C OMPOSITIO M ATHEMATICA
R OBERT E. J AMISON
Some intersection and generation properties of convex sets
Compositio Mathematica, tome 35, n
o2 (1977), p. 147-161
<http://www.numdam.org/item?id=CM_1977__35_2_147_0>
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147
SOME INTERSECTION AND GENERATION PROPERTIES OF CONVEX SETS
Robert E. Jamison*
Introduction
In this paper it is shown that in a real vector space of at most countable dimension any convex set whose
complement
is also con-vex is the convex hull of a countable set. This result is valid over any ordered field whose order
completion
is first countable in the ordertopology.
In an abstractsetting
ofconvexity
it is also shown that the abovegeneration property implies
that anyfamily
of convex sets withempty
intersection has a countablesubfamily
with empty intersection.In a 1956 paper
[10,
p.59],
V. L. Kleeproved
thefollowing
intersection theorem for convex sets:
KLEE’S THEOREM: Let V be a real vector space
of
countabledimension.
If
F is afamily of
convex sets in V with empty in-tersection,
then some countablesubfamily of
JF has empty in- tersection.Later in a lecture series
[12]
he observed that theproof
in[10]
wasapplicable
to vector spaces over any ordered field with a countable dense subset. Oneobjective
of this paper is to extend Klee’s result to the class ofcompletely sequential
ordered fields: every nonvoid subset contains anincreasing
cofinal sequence. This is thelargest
class of fields for which Klee’s theorem is valid. It may not be a
priori
obvious that
completely sequential
is astrictly
weakerrequirement
than order
separable. However,
we shall show that the real-valued function fieldgenerated by
the functionsx ~ xr,
r a realnumber,
is* The author is grateful to the Department of Mathematical Sciences of Clemson University for their support during the preparation of this paper.
Noordhoff International Publishing
Printed in the Netherlands
completely sequential
but not orderseparable. (The
"eventual" order-ing
is used here:f - g
ifff (x) g(x)
for allsufficiently large x.)
Another aim of this paper is to prove thefollowing
relative ofKlee’s theorem:
(A)
THEOREM: Let V be a vector spaceof
countable dimensionover a
completely sequential
orderedfield. If
H is a convex set in Vwhose
complement
V -- H is also convex, then H is the convex hullof
a countable set.
In
fact,
as we shallshow,
this result isformally
stronger than the extension of Klee’s theorem. This will be demonstrated in thesetting
of
general convexity developed
in[6].
As Klee’s result can be
interpreted
as a countable version ofHelly’s
celebrated theorem[5]
on intersections of convex sets and asTheorem A is a countable
analogue
ofCarathéodory’s
celebratedresult on formation of convex hulls
[1],
the abstract connection between these two results can beregarded
as a contribution to theunderstanding
of theinterplay
betweenHelly
andCarathéodory
pro-perties
of abstract spaces. A somewhat old but still valuable survey of results in this direction is[2].
In the course of thepresent develop-
ment, we shall also prove ageneral Helly
type theorem for compact(in
a combinatorialsense)
families of convex sets. Other intersection theorems related to this and to Klee’s theorem above are discussed in[11].
The main result
In this
paper F
will denote atotally
ordered field.(For
a discussionof ordered fields and some
interesting examples,
see[4, Chapter 13].)
If K is a subset of a vector space V over
F,
then K is convex if andonly
if03BBx + (1 - 03BB) y ~ K
for every x and y in K and each scalar À with 0 03BB 1. If both K and its
complement
V --- K are convex, then K is called ahemispace.
(The
open and the closedhalfspaces
determinedby
linear functionalsare
hemispaces.
But there are manyothers.)
A
totally
ordered set T will be calledcompletely sequential
if every subset of T has a countableincreasing
cofinal subset and a countabledecreasing
cofinal subset. Let us note inpassing
that if T is com-pletely sequential,
then so is the ordercompletion
of T. Infact,
T iscompletely sequential
iff the ordercompletion
of T is first countable under the ordertopology.
