Title: Projective convexity
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Permanent Link: http://ufdc.ufl.edu/UF00097983/00001
 Material Information
Title: Projective convexity
Alternate Title: Convexity, projective
Physical Description: v, 39, 1 leaves. : ; 28 cm.
Language: English
Creator: Hare, William Ray, 1936-
Publication Date: 1961
Copyright Date: 1961
Subject: Topology   ( lcsh )
Generalized spaces   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
Bibliography: Thesis -- University of Florida.
Bibliography: Bibliography: leaf 38.
Additional Physical Form: Also available on World Wide Web
General Note: Manuscript copy.
General Note: Vita.
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Bibliographic ID: UF00097983
Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
Resource Identifier: alephbibnum - 000574216
oclc - 13841471
notis - ADA1579


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June, 1961


In recent years much attention has been given to the

study of convexity. The application of this theory to the

solutions of systems of linear inequalities has further

intensified the research. Klee [3] has studied convexity in

the general setting of a topological linear space. It is

the purpose of this paper to define a generalization of con-

vexity and to make an analogous study of the generalization

in the setting of a topological linear space.

The topology will be as general as possible. That is,

no separation axioms will be assumed unless explicitly stated.

Thus, the results will hold in spaces whose topologies are

even weaker than To. (Recall that in a To space X the fol-

lowing property holds: if x,yeX, then there exists a neigh-

borhood U of x such that yfU, or there exists a neighborhood

V of y such that xtV.)

The author wishes to thank his supervisory chairman,

Dr. J. W. Gaddum, for suggesting the problem and furnishing

much valuable assistance and criticism during the research

and writing. Also, he wishes to express gratitude to the

other members of his supervisory committee--Drs. J. E. Maxfield,

W. R. Hutoherson, J. T. Moore, T. O. Moore, and C. W. Morris--

for their guidance during his graduate work and for their

assistance on this paper. Particular appreciation is also

expressed to Dr. Paul Civin for suggesting generalizations

of Theorems 9 and 10, and to Mr. T. C. Rogero for his assistance

on several occasions. Finally, he would like to thank

Dr. Jolm Kenelly and Mr. Shiu F. Yeung for their aid in the

geometric considerations of the last chapter and for their

assistance in the technical aspects of this thesis.


This Thesis is Dedicated

to P.M.H. and K.A.C.



PREFACE . . . . . . . . . . .

CONSIDERATIONS . . . . . . .





BIBLIOGRAPHY . . . . . . . . . .


Chapter I



Let X be a linear space over the real field R, where R

has the usual topology. Suppose, also, that X has a topology

superimposed. X is a topological linear space over R if and

only if (hereafter shortened to iff) vector addition xl+x2

and scalar multiplication ax are continuous functions on

X x X and R x X to X, respectively.

Let x and y be elements of X. The set of all points

Xx+(l-X)y, NCR, is called the line through Z and y and is

denoted by xy. Those points for which O<
internal segment which is denoted by ry. Finally, those points

for which _l make up the exteral segment, xy.

A subset S of X is convex iff, for each pair of points

x,y of S, it is the case that xyCS.

A subset S of X is proJectively convex (hereafter shortened

to p.c.) iff, for each pair of points x,y of S, either xycS

or xyC S.

Thus, we see that protective convexity is a generaliza-

tion of the notion of convexity. We state this formally as

the first theorem.

Theorem 1. A convex set S is p.c.

Proof. By definition, for each pair x,y of S, we have xyC S.

Thus, S is p.c.

It is obvious that the complement of a convex set need

not be convex. However, there is a more pleasant result for

p.c. sets.

Theorem 2. The complement of a p.c. set is p.c.

Proof. Let S be a p.c. set and suppose that its complement S'

is not p.c. Then there are points x,y of S' such that xy(S'

and xyQ/S'. That is, there are points u,v in S such that
u = Nlx+(l-Al)y, O
v = X2x+(l-X2)Y, A2<0 or X2>l.

If we solve for x and y in terms of u and v, we have that

x = 1-2 u+ -l1 v,

y = U2 u+ u I+ V.

Setting 81 = 1-X and 82 = -"X we have that
Xl-%2 l- 2

x = 61u+(1-61)v

y = 82u+(1-82v.

We now must consider two cases.

Suppose that O1 and 0<82<1, so

we have that xeuv and yeuv which contradicts the assumption

that S is p.o.

If O< ll, we have 82>1 and 0<81<1, making xeuv

and yeuv, again a contradiction.

Thus, the assumption that S' is not p.c. is untenable.

We conclude that S' is p.c.

Examples of p.c. sets other than convex sets and their

complements are furnished by the next theorem. Recall that

Q(x) is a quadratic form on X to R iff there exists a bilinear

form B(x,y) on X x X to R such that Q(x) = B(x,x).

Theorem 3. The set of points at which a quadratic form is

positive (non-negative, non-positive, negative) is a p.c. set.

Proof. Suppose that x and y are two points for which Q(x)>O,

Q(y)>O. Then

Q(AX+(1-X)y) = h2Q(x)+(1-1)2Q(y)+X(1-X)(B(x,y)+B(y,x)).

If (B(x,y)+B(y,x))_O, we have that Q(Xx+(l-X)y)>O for

Q(Nx+(l-X)y)>O for X1.

Thus, if the quadratic form is positive at x and y, it

is also positive at either xy or x.

We now consider several purely algebraic notions about

p.c. sets. First, recall that a linear variety is a translate

of a linear subspaoe of X. That is, a linear variety is a

set of the form x+S, where x is some element of X and S is a

linear subspace of X.

Lemma 1. If V is a linear variety and y,zeV, then yzcV. In

particular, a linear variety is convex, and hence p.c.

Proof. Let V = x+S, where x is some element of X and S is a

linear subspace of X. Then we may uniquely express y and z

as y = x+sI and z = x+s2, where s1,s2ES. Now any point on

yz has the form
Ny+(l-N)z = X(x+sl)+(l-N)(x+s2)
= x+(sl+(l-?)s2)

= x+3 ,

where s3ES. Thus, any point of yz is in V, so that yzcV.
It is an elementary theorem in convex set theory that

the intersection of a family of convex sets is also convex.
However, the intersection of two p.c. sets need not be p.c.

To see this, let
S = {(x,y)lx2+y2<},
T = {(x,y)Ix2y2>1}.

