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Projective convexity 

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Convexity, projective 

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v, 39, 1 leaves. : ; 28 cm. 

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Hare, William Ray, 1936 

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1961 

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1961 
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Topology ( lcsh ) Generalized spaces ( lcsh ) Mathematics thesis Ph. D Dissertations, Academic  Mathematics  UF 

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bibliography ( marcgt ) nonfiction ( marcgt ) 
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Thesis  University of Florida. 

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Bibliography: leaf 38. 

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Vita. 
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University of Florida 

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Full Text 
PROJECTIVE CONVEXITY
By
WILLIAM RAY HARE, JR.
A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
June, 1961
PREFACE
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 committeeDrs. 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.
iii
This Thesis is Dedicated
to P.M.H. and K.A.C.
TABLE OF CONTENTS
Page
PREFACE . . . . . . . . . . .
Chapter
I. ELEMENTARY ALGEBRAIC AND TOPOLOGICAL
CONSIDERATIONS . . . . . . .
II. HYPERPLANES AND PROJECTIVELY CONVEX SETS .
III. PROJECTIVE CONVEXITY IN TWODIMENSIONAL
TOPOLOGICAL LINEAR SPACES . . . .
IV. APPLICATIONS OF THE RESULTS FOR L2 TO
SPACES OF ARBITRARY DIMENSION . . . .
V. SOME PROPERTIES OF PROJECTIVE CONVEXITY
IN EUCLIDEAN SPACE . . . . . .
BIBLIOGRAPHY . . . . . . . . . .
BIOGRAPHICAL SKETCH . . . . . . . .
Chapter I
ELEMENTARY ALGEBRAIC
AND TOPOLOGICAL CONSIDERATIONS
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+(lX)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+(lAl)y, O
v = X2x+(lX2)Y, A2<0 or X2>l.
If we solve for x and y in terms of u and v, we have that
x = 12 u+ l1 v,
y = U2 u+ u I+ V.
Setting 81 = 1X and 82 = "X we have that
Xl%2 l 2
x = 61u+(161)v
y = 82u+(182v.
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 (nonnegative, nonpositive, 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+(1X)y) = h2Q(x)+(11)2Q(y)+X(1X)(B(x,y)+B(y,x)).
If (B(x,y)+B(y,x))_O, we have that Q(Xx+(lX)y)>O for
O
Q(Nx+(lX)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+(lN)z = X(x+sl)+(lN)(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
varieties.
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.
5
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+(xax+asl8x+8s2)
= x+s3
since (1a+a8+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 nonzero. We consider three cases.
Suppose that 0<8k+l<1. We know that
y = 81Y+...+8kkeS
81+...+8k
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+(lN)v = Xh(x)+(lX)h(y)
= h(Xx+(lX)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)+(1N)(yl,y2) = (x1l+(lX)yl,Xx2+(lN)y2) is an
element of SleS2 since Xk+(lA)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 nonnegative 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+(lX)v = a(a(x+s1)+8pl)+(l7)(a(x+s2)+8p2)
= a(x+(xsl+(lA)s2)+8( pl+(l1)p2).
Now x+(s l+(lX)s!eV for all real N. For O
or X>1 we have that ?pl+(lX)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 11 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 xyqS, a contradiction
to the assumption that S is p.c.
In a linear space it is wellknown 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
theorems.
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
MooreSmith Convergence, see Kelley [2], chapter 2.) Suppose
that this holds for internal segments. Let p be defined by
p = >x+(lX)y, O
Consider the net {pU;UeR} = {7xU+(lX)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[XA], 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
ScraU.
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
plication.
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
n
z = Xx ,
i=1 1 1
n
with i>_O and Z hi = 1. Let {ym be a sequence in the con
i=1
vex hull of S. We may write
n
Ym = Zimxi
i=1
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
bounded.
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+(l7)y, x y}.
Let { an = {l/n} and {\n} = {n}. We Imow that nx+(ln)yeL
for all positive integers n; thus {xn = {Xnx+(lNn)y} CL.
Clearly an is a null sequence. Now consider the sequence
{anxn} = {x+(l/n1 )y}. This sequence converges to xy 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
HYPERPLANES AND PROJECTIVELY CONVEX SETS
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
S.
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+(lX)y. We have that
f(p) = f(Az+(lA)y)
= Nf(z)+(lX)f(y)
= Xc+(lX) f(y)
< 7a+(lX)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+(lXn)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+(lX)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
f(z)f(y)
Az+(lN)y is a point such that