For an ordered field to becompletely
sequential
it is sufncient that every bounded set ofpositive
numbershave a countable
increasing
cofinal subset.(It
issurely
necessary but notsufficient
that 0 be the limit of a sequence ofpositive elements.)
Note that if the ordered set T has a countable order dense
subset,
then T iscompletely sequential.
MAIN THEOREM: Let V be a vector space over the ordered
field
F.Then the
following
areequivalent :
(i)
The dimensionof
V over F is countable and F iscompletely sequential.
(ii) Every hemispace
in V is the convex hullof
a countable set.(iii) Every family of
convex sets in V with empty intersection contains a countablesubfamily
with empty intersection.We first show that
(i) implies (ii).
From thegeneral theory
de-veloped
next, it will follow that(ii) implies (iii).
As it isquite
easy tosee that
(iii) implies (i),
this willcomplete
theproof.
One
might
wish to note that the Main Theorem is a resultphrased
in terms of the
countably
infinite cardinal. Ananalogous
result is truefor any infinite cardinal. In
fact,
theproof presented
hererequires only slight,
butjudicious,
modification to beapplicable
to thegeneral
case.
However,
thepresentation
is restricted to the countable case forconceptual
ease and concreteness.Hemispaces
as convex hulls of countable setsWe
begin
with aproof
of Theorem A((i) implies (ii)
of the MainTheorem).
For this result it will be useful to know that if two linearfunctionals .0 and Ji
are bounded(either
from above orbelow)
on anonempty hemispace H, then 0
andtp
arelinearly dependent.
To seethis,
first note thatby multiplying by - 1
if needbe,
one may assumethat 0 and Q
are bounded belowby
numbers a and03B2, respectively.
If~ and Ji
arelinearly independent,
there is apoint p
in V such that~ (p ) = -1
and03C8(p )
= 1. Let x be anypoint
of H. Choose a scalar Àlarger
thanboth ~ (x ) - 03B1
andI/J(x) - (3;
then setv = x + 03BBp
andw =
je - Àp. Then ((v)
a andI/J(w) 03B2,
so v and w are not in H. Asx is the average of v
and w,
this contradicts the assumedconvexity
ofV --- H.
PROOF OF THEOREM A:
Proceeding by induction,
we shall first establish the result when V is finite dimensional. Let H be anynonempty hemispace
in V.(The
result is trivial if H =0.)
If dim V =1,
then V may be identified with F. Since F iscompletely sequential,
there is a countable subset C of H which is both cofinal
increasing
and cofinal
decreasing. Evidently,
H = conv(C).
Now suppose dim V = n > 1. Since the linear functionals bounded on H are all
linearly dependent,
there is a linearfunctional ~
that isnot bounded on H. Thus for each À in
F,
the setH03BB = lx e H: 0(x) =,kl
isnonempty.
Since F iscompletely sequential,
there is a countable subset I of F that is both cofinal
increasing
andcofinal
decreasing.
For any subset S of F with I CS,
letOur first
goal
is to show that H =C(S)
for some countable subset S of F.To this
end,
for each subset Sof F,
letIf S is countable and I is included in
S,
thenR(S)
will be called aresidual subset of F. We shall now show in a series of
steps that 0
is residual.(i)
Each residual set is order convex in F.Check. For a
(3
y with a and y in the residual setR(S),
let x and zbe
points
in H -C(S)
with0(x)
= a and~(z)
= y. For somepoint
yon the segment from x to z, one must have
0(y) = 03B2.
Since thecomplement
of H is convex, either the ray from ythrough
x or theray from y
through
z mustbelong
to H.Suppose
the former obtains.Since ~(x ) ~(y)
and I contains adecreasing
cofinal subset ofF,
there is apoint p
on the ray from ythrough
x with~(p) =
À for someÀ in I. As I
C S,
thisimplies
that p EC(S).
