cor-0113117. let S CEm be a closed connected starlike set having prop-
erty P". Let S be a suspension of S constructed by choosing the suSpen-
sion points v1 and v2 to lie on a line orthogonal to Em in EMI such that

vlv2 intersects the kernel of S. Then S has property P“.

M. Let K be the kernel of 3'. Clearly, K is nonempty since
KC K. Let v e v1v2. Then v c K. This follows from the fact that the
suspension of a convex set M in Em, having as suspension points two
points such as v1 and v2 which lie on a line orthogonal to Em in 153m"1
and intersecting M, is quite evidently a convex set. Since K is known
to be the intersection af all maximal convex sets in S, and S is contain-
ed in the union of the suspensions of all maximal convex sets in S, it
follows easily that vlvzis contained in every maximal convex set in 3'
and hence in K.

Nov let x1, 1:2,...,xn be n distinct points of S' which are not
Joined in S. Since by Proposition 2.6, v18 and v28 have property 1’“,
not all ofthe points x1,x2,...,xn can lie in one of these two sets.
letxleVISandxzevzs. Thenxl-dx+ (l-a)vl,0:_a_f_l, for
sorrechandxz-By+ (l- Blv2,0_<.8_f_1, forsomeycS. Since
VIVZCK, if xyC S, then the Join of vlv‘.2
or a 2—simplex, (in either case a convex set), which contains xlx2 and

and xy would be a 3-simplex

lies entirely in S. This implies that x

1x2: 3', contrary to assumption.

Thus if xlx2 <2: S, xy C 8. By the proof of Proposition 2.6, the same'

conclusion may be drawn if x1 and x2 are both in v18 or in v28. Thus

if xl,x2,...,xn are not Joined in 3', there exist n distinct points which

are not Joined in S. This contradiction proves the result.

III‘IO{..|‘ 'Lur

 

 

-11-

Remarks: From Proposition 2.6 it is clear that one may obtain a
closed, connected set having property Pn which is starlike by simply
constructing the cone over a closed, connected set having property P”.

It also follows from.Proposition 2.6 that the suSpension of the
set S between two points v1 and v2 must have property P2n.1. It may,
of course, have property Pm for n _<__ m _<_ 2n-l, as illustrated by the
.Corollary to Proposition 2.6

Tb illustrate how the situation changes, (and consequently the
methods of’proof), as the dimension of the set S increases, and to
provide an example for the preceeding reSult, we consider the conven-

tional five-pointed star in E2.

This set obviously has property P3, may be written as the union
of three (and no fewer) convex sets, and has exactly five isolated
points of local nonconvexity. The cone over S is a three dimensional

set having property P3, but has no isolated points of local non-convexity.

 

 

.. 12 ..

The following result of Valentine [1?] is a clear indication
of how the situation changes: Let S be a closed set in a linear
topological space L where the dimension of L is greater than two.
Assume that S has property P3, and that S is not contained in any
two-dimensional variety of L. Then if S has one isolated point of

local non-convexity, S has at most two points of‘ local non-convexity.

Proposition 2.7 Let S C B“1 have property Pn and C be a convex set in

Em. ThenC+S={x+yIXcC,yeS}haSproper-tyPn.

Proof. Let c1 + x1, 1 = l.,,,.n, be n distinct elements of‘
C + S. Since xich 8 for some 1,3, and c1 cjc C for all i, ,1, we

haveO;o_<_l,a(ci+x1)+(l-a)(cJ+ J)=°‘°+(1'°‘)°J+

axi + (l - a)XJ is in C + S. Hence, C + S has property Pn.

Corollary: If under the hypotheses of Proposition 2.7, C = {x}, then
x + S has property P“.

Pr0position 2.8 rl‘he Cartesian product of two starlike sets is starlike.

Proof. Let A c ER and Bc: Em be starlike sets, and let a0 and

b o be elements of the kernels of A and B, respectively. In Ekhn con-
sider the vector expression
0(ao ,bo ) + (1 - a)(a,b) = (cao, b0) + ((1 - a)a,(l - c)b)
(cao + (1 - a)a., cbo + (l - 00b)

which is an element of AXB for all 0 _<_ a. _<_ 1.

since
aao+(l-a.)aeA,0_<_of_l

and
abo+(l-a)ch,O_<_o_<_l.

-13-

Corollary (to the Proof) The product of the kernels of A and B is

the kernel of the product A X B.

Proposition 2.9 The Cartesian product of a convex set, C, with a set

S having property P“, has property Pn.

Proof. let (c1,xi), i = l,2,...,n, be n distinct points of
C X S. Then cicJC; C for all i and J and xixJ CS for some i and ,1.

Thus no + (1 - a)c is an element of C and (xi + (1 - <::)xJ is an

i J
element of S for some i and J and O _<_ a _<_ 1. But this implies that
(cue1 + (l - “)CJ’ aux1 + (1 - a)xJ) is an element of C X S for some

iandJ, and foralloiail as was tobe shown.

Proposition 2.10 Let S cEm have property PH and e > 0. Then U(S,e),

the parallel body of S, also has property P”.

M. let x1,x2,...,xn be distinct elements of U(S,e). There
exist elements yl,...,yn of S such that llx1 - yill < e for i = l,...,n,
by the definition of U(S,e). Assume first that all of the yi are distinct.
Then there exist 1 and ,1 such that yiy‘1 c S by property Pn. Fixing i and
J,.consider xixJ. Let x 8 0x1 + (l - o:)xJ for somea between 0 and 1. Then
for the same a, y = ayi + (l - coyJ belongs to 8. Moreover,
le- y||=||ax1 + (l - °)xj - ayi + (l — (0%”

I I|c(x1 - y1)+ (l - m)(xJ - VJ)“

<ce+(l-c)c=c
which izrplies that x is an element of U(S,:) which in turn implies U(S,c)

has property Pn.

-11}-

Suppose now that the y1,...,yn are not all distinct. let y1 =
y 3’ i 7‘ .1. If it is possible to select an element 373 in S such that
y:l 9‘ 37d and «13.33) < e, and this is indeed possible for each pair
which are equal, then the above argument applies. If not, then
U(y1,e) n S = y1. ' Since xi’xj are in U(yi,e), which is convex,

xichU(y1,e)c U(S,e) which again implies U(S,e) has property P".

Proposition 2.11 If S is a set in Em, and X c S has property Pn rela-

tive to S, then cl(X) has property Pn relative to cl(S).

Proof. Let xl,x2,...,xn be distinct elements in cl(X), then for
any a > 0 there exist ul,...,un, such that lluill < c and x1 + ui,
i a l,2,...,n are distinct elements of X. Since X has property Pn
relative to S, the line segment (x1 + u1)(xJ + ud), say, is in S. Thus
fbr any a such that O :_a :.l, a(x1 + ui) + (l - a)(xJ + “3) is in S.
Now

||[c(x1 + u1) + (l - a)(xJ + ud)] - [0 xi + (1 - °)XJ]I'

==Huu1+(l--<x)uJ II<ae+(l—a)e=e

Since a is arbitrary, or xi+ (l - a) xJ is in cl(S).

Proposition 2.12 let n _>_ 2 be a positive integer. The limit 3 of a
sequence {Sk }of compact sets having property Pn is a compact set havino;

property Pn.

Proof. It is well known that S is compact [10]. Hence, let

x1,x2,...,xn, be n distinct points of S, and let p(S,Sk) a ck in the
Hausdorff metric. Since Skc: U(S,ck) and ck + O, we may find a sequence

- 15 -

k 1

{xi} (1 = l,2,...,n) such that xi is in S andlimxk = X1 (i = 1,2a---2n)-

Fbr each value of k, at least two points of xi,xi,...,x: are Joined by

a line segment which lies entirely in Sk.- Since the number of possible
pairs which may be Joined in SR fer each k is finite, we may choose a

subsequence {SJ} of {Sk} for which the sequences {xi}, {xi},...,{xg}
1 ,

,x2,...,xn, and such that fer some pair' of points x1

and x2, say, x§x§c33J fbr all values of J. Since S is closed and

converge to x

a:I + 0 we have x1 x2C S; that is, S has property PD.
Hbrn.and Valentine [10] have generalized the notion of a convex

set in the fbllowing manner: A set X.in,E2 is called an Rn set if for

every pair of points x and y in X, there is a polygpnal path, consisting,
of at most n segments, lying entirely in X, which Joins x to y.

Perhaps the most striking result obtained fer this class of sets
is the following result which was proved by Bruchner [3]. Theorem: A

necessary and sufficient condition that the set X in E be compact and

2
connected is that X be the limit of a sequence of compact Ln sets fer

some natural number n. This result has been generalized by J. w. MCCoy
to a set X contained in a complete, convex, locally compact metric space
[13]. Kay [1.] has shown that a closed, connected Pn set S in a Minkows‘

space is an Ph- set.

1
Thus, the class of all closed, connected sets having property Pn

is a subclass of the class of Ln— sets. The following example shows

1
that they form a proper subclass. For n.3_2, one needs only to take the

cone at (ébl) over the points (0,0), (0,%),...,(0,E§l9, (0,1) to obtain an

L2 set with property Pn+2.

-15-

As a.generalization of Definition 1.2 of property P“, the

following is given.

Definition 2.5 A set S in Em is said to have property P: if for each

 

n distinct points of S, at least r + l of the points, 1 :_r :_n - 1,

are Joined by line segments which lie entirely in S.

Proposition 2.13 Let S be a closed connected set in Em (m.; 1). Then

 

S has property P: if and only if S has property P3'1 5 Pn'l.

It is immediate that S has propertyPg whenever S has property
'2‘, and assume that {xk}, k = l,2,...,n-l

is a collection of n - 1 distinct points no two of which are Joined in S.

Pn-l. Hence, let 3 have property P

Let x c S, x # xk, k = l,2,...,n-l. Then by property P3, x x1 and x x2,

say, are in S. let {zi} be a sequence of points in (x xi) converging to

x Then since xl xk d: S, k = 2,...,n—1, there must exist a neighborhood

1.
of N of xi such that'for all 21

21 and xk, k . l,2,...,n-l, violate property ngl, a contradictidn.

in N, 21 xk¢ S, k g 2,3,009,n-lo Then

Corollary: Under the conditions of the Theorem, S is convex if and only

if S has property Pg.

CHAPTER III

PLANAR SETS HAVING PROPERTY P“

The results of this chapter were obtained in an attempt to
determine the number of convex sets which are required if a closed,
connected set S in 32 having property Pn is to be expressed as the
union of closed, convex sets. Two results of valentine are extend-
ed and several new results obtained..

Uhlike the results of Chapter II, many of the proofs of this
Chapter depend upon the prOperties of the containing space, E2.

The following notation and terminology will be standard through-
out the remaining chapters.

The letter S denotes a closed connected set in E2 unless other-
wise stated. K is the convex kernel of the set S. The letter Q al-
ways denotes the set of‘points of local non—convexity (lnc) of S. (Q
is evidently closed if S'is closed).