f(Nz+(lA)y) = Nf(z)+(lN)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+(1X)( ny+(1kn)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'+(lX)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'+(12X)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+(lX')z, where
= a!f(z)
1/2f(x)+l/2f(y)f(z)
Also, let r = A"y+(l?")z, where A" is defined by
n = CLf(z)
f(y)f(z)
Note that N' = X". The points s and r are, respectively,
pzAHI and yzAH1. Define the sequence {xn} by
xn = nr+(ln)s
and the sequence {yn by
y = Xn p+(1N )x ,
where Nn = ll/X'n. In a different form we have that
yn = [+(ln)(ln)X']p+[(ln)nh']y+[(1n)(lX')]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
PROJECTIVE CONVEXITY IN TWODIMENSIONAL
TOPOLOGICAL LINEAR SPACES
Our attention is now directed toward the study of pro
jective convexity in twodimensional 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 ndimenslonal topological linear spaces L and
in spaces of arbitrary dimension.
An open halfplane 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 halfplane.
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 2space, 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 noncollinear points of L2.
First, we will let al = p2P3, and a2 and a3 will be defined
analogously.
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 nonnegative. 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 nonnegative and the others be nonpositive.
(iii) The three regions Bi.
These regions have the ith ooefficient nonpositive and
the other two coefficients nonnegative.
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 halfplane 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 halfplane, will con
sist either of a single point, or a finite (open, closed, or
semiopen) 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 halfplane 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
[13
xz is contained in S. It is immediate that any convex set
has property P3. Also, the union of two convex sets has this
property.
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) A'AAi is convex. For example, let x,yeA'AA1; then
we have that
3
x = alP+l2p+a22+a3p, ai = 1, al>0, a2<0, a3<0,
i=1
3
y = 81P1+82P2+863P, Z = 1, 81>0, 82<0, 8 <0.
i=1
Any point on xy has the form 7x+(lA)y, OA<
qexy, we have that
q = (alPl22P+a2+3p )+()(81p +82P2+83p3)
= (Xal+(l)81)pl+(haX+(l)82 )p2+(Xa +(lX)8 )p3
= TiPni+r2P2+nTp ,
where
3 3 3
SwTi = Z \ai + E(1N)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+(lX)y = X Z api, + (lh) E 8ipi
i=l i=l
3
= 2 [Xai+(lX)Si3pi,
i=1
with X defined by
= 83
a383
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 xyxp. Thus, this set is p.c. and so is
our original set S.
25
(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 halfplane 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
APPLICATIONS OF THE RESULTS FOR L2 TO
SPACES OF ARBITRARY DIMENSION
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+(lX)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 twodimensional 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 twodimensional 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.
30
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
SOME PROPERTIES OF PROJECTIVE CONVEXITY
IN EUCLIDEAN SPACE
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 ndimensional 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 ndimensional 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
pqCA.
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 = 2x1y. 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 En1,
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 AAHn1. 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'
n+1
where E 8 = 1 and 0<86<1, i = 1,2,...,n+l. Now let us con
i=1
sider the region R defined by
n+l
R = {xjx = lX1+a 2x2+a 3x+.. .+'nx+'n+lz' ai = 1,
1=1
n
I ai 1, 0<.2, OISialSla^, i = 2,...,n}.
i=l
If yWA'AR, we have the following considerations. Let pl be
the point hy+(lX)p, where X is defined by
S= n+1
8n+l"n+1
Since an+lO, we have that 8n+lan+l>O. Hence 0
pleyp. If we write out the expression for pl, we have that
p = 83alSn+lCLn+ 81 x" +.' +nn+l 8nx+an+ln+la n+ +z
1 n+lan+1 n+lan+l n+ln+1
Each coefficient is nonnegative, making pleS(xl,...,x1).
That is, ypAS(xl,...,xA) = {pll.
Now let p2 be defined by p2 = Ny+(lX)p, where X = 1
Since we know that
811 = (81a2al82)+(83a183 )+..+(8* 1n+l'ln+l)
we have A<0; thus, p2eyp. Substituting the value of X, we
have
P 81 aal82x +...+81_n+1a18ntlz'.
p2 1=1 1lt
Each coefficient is nonnegative 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 n1 other points of A.
Proof. There are n+l points x1,x2,...,x n+ in A which generate
En. In particular, we have that
n+l
p = 81x+2x2 +.+n+xn+, 8= 1.
i=1
Suppose that 81 0. Then
1
x = 8 l(P82x2 n+ln+l).
Now {p,x2,...,xn+l} generates En, so there is a hyperplane
H(p,x2,...,xn) through p and nl 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 Enl. 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.
BIBLIOGRAPHY
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, 441464.
4. Klee, V. L., Jr. "Convex Sets in Linear Spaces II," Duke
Mathematical Journal, Vol. XVIII, 1951, 875983.
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.
BIOGRAPHICAL SKETCH
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:
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