If y were inC(S),
theconvexity
ofC(S)
would forceC(S)
to contain x. The choice of xprecludes
this. Thus y E H ---C(S), so (3
=~(y)
is inR(S).
The other alternative above is handledsimilarly. #
(ii)
The countable intersection of residual sets contains a residual set.Check. If
R(Sn)
n =1, 2,
... is a countable collection of residual sets, then S= ~ ~n=1 Sn
is countable and contains I sinceeach Sn
containsI. Thus
R(S)
isresidual,
and it iseasily
checked thatR(S)
ÇR ( Sn )
forall n.
yi
(iii)
If F is adecreasing
chain of order convex subsets ofF,
then there is a countablesubfamily W
of Fwith n %
=n
F.Check. For each set A in
F, define
and
Now let
and
Choose a cofinal
decreasing
sequence v1 > v2 > ... in U and a cofinalincreasing
sequence03BB1 À2:5 ...
in L. For each n, select setsAn
andBn
in F suchthat un
EU(An)
and03BBn
EL(Bn).
Let % be the(obviously countable)
collection of theAn’s
andBn’s
so chosen. Nowif 03B2
is anumber in F that is not
in ~ F,
then03B2 ~
C for some set C in 3i.Since C is order convex, either
j3
EU(C) or (3 E L(C).
In the first case,03B2
is inU, so 03C5n 03B2
for someinteger
n.Whence 03B2 belongs
toU(An),
so03B2
cannotbelong
toAn. Consequently 8
is notinnw.
The second case is handledsimilarly.
It followsthat ~ G = ~ F. yi
(iv)
There is a minimal residual set.Check. Assertions
(i), (ii)
and(iii) imply
that the intersection of anydecreasing
chain of residual sets contains a residual set. Zorn’s lemma thereforeapplies. #
(v)
A minimal residual set is empty.Check. If
R(S)
is a nonempty residual set, let 03BB be any number inR(S),
and take S’ = S U À. Then S’ still includes I and is countable.Obviously, 03BB~ R(S’),
butR(S’)
çR(S).
ThusR(S’)
is a residual setproperly
contained inR(S). yi
As a result of
(iv)
and(v) above,
theempty
set is residual. It follows that there is a countable subset S of F with I C S such thatC(S)
= H. Now for each À inF, H,
is ahemispace
in the(n - 1)-
dimensional flat
{x ~ V: ~(x)=03BB}.
The inductionhypothesis
thenguarantees that
H03BB
is the convex hull of some countableC03BB.
The setC =
U {C03BB : 03BB
eS}
is countable and for each À inS,
one obtainsH03BB
C conv(C).
Thus H =C(S)
= conv(C).
Thiscompletes
the in- duction and proves the theorem when V is finite dimensional.Now if the dimension of V is
countably infinite,
V can be ex-pressed
as the countable union of finite dimensionalsubspaces Ln.
IfH is a
hemispace
inV,
then H f1Ln
is ahemispace
inLn
withH ~
Ln
= conv(Cn). Letting
C =U :=1 Cn,
one caneasily
see thatH = conv
(C).
Moreover as the countable union of countable sets, C is countable. DAn
example
We shall
give
here anexample
of atotally
ordered field which iscompletely sequential
but does not contain a countable order dense subset. Thisexample
is based on thehyperreal
function fields in-vestigated by
L. Gilman and M. Jerison[4,
pp.171-189].
Theirterminology
isemployed
below. Let R denote the field of real numbers and set X ={r
~ R : r >1}.
Choose a free maximal ideal M[4,
p.54]
in thering C(X)
of continuous real-valued functions on X.The
quotient ring C(X)IM
is a nonarchimedean ordered field K[4,
p.70].