Both S and T are p.c., but their intersection is not p.c.
However, we are able to get a somewhat analogous result, which

also serves to characterize linear varieties. First, we need

a lemma which gives an alternative characterization of linear


Lemma 2. A nonvoid subset S of X is a linear variety iff we

have alxl+a2x2+...+anxneS whenever {x1,x2,...,xn}CS and

al+a2+...+n = 1. The index n is not fixed.
Proof. If S is a linear variety, then S = x+V for a subspace V.

Let {xl,x2,...xn}CS and write xi = x+vi, vieV, i = 1,2,...,n.
Using this representation we have

alxl+a2x2+...+anx = 1(x+v1)+a2(x+v2) +... +an(x+vn)
= x++al1+a2x2+.. .+nXn,

which shows that the condition is necessary.


To show sufficiency, let xeS and define T = -x+S. We

will show that T is a subspace. Let tl,t2 T and write

t1 = -x+s,

t2 = -x+s2'
for sl,s2ES. Now let a,8 be any two real numbers. We have

atl+6t2 = a(-x+sl)+8(-x+s2)

= -x+(x-ax+asl-8x+8s2)

= -x+s3

since (1-a+a-8+8) = 1. Thus, T is a linear subspace.

Theorem 4. A set S is a linear variety iff SAP is p.o. for

every p.o. set P.

Proof. First we will show the necessity of this condition.

Let x,yeSAP. From the definition of projective convexity we

know that xyCP or xyCP. Both xy and xy are contained in S,

by Lemma 1, so at least one of the two segments is contained

in SAP, making this intersection p.c.

Now we consider the somewhat more difficult problem of

the sufficiency of the condition. It will be shown that our

condition implies the condition of Lemma 2. To do this we

will use induction on the number of elements in the subset

of S.

For n = 1 we have {x,}CS and for al = 1 we have alx1ES

without applying our condition.

Now assume that for n = k {xl,x2,..Ixk)}CS and for

al+a2+...+ak = 1 we have alxl+a2x2+... +akxk6S. We will see

that this holds for n = k+l.

Let {yY2,..*.,yk+l)C S, and let {81,82',..,8k+1) be a
set of real numbers whose sum is one. If any one of the 8 's

is zero, the condition holds by the inductive assumption.
Thus, assume that all are non-zero. We consider three cases.
Suppose that 0<8k+l<1. We know that

y = 81Y+...+8kkeS

by the inductive hypothesis. Now yYk+l is a p.c. set and hence

SAyyk is p.c. Since yyk+lCS, we have that
k k+l
SiY+8k+lk+1 = j iYieS.
i=l i=l

Next, suppose that 86k+l. As in the first

case we have yeS by the inductive hypothesis. yYk+l is p.c.
so SAyy1 is p.o. Since yyk+CS, we have that
so S~YYk+l k+l
k k+l
Z 8iY+8k+lk+l = Z 8iYiCS.
i=l i=l

Finally, suppose that 8k+ = 1. We know that for some j,

68<0. Let 86 = 8n for n k+l, n a j, and let 816+1 = 6,
8, = 8k+1. Make a similar renumbering of the yi's. We now
have the situation of the second case, so the lemma holds.

In all three cases the condition of Lemma 2 is satisfied,

so we conclude that S is a linear variety.

A natural question of an algebraio nature is whether or

not projective convexity is preserved under the formation of
a quotient space or a direct sum of two spaces. That is, if

we form a quotient space of X modulo a linear subspace, is the

quotient space image of a p.c. set also p.c.? And, if we form

a direct sum of two linear spaces, is the direct sum of two

p.c. sets also a p.c. subset of the direct sum space?

The answer to the first question is in the affirmative,

but the second question receives an affirmative answer only in

a very special case. We now state these two results.

Theorem 5. Suppose h is a homomorphism of the space X onto

the space Y. Then if S is a p.c. (convex) subset of X, h[S]

is a p.c. (convex) subset of Y. In particular, the image

under a linear transformation of a p.o. set is p.c.
Proof. Let u,veh[S]. For at least one pair x,y in S we have

that u = h(x), v = h(y). Either xy or xy is contained in S,

so let us assume that it is the former, and the other possi-

bility will follow in a similar manner.
7u+(l-N)v = Xh(x)+(l-X)h(y)
= h(Xx+(l-X)y)

is an element of h[S] if O
To consider the question of direct sums let the spaces

X1 and X2 be E1. Let S1 = (0,1) and S2 = (l,oo)v(-c,-l). Both
S1 and S2 are p.c. subsets of X1 and X2 respectively, but

S14S2 is not p.c. in the direct sum space. Thus, if even one

of S1 and S2 fails to be convex, the direct sum need not be

p.c. However, we are able to state the following result.
Theorem 6. Suppose S1 and S2 are convex subsets of the spaces

X1 and X2, respectively. Then SlSS2 is convex in X1~X2.

Proof. Let (xl,x2) and (yl,Y2) be elements of Sl"S2. We have
that X(xl,x2)+(1-N)(yl,y2) = (x1l+(l-X)yl,Xx2+(l-N)y2) is an
element of SleS2 since Xk+(l-A)yicSi, i = 1,2.

It is a theorem in convex set theory that, if S and T
are convex sets and a and 8 are non-negative real numbers,
then the vector sum set aS+ST is convex. For a proof of this

theorem see Taylor [5], page 130. This need not be the case
for p.c. sets as the following example shows.

Let the space be E2 and define S and T as follows

S = {(x,y)Ix = O, y<-l or y>l},

T = {(x,y)ll By direct computation it is seen that

S+T = {(x,y)lll},
which is not p.c.

With a suitable restriction, however, we are able to ob-
tain a theorem.
Theorem 7. If P is p.c. and V is a linear variety, then the

vector sum set aV+SP is p.c. for any real numbers a and 8.
Proof. Let V = x+S, with x some element in X and S a linear
subspace of X. Let u,v be members of aV+SP. We may write

u = a(x+sl)+6pl,
v = a(x+s2)+6p2,
where sieS, pieP, i = 1,2. We have that

Xu+(l-X)v = a(a(x+s1)+8pl)+(l-7)(a(x+s2)+8p2)

= a(x+(xsl+(l-A)s2)+8( pl+(l1-)p2).
Now x+(s l+(l-X)s!eV for all real N. For O

or X>1 we have that ?pl+(l-X)p2eP. Hence, either uv or uv

is contained in aV+8P, making this set p.c.