The closed line segment Joining x to y is denoted by xy; the
corresponding open line segment is denoted by (xy). The line deter—
mined by the points x and y is denoted by L(x,y). By R(x,y) is meant
the ray emanating from x and passing through y. By W(x,y) we shall
mean the open half-plane determined by L(x,y) and lying to the left
of the line L(x,y) if L(x,y) is considered as directed from x to y.
While the meaning of the notation W(x,y) as given above is not
standard, the economy of words which it allows in that which follows

Justifies its usage.

- 13'-

- 18 -
The interior, closure, boundary, and convex hull of a set A

are denoted by intA, clA, bdA, and convA, respectively.

Defipition 3.1 Let S be a connected space. A point q of S is called
a cut point of S provided that S - {q} = A U B, where A and B are dis-

Joint, nonempty, open subsets of S.

Theorem 3.1 If’q is a cut point of the P“ set S, and all the components

of S - {q} are convex, then S is the union of n - 1 or fewer convex sets.

22992} If all the components of S - {q} are one dimensional the
proof is immediate so assume that at least one of the components, 8*, is
two dimensional. we proceed by induction on the number of components.

If there is but one additional component the conclusion is clear.
If not, q is a cut point of (S - 8*) L]{q} and we claim (S - 8*) L){q}
is a Pn.; set. For if (S - 3*) Lij} contains points xl,x2,...,xn__1
no two of which are Joined in that set, then since S” is not a subset of
the union of the lines L(q,xi), there is a point x in 8* not Joined in S
to any x1. Thus 3 is not P“. Hence, by induction, (5 - 5*) u {q} is the

union of[n - 2 or fewer convex sets and S itself the union of n - l or

fewer.

Theorem 3.2 If a closed, connected set S in Em has exactly one point,

q, of local noneconvexity, then S is starlike from q.

-19..

31293. By Proposition 2.3, it (suffices to show that q is con-
tained in every maximal convex subset of 8. let M be such a set and
suppose q t M. Since M is closed, there exists a hyperplane, L, such
that L n M = H with q in one open half-space and M in the other. Let
w be the Closed half-space containing M.

If for each y e M there exists a Sphere, a(y,p) c: M then M is
both open and closed relative to S and S is not connected. Thus for
some 2, z e M, each Sphere a(z,p) intersects S - M.

Since 8 is locally convex there exists at z, a sphere a(z,p1)
with o(z,pl) n S convex and furthermore for some 02 < p1,a(z,02)c w.
Thus, a(z,02) n S cw n S, is non empty, convex and is not a subset of
M. It follows that M is a proper subset of the component, K, of w n S
which contains M.

' Suppose, now, that x is any point of K. There exists a(x,03)
such that a(x,p3) n S nw is convex, and since x e K, this set is a sub-
‘set of K. That is to say, K islocally convex at each point and, being
closed and connected is by Tietze's theorem, convex. This contradicts

the maximality of M and shows that q must be an element of M.

Definition 3.2 If the rays R(q,p) and R(q,r) are not on a line they
bound a convex and a non convex sector of the plane. The closed convex
sector will be denoted by T(pqr) and the closed non convex sector by
T‘*(pQr). I

A sector of a circle which is non convex is a 59:195. circular

sector. The center of the circle is also called the center of the sector.

-20’.

lemma 3.1 If q is the only point of local non-convexity of the closed,
connected set S in E2, and q is not a cut point of S, then corresponding
to points p and r of S such that pr ¢ S there exists a circular disk
D(pqr) such that D(pqr) n T*(pqr) C: S.

w. Since S is starlike from q, the points q, p and r are not
collinear. Since S - {q} is connected, locally compact, and locally
connected, S - {q} is arcwise connected [20], and there exists an are C
in S - {q} containing p and r. let the distance from q to the compact
set C be a, and consider the circular disk D(pqr) with center q and
radius u/2.

If every ray R(Q,x) in T(pqr) intersects C, then q is a point
of local convexity of T(pqr) n S and the component of T(pqr) n S con-
taining qr U qp is convex. Then pr c S, a contradiction. Thus there
exists a ray R(q,x) in T(pqr) which fails to intersect C. Now if any
ray from q in T*(pqr) fails to intersect C, then C would lie in two
separated subsets of the plane. Hence every such ray intersects C and
D(pqr) n T*(pqr) C S as required.

lemma 3.2 If S and q are as in lemma 3.l, then S has property P3.

Proof. Suppose x,y,z are points of S no two of which are Joined
in S. If q ; conv{x,y,z}, then the smallest of the three disks D(qu),
D(yqz), and D(xqz), guaranteed by lemma 3.1, is a subset of S, and S is

locally convex at q, a contradiction.

-21..

Suppose then that q t conv{x,y,z}. One of the rays R(q,x),
R(q,y), and R(q,z) is in the interior of the convex sector defined
by the other two. Assume that R(q,y) c int(T(xqz)). Then D(qu) C
T(yqz) n S, and q is a point of local convexity of 8 n T(yqz). The
componentof this set containing y, q and z is thus a convex set and

yz is in S. This contradiction establishes the theorem.

Theorem 3.3 If q is the only point of local non-convexity of the
closed, connected Pn set S in E2, then S is the union of n - l or

fewer convex sets .

Proof. If q is not a cut point of S then S is P3 and it follows
from [17] that it is the union of two convex sets.

Suppose then that q is a cut point of S. If all the components
of S — {q} are convex, the conclusion follows from Theorem 3.1. We
consider now the remaining possibility that one of the components, 8*,
is non-convex. Now q is clearly the only point of non-convexity of
8* U {q} , and q is not a cut point of this set. The set 8* u {q} then
satisfies the hypotheses of lemma 3.1 and q is the center of a maJor
circular sector, D, lying wholly in s" U {q}. This means that the re-
mainirg components of S - {q} lie in the convex sector of the plane de-
fined by the rays which intersect D only in q. If M is such a component,
it is clearly convex, since its only possible point of non-convexity is q
and q is hardly the center of a maJorxcircular sector lying wholly in
M U{q}. So M U‘{q} is convex.

-22..

Since (8 - 8*) U {q} is now clearly Fri-2 and satisfies the
hypotheses of Theorem 3.1, it. is the union of n - 3 or fewer convex
sets while 8* is P3 and is the union of two convex sets. Thus 8 is
the union of n - 1 or fewer convex sets.

In the proof of Theorem 2 of [17] it may be observed that
the use of property P3 is unnecessary in the case where each point
of local non-convexity of the set 8 is in the convex kernel of 8 if
one introduces lemma 3.1} below which itself is independent of prop-
erty P3 and requires only that S be closed and connected. That is,
the following theorem, a generalization of the theorem cited above,

can be proved.

'meorem 3.14 let 8 have at least two points of local non-convexity.

If every point of local non-convexity is in the kernel, K, of S, then

S may be written as the union of three closed convex sets. The num-

ber three is best.

The proof of this theorem is a modification of that given by
Valentine [17] which avoids the use of property P3. Four definitions

and five lemmas are needed.

Definition 3.3 A cross-cut of a set Y contained in E32 is a closed

segment xy such that (xy) c: intY and such that x and y are in MY.

lemma 3.3 Each open segment (uv) of the convex kernel, K, of 8

contains no points of local non-convexity.

\

-23..

P__1'_o_gf_'_. let w be an element of Q n (uv). Clearly 8 C L(u,v).
let 2 be in S - L(u,v). Since uv C K, Auzvc 8. Hence, each suffi-
ciently small neighborhood of w contains no crosscuts of E2 - 8, since
such a crosscut xy would have to have its interior (xy) in one of the

open half-planes bounded by L(u,v).

Definition 3.1! A component of the complement of a closed connected

set 8 is called a residual domain of S.

lemma 3.1} let D be a bounded residual domain of S. Then, the bdD

contains at least three points of local non-convexity of 8.

M. Consider the set E12 - D which contains 8. E2 - D is
closed since D is by definition an open subset of an open set in E2.
Moreover, bdD = ch n (E2 - D) is closed and bounded, and hence compact.
let p be a fixed element of bdD and x be an arbitrary element of bdD.
As x varies over bdD the distance from p to x defines a continuous
mnction d from the compact set prdD into the reals. Hence, d must
attain its maximlm at some point ql of bdD. Consider the sphere
o(p,r) having center at p and radius r = d(p,ql). Evidently DC C,
where C is the open disk bounded by a(p,r). Thus, each point of o(p,r)
is contained in E2 — D. If ql is a point of local convexity of £2 - D,
then there exists an open spherical neighborhood N of q1 such that
N 0 (E2 - D) is convex. In particular, if x and y in a(p,r) n N are
such that pql n X)! # D, we have xy C E2 - D which contradicts the‘
assumption that ql is in the bdD. It follows that ql is a point of

local non-convexity for E2 - D. Moreover, ql is a point of local

-211-

non-convexity of 8. Let x and.y be elements of’E2 - D such that
xyfiEz-D. Thenxyanfl. SinceDisanOpen setxyanay

be eXpressed as a countable union of disjoint open intervals. Let
(uv) be one such interval. Then u, v c bdD C 8, and WC 8. More
specifically, if in every spherical neighborhood N' of ql there exist
points x and y of 82 - D such that xy C E2 - D, then there exist
points u and v in 8n xy 0 N' such that WC 8. That is, if ql is

a point of local non-convexity of’E2 - D, then ql is a point of local
non-convexity of S. In the same way we may next locate a second
point of local non-convexity q2 in bdD at a.maximal distance from

ql. (Which will not be p, in general). The third point q3 is ob-
tained in like manner by maximizing the sum.of the distances d(ql,q3)
and d(q2,q3) to obtain an ellipse with foci at q1 and q2 passing
through q3. Because the ellipse, like the circle, is a convex curve,

the very same argument gives the desired reSult.

Remark: The boundary of the triangle indicates that the number 3

is best.

lemma 3.5 Under the hypothesis of Theorem 3.1}, 8 has at least one

isolated point of local non-convexity.

Proof. let xy be a crosscut of a residual domain of 8. The
set D - (xy) is the union of two disJoint open sets, denoted by D1

and D2 [11!]. Since 8 is starlike, D , say, is bounded while D is not.

2
contains a point q

1

Then bdD is a continuum [114]. By lemma 3.1}, bdD

1 l

-p 25-

different from x or y which is a point of local non-convexity of 8.
Since S is starlike from q, xq and yq are in 8. This implies that
ch quy since D was a residual domain and q is in ble.
Consider the lines L(x,q) and L(y,q) or more specifically, the
v - shaped domains V1, 1 a l,2,3,l}, that they determine. Order the
V1 in a clockwise direction about q so that Vl DDl. Suppose ql is
an element of (clVl-q) n -Q. Then since (qlx U qu) C 8 we have
D1 (3 Axyql. But this contradicts the fact that q is in ble. Suppose
next that there exists an element q1 in V2 0 Q. Then quql and quq1
are contained in 8 and once more we would contradict the fact that q

is an element of ble. Similarly for V“.

