0047-2468/86/020105-1051.50+0.20/0 9 Birkh~user Verlag, Basel

Journal of Geometry Voi.26 (1986)

A KRASNOSEL'SKII

Marilyn

THEOREM FOR NONCLOSED

SETS IN

Rd

Breen

This w o r k will be concerned w i t h a K r a s n o s e l ' s k i i theorem for nonclosed bounded sets in R d , and the following theorem will be obtained: For each d ~ 2 , define f(d) = d 2 - 2d+3 if d # 3 and f(d) = 2d + 1 if d = 3 . Let S be a nonempty bounded set in R d , d ~ 2 , and assume that cl S ~ S is a finite union of convex components, each having closure a polytope. If every f(d) points of S see via S a common point, then there is a point p in cl S such that Bp ~ {s:s in S and (p,s]~ S} is nowhere dense in S.

i.

INTRODUCTION

We begin w i t h some definitions points

x

and

y

corresponding

in

sees

x

kernel

via of

[x,y]

is possible

Without

by examples

be a n o n e m p t y

in

S

bounded

of locally

and

throughout

in

S

be a subset

via

Set

S

S

of

Rd .

For

if and only if the

is said to be starshaped

such that, p

for every

is called

(p,s] ! S} the planar

in [3].

relative

respectively,

R2

the

components.

S

x

in

if

S,

p

(convex)

S .

dense in

Rd .

way:

The purpose

that

Let

S

in

S

see

B m {s : s

of this w o r k is

terminology

will be used

cl S , int S , rel int S , bdry S ,

the convex hull,

boundary,

relative

affine hull,

Boundary,

S

of a finite

in cl S , the set

The following

Conv S , aff S,

S .

consists

set

see via

no such theorem

If every 3 or fewer points

is nowhere

will denote

S

[2] it was shown

in the following

complement

a compact

of

be compact,

p

to

S

points

in [i] and

starshaped whose

that for

d+l

then for some point

interior,

for set

that

However,

to being

result

the paper.

[4] states

the requirement

compact

rel bdry S , and ker S interior,

p

y

if and only if every

set in

a common point,

to extend

S .

theorem by Krasnosel'skii

such sets may come close

S

sees

S

The set of all such points

is starshaped

a common point.

number

Let

S .

Rd , S

via

x

lies in

is some point

S .

A well-known in

S , we say

segment

and only if there

from [i].

closure,

and kernel,

106

Breen

If

S

is convex,

y , dist(x,y) L(x,y)

S

will be the dimension

will denote

will represent

referred

2.

dim

the distance

from

the line determined

to Valentine

of

x

by

S .

to x

For points

y , and when

and

y .

[6] and to Lay [5] for a discussion

x

~nd

x # y ,

The reader

is

of these concepts.

THE RESULTS

DEFINITION. points

Let

S

x I, .... x k

in

common point of that

xi

T

Let

If every

j(n+l)

common point

in

convex

by induction

that every in

6.18]).

n+l

T , where

points

T

for

assume k+l

t ~ T

We say

S

sets a

S 1 .... ,S k

in

(depending

on

~

such

xi

and

. compact

sets in

R d ~ j ~ i.

see via their respective

T , dim T ~ n , then

S

sets a

ker S I N oo. N k e r

S

convex

Let

every

T

For convenience

k(n+l)

the disjoint distinct

compact

sets

open halfspaces

Sk+ 1 N T ~ H+ and to see via

by our induction

in

T' in

T' c H

fails

(n-l)-flat and ker

hypothesis,

ker

S

k

n .

Then

(See [6, Theorem

or fewer

convex

sets,

of compact

, to

sets such that

see via

S

sets

at most

their

hypothesis,

T~

ker Sk+ 1 n T # ~ .

to meet

T'

to reach a contradiction.

loss of generality), H

1

set of dimension

S I U ... U S k

~n

aff

T

Sk+ 1 n T ~ let

determined

If every

their respective

at most

a common

see via their respective

in

Similarly,

aff T

j = ! and S1

T , so by our induction

(and without

Choose an

see via

theorem.

be a family Sk+ 1

points

ker Sk+ 1 n T

of notation

dim T = n .

is true for

is a closed

ker S I n ... n ker S k n T # ~ 9 that

S1

Let

j = 1 9

S I U ..~ U

sets a common point

We will assume

set

set of dimension

SI,.~O,Sk+ 1 in

of sets.

of Krasnosel'skii's

is true for

T , where

Then certainly

respective

of the compact

is closed

points

in

j , the number

that the result

sets.