A motivation for the terminology projectivee" convexity

is given in the next theorem. Recall from protective geometry

that if a,beX such that aexy and bexy for x,yeX, then the

pair (a,b) is said to separate the pair (x,y). If (a,b)

separates (x,y), then (x,y) separates (a,b).

Theorem 8. Let T be a 1-1 transformation on X onto X which

maps points into points, lines into lines, and having the

property that

(a,b) separates (x,y) iff (Ta,Tb) separates (Tx,Ty).

Then, if S is p.c., so is T[S]. That is, the image of a p.c.

set under a projective transformation is p.c.

Proof. Let Tx,TyeT[S] and assume that TxTyS, TxTytS. Then

there are points Tu,Tv such that TueTxTy and TveT'Xy and

Tu,TVE(T[S])' = T[S]'. Then (Tu,Tv) separates (Tx,Ty), so

(u,v) separates (x,y) Thus, xyCS and xyq-S, a contradiction
to the assumption that S is p.c.

In a linear space it is well-known that there is a unique

minimal convex set containing a given set, and this set is

called the convex hull of the given set. For a proof of this

statement see Taylor [5], page 131. However, if our set is

simply two distinct points, there are two minimal and non-

comparable p.c. sets containing the two points, namely, the

two segments determined by the two points. This situation

prevails in general as the following theorem states.

Theorem 9. For a subset S of a topological linear space X

there exists a minimal p.c. set containing S.

Proof. Let A be the family of all p.c. sets contained in the

complement S'. A is partially ordered by set inclusion. For

B a chain in A consider VB. Let x,yeVB; there are sets C,D

in B such that xcC, yeD. Either CCD or DCC, so let us

assume that the former holds. Then x,yeD and since D is p.c.,

we have that xy or xy is contained in D. This means that xy

or xy is contained in VB, making VB a p.c. set. Now clearly

VBC S', so we have that each chain in A has an upper bound

in A. Thus, by Zorn's Lemma, there is a maximal member F of

A. F' is a minimal p.c. set containing S. For, suppose that

there is a p.c. set M such that SCMCF' and M / F'. Then

M' is a p.c. set with F N' and FCM'CS', contradicting the

maximality of F.

Such a minimal p.c. set containing the subset S will be

called a D.c. hull of S. A problem that remains open at the

present is the determination of the class of sets for which

there exists a unique p.c. hull.

Before proceeding with our development, it seems advis-

able to make an observation about one property in a topologi-

cal linear space. Since translations are homeomorphisms, the

neighborhood system at any point xeX is obtained by translat-

ing the neighborhood system of the origin by the vector x.

Thus, if S is the neighborhood system of the origin, then

{x+VIVeS} is the neighborhood system at x. The importance of

this observation lies in the fact that the neighborhood system

at any point ycX can be indexed by means of the same index

set S. It should also be remarked that S is directed by set

inclusion. Both of these facts will be used in the next two


Theorem 10. The closure of a p.c. set A is p.c.

Proof. Let x,yeA". For each neighborhood rI, VeS, of x, there

is a point xv~AAM Likewise, for the corresponding neighbor-

hood NV of y, there is a point y vAANv. Now for each V either

XyYV or xvYV is a member of A, since A is p.c. Thus, for at

least one of these two cases there is a subset R of S with the

property: if VeS, there exists UeR, with UCV, such that

Xu U (or )U J) is contained in A. (Such a subset R of S is

said to be a cofinal subset. For a detailed discussion of

Moore-Smith Convergence, see Kelley [2], chapter 2.) Suppose

that this holds for internal segments. Let p be defined by

p = >x+(l-X)y, O
Consider the net {pU;UeR} = {7xU+(l-X)yU}. It is eventually

in each neighborhood of p, so p is a limit. Since all of the

values of this net are in A and p is a limit of the net, we

have that peA". Since p was any member of xy, we conclude

that xyCA. The foregoing argument can be modified slightly

for the other case.

Theorem 11. The interior of a p.c. set A is p.c.

Proof. From A = X-[X-A]-, the preceding theorem, and the fact

that the complement of a p.c. set is p.c., the theorem follows.

We will now digress somewhat to develop some properties

of boundedness that will be needed in the sequel.

A subset S of X is bounded iff for each neighborhood of

the origin, U, there exists a positive scalar a such that


An equivalent condition is that S is bounded ifft if

a is a null sequence and {x}) is a sequence in S, {anxn)

converges to the origin. See Taylor [5], page 129.

Lemma 3. The union of two bounded sets is bounded. (A proof

can be found in Bourbaki [1].)

Lemma 4. Any finite set is bounded.

Proof. We will use induction on the number of elements in

our set.

For n = 1 let S = {xl. If an is a null sequence, then

{o:nxn converges to the origin by continuity of scalar multi-


Assume that any set of n elements is bounded. For n+l

let Sn+ = ,x ,..., ,x n+}. This set can be expressed

as the union of a set of one element and a set of n elements,

both of which are bounded. Thus, Sn+1 is bounded by Lemma 3.

Theorem 12. The convex hull of a finite set S is bounded.

Proof. Let S = {x1,X2,...,xn}. The convex hull of S consists

of all points of the form

z = Xx ,
i=1 1 1

with i>_O and Z hi = 1. Let {ym be a sequence in the con-
vex hull of S. We may write

Ym = Zimxi
Now let {an} be any null sequence, and consider the sequence

{anYn}. Since O< in n. Hence, each term in the expression of ym approaches the

origin as a limit by the continuity of scalar multiplication

and thus, by the continuity of vector addition {a nYn con-

verges to the origin. Therefore, the convex hull of S is


The next result, although elementary in appearance, is

quite important in our next theorem.

Lemma 5. A line L is an unbounded set.

Proof. Let the line L be defined as the set

{zlz = Nx+(l-7)y, x y}.

Let { an = {l/n} and {\n} = {n}. We Imow that nx+(l-n)yeL

for all positive integers n; thus {xn = {Xnx+(l-Nn)y} CL.

Clearly an is a null sequence. Now consider the sequence

{anxn} = {x+(l/n-1 )y}. This sequence converges to x-y which
is not the origin. Thus, L is not bounded.