Now if ql is in V3 n Q. (quql U quq1)c 8 which implies q is
isolated since V1 contains no points of Q. Finally, since no open
segment of K contains a point of Q, there does not exist a sequence of

points of Q along L(x,q) n clV or L(y,q) n clV with q. as a limit

3 3
point. Thus, q is an isolated point of local non-convexity of 8.

Corolla. let Q' be the set of isolated points of local non-convexity.
Then Q . cl(Q' ).

M. let q s Q. If q c Q', then q s cl(Q'). If q L’ Q', let M
be an open Spherical neighborhood of q. Since q C Q, there exist x,
y c N n 8 such that m at 8 and xy defines a crosscut x'y' of a residual
domain D of 8. As in the proof of lemma 3.5, the boundary of the bounded

component D of D—(x'y') contains an element q' of Q'. Since q a K,

1

.. 26 ..
chl C Ax'qy' C N. Thus every neighborhood N of q contains an element
q' of Q'. That is, Q C cl(Q'). Since Q is closed, cl(Q') C Q, and we

have Q = cl(Q').

lemma 3.6 The boundary of coan is connected, and contains at most one

ray.

25931;. Since H coan is convex, if de were not connected, it
is well known that it would consist of two parallel lines. Then
lemma 3.3 would imply that each of these parallel lines contains at
most two points of local non-convexity. But then Q would be bounded
and de would be connected. If de contains two rays, then lemma 3.3

would again imply that Q is bounded, a contradiction.

Definition 3.5 An edge of bd(coan) is a closed segment xy or a closed

ray x a whose endpoints are elements of Q.

lemma 3.7 let x and y be successive points of local non-convexity in

bd(conVQ). and W be an open half-plane of support to coan which abuts
on the edge xy (or x n). Then (coan) U (w n S) is a convex subset of
S.

Proof. If u is in coan and v is an element of 8 n w, then
uVC 8, since coan C K, K being convex. Now, to show that

uvc(coanU (8 nW)) we Show that xynuvylfl (or uvnx “7425).

-22..

Recall that x and y are in ms. Suppose uv n xy = {2}. Then
xv and yv are in S which implies that x is in int(Am/v) n S ‘or y
is in int(Auxv) n S which implies x or y is not in Q, a contradiction.

\

Suppose next that u and v are elements of S H w. let 2 be an
element of (xy) (.or (x 6)). Suppose uv ¢ S n w. Since uz, va S,
Auvz would contain a point of Q, by lemma 3.1!. This is a contradiction
since W n Q = fl and by Lemna 3.3 (xy) (or x ~) contain no points of Q.
Hence (coan) U (W n S) is convex. (If coan = xy, then (coan U (w n 8))

may or may not be closed).

lemma 3.8 Let {xiyi} be a countable number of pairwise disjoint edges
in bd(coan). Assume that bd(coan) contains at least three edges,
and let w be the open half-plane of support to coan whose boundary
\ ' a ' " = U "
contains (xiyi), (xiyi may be x1 ). Then the set X - (coan) U (8 fi( 1 '11))

is a closed convex set.

Proof. Choose an order on the boundary bd(coan), and assume

that in this ordering Xi is the beginning of the edge x and that

1Y1
yi is the end point of xiyi.

let xiyi and nyJ be two disjoint edges, and consider the convex

region V bounded by the lines L(x1,yJ) and L(xJ ,yi) and containing the

quadrilateral xiy 1ny3. let V1 and V J be the portions of V adjacent

to xiy:l and nyJ , reapectively. (These two sets may not be bounded).

If
"3

Now, S n w c: V since otherwise x or yJ are not in Q, a contradiction.

J J J

5"} " Xffiéw, then L(x1,~.)_ is a line parallel to the ray xjw.

-28...

let uandvbe elements of S. Ifuand vare in (conVQ) U(S nwi)
thenbylemna 3.7, uVCX. IfuisinSnwiandvisinSan,
iiJ,thenuisinV1andvisinVJ.
V1“ xiyi nyJB xiyi, and we have uvn xiy1 ’4 H which implies uVCX.

Since V is convex,

Finally, X is closed. The finite case is immediate since
“”1 n Sc Vi implies cl(wi n S) C (W 1n S) U bd(coan). If there
are an infinite nwnber of disjoint edges, let r be a°limit point of

the sequence of sets W n S. Since w n S cv in , by fixing (x y )

1,1,, 3.1

of the preceeding paragraph it follows that (x1 yi ) + q, a fixed point
n n

of bd(COnVQ), as in " ”0 Since then V1 '> q as in + O we have I! = q’

n
an element of coan. Hence X is closed since coan is closed.

Proof of Theorem__3_:_h_ First assume that Q I {q1 U qz}. The line L(ql,q2)
divides the plane into the two open half-planes Ni (1 - 1,2). By lerrma
3.7, W1 0 S is convex (i - 1,2). Hence, S = cl(wln S) U cl(wzn S) U
L(q1,q2) n S is the desired decomposition.

Next assume that Q = {q1 U q2U "‘UQZm} where m > 1. Order the

edges of bd(coan) in a counterclockwise manner so that q1 .=. q2m+1.
let W,L denote the open half-p] ane of support to coan adjacent to
qiqi+1‘ By Lemma 3.8 each of the sets

m
31"(conv5a2)USf)(iUl M211)

= (conVQ) U S n (1:31 W21)

2m
. isa closed convex set. Since S C (coan) U S n (121 W1) we have S =

31 u 52.

-29..

Next. if Q =liql U <12 U ...Uq2m1} m _>_ 1, we add 53 =

(conVQ) U (s n wzmfl) to the sets 51 and 3 Finally, 11‘ Q is

2.
infinite, we need the following definition:
Definition 3.6 A closed, connected subset I of bd(coan) is called
a polygonal element if the following conditions hold:
1. It is the closure of the union of edges of bd(coan).
(An edge of bd(coan) is a closed segnent xy or a closed
ray x-o whose endpoints are contained in le.)
2. Its endpoints are limit points of elements in Q.
3. If I - bd(coan), then I contains at most one limit point
of elements in Q. If I ’5 bd(coan), then only its endpoints
are limit points of elements in Q.

Note that a polygonal element is maximal in the sense that it
is not a proper subset of a larger polygonal element. The number of
polygonal elements of bd(coan) is countable. This. follows from the
fact that they are convex subarcs of the boundary of a convex set,
cva, which do not overlap. By definition, each I contains at least
one segment. Hence, relative to bd(coan) , each polygonal element
has a non-empty relative interior, and the non-overlapping of the poly-
gonal elements implies countability. If coan is bounded, it is clear
that there can be at most finitely many polygonal elements of length
at least l/n times the perimeter of bd(coan). In the unbounded case,
we may simply consider a (countable) monotone increasing sequence {01}
“of closed disks concentric about the origin. Then bd(ai n conVQ) con-

tains at most a countable number of polygonal elements for each value

-30..

of i. I Since {01} is countable, we have bd(coan) contains at most

acountable number of polygonal elements.

let 11’ IZ”"’Ik"” be a well ordering of these elerrents. F r

each polygonal element I divide the edges it contains into two

k!
classes bill, and ME such that no two edges of Mi (1 =. 1,2) are adjacent,
that is, have an endpoint in common. It may happen that one of the m:
may be empty. For each edge e e m: we let w: denote the open half-

plane of support to bd(coan) whose boundary contains e. Define

1,2) and let

F113 gm: (Win 8) (1

II

S = confiU<EFb (i 1.2)

1
Since each edge in Mi is separated from each edge in M2; (14 54 m), LemT‘a

3.8 implies that S and S are closed convex subsets of S. Moreover,

I 2
since for each point x c S, either x e coan, or x is in some w: n S,

we have S = Sl U .32 and S1 n S is non-empty.

2
Corollary. If Sins property P3 and two or more points of local non-
convexity, then S ray be expressed as the union of three or fewer

closed convex sets.

Proof. If q is a point of local non-convexity of S, then S is
starlike from q, [17]. The result now follows, since the hypothesis

of the theorem is satisfied.

If the set S bounds a bounded residual domain of itself, the

following two results may be obtained.

- 31..

Lemma 3.9 Let 8 have property P“. Then S can bound at most one
bounded residual domain of S.

Proof. Let Kl be such a domain. By Lemma 3.“, bdK1 contains

three or more points of local nonpconvexity of S. Let q1,q2,q3 be
three such.points. Then, by property P” and the closure of S, these
points are Joined by line segments lying entirely in 8. Thus, by the
definition of K1, K1 C Aq1q2q3. Moreover, there can not exist a
fourth point Q, of local non-convexity in Aq1q2q3 since then quqfc:s,
i - 1,2,3, which would either contradict the definition of K1 or the
fact that qi is an element of bdKl. For the same reasons if x1,yl,
and 21 are elements of (qlqa), (q2q3) and (qlq3), respectively, then
the line Joining any two of these points does not lie entirely in S.
Hence, by preperty Pu, every point of S is Joined to at least one of
the points xl,yl and 21 by a line segment lying entirely in S. If a
second residual domain, K of S exists, then there exist points of

2

bdK2 corresponding to ql,q2 and Q3 of bdK at least one of which is

1
distinct from ql,q2, or q3 and points x2,y2, and 22 corresponding to

x1,y1 and zl, reSpectively of Aqlqzqa such that the three line segments

and z' to x are not in S or the three line

1”1* 1 2’y2 °r z2
segments Joing x2,y2 and 22 to x1,yl, or 21 are not in S. In either

Joining x

case prOperty P1‘ is violated. Thus S can bound at most one bounded
residual domain of 8.

Theorem 3.5 Let S have property P“. If S bounds a bounded residual
domain of S, then S may be expressed as the union of five or fewer

closed convex sets. The number five is best.

-32..

M. let B be the bounded residual domain which is bounded
by S. Then B is entirely contained in some triangle whose vertices
ql,q2 and q3 are points of local non-convexity of S, as was shown in
the proof of lemma 3.14. We denote the six unbounded domains into

which the plane is divided by extensions of the sides of Aqlqzq as

3
shown in Figure 3.2. By V0 we shall mean the set cl(Aqlq2q3) - B.
The component of int(S 0V0) whose closure contains (1qu will be de-

noted by JiJ' Denote (S nVfl)U JiJ U ((11613) by DiJ .'