(k+l)(n+l)

a common point

were

U Sj

set

on

by a version

Thus the result

Inductively,

that

a

of nonempty

SIU ...

in some closed

(ker S I) n T # ~

n .

be a family

points

T c Rd

in S} see via their respective

[xi,t] ~ S i , 1 ~ i ~ k

S 1 ..... Sj

We proceed

assume

every

and let

Tr

PROOF.

prove

: S

Rd

if and only if for every choice of

with

LEMMA i.

point

U{S

of sets in

E S i , 1 ~ i ~ k , there corresponds

S i , i ~ i ~ k)

S.n ]

be a family

k(n+l)

by

Separating

H+

~U

and

H , labeled

points

of

sets a common point

S IN:,.. N k e r

we may assume

strictly

so Lhat

S1 U ~ in

denote ker

U Sk

T n e! H+

S k n (T n cl H+) r ~ ,

, then

Breen

107

impossible since points of

ker S I N ... N ker S k N

S I U ... U S k

common point in w h i c h fail

Similarly,

Sk+ I

(k+l)(n+l)

fail to see via their respective

contradiction,

However,

COROLLARY. (n+l) 2

Let

2

S

n+l

points in

By induction,

U{S : S

in

2~

, every

in

2

S

(n+l) 2

THEOREM i.

in

Hence every

2} ~ (n{ker S : S in

For each

f(d) = 2d + i

if

and assume that

d ~ 2 , define

d = 3 . cl S ~ S

closure a polytope.

Let

n+l

p

S

S

in

T .

~.

S I U ... U Sn+ I ker

of the compact convex sets in By Helly's theorem,

2})N T r ~ .

f(d) = d 2 - 2d + 3

f(d)

If every

By Lemma i,

if

d # 3

be a nonempty bounded set in

points of

el S

j ~ i.

sets a common

S

such that

see via

B

and

R d , d ~ 2,

is a finite union of convex components,

If every

then there is a point

Rd .

points o f

{ker S N T : S in 2} have a nonempty intersection. N{ker S n T : S

for all

(N{ker S : S in 2 } ) n T #

sets a common point in

S I n o.. N ker Sn+ I N T # ~ .

U

We have a

see via their respective

PROOF.

S I .... ,Sn+ I

which

(T N cl H+)

the lemma is established

T, dim T ~ n, then

For

Sk+ I

ker Sk+ I N T N T' ~ ker S I N ...

point in some closed convex set

see via their respective

sets a

S I U ... U Sk+ I

be any family of nonempty compact sets in

points of

k(n+l)

Combining these

sets a common point of

is false, and

S

points of

T n cl H

this violates our hypothesis.

our a s s u m p t i o n

N ker Sk+ I n T # ~ .

there are

a common point in

points into one set, we obtain

(T N cl H_) = T.

Thus there are

w h i c h fail to see via their respective

T N cl H+ o

to see via

T ~ T' ! H_

each having

S a common point,

~ {s : s in S and

(p,s]

P S}

is nowhere dense in

PROOF.

S .

To begin, we dispose of the case in w h i c h

Observe that by [i, Lemma 2], cl

S

d = 2

and

of

cl S ~ S

are either isolated points or segments.

and

[2] hold~ and the result fellows from [2, Theorem I].

For the remainder

f(d) = 3 .

is starshaped and hence the components

of the argument assume

d > 3 .

Thus the lemmas in [i]

For convenience,

the proof

is divided into three parts. PART i.

We define families of convex sets

consider the collection of all points true:

There is some n e i g h b o r h o o d m ~ d-l, w h e r e

in

of N

x

m-flat

P

Let

be the collection of all the flats

P

for some

x

N

P,

@ , S

and

C .

for w h i c h the following

such that

N n s

is selected so that P .

First, is

lies in an m

is minimal.

108

Breen

Next, for each

P

in

cl(S N P) ~ (S N P)

P , let

O

denote the family of components A of P which are a full ((dim P)-l)-dimensional and which

satisfy the following property: sequences

{d n} , {d~}

halfspaces of e

in

in

S N P

determined by

a

in

rel int A , there exist

converging to aff A ~

Let

a

and in opposite open

@ = {A : A

in

0p

for some

P} .