Theorem 13. A bounded p.c. set S is convex.

Proof. Since {x,y} is bounded, xy is bounded, being the con-

vex hull of a finite set. If xy were bounded, xy =xyVxy

would be bounded. Hence, xy is not bounded. Therefore, if

x,yeS, we see that since xy(S, we must have xycS.

Chapter II


A hyperplane H is a set of the form {xlf(x) = a} for a

linear functional f and a scalar a. H is said to separate

the set S iff there exist points x,yeS such that f(x)
f(y)>a. Clearly, if H separates T and TCS, then H separates

We now give several theorems in which separation by

hyperplanes plays an important part.

Lemma 6. Suppose that the hyperplane H fails to separate the
set S. If x and y are points of S such that xCS, then either

xyc H or xyAH = 0.

Proof. Let H be represented as {xlf(x) = a} for a linear func-

tional f and a scalar a. If H does not separate S, either

f[S]>a or f[S] suppose that xy
in view of Lemma 1. Suppose that it is y. Let zexyAH; for
X>1 consider the point p = Az+(l-X)y. We have that

f(p) = f(Az+(l-A)y)
= Nf(z)+(l-X)f(y)

= Xc+(l-X) f(y)
< 7a+(l-X)a = a.

This says that H separates xy since f(y)>a, so we have that

H separates S, a contradiction.

Lemma 7. There is no open set contained in a hyperplane.

Proof. Let H be a hyperplane and let x be a vector such that

xtH. If S is open and SC H, then consider xs for seS. No

point of xs is in S except for s, and s is an accumulation

point of S'. To see this, let {\n} be a sequence of real

numbers with O< n<1 and such that the sequence converges to

zero. Then {Xnx+(l-Xn)s} is a sequence of points in S' which

converges to s by the continuity of scalar multiplication and

vector addition. S' is closed since S is open, so we have

that seS', a contradiction to the assumption that seS.

We may remark that, in view of the preceding lemma,

if S is open in Lemma 6, then SAH = 0 so that yAH = 0.

Theorem 14. If A is an open or closed p.c. set and if there

is a hyperplane which does not separate A, then either A is

convex, or A is contained in a hyperplane.

Proof. Suppose that A is a closed p.o. set and H1 is a hyper-

plane which does not separate A. Then there exist a linear

functional f and a real number al such that H1 = {xIf(x) = al},

and f(x)>2a or f(x)al for all xGA. Assume that the former

holds. If A is not convex, then there are points x,yEA such

that xycA. That is, there exists a point peA' such that

p = Xx+(l-X)y with O
xyC A. By Lemma 6 either xyC H1 or xyAH1 = 0. Thus, in the

first case xy is contained in H1 and in the second xyC H2,

where H2 is the hyperplane defined by H2 = {xlf(x) = a2 for

a2>al. Let us consider the case xyCH2 and the other follows
similarly. If there are no points in A xy, the theorem is
established. Hence assume that there exists a point zeA such
that f(z) a2. Either alEf(z)a2. Consider the
first case; yztA since for any 7 where

Az+(l-N)y is a point such that
f(Nz+(l-A)y) = Nf(z)+(l-N)f(y) Similarly in the other case yz' A. Thus, we must have that
yzCA in either case.
Now let qeyz; in a manner similar to the foregoing, we
can show that xqCA. That is, the point
[X +(1-?)(X'y+(1-') z)]EA
for all N' for which O<^'<1 and X has the same value as before.
Let {\n} be a sequence such that O one. Then

{ x+(1-X)( ny+(1-kn)z)}
is a sequence of points in A which has p as a limit due to
the continuity of scalar multiplication and vector addition.
Since A is closed, peA, a contradiction. Hence, if A is not
convex, it is contained in a hyperplane.
Let us consider the case in which A is an open p.c. set
and H1 is a hyperplane which does not separate A. We again
have that H1 = {xif(x) = al} for a linear functional f and

a scalar al, and f(x)al for all xeA. If A is not
convex, there are points x',y'EA such that for 0<<1
p = [xx'+(l-X)y']eA.
Either 0 p is the midpoint of xy. Suppose that the third case holds;
let x be defined to be the point 2Xx'+(1-2X)y' and y' be called
y. Since 1<2N and A is p.c., we have xcA. Also, we have that
p = 1/2x+l/2y. That is, in this case we can find x,yeA such
that the midpoint of xy is not in A. This can also be done in
the other case, so for the remainder of the proof we will
assume that this has been done. Since A is open AAH1 = 0 so
xyAH1 = 0; hence, xyCH2 = {xlf(x) = a2} for alc Assume that the former holds. If A H2, there is a point
zEA such that f(z) a2. Either ala2.
We will consider in detail the first possibility and the other
will follow in a similar manner. As in the first half of the
proof we have xzCA. Let s = A'p+(l-X')z, where

= a!-f(z)
Also, let r = A"y+(l-?")z, where A" is defined by

n = CL-f(z)
Note that N' = X". The points s and r are, respectively,
pzAHI and yzAH1. Define the sequence {xn} by
xn = nr+(l-n)s
and the sequence {yn by

y = Xn p+(1-N )x ,
where Nn = l-l/X'n. In a different form we have that

yn = [+(l-n)(l-n)X']p+[(l-n)nh']y+[(1-n)(l-X')]z.
pxn contains points of A so it cannot be contained in A'; thus
p CA'. Since ynepxn, we have that YnA'. It is seen that

{yn) converges to y; since A' is closed, yEA'. But yEA, so
we have reached a contradiction. Thus, if A is not convex,

it is contained in a hyperplane.

Corollary. If A is an open p.c. set and if there is a hyper-

plane not separating A, then A is convex.

Proof. By Theorem 14 A is either convex, or is contained in

a hyperplane. By Lemma 6 A cannot be contained in a hyper-

plane; hence, A is convex.

Chapter III



Our attention is now directed toward the study of pro-

jective convexity in two-dimensional topological linear spaces

which will be denoted by L2. Such a space will have a basis

of two vectors. In particular, a connection will be established

between projective convexity and a certain convexity property

which is called property P Further information about the

latter idea can be found in Valentine [6]. The theorems of

this chapter will be quite useful in helping to characterize

p.c. sets in n-dimenslonal topological linear spaces L and

in spaces of arbitrary dimension.