Figure 3.;

 

 

v2 v23 \V3

Observe first of all that S n D and S n D are each convex

12
, say, such that

13' 5 ””23

sets since if x and y are elements of S n D23

xy (2: S n D then, a and b (as shown in Figure 3.2) together with x

23’
and y are four points which violate property P“. The same argment

applies to S n DJ. and S n D12. Observe next that the points c, a,

3

and b are not Joined to any point in S1 = (S n W(q2,q3)) - J23, 82 =-
(3 n‘-‘l(q3.ql)) ' J13: and S3 = .(S n w(qlnq2)) "' J12, r'eSPeCtiveJ-Y!

which implies that these sets haVe property P3. Hence, by Proposition 2.11

-33-

and the Corollary to Theorem 3.14, the closure of each of these sets
may be written as the union of three or fewer closed convex sets.

let Q1 denote the set of points of local non-convexity of S Since

10
.81 has property P3, Q1 is contained in the kernel of 51' Clearly,

Q1 is contained in cl(V1 n S), and is non-empty, since q1 is contained

in Q1. We denote the cardinality of Q1 by N Since 8 = 81 U 82 U S

1.
the remainder of the proof consists of showing that under all possible

39

circumstances these three starlike sets may always be written as the

union of five or fewer closed convex sets.

Order the points of Q1 and Q 2 in a counterclockwise direction
“1

5 Q2 1, reSpectively. let Q1 8 U
0

J=l {£11,J }

start with :-: and
ins N2 <11 qlfl <12

. U
and Q2 kai .(q2,k}.

CASE I Assume first that N1 = N2 = 1. We consider a closed disk G1
with center at qi, i 8 1,2, and with radius sufficiently small so that

Gin SC S (In the proof of Theorem 3[l6] it is shown that the number

1.
of components of S fibd(conv(Cin S)) is 1,2, or 14. If the number of

components is 1!, then it is shown that cls is the union of two distinct

i

rays or line segments intersecting at C11. Here this implies that S1 is

contained in L(ql,q2) U L(qi,q3), i = 1,2. For future reference, we de-
1

note the convex subset S1 n L(q l,q2) of Si by X12 and denote the convex

1
subset 51 n L(q1,q3), by X13.

If the number of components of S n bd(conv(Cin S)) is 2, then

S is known to be expressible as the union of two convex sets each of

i
which is determined by one of the two distinct components of

-3“-

s nbd(conv(C1 n 3)). (By the proof of Theorem 3[17J). The inter-

section of these two convex subsets of S is {qi}' let B be the

1 1

component of S n bd(conv(Cin S)) which intersects 8!] D12. let X1

12

be the unique convex subset of 8 determined by B . By the above

1 1
discussion of the proof of Theorem 3[17], q:L is a cut point of S1 and

X12 - {C11} is a component of S1 - {qi}. Since Xiz - {qi} intersects

the convex (connected) set S n D12, the latter must be contained in

Xi? - {Q1}, and consequently in X112, 1 =- 1,2. By the same argument
1 - 1

If S n bd(conv(Ci n S)) has exactly one component, i = 1,2, then the

is a convex subset of S containing S n D
support line to the convex set J12 through qi, distinct from L(q1,q2)

if J12 # fl and L(q1,q2) if J = {5, determines two convex subsets X1

12 12
i
and X113 of Si such that qic(ch12 n ClXiB)’ Si c1X12Uc1X13,
i i
X1235 0012 and X13DSfli D 13.
CASE II Assume now that N1 = 2211, i = 1,2. Consider the edges

gl’Jql,J+l and q2,kq2,k+l .1 31,000,231, k ' 1,000,21le Where

qlfinfl 5 q1,1 and q2,2rm2+l E q2'1. let the open half-plane of

support to coanl and coan2 adjacent to ql. Jql, J +1 and q2,kq2’k,1,
respectively, be denoted by w1 and Vi, respectively. By lemma 3.8,

J:
J=lw21

are convex sets, having S n D12 in common. Moreover,

1

X12

and (coanz) U cl(Szn

the sets (coanl) U cl(Sl n (U ))

(kt-91 wax-1” EX12

by the proof of lemma 3.8, X12 is lmown to be one of the two convex sets

whose union is S i = 1,2. The closed convex sets

1'

.. 35-
ml

(convol) u cl(sl n <ng ”23-1” 5 x13

and m2

(conVQ2 ) U cl(S2 n (kr-l W-2k)) 52X3

sxl 2Ux1 andsno cx13 fori=1,2,asin

113 13 13
the proof of lemma 3.8.

are such that S

CASE III If one (or both) of the Ni’

by considering polygonal elements of bd(Coani), as in the proof of

i = 1,2, is infinite, then

'Iheorem 3.11, we may express the set 81 as the union of two distinct

1 contains 8 n 012 and X1

convex subsets X1 12 13

12
113,

and X13, such that X

contains 3 n Di =1,2.

CASE IV We next assume that Ni = zni+1 for mi > 1. Again, by

lemma 3.8, the sets
- 1
(conVQ1)U cl(Sl n (3U 81 WZJ U wl : 1712
(comel) U cl(sl n (39,1 ”n+1” - 12

II
N

are closed convex sets. Likewise,

. m2
(conVQZ) U cl(S2 n (#1 Wad”

m
N

m2
,2 = 2
(convoz) u cl(82 n (kgz 5’ku win .. YR

are closed convex sets. The same lemma 3.8 shows that

1

(conVQ1) U cl(Sl n (W1 U wl )) s X13

and

(coanz) U cl(S2 n ('ng U wgmlfln .=. X

and
2

23

-35..

are closed convex sets containing S 0 D1 13 and S 0 D23, respectively.

1 i
=X13U Y1 2U212, and the sets leandz

Moreover
’ 12

Si each contain

3 n 012,1:- 1,2.

In the four cases considered we have selected convex subsets

1 2 1 2' l 2
1 and $2 denoted by X12, 12, Y12, Y12, 212, and Z12

which contains the convex set S n D12. At the same time we have

selected the convex subsets Xh and X23 of 81 and 82 respectively,

2
in such a way that X13 contains S n D13 while X23 contains S n D23.

Moreover, in each case these subsets have been chosen so that S

1 B
X13 U Y12 U 212 or 81X13 u X1? It is clear that several other

of S X , each of

cases arise as combinations of the four cases considered; for

example, N1 might be an even positive integer while N is an odd

2
positive integer. However, in each of these cases, the decomposition

of S and 3 may be taken to be exactly the same as in the four cases

1
considered.

2

To reduce the number of convex sets from 9 to 5 we first show
that cl(S:l U 82) may be expressed as the union of it or fewer closed
convex sets. This is accomplished by showing that cl(Gi2 U Hi2) is a

closed convex subset of cl(SlU 82) where X,Y, and Z my be substituted

1
1?.

2 a H is a convex subset of 32 also con—

for either G or H. Assume then that G

2
1

taining S n D12. Consider the closed subset cl(G U H) of the closed

5 G is a convex subset of S1

containing S 0 D12 and that H

set S. Since G and H are convex sets, each is connected. Since G n H =

S n D12, a non-empty, convex (connected) set, C U H is a connected set.

-37...

Hence, cl(G U H) is a connected set. By Tietze's Theorem, if

cl(G U H) is locally convex, then cl(G U H) is convex. Assume, to
the contrary, that cl(G U H) has a point of local non-convexity p.
Since ch and clH are convex subsets of c181 and c182, respectively,
p is not a point of local non-convexity of ch or clH. Thus, if a
is a Spherical neighborhood of p of radius r, and x,y c a n cl(G U H)
such that xy ¢ cl(G U H), then x e (ch - ell-I) while y c (clH - clc).
This implies x e cl(S1 n V1) while y c cl(Sz n V2). If we take r to
be less than one-half the distance from ql to q2, then no such x and y
can exist. Since this contradicts the definition of p as a point of
local non—convexity of cl(G U H), it must be true that cl(G U H) is

locally convex, and hence convex.

For the cases discussed, it has thus far been shown that the
6 possible closed convex sets which could arise from cl(S:L U 32) may
always be reduced to l! or fewer closed convex sets. The arguments
given for the four cases above when applied to clS3 show that clS3
may be expressed as the union of 3 or fewer closed convex sets in such
a way that one of these sets X33 contains S 0 D13 while another X33
contains S n 023. Then the same argument given above applies to show
that cl(Xi3U X23) and cl(Xg3 U X33) are closed convex subsets of S.
Recalling the arguments given in the four cases above, it is clear
that this can be done in such a way that there will remain in clS3 at
most one closed convex set. It has now been shown that the seven

possible closed convex sets which remained may be expressed as five

or fewer closed convex sets, the desired result.

CASE V. There is one case which remains to be settled; namely, the

case which arises when N = 3 fOr one or more values of i.

i
If Ni = 3 fbr exactly one value of i, then without loss of

generality we may assume that N = 3. As in Cases II, III, and 1V

3
we consider the edges q3l,q32, q32q33, and q33q31 of coan3. we

again denote the open halfeplane of support to coan3 adjacent to

1,2,3 letting q3u 5 By Lemma 3.8 the

q3.1"3.J+1by ”1’ J q31'
set (coan3) Uc1(S3 ruw3)3 = 1,2,3, is a convex set. Mbreover,
(coan3 ) Ucl(S3 nw3) a X33 contains 023 while (coan3)lJ cl(S3 [1 W3)
=X§3 contains 013.1Finally, S3= X33U X3 3U ((coanB) U cl(S3 n L13))
By the same arguments given for the preceeding three cases
cl(Xi 3U X3 3) and cl(Xé UX33) are convex sets. The only distinction
to be 3made now between this case and previous cases is that the re-

training convex set in cls3 which is the fifth convex set in that de-
composition is the set (coan3)lJ cl(83fl Hg).

we next assume that N1 = 3 fer i = 1,2 and that N is one,

3
even or infinite. In this case, as in Cases.I, II, and III, the set

013 may be expressed. as the union of two closed convex sets cl(X23)

3

and cl(XlB), containing D and 013, reSpectively.

23

1 = 3 for i = 1,2, S1 may be expressed as the union of

the three convex sets, (coani) Llc1(Sil) Hi), 1 = 1,2,3, 1 = 1,2,

Since N

with the property that

~39-

xiz s (conVQl) u cl(Sln wfi) 3012
Xiz e (coanz) u c:L(s2 n wi) 31312
and Xi‘B 5 (coanl) U cl(Sl n Wi) 31313
x33 5 (conVQZ) U cl(s2 n wg) 31323
By the argument given after Case IV, cl(lel.2 U Xiz),
cl(Xi3 U XiB), and cl(Xg3 U X33) are convex sets. Since S is evidently

the union of these three sets together with the convex sets (coan1)U
cl(Si n Wé) i a 1,2, we see that S is once again expressible as the
union of five or fewer closed convex sets. Finally, we consider the

casewhereNi=3,i=l,2,andN =2m+lform_>_l. Itiscon-

3

fvenient to consider the case m s 1 first.