InK,
let F be the subfieldgenerated by
thequotient
mapimages
of the functions x ~ xr where r is any real number. The
ring generated
in Fby
the functions x" can be identified with the"pseudo- polynomials"-functions
of the formp(x) = ~i=1n c(i)xr(i)
where n is apositive integer, c(i) E R,
and ther(i)
are distinct real numbers. Thelargest r(i)
such that thecorresponding c(i)
is nonzero, will be called thedegree
ofp(x)
and thecorresponding c(i)
theleading coefficient.
Any
element k of F can berepresented
in the form À =p(x)lq(x) where p and q
arepseudopolynomials
and theleading
coefflcient of q is 1. Thus 03BB ispositive only
when theleading
coefficient ofp-which
will also be called the
leading
coefficient of 03BB- ispositive.
We takethe
degree
of À toequal
thedegree
of p minus thedegree
of q.We shall now outline an
argument
that F iscompletely sequential
but possesses no countable dense subset. The latter is easy to see.
Note that for any real numbers r s, if p E F has
degree
r and o- e Fis
positive
ofdegree s,
then p u. Hence the open order intervals of the formare
mutually disjoint
and uncountable in number.To show that F is
completely sequential,
we shall assume theopposite
and derive anabsurdity. Suppose
that S is a bounded set ofnonnegative
elements of F with noincreasing
cofinal sequence. For each countable ordinal q, we shall construct an elementÀll
of S and areal
number dn
such that whenever yprecedes
q, we have(i) 03BBy An
and
(ii) d, - d,
with strictinequality
if y is a nonlimit ordinal.(For
a discussion ofordinals,
see[8,
p.266], [4,
p.74].)
Let Ào = 0 and take
do
to be any real number such thatdeg cr do
for all 03C3 in S.(Such
exists since(x r)
r E R is cofinal in F and S isbounded.)
Now suppose
03BBy
anddy
have been chosen for all ordinals y less than aparticular
countable ordinal q. If q has an immediatepredeces- sor 03B2,
letand consider the set D of
degrees
of elements in T.Define d."
= sup D.If
dn.
is not inD,
select a sequence(Tn)
from T such thatSuch a sequence is cofinal in
T,
whence(T + 03BB03B2)
is a cofinal sequencein S. As this contradicts the
assumption
onS,
there must be an element of T with maximumdegree.
Let C be the set of allleading
coefficients of elements of T which have
degree d",.
Set c = sup C. If cis not in C
(or
isinfinite),
select a sequence of elements in T of maximaldegree
whoseleading
coefficients have c as their supremum.This sequence is cofinal in
T, again
a contradiction. Hence c is finite and in T so there is someelement 1£
ofdegree dn,
with c asleading
coefficient.
Taking 03BBn
= IL +03BB03B2,
weclearly
get03BB~
>03BB03B2
asrequired
in condition(i).
Now note that ifthen also
From the choice
of u,
it follows that every element of T’ hasdegree strictly
less thand~.
Inductive condition(ii)
can be inferred from this.If q
is a(countable)
limitordinal,
the construction is easier. LetThe countable set
{03BB~ :
y~}
cannot be cofinal in Sby hypothesis
onS,
so there is a03BB~
in S such that03BB~ > 03BBy
for all ypreceding
q. Bothinductive conditions
(i)
and(ii)
areeasily
verified in this case.Now for each countable ordinal q, we have an open interval of real numbers:
(Here
q + 1 denotes the successor ofq.) By
inductiveproperty (ii),
these are
nonempty
anddisjoint.
But this isimpossible
since there areuncountably
many countable ordinals and every collection ofdisjoint
open sets in R is countable. Hence F must be
completely sequential.
Two
Helly-type
theoremsIn this section we shall establish the second
implication
in the MainTheorem. This will be done in the
general
context ofalignments,
asdeveloped
in[6]. Only
a brief review of the essentialsrequired
for thedevelopment
will begiven
here. For more details andexamples,
see[6].