Finally, a full c

P

For some

let

C

be the collection of components

(d-l)-dimensional

in

rel int

(N{P : P

that

N{P : P

in in

P}) N (n{aff P} # @ .

P # @ . Select a subfamily

B

If P'

Rd

a minimal collection

there are at most

{bn}, {b~} in

determined by

:

B

of

PI,...,Ps

P

for which N{P d'k .

from

P

which are

of the sets

PI,...,Ps

For some

converging to

aff C .

: P

Clearly

such that

Since

We assert that for each

S

in 0UC}) # @ .

PI .... 'Ps ~ P ' ~ k

cl S ~ S

c

We will Show

We begin by showing

P = @ , the result is immediate,

and has smallest possible dimension

We may assume that

of

and which satisfy the following property:

C , there exist sequences

and in opposite open halfspaces of that

C

in

so we assume

~}

is nonempty

0 ~ d-k ~ d-i

Choose

dimn{P i : i ~ i ~ s} = d-ko

dim Pi ~ d-I , so

for

i ~ i ~ s,

I ~ s ~ k ~ d o

in P P0 contains K ~ n {Pi : i ~ i ~ s} . o Of course, there are only three possibilities for flats PO and K : Either P0

meets

K

P0

contains

P

in a nonempty set of dimension less than K .

P0 n K = ~ ~ or

The first of these cannot occur by our choice of

see that the second cannot occur, proceed as follows: {P0,PI,...,Ps } , select a point of

xi

xi

in

such that

S n P.1

neighborhood

Ni

Xo,Xl,...,Xs

yield a collection of at most

By our hypothesis,

d-k ,

they see via

S

Ni N S !Pi

For each

P. i

P~ .

To

in

and an associated ' 0 ~ i ~ s .

s+l ~ d+l ~ f(d)

a common point

The point s points in

z , and clearly

S o

z ~ O{Pi:

0 ~ i ~ s} ~ P

N K . Hence p n K # ~ , and the second possibilit~ cannot p o The only remaining possibility is that K ~ P0 for each P0 in ~ ,

occur.

and the assertion is established. ... N p

= N{p :

Moreover,

it is easy to see that

K ~ PI N

p in P} # ~.

s

Next we will show that that

OUC

that

P # ~ , and let

Q{aff B : B in

P1 .... 'Ps

and

associated points selected previously. empty) of

OU~}

Meets

# @ , for otherwise the result is trivial.

OUC

for which

Xl,...,x s

N{P : P

in

P}.

be the members of

Choose a subfamily

K N (N{aff B : B

in

B'})

Assume

For the moment, assume

B'

~

and

(possibly

is nonempty and has

Breen

109

smallest

possible

d-j < d-k that

dimension

, choose

d-j ~ d-k

a minimal

.

Define

co]lection

Define

J = K

1 4 s+r ~ j ~ d We assert soning,

that

{d~n}

and

in

S N P0 by

in

rel int C O

to

cO

r 4 j-k

for each

B0

a0

0UT

to

select

associated

sequences

above,

accordin~

to whether

a0

B0

sequences

{din} B.z

{don}

of 9

and E0

If

Rd

opposite don

sees via

S

ds+l, n , ds+l,n'

If

B.I EC

we have near

in

to point

said that

J .

for

near

J .

of

P0

For each

' ds+r,n}

If

same contradiction, The case for

r

and

d+l

S q

in q

points.

J ~ aff B 0 .

point

cl S .

P0 cO

converging aff C O .

that

J N aff B 0 =

J !P0

' J

must

by

aff A 0

For an appro-

n , {x I ..... Xs} U {d

qn

points

s+l + 2r = S .