For m = l, we define the following sets
A1 = cluszu 83) n (14(q22.q23) n wcq32.q21) n W<q32.q33>>1
A2 a c1[(SlU 83) n (W(q12,ql3) n W<qn,q33) n W(q32.q33))]
A3 = c:l[(S2 U 83) n (w(q21.q22) n W(q23,q32) n w(q33.q31))]
Au ' Cl[ (81 U S3) 0 (w(ql3’qll) n w<Q319qB2) n w<Q331q12DJ

Finally, let
131 - ((51 n “’35) n W(q12.q33))

132 = (<82n w‘g) n w(q32,q23))

Since each is the closure of the intersection of convex sets, each is

convex. We define A5 to be the set cl(conv(B‘.L U I32 U 812)). Consider

.40...

the set A1. Since 82 and 83 have property P3, each point of local
non-convexity of $2 and S3 is in the kernel of 82 and S3, respectively.
By Lemma 3.7, cl(conVQ3 u (33 n win and cl(coanz up (32“ 14%)) are
convex sets. The argument following Case IV may be applied to these
two sets to show that their union is a convex set. Thus, the inter-
section X of the closed half—plane cl(W(q32,q21)) with this set

is convex. From Lemma 3.7, it is also known that cl(S2 n Ni) U coan2
is a convex set. Thus, Y a (cl(S2 n W?) U coan2) n cl(W(q32,q21)) is
a convex set. From its definition, A

=XUY. Letx,y€A Ifx'

l l'
ychrx,er, thenxycAl sinceXandYareconvex. Ifx eXand

y c Y, then xy n coan2 # 25 which implies xy c A1, since coanzc X n Y.

In exactly the same manner it can be shown that A is a convex set.

2

We next consider A3. By lemma 3.7, X s coan3

Y- .=. coan2 U cl(82 n W3), Z1 5 coan2 U cl(S2 n w?) and 22 conVQ3 U

3
U cl(S3 n w2),

cl(S3 n W?) are convex sets. By the argument following Case IV, '2- a

Z1 U 22 is a convex set. By definition, A3 is the intersection of

X UY U '2' with the closed convex region bounded by L(q33,q3l),

L(q21.q22) and L(q23.q32). Thus, if x, y c A30 3?, A3 n Y, or A30 '2',

3. If x 5 A30 35, y 2 A30 Y, then xy n conVQ2 and xy 0 coan3

and xy ('1 D23 are non-empty. Since coan2 and coan3 are in the kernels

If x e Xand z e 7, then xz n coan3

then m C A

of $2 and S3, respectively, xyc: A3.

is non-empty which implies xz c: A3, since coan C in '2'. Similarly,

3

if y a Y and z e '2’, then yzc A , since yz n coan2 is nonempty and

3
coan2 C Y n '2'. The set Au is defined in precisely the same way as A3,

and may be shown to be convex inexactly the same manner. Lastly, we

-l|1-

consider the set A5. By construction, each element of B1 U 812 is

not Joined to the point of local non-convexity q33; otherwise C112,
qn or q21 would be a point of local convexity. Similarly each
element of 82 U 812 is not Joined to Q32 by a line segnent lying
entirely in S , since then qn,q21 or q23 would not be in Q. There-
fore, by Proposition 2.2, Bl U812 and 82 U B12 are convex relative
to 8. By Bunt's refinement of Fenchel's Theorem [5](see Theorem 3.6

for statement) the convex hull of the two components B and 812’ and

l

the convex hull of the two components B and B are contained in S.

2 12
Next consider the set conv(Bl U B2). By Bunt's refinement of
Fenchel's Theorem conv(B1 U 82) C S if B1 U B2 is convex relative to

S. Since by definition B and B are convex subsets of S, it suffices

l 2

to show that for x c B1 and y ‘3. 82, xy C 8. By the definition of B1
1
andBZ, x swanslandy ewgnsz. Sinceqn c KlandqzleKz,

x, y c W(qn,q12) n W(q23,q21). Evidently, if xy C S, then either
x or y is contained in E‘J(q22,ql3). Assure that x e W(q22,ql3). By
property Pu and the closure of S, we know that q22q'13C S. Since

c clV there exists an element 2 in w: n Q22q13 UW(q33,qll)n

q33 3'
W(q21,q33) such that z Q33 at S. Since x : W(q22,q13) if xz C S, then

q13 t Q3, a contradiction. But if xz ¢ S, then property P" is violated,

since xq33tS and c Q implies that each point which is not Joined to

q33 ,
q33 by a line segnent lying entirely in S, is Joined to x by such a line
segment. (See the proof of Proposition 2.2). This contradiction implies
xyC S. That is, B1 U 82 is convex relative to S. The same argument
applies if y c w(q22,ql3), or if x, y c W(q22,ql3). It follows that

conv(Bl U 82) C S.

.133.

1, B2, and 1312 with the

property that conv(Bl U 812), conv{B2 U 812) and conv(B:l U B2) are

We now have three convex components B

contained in S. We wish to show that conv(Bl U B2 U 812)C S. Assume
that this is not the case and let w e conv{Bl U 82 U 812) - S. It is
well known that this would imply the existence of three or fewer
points x1,x2, and x3 of B1 U 82 U 812 such that w e conv{xl,x2,x§}- S.
Since conv(Bl U 812), conv(B2 U. B12) and conv(Bl U B2) are contained
in S, the existence of such a w is impossible unless there exists
such that x e B

3 1 l’

2 3} - S= Axlx2x3-S.

)C 8 while int(Axlx2x3) contains w t S. Since

exactly three non-collinear points x1, x2, and x

x2 e 82' and x3 e B12, say, and w e conv{x1,x ,x

We now have bd(Axlx2x3

S is closed, there exists an open spherical neighborhood a of w such

that on S = fl and o C int(Ax Thus, bd(Ax ) bounds a

1x2x3). lxzx3
residual domain of S. From Lemma 3.9, it. follows that Aqllq21q31 C

Axlx2x3. But, since x1 e clVi, i = 1,2, and x c D while qll e CW

3 12
t Axlx2x3, a contradiction. Thus, conv(Bl U 82 U 812): S, and

3 3
qll

since S is closed, cl(conv(Bl U B2 U B12) C S. That is A5 is a closed

, convex subset of S.

5
It must now be shown that S = kgl Ak' Consider the set S
3

(coani) U (ng (S1 0 W?» i = 1,2,3, and recall that Q1: clVi,
1,2,3. Evidently, (sl - WE): A2 u A“ U 812 U A1, (S2 — w?) c A1 U A3

1
is:

3 ' _
U 812 U A2. and (S3 - NZ): Al U A2 0 A3 u A“. The claim is that
1
($10 wz)c A“ u A5, (32 n wg) c: A3 U A5 3
important to recall that (S 1 0 Ni) is a convex set whose closure inter—

.3 ‘
and (330 112): A U A“. It is

.e 113 ..
__ i ' i
sects (S1 Wz) in the line segment q12q13, and that (Sin Wz) c
A Cl(w(qil’q12) n w(qi3’qil)) Since qil 8 K1’ and in’qi3 E Q13 1 a 1:203.

With this in mind, we direct our attention first to (S3 n W Since by

14 2)
property P and the closure of S, Q is convex relative to S, the line
3
segments q32q23 and c133q12 lie in S. If (S30 W2) C A3 U A“, then
3 .
there would exist three points x e (S3 0 W2) n W(q33,q12) n W(q23,q32),
y c 33 n (q3z,q23) and z 5 S30 (q33,q12) such that x, y and z violate

property P3 of S Tnus, (S3 n Wg) c A3 U Au.

3.
It is clear that the convex sets (810 szl) and (S2 n W2) are
each separated into two convex sets by the lines L(q33,q12) and
L(q32,q23), respectively. One of the convex subsets of S1 is Bic As,
i - 1,2. Thus, it suffices to show that (sin wé) - 131 is contained in
_ ~ . ,1 _
A3 U A“. This in turn reduces to snowing that (81“ WZ) Bl CW(q33,q12)n
W(q13,qll) and that (San wg) - 82 c W(q21,q22) n W(q23,q32). This,
however, is evident from the fact that (810 W3) C cl(‘.~!(qil,q12) n

w<q13.qfl) ) .

Consider now the case N1 = 3, i = 1,2, and N3 = 2m + l for m > 1.

We first define the sets

("U 3
A1” (53 ”(Li wan”
and
= (s3n (301%))

The set S is expressible as the union of the following convex sets.

-uu-

A2 a cl[(Sl U 83) n (w(q12.q13) n W(qn.q3’2m+l) n W(q32.q3,2m,l))3
K3 e cl[(82 U 83) n (W(q3,2ml,q3l) n w<q21,q22) n ”(923'q3,2m))3

I“ = cllI(Sl U 83) n (‘:I(q3,2m,1.q12) n W<ql3.qn) n W(q3l.q32) n

‘
”‘q3.2rn'q3.2m+1” " A1

and

I5 = cl(conv(BlU B'ZU B12».

_ g ‘1 ‘-

52 a (32 n vi) n \~I(q3,2m.q23).

and a '.-. ‘
B‘12 Dl2n "(q21'q3,an) nw(q3,2n+l’qll)°

The sets 172,133 and K5 are defined in the same way that A2, A3,

and A5 were defined, respectively. The proof that each is convex re-

quires only the obvious change of subscript; that is, the replacement

of Q32 by q3,2m and of q33 by q3,2m+l in the above argument.

111-1

’1. By Lemma 3.8, coan3 u (s3 n (Jgo WSW»

is a convex set. Thus, W(q3,2m,q21) n W(q32,q33) intersects this set

Consider the set

in a convex set. Moreover, Ai is contained in this intersection since

Q21Cl3 2m and q32q33 intersect coan3 in line segments and if A1 were

’ a

not so contained, either q or q would be a point of local
3,2111 33

convexity, contrary to assumption.
The remainder of the proof that K1 is convex is the sat?“e as

that for Al.

.. “5 ..
The set A“ is defined in the same way that A" was defined

except for its intersection with W(q3 2m'_1,q32). By Lemma 3.8,
’

coanBU u(83 n W31) U A," is a convex set. Thus, the intersection

of this set with W(q3,2m,q3’2m+l) n W(q3'2m+1,q12) n W(q31,q32) is
a convex set. Again, A," is contained in this set since q31q32 and

I
q3,2mq3,2m+1 are contained in the kernel of S3. If A,4 were not so

3.

the proof that A“ is convex is the same as that for Au.
5

The proof that S=1U1K1 is identical to that of N3: 3

This case differs for the set (83 n

contained then q32 or q3 2m would not be in Q The remainder of
l

with the exception of 33.

W(q3.an+l,q32)), which is seen to be the union of A1,A,;, and
(83 n Wan) by checking the indices of wtion. It is known from
Theorem 3. A that w383 = coan3 U(S3 n (JU1 W3». Since (S30 W3 )C
KIBBUUA’,(S éHl)CAUAuandcoan3123CAUKUKUmee

U
see that S3 is contained in i 1 A1.

Thus, in every case, S is expressible as the union of five

or fewer closed convex sets.

To see that the number five is best consider the following
example which satisfies the hypothesis of the theorem but which can

not be expressed as the union of fewer than five convex sets.