An open half-plane is a set of the form {xlf(x)
{xlf(x)>a) for a linear functional f and a scalar a. If less

than and greater than are replaced by less than or equal to

and greater than or equal to, respectively, the resulting set

is called a closed half-plane.

An open strip is a subset S of L2 suoh that there exist

a linear functional f and scalars a and 8 with c<8 for which

S is the set {xla
f(x) = 8 is a boundary point of S.

Following the terminology of Euclidean 2-space, a hyper-

plane in L2 will be called a line. This is also in agreement

with our earlier usage of the term.

Before starting with our main project of this chapter,

we will establish a system of terminology and notation which

will be very useful in the sequel.

Let pl' p2, and p3 be three non-collinear points of L2.

First, we will let al = p2P3, and a2 and a3 will be defined


We next wish to define seven regions in a manner which

agrees with the standard terminology used for the Euclidean

plane. We will consider a point of the form

3 3
Z aiP1, where Z ai = 1.
i=1 i=1

(i) The triangle P.

Here each of the coefficients is non-negative. If one

of them is zero, as for example al, we have a point on one

of the three internal segments, in this case p2 p. If two

of the coefficients are zero, the third is automatically one,

so we have one of our three points.

(ii) Th three reiaons Ai.

These regions are defined by requiring that the ith

coefficient be non-negative and the others be non-positive.

(iii) The three regions Bi.

These regions have the ith ooefficient non-positive and
the other two coefficients non-negative.

We first prove a lemma which will be quite useful in

establishing our main theorem of the chapter.

Lemma 8. If A is a p.c. subset of L2 and a line L fails to

separate A, then A is either

(i) an open half-plane plus part of its boundary; or

(ii) an open strip plus part of its boundary; or

(iii) contained in a line; or

(iv) convex.

Proof. Suppose that A is not convex; then there are points

x,yeA such that xy(A. We know that xkyA, so we have that

xy is parallel to L. We may represent L as {xlf(x) = a1}

for a linear functional f and a scalar al. Then xy can be

represented as {xlf(x) = 21}. If ACtxy, let zEA xy. Then

f(z) a2. The line L' = {xlf(x) = f(z)} bounds, with xy,
an open strip whose points all belong to A. If we can find

a sequence z n} in A such that the sequence {f(z n)} is strict-

ly monotone increasing and unbounded above, A is the set of

case (i). If no such sequence exists, A is the set of case (ii1

It might be remarked that the part common to A and the

boundary of the open strip, or the open half-plane, will con-

sist either of a single point, or a finite (open, closed, or

semi-open) line segment or the union of two infinite rays (with

two, one, or no endpoints included). Any other collection of

boundary points would fail to be p.c. and, therefore, A would

not be p.c. It, of course, might happen that the entire

boundary of the open strip or open half-plane is in A.

A subset S of the topological linear space X is said to

have property PE iff for x,y,zES, at least one of xy, yz, and
xz is contained in S. It is immediate that any convex set

has property P3. Also, the union of two convex sets has this


Theorem 15. Suppose that A is a p.o. subset of L2 which does

not have property P3. Then

(i) A' has property P3;

(ii) A has at most two components; and

(iii) if A is open or closed, then it is connected.

Proof. Let pl, p2, and p3 be three points of A such that none

of their internal segments is contained in A. Since A is

p.c., all three external segments are contained in A. We

will first determine some general structural properties of

(a) A'AAi is convex. For example, let x,yeA'AA1; then
we have that

x = alP+l2p+a22+a3p, ai = 1, al>0, a2<0, a3<0,

y = 81P1+82P2+863P, Z = 1, 81>0, 82<0, 8 <0.

Any point on xy has the form 7x+(l-A)y, OA<
qexy, we have that

q = (alPl22P+a2+3p )+(--)(81p +82P2+83p3)
= (Xal+(l-)81)pl+(haX+(l-)82 )p2+(Xa +(l-X)8 )p3

= TiPni+r2P2+nTp ,

3 3 3
SwTi = Z \ai + E(1-N)8i = X+(1-%) = 1,
i=1 i=1 i=1
and n1>O, r2<0, and m3<0. Hence, qcA1 which makes Al convex.
Likewise A2 and A3 are convex.
Since A' is p.c., either xy or xy is contained in A'.
We will see that xytA'. If a = 81 and a2 682, then also
a3 = 68 and x = y. Thus, let us assume that a3>6 The same
argument applies to the other two pairs of coefficients. Con-
sider the point
3 3
Xx+(l-X)y = X Z api, + (l-h) E 8ipi
i=l i=l

= 2 [Xai+(l-X)Si3pi,

with X defined by
= -83

We have that X>1, so the point is on xy. If, however, we
substitute this value of X in the above representation of
our point, we obtain a point on pPi2, which we have already
observed to be in A. We have reached a contradiction, so we
a must have xycA'. Hence xyCA'AA1, making this intersection

convex. A similar argument holds in the other two cases.
(b) ILf A'A4, then A' contains no unbounded subset
In either of th_ Bj4' adjacent to Ai. For example, let
xeA'AA1 and yEB2AA'. If y lies outside the triangle which is
determined by pl, p3, and x, then xy intersects both p2P3 and

pip3. Regardless of where y lies within its specified region,

xy intersects plP2, so that A' would not be p.c. Hence, B2AA'
is contained in a triangle and is thus bounded by Theorem 11.
Similarly B AA' is bounded.

(c) A-' intersects fa mDt one A. For, suppose xeA'AA
and yeA'AA2. xy contains a point on each of P3 and P13, so
xy must be contained in A'. Now xy is parallel to p 2, for
otherwise xy would contain a point of p1P2. Thus, A'A(A VA2)
is contained in a line parallel to pl 2. Since plp 3A, there
is a point z 9epp AA'. But since the lines yz and xy are not
identical, we have reached a contradiction.
(d) A'A(B1VB2VB3VP) = S j p.c. Consider the set

AV(A1VA2VA3 ),
and let x and y be elements of this set. We must take into
consideration three cases.
If xEAi, y6A1, we have xyCAi since Ai is convex.
Suppose that xeA yeA with i / j. For example, let
xeAl, yeA2. xy contains points of A, say ql and q2, where
q 1p2p3 and q2Eplp3. Now either q q2cA, making xy in our

set (since xqcA, yq2cA2, and xy = xqlVqlq2Vq 2y), or

q1q2CA, making xkyCAV(AlVA2VA3), since xycq" 2
Finally, suppose that XEA, yEAi. If XEAj for any j, the
case has already been considered. Thus, let xe(B1VB VB3VP).
We know that x contains a point p of A. Either xpCA, which
makes xYCA (since ypcAi and xy = xpVy), or xpCA, which

makes xycA, since xy-xp. Thus, this set is p.c. and so is
our original set S.