By

sees via

our assumption

is false,

of

: 1 ~ i ~ s} =

N aff Bs+ r = J 9

lars

changes

of

{qn} _c N{P.I

don

B 0 = C O , minor

S and

in the argument

However, no point J N aff A 0 =

above yield

the

J N aff B 0 # fl .

is a simplified

The details

of at most

,

on

, and some subsequence

Clearly

( K N aff Bs+ 1 N " "

sufficiently

and again

d-j = d-k

S

by

of

Select

determined

is a collection

a common

We have a contradiction 9

J Q aff B 0 r fl .

s+l ~

n

in

,

{d' } sequence {d } belongs to the on ~ on (aff A0) - . Observe that for n sufficiently large,

'''''ds+r,n

p } z K ~ and

{don}

B0 = CO ~

both are flats) 9

(j-k) + 1 = 2j - k+l < 2d+l ~ f(d)

converges

B0 =

open halfspaces

Assume

see via

: P

~0, let

sequences

B 0 = A 0 , then since

A0

rea-

B. ~ s+l ~ f < s+r I in the manner described

(s+r) + r+l < j +

{qn }

B0

, {d~n}

hypothesis,

N{P

case we have

and

no point

these

J = K

Using previous

determined

{din}

or

(since

}

define

For each set

(aff A0) +

from

on open halfspace

such

Bs+l,...,Bs+ r

in either

(C 9 let

lie in one of the open halfspaces

{d

Then

If

OUC

d-j = d-k,

and in opposite if

and a positive

of

that

with associated

d-j < d-k

r 9 to reach a contradiction.

distance

from

, .7 ~ aff B 0 .

open halfspaces

the case for

labelin~

If

J N aff B 0 # ~ o

Similarly,

associated

and in opposite

priate

Assume

r = 0 ,

rel int A 0

converging

with

as follows:

. in

in

aff A 0 ,

consider

let

to show that

Select

determined

First

of notation,

it suffices

A0 (O P0

,

N aff g s + I N ... N aff Bs+ r .

and for convenience

J

Bs+ 1 .... ,Bs+ r

d i m ( K N aff B s + i N --o N aff Bs+ r) = d-j

B'.

set

version

are omitted.

The assertion

of the argument

We conclude

is established.

that

Moreover,

above,

using

J N aff B 0 #

ii0

Breen

J = K n {aff

: B

in

0UC

} =

(n{P

: P

in

P }) .q (N{aff

B : B

in

}) ~ ~ .

ouc If

B

P = ~ , then

n{aff

C : C

(n{aff

O = ~ , and an argument

in

B : B

C } r ~ ,

iu

PART

2.

seen

that

T # ~ ,

PIn

''' n P s

0 aff

chosen

Define

@UC set

Hence

}) ~ ~

T = For

(n{P

: P

in P }) n

convenience

Also

the one above

that

(N{P

shows

: P

in

that

P }) N

in general,

Bs+ I n ,.-n aff

previously.

like

we c o n c l u d e

assume

(n{B

of n o t a t i o n ,

Bs+r that

where

: B

the

points

in

when P

x.

@UC

P # ~ and

are

})

B

sets

chosen

,

just d-j

as they w e r e

{din}

earlier.

Recall

for

P.

that

this

by w r i t i n g

those

,

l

chosen

for

B i ' s+l ~< i ~< s + r ~

dim(P 1 n . . , n P ) = d-k

dim T =

0 ~< r ~< j-k

T = aff Bs+ 1 R . . . N a f f

,

G/hen

Bs+ r , w i t h

s = 0 .

Define will

I~

consider,

X(P) P =

CASE

to be the r

prove

define

n {ker

determined % - %(P)

= r

that

I.

=

Assume

cl W

If

P

that

P

contains

We will

show

that

every

respective

Let

.

Parts

{Vl,.,.,v%}

arbitrary

W.

of sets

If

S)

P

has

in

P

: P

in P

There

are

Hence

at least

one member,

} U {S} three

for

We

cases

convenience

two d i s t i n c t ~(P)

.

in

IV

at least

and d i m ( n { P

~ E (d-j+l) 2

cl W

sets

= 2(d-j+l)

W

with

v.

: W

( cl W.

I

I

members,

in I~ }) ~< d-2

points

of U {cl W

are

_cU{cl

two distaoct

: P

a common

of the a r g u m e n t

I

a collection 2) =

However,

we

members, If

point

: W

in the closed

numbered

in ~ } .