 

_ u5 -
Theorem.3.§_ If S has at most one point of local non-convexity which
is not in the kernel K of S, and S has property P”, then S may be

expressed as the union of four or fewer closed convex sets.

22222, If Q, the set of points of local non convexity of S,
is empty, then S is a closed, connected, locally convex set, which,
by Tietze's theorem is convex. If Q has exactly one element, then S
is expressible as the union of three or fewer closed convex sets, by
Theorem 3.3. If Q has two or more elements, each of which is in the
kernel, K, of S, then S may be expressed as the union of three or
fewer closed convex sets, by Theorem.3.U. Thus, we may assume that

Q has two or more elements, exactly one of which is not in K.

Let q a Q-K. It was shown in the proof of Proposition 2.2 that
S is expressible as the union of a starlike set whose kernel contains
q, and the set X of all points of S which are not Joined to q by a
line segment lying entirely in S. By Proposition 2.2, since S has
property P“, the set X has property ?2 relative to S. That is, X is
convex relative to S. We show that coan C S, and that S-X may be

expressed as the union of three or fewer closed convex sets.

It is first shown that X has at most two components. Assume that
X has three components Cl’ C2, and C3 in S. Let c1 e Ci’ 1 = 1,2,3.
Since X is convex relative to S, we have that clcz, c2c3, and clc3 are

entirely contained in S. Since, however, Cltl CZ’ CZIW C and lel C

3!
each intersect S - X. Let

3

are empty, we have that clc2,c2c3, and clc3

a12,a23, and al3 be elements of clc2 r](S-X), c2c3 {T(S-X), and

clc3fl (S-X), respectively. Then qa12,qa23, and qa13 lie entirely
in S-XC S. The elements cl,c2,c3, and q are not collinear since

the fact that qa12,qa23, and qa lie in S would imply qc 1C S for

13
2, and c3 are collinear in that order, say, then

qa12,qa23, and 812823 are contained in 8 while qc2 I S. Since S-X

some i. If cl,c

is closed, there exists an open spherical neighborhood N of e such

2
that x e N {TS implies xq CS. Thus bd(Aaqua23) C S bounds one or

more residual domains of S. Since S is starlike, this is impossible.

If c1,c2, and c are not collinear, then they determine the triangle

3

Ac1c2c3 a '1‘, whose vertices we may assume to be in that counterclock-

wise order. Then q a intT, or S-(S n T). (For the same reasons as
given above, q can not lie on an extended side of T.) If q a intT,

bd(qa ) bounds one or more residual domains of S, a contradiction.

12°2a23
If q c S—(S n T), then without loss of generality, we may assume that q
is in the v-shaped region detemined by L(c2,c3) and L(02,cl) and con-
taining (c1c3), or that q is in they-shaped region determined by these

lines which intersects T in the point e In the former case

2.

bd(qa ) bounds a residual domain of S, and in the latter. case

23°2312
bd(qal2clc3a23) bounds a residual domain of S. Since in each case we
are led to a contradiction of the starlikeness of S, we have that X has

at most two components C and C2, say, in S.

l
w. Fenchel [7] showed that a necessary and sufficient condition

for a point x to be contained in the convex hull of a compact, connected

set X in Em is that there exist m (or fewer) points pl,p2,... ,pm in X

such that x e conv(plU p2U ... U pm). Later L.N.H. Bunt [5] showed

-118-

that compactness was unnecessary and that "connected" may be replaced
by "having at most m components". For m = 2, this theorem may be
stated: A necessary and sufficient condition for a point x to be '
contained in the convex hull of a set having at most 2 components

is that there exist 2 (or fewer) points pl,p2 in X, such that x c
conv(pl U p2) .=.' plpz. Since in this case X = {z | zq 228} has the

property that for all 21 and 22 in X, 2 C S, if we considered the

1Z2
set Y of all points of S which are in X or are contained in a line
segment, the end points of which are in X, Bunt's refinement of

Fenchel's theorem applies to give Y C S is the coan. That is, X

is contained in a convex subset of S.

We now direct our attention to the closed subset S-X of S.
Let Q and ‘1? denote the set of points of local non-convexity and kernel
of S-X, respectively. Let p c Q and assume that p t X. Then p 7! q
since q a K’. Let x c S-X be such that px ES-X. If p c Q of S, then
px C S, which then implies that px n X 7‘ E. If x e L(p,q),px CS-X
since qx and qp are in S-X, a contradiction. If x p! L(p,q), then we
may consider Apxq. Since p c Q, pxCS. Since q a K, and x, p e S—X,
qx and qu S—X CS. Also, since px f) X 7‘ fl, and S—X is closed, there
exists an Open subinterval (uv) of (px) such that (uv) C X which implies
that bd(Aqu) C S bounds a bounded residual domain of S , a contradiction,
since S is starlike. Finally, we show that the case where p i Q is
impossible by showing pe Q irplies p c Q. Assume, to the contrary,
that p a Q, and p L’ Q. Then there exists a Spherical neighborhood N of

p of radius r such that N f) S is convex. Since p c Q, there exist

.19..

elements y and z of N n (S-X) such that yz «ES-X. Since N n S is

convex, yz C S which implies that yz n X 7‘ fl. Consider the points

y,z, and q. If q 6 L(y,z), then yzC S—X since qy and qz are con-

tained in S-X. If q t L(y,z), we may then consider quz. As before,
since yzn X 5‘ f5 and S-X is closed, there eixsts an open interval

(uv) C (yz) such that (uv) C X. This implies that bd(quz), which

is contained in S, bounds a bounded residual domain of S, which again
contradicts the starlikeness of S. Hence, each point of local non-
convexity of S-X is in the kernel of S-X. Since S-X is a closed, A
connected set, if S-X has two or more points of local non-convexity,

then S-X may be expressed as the union of three or fewer closed convex
sets, by Theorem 3.1!. If S-X has exactly one point of local non-convexity,
then S-X is expressible as the union of three or fewer closed, convex
sets, by Theorem 3.3. If S-X has no points of local non-convexity,

then S-X is convex, by Tietze's theorem._ Thus, in each case, S-X may

be expressed as the union of3three or fewer closed, convex sets, Mi’
1 =- 1,2,3. Then S = clY U ([1J Mi) is the desired decomposition of S into

closed convex sets.

Lemma 3.10 Let S have property 1’“, and 2 be a point of local convexity
of S. If S is one-dimensional at 2, then there exists a line X through
2 such that cl(S-X) has property Pn'l.

Proof- Since 2 is a point of local convexity, there exists an
open spherical neighborhood, o(z,t) of 2 such that o(z,c) n S is convex.

Since S is one-dimensional at 2, there exists an 1:- _<_ e such that

-50..

bd(o(z,?) n S) is zero-dimensional. That is, cl(o(z,'e-) n S) is
a line segment, Y. Hence, xz CS implies x c S n X, where X is the

line determined by Y. Thus if xl,x2,...,x are n-l points of

n-l
S-X which are not Joined in S, the points xl,x2,...,xn.1, 2 would
violate property Pn. Hence S-X, and consequently cl(S - X) s A,
have property Pn'1 relative to S. Let w1 and V12 be the open half-
planes determined by X. Then A = c1[(w1n S) U (W2 n 8)]. Assume
now that A does not have property Pn'l (relative to A). That is,
letx x1,x2,...,xr1 _1 be n-l distinct points of A which are not Joined
in A. Not all of these points can be in cl(wln S), or in cl(W2 n S),
since these sets have property Pn-l. Consider first the case where
1‘ X n S for i = l,2,...,n-l. Since 2x1 CS for i = l,...,n-l,

C S for some i and 3 because A has property Pn-1 relative to S.

”1.1
SincexixJCA, x:l exrllnSwhile xJ ewzn S, say. Thus xifoHan)
is non-empty. Let u= xixJn (X n S). Since A is closed and xix.1 -{u} C

A, u c A. That is, A has property Pn—l. Next consider the case where

x1 c (X n S) for some 1.. Without loss of generality we may let i = 1.
Since x1 5 cl(l-Il n S) or xl c cl(w2 n S), for each neighborhood N of
XI, there exists an element Elin ian1 n S or W2 0 S such thatx x1,x2,...,

_ xn-l are not joined in A, since A is closed. Then x1,x2,...,xn_1, z
violate property P". Thus in this case too, A has property Pn'l. That
is, cl(S - X) has property Pn-l.

Corollary If S is as in the lemma and n = ll, then S is expressible as

the union of four or fewer closed convex sets.

—51—

Proof. Since S satisfies the hypothesis of the lemma,
cl(S - X) = A has property P3 and is expressible as the union of
three or fewer closed convex sets. If X n S has exactly one (convex)

component, the result follows.

1 and X2 of X“ S with the

preperty that neither is a subset of A, then there exist points 21

If there exist two components X

c 11 and 22 c X2 with the property that 2122 C 8. Moreover, the con—

ditions of the lemma are satisfied at 21 and 22. It is clear that A

must be a convex set and the result follows.

Finally, the set X n S can have at most two such components.

For then cl(S - X) = fl and S is not a connected set.

Theorem 3.7 Let S have property P“. If S has a cut-point q, then

S may be expressed as the union of U or fewer closed convex sets.

Proof. SinCe q is a cut-point of S, S - {q} has two or more

canponents .

If one of the components, C, of S - (q) is convex, we consider
the two cases ,' C is one-dimensional and C is two-dimensional. If C
is one-dimensional, then clc is evidently a line segment qy or a ray
R(q,y) emanating from q and containing qy. In either case the hypothesis
of Lemma 3.10 is clearly satisfied at y c C, and the result follows from

the Corollary to Lemma 3.10.

Assume next that the convex component C of S - {q} is two—

dimensional. Since chn cl(S - C) = {q}, if c s cm and b e cl(S - C),

-52—

c a! q 7! b, then cb C S implies q a (Cb). Moreover, if a, b c cl(S - C)
and ab C S, then ab C cl(S - C). Assume, to the contrary, that ab ¢ _
cl(S - C). Since q c cl(S - C) and abC s, ab n (010 — (<11) y! a.

That is, ab n C 7‘ C which implies C is not a component of S - {q} ,

a contradiction. Now, let xl,x2,x3 be three distinct points of cl(S - C)
which are not pairwise Joined in cl(S - C), then by the above remarks,
if c c ClC, c 75 q, then cx1C S implies q a (exi) and the points xl,x2,x3
are not pairwise Joined in S. Since ch is two-dirrensional at c, and
{x1,x2,x3}is a finite set, there exists in some neighborhood N of c, a
point c' of ch such that c' x101 S, i = 1,2,3. The points c', x1,x2,x3
new violate property P3. rmus, S is once again expressible as the union

of three or fewer closed convex subsets of cl(S - C) and the closed

convex subset ClC.