If X is any set, an
alignment
on X is afamily L
of subsets of X(which
will be called convex subsets ofX)
such that(Al) 0
and X are convex,(A2)
thearbitrary
intersection of convex sets is convex, and(A3)
the union of a chain of convex sets(that is,
afamily
ofconvex sets
totally
orderedby inclusion)
is convex.The
pair (X, L)
is analigned
space. Analignment Y
on X determinesa convex hull
operator
on subsets of XThis
operator
can be shown tosatisfy
the usefulfinitary
property:(*)
: E is a finite subset ofS}.
A direct
proof
of this may begiven using
transfinite induction[6,
p.5].
An
alignment
on Xis,
of course, a subset of the power set Pow(X)
of X.
By identifying
each subset of X with its characteristicfunction,
Pow
(X)
may be identified with thecompact
Hausdorffproduct
ofdiscrete two
point
spacesfO, 11,
indexedby
X. Theresulting topology
on
Pow (X)
will be called the inclusion-exclusiontopology.
Familiesof the form
f A
C X : I C A and E rl A =~}
where I and E are ar-bitrary
finite subsets ofX,
form a base of open(and closed)
sets forthis
topology.
Thefinitary property implies
that anyalignment Y
on Xis
actually
a closed subset of P ow(X)
under thistopology.
For if S isa subset of X that is not convex, there is a
point
p inL(S)
that is notin S.
By (*),
there is a finite subset I of S with p inL(I).
Thefamily f A
C X : I C A andp e A}
is then an openfamily
in Pow(X)
contain-ing
S butdisjoint
from 5£.WARNING: The inclusion-exclusion
topology
on Pow(X)
in no way reflects anytopological
or other structure X may possess. It ispurely
combinatorial in nature,
depending only
on thecardinality
of X.A set H in an
aligned
space(X, L)
is ahemispace provided
H andits
complement
X - H are both convex. Since thealignment Y
isclosed in
Pow (X)
and thecomplementation
map A-X - A is ahomeomorphism
ofPow (X)
ontoitself,
thefamily
5£’ of sets whosecomplements
are convex is also closed. Thus thefamily
of he-mispaces
inX,
as the intersection of the closed families 5£ and5£’,
is closed.Hemispaces provide
a natural notion to be used indescribing separation properties
of analigned
space. Werequire only
one suchproperty
here(see [6,
p.24]).
SEPARATION Axiom
S3:
Given any convex set K andpoint pe K,
there is ahemispace
H such that K C Hand p ~
H.The
ordinary alignment
of convex sets in a vector space over atotally
ordered field satisfies this axiom. To seethis,
take H to be any maximal convex setcontaining
K butmissing
p. It is not hard to check that such an H must have a convexcomplement.
An
alignment Y
is said to haveHelly
number hprovided
(t)
for any finitefamily 3i
of L-convex sets withempty intersection,
some
subfamily
of at most h sets from 3i hasempty
intersection.This is a
compactness type property.
Infact,
we shallshortly
prove that the "finite" above can bereplaced by "compact"
in the sense ofthe inclusion-exclusion
topology
on Pow(X). Helly’s
theorem[5]
asserts that n + 1 is the
Helly
number for theordinary
convex sets inan n-dimensional real vector space. A standard
proof
ofthis,
such asgiven
in[3,
p.33],
iseasily adapted
for vector spaces overarbitrary
ordered fields. An
aligned
space(X, L)
hasu-finite Helly
type if X is the union of a countablefamily
of subsetsYn
suchthat,
for each n,the relative
alignment
has a finite
Helly
number. Thus any vector space of countable dimension over an ordered field has 03C3-finiteHelly type.