.

for

For each

, 1 ~< i ~< % ~

Hence

Thus

k ~> 2

in W } couvex

future

reference.

point

v.1 ' select

Observe

that

,

see set

T ,

an

either

W. = P n S for some 1 i in W. c o n v e r g i n g to l {Vln .... ,V%n}

to

d-j+l

}) # d-i

dim T = d-j

{(Pn

W } n T # r .

has e x a c t l y

in P

their

iu

by the number

: P

via

: W

as follows:

(d-j+l) 2 , %(P)

of sets

dim(n{P

i)

are

, 1 ~< s ~< k ~< d , I ~< s+r ~< j ~< d , k < j , and

P = r , we m o d i f y k=

' {d', In }

T =

are

I

i ~< i ~< s , and s e q u e n c e s

We have

assume

P. in P or W. = S . T h e r e is a s e q u e n c e {vin} I I " v. , 1 ~< i ~ % . For every n , the points 1 ' ~ } yield U {x I .... ,x s} U {ds+l, n , d s+l,n .... 'ds+r,n ' d ser,n

of at most %+s+2r

(d-j) 2 + 2(d-j)

%+s+2r

points

= (d-j+l) 2 +

+ ! + 2j-k =

(s+r)

in

S .

+ r ~< (d-j+l) 2 + j +

(j-k)

(d-j) 2 + 2d - k + 1 ~< (d-k) 2 + 2d - k + 1

Breen

The

Ill

function

1 < 0 o

g(k)

Thus

is as small

=

g

(d-k) 2 + 2d - k + 1

is d e c r e a s i n g

as possible.

That

2d - 1 = d 2 - 2d + 3 ~ f(d) 3)

Thus

we h a v e

pothesis, n{P

: P

q

in

they in

see v i a

For each then

S

is, w h e n

a common

q ( P1 n . . . n Ps

6 Pin

assumes

k = 2 .

f(d)

point

qn

a subsequence

i , 1 < i ~ % , we

Vin,q n

and

of at most

Moreover,

cl S , and

k

derivative

c S ~ W.

n aff

show

that

S , so

[Vin,q n ] c

Hence

by a simple

--

that

point

4) B y

= -2d + 2k value

g(k) ~

k

points "

of

in

Clearly {qn }

qn (

converges

Bs+I n -.. n a f f [Vin,q n] c W i :

Pi N

S .

S ~ W.

By our hyPI N... N P s to some

Bs+ r ~ T If

If

o

W i = P1. n

W. = S

1

1

convergence

point

S '

clearly '

argument,

[vi, q] c c l

W.

--

points

of

when

(d-2) 2 +

i

We c o n c l u d e common

.

its m a x i m u m We have

--

[Vin,q n]

g'(k)

.

a collection

P } .

in

has

Vl,...,v %

see v i a

their

respective

cl W

o

i

sets

a

T o

the c o r o l l a r y

to L e m m a

i, n { k e r

cl W

: W

in

W } n T r ~ , the d e s i r e d

result. CASE

2.

W~

{PIn

Assume

that

S , S}

and a s s o c i a t e d argument S . (j-k)

contains

i, Part

%+s+2r

Vl,...,v %

see via

Finally,

we m a y

by L e m m a

apply

their i, ker

cl

the d e s i r e d

CASE

that

Again

Assume

select

via W

cl S in

PART in

at most

+ 0 + 2r ~

a common

W } n T # r . 3.

Finally,

N } n T .

(PIn

This

choose

Then

(d-j+l)

Then

1 = s ~ k

points

,

v. i of

+ r ~ 2d - 2j + 2 + j + contains

a common

at most

that

f(d)

points

point

cl S R T ~ N { k e r

s = k = 0 , ~ = {S}

of

T

point

sequences

of

cl W

T . : W

%+s+2r

i, Part

, and by L e m m a

p

point

Case

in W

}

~ = d-j+l

3),

I, ker

~

i, Part time,

, so the c o l l e c t i o n

points

Vl,...,v %

see

cl S N T ~ n {ker

cl W

3.

in the n o n e m p t y p

, and

{v~ } as in Case In p o i n t s of S . This

+ 2j ~ 2d + 1 ~ f(d)

By Case

finishes

that

(s+r)

sets

S) n ker

points.

point

We assert

"

Select

result.