Let us now assume that the closure of each component of S -— {q}
is not convex. Let C and C' be two such components. Let x, y a C10
and a, b c CIC' such that xy C ch and ab (Z ClC'. By a‘previous argu-
ment this implies )qtcz S and ab d S. We first assume that q is not one
of the four points x, y, a, b. By property Pu, two of the four points
x, y, a, b must be Joined by a line segment lying in S. Without loss
of generality we may assume that xa C S. Once again, this implies
q s (xa). Since C is a component of S - {q} and x a! q, there exists a
spherical neighborhood a of x such that C n SC C. If there exists a
spherical neighborhood a' C a of x such that for all x' e a' n S,

x' aC S, then q C (x'a) for each x' and we have cl(a' n S) is a line
sesnent containing x. ((a' n S) 7‘ {x} since 5 is connected) Clearly

x Satisfies the typothesis of Lerrna 3.10 and the result follows from

.- 53 ..
the, arguments used in the one-dimensional convex case above. Assume
then that for all spherical neighborhood 0' of x there exists x' )4 x
in an S such that x' aC 8. Since 8 is a closed set and xy (2 S,
there exists an open spherical neighborhood a' C C such that for all
x". c a' n S, x'yCS. Let x' c c' n S such that, x' a CS. Then,
since x'a, x'y, and ab are not contained in S, by property Pu, x'b,
yb, or ya is contained in S. If ya C S, then q, y and a are collinear
since ch n CIC' = {q}. Since xaC S, q, x, and a are collinear for
the same reason. Thus, if ya C S, xy C xa U ya C S, a contradiction.
Assume then that x'b C S. Since x' c o n S CC, x' a! q and x'b C 8

implies q a (x'b).

If there exists a spherical neighborhood 1 of x' such that for
all z e 1 n S, sz S, then q a (zb) for all z e T which implies
cl(r OS) is a line segment containing x'. Then x' satisfies the
hypothesis of Lemma 3.10 and the arguments of the one-dimensional
convex case apply to give the desired result. Therefore, we may
assume that there exists in every spherical neighborhood ‘1' of x' an
element 2 such that zb C S. Rt 1 C o' be a Spherical neighborhood
of x' which does not intersect the component of L(x,a) n 8 containing
xa. Let 2 e r such‘that szZ S. Since 2 e C', zyCS. Since 2 e T
and 1 does not intersect the component of L(x,a) n S which contains
xa, z a ¢S. We now have a point z e C such that za, zb, zy are not
Contained in S. Since ab and ya are not contained in S, it must
happen that yb is contained in S, by property P". If, however, yb C

S, then q, y, b are collinear, since c1C n ClC' = {0.}. Since x' bCSa

-5u-

q, x', b are collinear'for the same reason. Thus, if yb c S,

x'yC x'b U ybC S, a contradiction.

Thus, if xa and x'b are contained in S, x, y, a, b are
different from q, and neither x nor x' , as defined above, has a
spherical neighborhood whose closure intersects S in a line segment,

we are able to exhibit four points z, y, a, and b which violate

property Pu .

We now consider the case where xa and yb are contained in S,
and x, y, a, and b are each difference from q. Without loss of
generality we may assume that none of the elements x, y, a, b has
a neighborhood whose Closure intersects S in a line segment. If x'
is defined as above, then x'b C S, for if x'b C S, then ch n CIC' =
{q} implies q, x', b are collinear. Since ybC S, q, y, b are collinear
for the same reason, and we have x'y C x'b U yb C S, a contradiction.
Let the a spherical neighborhood of b such that IT) S C 0'. Since S
is closed, and neither x'b nor ab is-in S, there exists a spherical
neighborhood 1" of b, I" C I', such that b' e 1" implies b'x' and b'a
are not contained in S. By our original asswnption each spherical
neighborhood of b contains an element which is not Joined to y by a
line segrent in S. Let b' e I" n S have this property. Again, since
ybC S, xaC S and xy C 8, ya C S. Thus x', y, a, b' violate property P”.
In the case where xa, yb C S, it has been shown that yaC S. The same
reasoning shows that xa, yb, be S can not occur. Finally, assume that
one of the points x, y, a, or b is q. Without loss of generality we may

assume that a = q. Then since S is a closed set and qb C S, there exists

-55..

a spherical neighborhood A of q such that for all a' e A n S,
a'bCS. Since q a ch', A 00' i‘ U. Let a' e A nC'. The argu-

ments given above are now applicable to the four points x, y, a' , b.

Hence, in each case, we are led to a contradiction unless the
closure of some component of S - {q} is convex or unless there exists
.a point having a neighborhood the intersection of whose closure is a

line segment .

Since the result has been established in each of these cases,’

the proof is complete.

Theorem 3.8 Let S be a closed. starlike set and have property P”. Let

the kernel K of S be a one-dimensional set. Then S may be expressed as

the union of four or fewer closed convex sets.

m. Since 8 is closed, by the corollary to Proposition 2.3,
the kernel K of S is closed and convex. Since K is one-dimensional, K
is a line, ray, or line segment. Clearly, K is not a line unless SC K.
Since S C K implies that S is convex, we may assume that K is a line
segment xy, or a ray R(x,y) containing xy. Let W(x‘,y) and W(y,x) be
the Open half-planes determined by L(x,y). Since S n L(x,y) contains
K, S n L(x,y) is a convex set. Thus, if SC S n L(x,y), S is convex,
and we are done. Moreover, if S n L(x,y) ct cl[(W(x,y) u W(y,x)) n S],
then there exists a point of local convexity, z, in S n L(x,y) such
that S is one-dimensional at 2. By Lemma 3.10, clES-(S nL(x,y))] has

property P3. Thus, the set S may be expressed as the union of the 3

-55..

or fewer closed convex sets which arise from C1[S~(S fl L(x,y))] and
the closed convex set S n L(x,y). Therefore, we may assume that

S n L(x,y) c cl[(W(x,y) u W(y,x)) n S], and that W(x,y) n S and
W(y,x) n S are not both empty. If W(x,y) n S and W(y,x) n S are con-
vex, we are again done, since then S is seen to be the union of the
two closed convex sets cl(W(x,y) n S), and cl(W(y,x) n S). Assume
then that W(x,y)n S is non-empty, and that cl(W(x,y)n S) is not

. convex. Then W(x,y) n S must contain a point of local non-convexity
of S. Assume, to the contrary, that (W(x,y) n S) contains no points
of local non-convexity of S. We show that (W(x,y) n S) is a convex
set. For if u, v e (W(x,y) n S) such that uv C (W(x,y) n S). Then
evidently qu S. Since x, y e K, and WC S, x, y i’ L(u,v). Thus,
we may consider the quadrilateral xyvu whose vertices may be consi-
dered to be ordered ina counterclockwise manner x, y, v, u, as indi-
cated. Let p = xv n yu, and consider the component C of the closed
set S n Aupv containing the connected subset of S, up U pv. Since C
is a closed, connected subset of S, if C is also locally convex, then,
by Tietze's result, C is convex. Since CC W(x,y) n S, if W(x,y)n S
contains no points of local non-convexity then C is locally convex in

which case u, v e C implies uv C C C S, a contradiction.

If y a e, the quadrilateral xyvu becomes an infinite convex
strip. The same arguments apply, however. Thus, in the case being
coriSidered, if W(x,y) n S contains no points of local non-convexity,
then W(x,y) n S and, consequently cl(W(x,y) n 'S) is a convex set.

. We see that if neither W(x,y) n S, nor W(y,x) n S contains a point

of local non-convexity of S, then S is expressible as the union of

the two closed convex sets cl(W(x,y) n S) and cl(W(y,x) n S).

We next show that each point of local non-convexity of W(x,y)
S is in the kernel of W(x,y)n S. Let q be a point of local non-
convexity of S in W(x,y) n S. Let z 7‘ q be a point of W(x,y) n S
such that zq at S. By the proof of PrOposition 2.2, the set S is star-
like from q U 2; that is, for each element u of S, either un S or
WC 8. We first assume that K = xy, y a! on. Evidently, L(z.q)n xy =
fl since otherwise zq C S. Since quy C S and szy C S, we have quyn
szy 5 TC X. Let u c intT. Then since u L’ K, there exists an element
v of S such that uv (2t 8.. Order the four v-shaped regions V1, 1 = 1,2,3,“
determined by L(x,u) and L(y,u) in a counterclockwise manner starting
with that region which contains (xy). Evidently v t clV1 U clV3 since
xy = K and there would then exist an element w in xy such that w, u,
and v are collinear which. would imply uv C S, a contradiction. Let
v e V2. Since S is starlike from q U z, vq or vz is contained in S.
We have, qu quy n S and uz C szy n S. The segments vy and yu are
also contained in 8 since y c K. Thus uv is contained in the quadri-
lateral uqu or uzvy, each of which lies entirely in S, and we have
uVC S, a contradiction. If v. e V“, we have uvc uqvxc S or uv C
uzvxc S , again a contradiction. This contradiction implies that
every point of W(x,y) n S and hence of cl(W(x,y) n S) is Joined to q
by a line segment which lies entirely in S. Thus each point of local

non-convexity q of S in W(x,y) n S is in the kernel of cl(W(x,y) n S) -

- 53 -
If y = a, the same argument applies to the regions .clvl, clV3,an
V“. If, however, v c V2, we have uv is contained in the convex strip
uqv .. if qvc S or in uzv . if zy C S, which again leads to the same

conclusion.

let 5 be the set of points of local non-convexity of cl(W(x,y)
n3) 5 A. The set 5 is non-empty, by assumption. If q a Q’ nL(x,y),
we show that q is the limit of a sequence of points of 5 n W(x,y).
Since each element of 5 n W(x,y) is in K, the kernel of A, it will
follow from the closure of S, that q s 3?. Assume that q s 5 nL(x,y).
If q is an isolated point of 5, then there exists an open Spherical
neighborhood a of q such that an 5 = {q}. Let u, v c a n A such
that uv¢ A. Evidently uv¢ S. If u, v e L(x,y) n S, then uvc S,
since L(x,y) n S contains xy, the kernel of S. Thus, at most one of‘
the points u and v is in L(x,y) n S. Assume that u c L(x,y) n S.
~Then since u e c1(W(x,y) n S), every spherical neighborhood 1 of u
intersects W(x,y) n S. Moreover, since 8 is closed, there exists a
Spherical neighborhood x' of u such that u' e 1' n A implies u' v¢ S.
We may assume that A' C a. The above argument indicates that we may
assume u and v to be in W(x,y) n S, whenever it is desirable to do so.
If q a K, the kernel of A, then qu and qv are in S. If q t! K, then
there exists a point z e W(x,y) n S, such that qz C S. Since S is closed,
there will then exist a Spherical neighborhood 1 of q, T C a such that
we r n A implies wz¢ S. Let u, v e r n (W(x,y) n S) such that uv¢ S.