The next
Proposition
will be used in theproof
that(ii) implies (iii)
from the Main Theorem. It also is of
independent
interest.(B)
PROPOSITION: Let(XY)
be analigned
space withHelly
number h.Suppose
Je is afamily of
L-convex sets that iscompact
in the inclusion-exclusiontopology
on Pow(X). If
the sets in K haveempty
intersection,
then somesubfamily of
at most h setsfrom
K has empty intersection.PROOF: First we show
by
a Zorn’s Lemmaargument
that among the closed subfamilies of K which haveempty intersection,
there is aminimal one. For any
decreasing
chain(Ky)
of closed subfamilies ofJe,
each withempty intersection,
letAs an intersection of closed families in Pow
(X),
.Ji is closed. Now consider anypoint
p in X. For each y, thefamily
is
nonempty
since the sets inKy
have void intersection and hencesome set in
Ky
excludes p.Furthermore,
asZ,(p)
isjust
the in-tersection of
Ky
with the basic closedfamily {S ~ X :p ~ S},
eachX-y(p)
is closed and the collection(Ky(p )
is adescending
chain. SincePow (X)
iscompact,
this chain must have a nonvoid intersection.Any
set in this intersectionbelongs
to .;U but misses p. Thus the intersection of the sets in « isempty.
One can now conclude from Zorn’s lemma that there is a minimal closedsubfamily
of 3t withempty
intersection.To prove the
Proposition
it now suffices to show that a minimalclosed
family
of 0-convex sets whose intersection isempty
must be finite.Proceeding by
reductio adabsurdum,
assume that « is such afamily
but that « containsinfinitely
many sets.Being
bothcompact
andinfinite,
JK must contain a set A which is an accumulationpoint
of the other sets in «
(with respect
to the inclusion-exclusiontopology). Working
toward acontradiction,
weinductively
chooseh + 1
points
from A as follows:Let xo be any
point
in A.Having
chosen xo, xi, ..., Xn fromA,
consider the closed f amilies
defined for each i from 0 to n. Now define
Clearly 16
is a closedsubfamily
of .;U. We argue that it isproperly
included in .;U. In Pow
(X),
is an open
family containing
A but no set from anyAli,
ThusCf1 ôm =
{A}.
But since A is not isolated relative toAl,
there must beinfinitely
many sets in « fl ô/1. Thus C is not all of Al.Therefore,
thesets in % must have
nonempty intersection, by
theminimality
of Al.Let xn+1 be any
point
in this intersection. Then x,,,, e A since A E C.Furthermore,
this choice also forces xn+, tobelong
to every set inAli
for each i n + 1.
With xo, xi, ..., xh selected as
above,
define for each i from 0 to han i%-convex set:
Clearly any h
of these h + 1 5£-convex sets havenonempty
in- tersection.However,
we shall show that all h + 1 of these sets haveempty intersection, contradicting
thehypothesis
that h is aHelly
number for 5£.
Let I =
L0 ~ L1 ~ . . .
nLh.
If M is inM,
we aver thatLi
C M forsome i and hence that I C M. In
f act,
if M contains all thepoints
xi, then M contains all the setsLi
and hence I. If M misses some x;,let j
be the least index such that
xj ~ M.
Then Mbelongs
to.Âlj.
Asobserved in the construction of the
xi’s,
for eachi > j,
thepoint
xi must lie in each set in.Âlj. Thus xi
E M when i >j. However,
for ij,
we have xi E M
by
the minimal choice ofj.
Hence M contains all of thepoints xi except for xi
and hence includes thecorresponding
convex hull
Li. Consequently
I C M for each M in .Âl. Thus I is contained in the intersection of all the sets inM,
which was assumedto be
empty.
This leads to the contradiction stated above andthereby completes
theproof.
DOne
might
wonderif,
for anyalignment,
a closedfamily
of convexsets with
empty
intersection mustalways
contain a finitesubfamily
with empty intersection. Were this so, the aboveProposition
could bededuced as an easy
corollary. However,
as shown in theexamples below,
this need nothappen
in the absence of a finiteHelly
number.(C)
EXAMPLE: Let X be an infinite set with thefree alignment:
allsubsets of X are convex. Let F be the
family
of all subsets of the form X - p where p is anypoint
in X.Any
net(not eventually constant)
of sets in F converges to X. Thus thefamily F
U{X}
isclosed and has
empty intersection,
but no finitesubfamily
hasempty
intersection.(D)
EXAMPLE: Thepreceding example
can be embedded in a vector space. Let V be an infinite dimensional real vector space and let X bean infinite
linearly independent
subset of V. Foreach p
inX,
letCp
= conv(X - p)
and set C = conv(X).