(e) tLet M be *h family lines parallel to 2i which
intersect B, I = 1.2s, Then some member of some 1J does
=o separate S, where is 11 same a. In 5~1 rt (d)l. Let us
first assume that all of the members of M1 and M2 separate
S. Let xEA'AB1 outside the closed strip bounded by a3 and
the line through p3 parallel to a Now suppose that yeA'AB2
such that y lies outside the closed triangular region deter-
mined by a2Aa3, a3Axp and,a2Axp.3. Such a y exists since
A'AB2 must be unbounded by the hypothesis about M2. xy is

not in S, since it contains a point of A. By (d) x" must be
contained in S, so it follows as in our previous work that
xy is parallel to a3. Since plp 3A, let zepp AA. There
are points on both yz and yz which lie in B2 but not on xy,
since xy # xz. One of these segments must be a subset of A';
but we have already observed that A'AB2 must be a subset of
xy. Hence, we have reached a contradiction and it follows
that there is no x outside the closed strip and in A'AB1.
(f) L. Is convex or is open strip plus part o1 if
boundary. By (d) and (e) S satisfies the conditions of
Lemma 8. S cannot be an open half-plane plus part of its
boundary since there are points from an Ai in every half-
plane. It cannot be contained in a line since the three points
of S on PP2, P2P 3, and pP3 cannot be collinear. Hence, the
conclusion holds.
Having established these preliminary results, we can now
proceed with the proof of our theorem.

A' has property Z. We must consider two possible cases.
Suppose that A' does not intersect any Ai. By (f) A' = S
is convex or is an open strip plus part of its boundary. In
either case A' has property P .
Suppose that A' intersects one Ai, say Al. (It cannot
intersect more than one by (c).) Then A' = (A'AA1)VS. Both
A'AAI and S are convex. The former is convex by (a) and the
latter follows from the fact that both A'AB2 and A'AB3 are
bounded, by (b), and hence S cannot be an open strip plus part
of its boundary. We have remarked earlier that the union of
two convex sets has property P .
A hs at most two components. Let xeAAA1 and y2, Y3 be

points on a2, a3 such that y2Y3 is parallel to a, and xey2y3.
Either xy2CA or xy2CA so that, in either case, x can be
joined to the boundary of A1 by points of A. Thus, AAA1 is
connected. From (e) we have that some A'AB is bounded. Let
it be A'AB There is some member of M which does not pass
through A'AB Then the internal segment K between al and a2
lies entirely in A. Since any point in AAA1 and AAA2 can be
joined segmentally to K, we have that AA(A1VA2) is connected.
Let peAAB1, and consider the line through p parallel to a1.
There are points q2ea2 and q3ea3 on this line. Either pq2

or pq3 is contained in A; in either case, p can be joined to
one of the two As adjacent to Bi. Finally, if rePAA, let s
be a point of K such that rs separates A If rsCA, r is
connected to K by a segment lying in A; if zsCA, then r is

connected to AAA Thus, A consists of at most two components,

namely, the ones containing AAA3 and AA(A VA2).

IL A ia either open or closed, it is connected. Suppose
that A is open; then we know that A' is closed. The assumption

that A'AB1 and A'AB2 are unbounded is equivalent to saying

that all members of M1 and K2 separate S. We have seen that

this requires A'A(B1VB2) to be contained in the strip bounded

by a and the line through p parallel to a Since plp 2A,

we know that there is a point x such that xeA'ApIP2. Now if

both A'AB1 and A'AB2 are unbounded, we can get a sequence of

points in A' having p as a limit which says that p1eA'. But

this is a contradiction to the fact that p EA. Hence, at most

one of the sets A'AB is unbounded. As before, let us suppose

that AA(A1VA2) is contained in a component. AAA can be joined

to AAA, or AAA2, since either A'AB1 or A'AB2 is bounded, by

points of A. Thus, A is connected.

If A is closed,A' is open. We can use a point of A'AB

in some neighborhood of x to show that at least one of A'AB

and A'AB2 is bounded. The remaining part of the proof is

the same.

Corollary. A p.c. set A in L2 has at most two components.

Proof. If A does not have property P the preceding theorem

says that A has at most two components. If A has property P ,

suppose that x, y, and z are points in three distinct com-
S r1%f l
ponents. All three of xy, yz, and xz fail to lie in A, con-

tradicting the assumption that A has property P .

Chapter IV



In this chapter we will make use of the results of the

previous chapter to establish the analogous results in arbi-

trary topological linear spaces. The somewhat slow and

tedious proofs for the L2 case will ease the way for shorter

proofs in our present considerations.

First we have the analogue of Theorem 15, part (ii).

Theorem 16. A p.c. set A has at most two components.

Proof. Suppose that Ci, i = 1,2,3, are three components of

A and xieCi, i = 1,2,3. First, suppose that the xi's are

collinear. Then one of them is on the internal segment of

the other two. Suppose that x2Exlx-. Then

x2 = Xx1+(l-X)x3,

where 0<\<1. Now x x(A, for then C and C would be seg-

mentally connected. Thus, since A is p.c., we have that

x X3CA. Since Xx2C:xx 3, we have that C1 and C2 are seg-

mentally connected, once again a contradiction. Thus, in

this case, A can have at most two components.

Now suppose that the xi's are not collinear. Then they

generate a two-dimensional linear variety V which is homeomorphi

to L2. It is clear that VAC1, VAC2, and VAC3 are components

of VAA, a contradiction to the corollary to Theorem 15.

Theorem 17. If an open or closed p.c. set A is not connected,

then each component is convex.

Proof. Let C1 and C2 be two components of A and suppose that

C1 is not convex; then there exist points x,yeC1 such that

xy(CCl. Let z be any member of C2. First, suppose that x,

y, and z are collinear. Then one is on the internal segment

of the other two; suppose that yExz. Now yztA and xz' A.