“Then, by Proposition 2.2, each point of S is either Joined to q or to z

- 59 ..

by a line segment lying entirely in 8. Since uz and vz are not con-
tained in S, uq and vq are contained in S. Thus, in either case
(chandth)weareableto findpointsuandvin o n
(W(x,y) n S) such that uv CS while qu, qvc S. By Lemma 3.3, q a
(x,y). let us assume that q, x, y is the ordering of‘ these three
points along the line L(X,Y). (We do not exclude the possibility
that x = q). Since y e K, y e K. Since quS, u, v and y are not
collinear. Thus, we may consider quu and quv. Since qu, qv C S
and y c K, each of these triangles lies entirely in A. Since qu and
qv are in S and uv¢ S, u, v, and q are not collinear. Hence either
yu n qv or yv fl qu defines a unique point p in W(x,y) n S with the
property that pu and pv are contained in a n (W(x,y) n 8). Thus,
Apuv lies entirely in o “(W(x,yn. Let C be the component of Apuvn
(W(X,Y) n S) which contains the connected set up U pv. Since Apuvc
a n (W(x,y), C is a closed, connected, locally convex set which must
then be convex, by Tietze's theorem. Since u, v e C, uv C C C S, a
contradiction, unless C contains a point of local non-convexity of

s in o n W(x,y) distinct from q.

If y = a, then the same argument holds for a suitable choice
of y' along the ray x c». It follows that every open spherical
neighborhood of q contains a point of local non-convexity of S lying
in W(x,y) n S. That is, q is a limit point of a sequence {pn} of
points of local non-convexity of S in W(X,y) n S. Since pn : K- for
each n, pn a C S for all a a A. For each a e A, the set {pna} con-

verges to the set qa which is in A, since A is a closed subset of

-60..

the closed set S. Thus, q E K, the kernel of A. We now have that
every point of 5, the set of points of local non-convexity of A is
in K, the kernel of A. Moreover, if 6 n L(x,y) is non—empty, then
the cardinality H of 22' is infinite.

Evidently the set A is a closed, connected set each of whose
points of local non-convexity is contained in its kernel K. That is, -
the set A satisfies the hypothesis of Theorem 3.1%. It follows that
if FT is one, even, or infinite, that A is expressible as the union

of 2 convex sets, as was shown in the proof of Theorem 3.1:.

If fi' = 2n + 1 is an odd integer greater than one then
5 nL(x,y) = fl and there exists an edge qoql of convfi and a support

plane No of coan which abuts convfi' along qu1 such that xy C cl(W0 n

(W(x,y) n S)). As in the proof of Theorem 3.14, on (W(x,y) f) S),
and, consequently, cl(Wo n (W(x,y) n 8)), are convex. Starting
with qoql we order the remaining edges of convfi' in a counterclockwise

manner. If we let W:l be the Open half-plane of support to conle'

whose closure contains qiqi+l’ we have the convex sets
mu
81 8 cl {(convfi') U [(S n W(x,y)) n 1_1W_21 1J}

. mU
szacluconv'e') uusn nW(x.y))n 19,1 ”213}

and S ' 61040 n (W(x,y) n 8)) as three convex sets containing

3
cl(S nW(x,y)), as in the proof of Theorem 3.“. let M1 and M2 be
maximal convex sets of cl(W(x,y) n S) containing 81 and 82, respective-

ly. We show that 83C MlU 9’12. let w e 83 such that w a/ Ml U M2. Then

-51..

there exists an element 3 in M and t in M such that sq of S and

l 2

WC S. By Lemma 3.8, s c cl(Wl n (W(x,y) n 8)) and t e cl(men
(W(x,y) n 8)). Let x,y,ql,q0 be the natural ordering of the ver-
tices of the quadrilateral xyqlqo. Then S must be contained in

3
the union of the two convex regions C2 and Cl bounded by L(x,qo)

and L(ql,q2m), and L(y,ql) and L(q0,ql) respectively, and contain-

ing the quadrilateral xyqlqo. Moreover, cl(Wl n W(x,y)n S))CCl

and cl(i-I2m n (W(x,y) fl 8)) C C If this were not the case either

2.
q0 or ql would not be in Q. Since the quadrilateral xyqlqOC K
and the above mentioned sets are contained in C1 U CZ’ we have swn
K or twn K is non-erpty. That is, sw or tw is contained in S.

This contradiction shows that S C' Ml U I42. As in the proof of

3
Theorem 3.11, the same argwnents hold for y = on, with only the obvious
changes. ' Since the above arguments may also be applied to cl(W(ihxm
S), we see that S is the union of 14 or fewer closed convex sets. The

following example is representative:

 

 

 

- 52 -

An examination of the proof of Theorem 3.6 discloses the

following:

1. If S has property Pu, and q is a point of local non-
convexity of S, then the convex hull of the set X of
points of S which are not Joined to q by a line segment
lying entirely in S is contained in S.

2. If w is a point in the kernel of S, then w is a point
in the kernel of S - X.

3. If w is a point of local non-convexity of S - X, then w

is a point of local non-convexity of S.

If S, X, and q are defined as above, and S does not bound a
bounded domain of its complement, then S—X has prOperty Pu. Assume
to the contrary, that x1, x2, x3, and x1I are four points of S-X
which are not Joined in S-X. Then X n xlxz, say, is non-empty
since 8 has prOperty 2’“. Clearly, q, x1, and x2 are not collinear
since qxl and qx2 are contained in S-X. Thus, we may consider
Aqxlxz. Since S-X is closed, there exists an open subinterval (uv)
in (x1x2)n X which implies bd(Aqxlxz) C S bounds a residual domain

of S, contrary to assumption.

Since by Theorem 3.5, a closed, connected set in E2 having
property P“ which bounds a bounded domain of its complement may be
expressed as the union of 5 or fewer closed convex sets, the problem
of determining the number of closed convex sets which are required if

2

a closed, connected set in E having property P“ is to be expressed

as the union of such sets, may be reduced to the case where the set

-53..

S does not bound such a domain. Since S is either convex or contains
a point of local hen-convexity q, (by Tietze's Theorem), we may assume
the existence of such a point. By the above remarks, S may be con-
sidered to be the union of a starlike set having property P“ with q
in its kernel, and a closed convex set. Thus, if the case where S is
starlike with one or more points of local non-convexity in the kernel
of’S were settled, the general case would then fbllow. Several of the
cases settled by the theorems of this chapter allow one to place even

greater restrictions upon the nature of the set S.

CHAPTER IV
FURTHER CONSIDERATIONS AND EXAMPLES

While it may be difficult to determine how many convex sets
are required to express a closed, connected set 3 having property
Pn as the union of Closed convex sets it may not be.as difficult
to establish an upper bound on the number required. The fellowing

theorem is a step in that direction._

2 and the set

Theorem.fl.l If S is a closed, connected Pn set in E
Q of points of local nonpconvexity of S is a finite set, then S is

expressible as the union of a finite number of convex sets.

Proof. Let N be the cardinality of Q. If N = l, the result
fellows from.Theorem 3.3. Hence, assume the result to be true fbr

N :_k - l and consider the case N = k.

Since coan is a compact, convex set in Em, there exists an
element q of Q such that q t conv(Q - {q}).' Thus, there exists a
line L which strictly separates q and the compact convex set

conv(Q - {q}).

Let U be the closed halfespace determined by L and contain-
ing conv(Q - {q}) in its interior, and'V be the closed halfespace
containing q in its interior. It is clear that U n s and v0 8
are closed sets having property Pn. 'Thus each has at most n - 1
components, and each component is a E” set containing at most k - 1
points of Q. Thus, by our induction hypothesis, each component is
expressible as the union of a finite number of closed convex sets.

The result now fellows.

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-55..
Remark: An example given by Kay [11] shows that a finite
number of convex sets may not suffice if S is not closed. The case

where S is closed remains unsettled.

Considering the results obtained in Chapter III for closed,
connected sets in the plane having property Pu, one might conJecture
that a closed, starlike set in E2 with property P“ is expressible as
the union of n or fewer closed convex sets. The fbllowing examples

show that this is not the case for n > H.

The first example given has property P5, but can not be ex-
pressed as the union of fewer than 6 closed convex sets. (Each
star is eXpressible as the union of no fewer than 3 convex sets,
and no point of one may be entirely included in convex subsets of
the other).

 

.. 55 ..

The second example given indicates how one might subdivide the
circumference of a circle to obtain a starlike set having property
P2k+1 which is expressible as the union of no fewer than 3k convex

sets, where k is the number of stars.

This particular example has property P17 and is expressible
as the union of 2h closed convex sets. Its kernel is the center Of

the circle.

 

 

 

 

 

 

 

 

 

-57..

Some of the topological properties of a set having property
Pn have been established. For. example, if S is closed and connected,
then S is arcwise connected by Proposition 2.2; by Theorem 7 of Kay
[11] which is stated without proof, S is locally starlike, and hence

locally connected.

For n _>_ 1|, a closed connected set need not be simply connected
since the boundary of a triangle in E2 has property P“. The question
of the connectivity of a closed, connected set in E2 is of some in-
terest.

For n = it it was shown in Lemma 3.9 that a closed, connected
set S in E2 with property P“ can bound at most one bounded domain of
its complement. For n a 5, each of the following examples having
property P5 bounds 3 such domains. (In these examples, 16, 5, and 6
convex sets have been used to bound the three domains).

 

 

For n > 14, the number of bounded domains which a closed,
2

connected set in E having property Pn can bound is an Open question.

- 53 _

The following class of examples of sets having property Pk+l which
bound 62-1 domains indicates that the number is greater than or equal
to'Clg-l. Let the k vertices of a convex polygon be chosen so that
the extension of each side intersects the extension of every other
side. The configuration consisting of the k extended sides will
bound cg‘l domains and will have property Pk+1. Given below are
examples for k = 3,14,5,6,7,8. Several other configurations have
been considered, but each has resulted in Gig-1 or fewer bounded do-
mains

 

 

 

 

-59..

The nature of P” sets in higher dimensions has received very
little attention. If the proofs given in E2 are to be generalized,
it would seem that the generalization of Valentine's results, Theorem
3.3 and Theorem 3.14 should first be considered, and the nature of

the set of points of local non-convexity understood.

For n = 3, Valentine [18] has given the following result:

let S be a closed set in a linear topological space L where the di—
mension of L is greater than two. If S has property P3, is not con-
tained in any two-dimensional variety of L, and has one isolated
point of local nonconvexity, then S has at most two points of local
non-convexity. The proof uses strongly the fact that for n = 3 the
set of points of local non-convexity of S are in the kernel of S,
which is not true for n > 3. The following example is a P3 set in

E3 having exactly. two points of local non-convexity.

By adding lines which pierce the sphere one may obtain a Pn set

having 2(n - 2) isolated points of local non-convexity.

- 7o _

It is clear from the example given after Proposition 2.6,
and the boundary of a 3-simplex which has property P5 that S need

not have any isolated points of local non-convexity.

l.

2.

3.

9.
10.

11.

12.

13.

IN.

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