Thefamily
3t of all theseconvex sets has empty
intersection,
but any finite numberalways
have a
point
in common.Furthermore,
thefamily 3t
is closed in Pow(V). (For
any convergent net from3t,
look at thecorresponding
net of extreme
points.
Note that ext(Cp )
= X - - p, and that a limit ofconvex sets must be
convex.)
In
[9]
V. L. Klee gave a result for open convex sets in Euclidean space which is similar toProposition
B. His notion of compactness,however,
seems todepend
on the Hausdorff metric. In an earlier paper[7]
Karlin andShapley proved
a related result forhemispheres
in
spheres. (Their
result isphrased
in terms ofcoverings.)
Both ofthese results can be deduced from
Proposition
B. We shall not do sohere since the
original proofs
arequite
easy and it would lead us too far astray. For a survey of other relatedresults,
see[11].
(E)
THEOREM:Suppose
that(X, L)
is analigned
spaceof u-finite Helly
type, and(X, Y) satisfies separation
axiomS3. If
every he-mispace
in X is the convex hullof
a countable setof points,
then anyfamily of
convex sets with empty intersection has a countable sub-family
with empty intersection.PROOF: Let 3i be a
family
of convex subsets of X with empty intersection. For eachpoint
x ofX,
letCx
be a set in 3i with xE5Cx.
By
axiomS3
ahemispace Hx
may be chosen such thatCx
CHx
andxe Hr
Let W denote the collection ofhemispaces
so chosen as xranges over X. To prove the theorem it suffices to show that some
countable
subfamily
of W has void intersection. To thisend,
define 3t to be the closure of W inP ow (X)
with respect to the inclusion-exclusion
topology.
Note that since W is included in thefamily
of allhemispaces and,
as observedearlier,
thisfamily
isclosed,
it follows that each set in 3t is ahemispace.
Now the collection llt of
hemispaces,
as a closed subset of Pow(X),
is acompact
Hausdorff space under the inclusion-exclusiontopology.
It is also first countable. To seethis,
consider any he-mispace
H.By hypothesis,
there are countable sets S =(s1,
s2,* *1
and T =
{t1, t2, ...}
such that H =Y(S)
and X - H =2(T).
For eachj,
thefamily
is a compact
neighborhood
of H relative to llt. If H’ is any set in the intersection of this sequence of compactfamilies,
then H’ is ahemispace including
S butmissing
T. Thus H’ ~.2(S)
= H since H’is convex, and
X ~ H’ ~ L( T ) = X ~ H,
since X - H’ is convex.Consequently H’ = H,
so H is theonly point
in the intersection of thedecreasing
sequence ofcompact neighborhoods Nj
It follows that llt is first countable. The information ofimportance
to begleaned
from this is that each set in 3t is the limit of a sequence of sets in G.Now write X as a countable union of subsets
Yn,
where for each nthe relative
alignment .21 Yn
has a finiteHelly
number. Foreach n,
is a
family
ofrelatively
convex sets inYn
withempty
intersection.Since intersection is continuous on Pow
(X)
under the inclusion- exclusiontopology [6,
p.43], K|Yn
is also closed inPow ( Yn ).
Sincethe
Helly
number ofcPl Yn
isfinite, Proposition
Bimplies
thatK|Yn
contains a finite
subfamily
with void intersection. Let the sets in this finitesubfamily
ofK| Yn
be the restrictions toYn
of the sets in a finitesubfamily Kn
of K. If Abelongs
toKn
for some n, we may select asequence of sets
(Ai )
from W which converge to A. Let 16 be the collection of all sets in W which occur in some such chosen sequence.As a countable union of
sequences, (6
is countable.We aver that the intersection of the sets in 16 is
empty.