Hence, yzCA and xZCA. Since xycyz, we have that xyCA.

Thus, there is a point pEC2 such that pExy. Since pycxy,

it follows that p can be segmentally connected with y, which

is a contradiction. The case of xsyz is similar to the fore-

going. The case of zexy is impossible, since z would then be

segmentally connected with x from the fact that xzCxy.

Thus, if x, y, and z are collinear, the theorem holds.

Now suppose that x, y, and z are not collinear. Then

they generate a two-dimensional linear variety V which is

homeomorphic with L2. Also, VAC1 and VAC2 are components in

the open (or closed) p.c. set VAA. But then VAA must either

have property P or be connected. If it has property P then

we have that xyC A since xzqA and yzCtA. This is a contra-

diction. Now VAA cannot be connected since VAC1 and VAC2

are both open and closed in the relative topology and neither

is void.

Hence, the theorem is established.


Theorem 18. If A is p.c., then any component of A is also p.c.

Proof. Suppose that we express A as the union of two components

C and C2. If x,yEC we know that xyCA or xyCA. Suppose
that xyCA; then xyC C1 for otherwise A would be connected.

A similar argument holds for the case of xyCA.

Chapter V



Several of the results which were desired require a more

restrictive topology. It might appear that merely assuming

that our topology is T1 or perhaps T2 would not be so special-

ized as restricting our attention to En, but a theorem in

topological linear space theory tells us the contrary. The

theorem states that all n-dimensional topological linear

T1 spaces with the same scalar field are linearly homeomorphic.

This theorem may be found in Taylor [5]. Thus, in particular,

the study of real n-dimensional topological linear spaces

"reduces" to considerations of En, since we are only concerned

with algebraic and topological properties of p.c. sets.

We will first prove a lemma and then a theorem for the

plane. The lemma holds in L2 and the proof is the same as is

given below. The theorem does not hold even in E .

Lemma 9. If A is a connected set in E2 which has property P3

but is not convex, then there exist points p,q,reA with pq

and pr contained in A but qr not contained in A.

Proof. Since A is not convex, there exist points t,reA such

that trh(A. Since A has property P every point of A can be

joined to t or r (or both) by an internal segment lying entirely

within A. If some point can be joined to both, the lemma

holds. Hence, suppose that no point can be joined to both.

Let T be the set of all points in A which are joined to t and

R be the set of points in A which are joined to r. Suppose

that T is not convex; then if x,yeT such that xyQIT, there is

a point zexyAR. By property P we must have that xzCA,

rzCA, and rxCA. Hence the lemma holds. Thus, let us sup-

pose that both R and T are convex. Since RAT = V and A is

connected, we know that some point peR is an accumulation point

of T. But this means that p is linearly accessible from T.

(See Klee [53. A point y is linearlv accessible from a .sb-

set S of X iff there is a point xES {y} such that Cx,y)cS.)

Thus, there is a point qeT such that qrtA, but prCA and


Theorem 19. If A and A' are complementary connected p.c. sub-

sets of the plane, then one of them is convex.

Proof. Since either A or A' has property P let us suppose

that A does. If A is not convex, by the preceding lemma there

exist x,y,zeA, with xz and yz contained in A, but with a point
% I \ r '*
u of A' on xy. Then xyVxzVyz lies in A and separates the

plane with A' on one side since A' is connected.

Now if A' is not convex, there exist p,qeA' with pq not

in A'; that is, there exists a point r such that reAApq. But

pu and qu must lie in A', and pqVpuVqu separates r from x and y,

contradicting the fact that A is connected. This completes
the proof.

That this theorem does not extent even to E can be seen

by considering the two p.o. sets bounded by a hyperboloid of

one sheet.

For notational purposes we make the following definitions.

Suppose that {x1,X2,...,Xn+l) is a set of completely indepen-

dent points in En. The simplex which this set determines

will be denoted by S(xl,x2,...,xn+1) and the hyperplane through

the first n of these points will be denoted by H(xl,x2,...,xn).

It might be remarked that Eis the same as H(x1,X2,...,xn+l).

Lemma 10. Suppose that A is a p.c. subset of En which is not

contained in a hyperplane. Let xl,x2,...,xn be vertices of a

simplex S(x1,X'1...,xn) which is contained in A. Then this

simplex contains a simplex S(Yl,y2,...,"n) which is a face of
a simplex S(yl,y2,...yfn+1) lying in A. Moreover, for some

arbitrary i, we have xi = y

Proof. Let S(x1) be a simplex in E1, with S(xl) = {x1}CA.

Since A is not contained in a hyperplane (a point in this

case), there is a point ycA distinct from x. Either xly or

xly is contained in A. If the former is the case, the lemma

holds. In the latter case let y2 = 2x1-y. Then we have

S(x1lY2) = xlY2Cxly C A,
so that the lemma holds in E1.

W now make the inductive assumption by assuming that

the lemma holds in En-1,

Suppose that x1,x2,...,xncA and S(x1,x2,...,x )C A. Since
A is not contained in a hyperplane, we know that there is some

point zeA which is not in H(xl,x2,...,xn). Consider the hyper-
plane H(x2,x3,...,. ,z) = Hn-. S(x2,x3,...,xn) is a simplex
in AAHn_1 and hence there are points xi,x4,...,x',z'6Hn.1 such
that S(x2,x ,...,vx)CS(x2,x,...,x) and S(x2,x,. .. -,,zl)

is a subset of AAHn-1. Now S(xlx2,x ,...,x), which is con-
tained in S(x1,x2,...',), and S(x2,x ,...,x',z') are faces

of S(x1,x2,x4,...,xA,z') which are contained in A. If no point
of A' lies in S(xl,X2,x ,...,z'), the lemma holds. If some
point p of A' lies in this simplex, it can be represented by

p = 81x +82x2+83x...+58n +Sn+lz2'
where E 8 = 1 and 0<86<1, i = 1,2,...,n+l. Now let us con-
sider the region R defined by

R = {xjx = lX1+a 2x2+a 3x+.. .+'nx+'n+lz' ai = 1,
I ai 1, 0<.2, OISial-Sla^, i = 2,...,n}.
If yWA'AR, we have the following considerations. Let pl be

the point hy+(l-X)p, where X is defined by

S= n+1-
Since an+l-O, we have that 8n+l-an+l>O. Hence 0 pleyp. If we write out the expression for pl, we have that

p = 83alSn+l-CLn+ 81 x" +.' +nn+l- 8nx+an+ln+l-a n+ +z
1 n+l-an+1 n+l-an+l n+l-n+1

Each coefficient is non-negative, making pleS(xl,...,x1).
That is, ypAS(xl,...,xA) = {pll.
Now let p2 be defined by p2 = Ny+(l-X)p, where X = 1
Since we know that

81-1 = (81a2-al82)+(83-a183 )+..+(8* 1n+l'ln+l)
we have A<0; thus, p2eyp. Substituting the value of X, we
P 81 a-al82x +...+81_n+1-a18ntlz'.
p2 1=1 1l-t

Each coefficient is non-negative and less than one, so we have

P2cS(x2,...,xn,z'). Hence, we have that ypAS(x2,...,xA,z') is

not empty.
Thus, we see that no point can be in A'AR, making the

entire region R lie in A.

For convenience we will let x1 = ul, x2 = U2,..., z' = un+1.
Then for i = 2,3,...,n define y. by
+y = i -!n_-lu +...+ + u
S 681+2+...+8n 8 1+62+ ..+n
and Yn+l by
281 Ul +...+ 28n un+(-l)un+.'
y ,, = 8 + 1 Y ..- + i
1n+ 2 n 1 2 n

Clearly for i = 2,3,...,n we have yi6S(x1,X ,...,x4) so that

S(YY2",...,)C S(xl,X,"...',x) and x = y1 Finally, the
coefficients of yn+1 are such that yn+ R, but yn+l is not in
S(x,x',...,x'). It is clear that we could have modified the
foregoing argument so that xi = yi for any i = 1,2,...,n.
Clearly S(yl,Y2,..., n+1 )CA.

This completes the induction.

Lemma 11. If A is not contained in a hyperplane, then for any

p6A there is a hyperplane through p which is generated by p

and n-1 other points of A.

Proof. There are n+l points x1,x2,...,x n+ in A which generate

En. In particular, we have that
p = 81x+2x2 +.+n+xn+, 8= 1.
Suppose that 81 0. Then
x = 8 l(P-82x2- n+ln+l).

Now {p,x2,...,xn+l} generates En, so there is a hyperplane

H(p,x2,...,xn) through p and n-l other points of A.

Lemma 12. Suppose that A is a p.c. subset of En which is not

contained in a hyperplane. Then every point of A is a vertex

of a simplex lying in A.

Proof. For n = 1 the proof is trivial. Let us assume the

lemma for En-l. Let pl6A; there are points x2,x3,..., x6A

such that {pl,X2,...,xn} generates a hyperplane. In this

hyperplane the inductive assumption says that there is a sim-

plex S(plP2,...,p )CA. By Lemma 10 there are n+l points

pl'q 2" '' q1 such that S(pl,q2" ... qn+)CA.
Theorem 20. If A is a p.c. subset of En, then either A is

contained in a hyperplane or A- = (Int A)-.

Proof. Suppose that A is not contained in a hyperplane. Let

xA; it is a vertex of a simplex whose interior lies in A.

Hence, there is a sequence {xn} in (Int A) which converges to

x. Thus, AC (Int A)-. If xeA A, there is a sequence {x }

in A which converges to x. For each n there is a sequence

{x } in (Int A) which converges to x'. Since we have
ni n
Lir (L4m x ) = x,
n ni
we have that

Lim x = x,
i xi
and thus xe(Int A)-. Therefore, we conclude that AC(Int A)-.

Since (Int A)CA, we know that (Int A) CA so that

A- = (Int A)-.

Corollary. If A is a p.c. subset of E then either A is con-

tained in a hyperplane or A has an interior point.

Proof. If A is not contained in a hyperplane, we have already

seen that AC(Int A)-. Since 0 = 0, we have (Int A) $ 0.


1. Bourbaki, N. Espaces vectoriels topoiogaues. (Elements
de mathematique, Actualites Sci. Ind., Number 119).
Paris: Hermann, 1953.

2. Kelley, J. L. General Topology. New York: D. Van Nostrand
Company, 1955.

3. Klee, V. L., Jr. "Convex Sets in Linear Spaces I Duke
Mathematical Journal, Vol. XVIII, 1951, 441-464.
4. Klee, V. L., Jr. "Convex Sets in Linear Spaces II," Duke
Mathematical Journal, Vol. XVIII, 1951, 875-983.

5. Taylor, A. E. Introduction to Functional Analysis. New
York: John Wiley and Sons, Inc., 1958.

6. Valentine, F. A. "A Three Point Convexity Property,"
Pacific Journal of Mathematics, Vol. VII, No. 2,
Summer, 1957.

7. Veblen, 0., and Young, J. W. Protective Geometry. New
York: Ginn and Company, 1910.


William Ray Hare, Jr. was born June 29, 1936, at

Murfreesboro, Arkansas. In May, 1953, he was graduated from

Delight High School, Delight, Arkansas. His undergraduate

studies were completed in June, 1957, at Henderson State Teachers

College, Arkadelphia, Arkansas, where he received the degree

of Bachelor of Science summa cum laude. The major fields were

mathematics and chemistry and the minor was physics. He was

awarded a graduate assistantship in the Department of Mathe-

matics of the University of Florida where he enrolled in June,

1957. He received the Master of Science degree in January,

1959. From June, 1959, until graduation he was a Graduate
Council Fellow in the Department of Mathematics. He accepted

a position in the Department of Mathematics of Duke University

beginning in September, 1961.

Honorary organizations in which he holds membership are

Alpha Chi and Phi Kappa Phi. He was president of the former

organization during his last undergraduate year. He is a mem-

ber of the American Mathematical Society and the Mathematics

Association of America.

This dissertation was prepared under the direction of

the chairman of the candidate's supervisory committee and has

been approved by all members of that committee. It was sub-

mitted to the Dean of the College of Arts and Sciences and to

the Graduate Council, and was approved as partial fulfillment

of the requirements for the degree of Doctor of Philosophy.

June 5, 1961

Dean, College of Arts and sciences

Dean, Graduate School

Supervisory Committee:

C -chaman

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