For if x isany
point
inX,
then x lies in someYn.
Hence there is a set A inKn
with
x ~
A. Since the chosen sequence(Ai )
converges toA,
for somei,
x must be excluded fromAi.
Thus x cannotbelong
to the in-tersection of all the sets in C. Hence (6 is a countable
subfamily
of Gwith
empty
intersection. 0As is inherent in the Main
Theorem,
the converse of Theorem E is valid in the case the space X is a vector space over an ordered fieldand Y is the
ordinary alignment.
Ingeneral, however,
the converse is not true. Forexample,
take the space to be the closed unit disc D in the Euclideanplane
with thealignment
ofordinary
convex sets. ThenD is
certainly
ahemispace
initself,
but D is not the convex hull of a countable set since D hasuncountably
manyextreme-points.
However,
since Klee’s theoremapplies
to anyfamily
of convex setsin the
plane,
itcertainly applies
to the convex sets contained in D.Consequently,
in thegeneral setting,
thegenerating property
con- cluded in Theorem A isstrictly stronger
than the intersection pro-perty expressed
in Klee’s theorem.Conclusion of the
proof
of the main theoremFrom Theorem A it follows that
(i) implies (ii)
in the MainTheorem. That
(ii) implies (iii)
is aspecial
case of Theorem E. Now in any vector space of uncountabledimension,
one can select per definitionem an uncountablelinearly independent
set X. One sees atonce that the
family
of convex sets associated with this X as inExample
D hasempty
intersection but every countablesubfamily
hasnonvoid intersection. Hence if
(iii) obtains,
the vector space V must be of countable dimension.Furthermore,
if(iii)
obtains inV,
then(iii)
obtains in any linear
subspace
of V. Since a one-dimensional sub- space can be identified with theunderlying
fieldF,
it follows that(iii)
obtains in F. From this it follows
easily
that F iscompletely
sequen- tial. Thiscompletes
theproof
of the Main Theorem.Note: The referee has
pointed
out that anupdated
Russian translation of[2]
is available:Danzer, Grünbaum, Klee,
andZalgaller, Helly’s
theorem and its
applications (Russian).
"Mir"Publishers, Moscow,
1968.REFERENCES
[1] C. CARATHÉODORY: Über den Variabilitätsbereich der Fourierschen Konstanten
von positiven harmonischen Funktionen. Rend. Circ. Mat. Palermo 32 (1911) 193-217.
[2] L. W. DANZER, B. GRÜNBAUM and V. L. KLEE: Helly’s theorem and its relatives. AMS Proc. Symp. CONVEXITY VII (1963) 101-180.
[3] H. G. EGGLESTON: Convexity Cambridge, Univ. Press, 1966.
[4] L. GILLMAN and M. JERISON: Rings of Continuous Functions. Princeton, Van Nostrand, 1960.
[5] E. HELLY: Über Mengen konvexen Körper mit gemeinschaftlichen Punkten. Jber.
Deutsch. Math. Verein. 32 (1923) 175-176.
[6] R. E. JAMISON: A General Theory of Convexity. Dissertation, Univ. of Washing-
ton, 1974.
[7] S. KARLIN and L. S. SHAPLEY: Some applications of a theorem on convex functions. Annals of Math. 52 (1950) 148-153.
[8] J. L. KELLEY: Topology. Princeton, Van Nostrand, 1955.
[9] V. L. KLEE: Critical set of a convex body. Amer. J. Math. 75 (1953) 178-188.
[10] V. L. KLEE: Structure of Semispaces. Math. Scand. (1956) 54-64.
[11] V. L. KLEE: Infinite-dimensional intersection theorems. AMS Proc. Symp. CON- VEXITY VII (1963) 349-360.
[12] V. L. KLEE: Mimeographed lecture notes, Univ. of Washington, 1973.
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