~ = ~ .

f(d)

P1

3) to c o n c l u d e

cl W

v. and a s s o c i a t e d 1 a c o l l e c t i o n of at most

= (d-j+l)

contains

i, Part

points

and o b t a i n

%+s+2r

+

, so the c o l l e c t i o n

Case

respective

n T r r , again 3.

one m e m b e r

% = 2(d-j+l)

= 2(d-j+l)

~ 2d + 2-1 = 2d + 1 ~ f(d) Thus

and

{v. } as in Case i, Part i). The s u b s e q u e n t In i) yields a c o l l e c t i o n of at most %+s+2r points

n o w we h a v e

elements.

i),

exactly

two m e m b e r s ,

sequences

in Case

However,

P

has

satisfies

intersection the

theorem.

N {cl W Let

: W

:

112

Breen

B = {s : s E S

and

(p,s] ~ S

S ~ cl B c S , cl(S ~

choose

x

in

S

the case for

cl B) ! c! S ~

to show that

p

N

N

is selected

E ker cl S , [p,x]

of

k = 1 , then

(p,x) cl(P

k

P E P

, and

the reverse

.

p

in

conclude

that

N Q S .

Hence

N N S

is m i n i m a l

of

Since

inclusion,

It suffices

to censi'der

k ~ 1 .

z .

(p,x]

in

c S .

P ~ S , and

6 aff{z}

= {z}

Similarly,

N n B = @

and

.

Since

could not lie in a segment

p

k-flat

p ~ N

.

P

Since

We p r o c e e d by in--

P , 1 ~ k ~ d .

some point

T ~ then

lies in a

and so that

p ( ker cl(P N S)

have to be an isolated point choice of

such that k

and

at

z

x

c c] S , L(p,x) c p , and

would m e e t ~ S n S)

To e s t a b l i s h

x ( cl(S ~ cl B)

so that

d u c t i o n on the dimension If

cl(S ~ cl B) = cl S o

x # p .

Select a n e i g h b o r h o o d where

to show that

(p,x]

~S

, then

p E ker el(P A S) in

P ~ S .

Hence

z

However,

would

(@

and

p = z ~ impossible~

cS

o

~ [p~x]

{z}

(p,x']

x E S~cl

If

for all

B ! cl(S ~ cl B)

x~

by our

in

We

N NP AS =

, the desired

result. Assume

that the result

There are two cases j+l < d

or

CASE i.

Suppose

is not

sion at most

j

to consider.

j+l = d .

contained

We will

is true for

k , i ~ k < j < d , to prove Each argument

Of course~ w h e n

is independent

j+l = d , then

that for every n e i g h b o r h o o d in a finite union of flats

N' in

and c o n t a i n i n g

F

p , such that

of flats, B n P c U{F

: F

I)

For component

E

j-!

, aff(p

U E) to

with

N ~ c N ~ N~N S

of

cl(P

is a flat

N S) ~

in

P

(P N S)

in

F } .

Members

of

F

w h i c h has d i m e n s i o n at most

of d i m e n s i o n

at most

j o

Assign

F

For component

which

R = Rd

each h a v i n g d i m e n s i o n at most

in one of the following ways;

2)

of w h e t h e r

j .

define a finite family

U E)

x

j+l

P , each of w h i c h has dimen-

will be d e t e r m i n e d

aff(p

of

for

E EOUC

E

of

, assign

cl(P N S) ~ aff E

(P N S)

to

F

9

whose

Recall

dimension

p ( aff E

is

j

and for

for every

E

in

0UC 3)

Fqr component

which

E

~@UC

E

of ci(P N S) ~

(P N S)

, p r o c e e d as follows:

Since

whose E

dimension

is a

is

j

j-polytope,

and for each facet

Breen

G

113

of

E

or a

determines

(j-l)-flat

We assert

that

a

in

P .

P E ker cl(p

(P N S)

,

If

above~

Since

Thus

E

and

E

~0UC

If

c aff(p U E)

~OUC

, then

p E aff E ~ then

of

E , aff(p

c

U{F

: F

If

and

E

{F

L(p,s)

U G) = aff E

s E B n P ~

a component

j-i

, then

E aff E

: F

in

c aff E ~

and

aff E

Then of

rel bdry E .

If

E

~ S,

by I)

If

E

has

by 2) above.

has dimension

j

to examine: for an oppropriate

by 3) above.

Again

facet

s E aff E

(p,s)

meets

aff E

Let

G

be a facet of

U G)

(~

by 3) above,

at a singleton

point

E

e

containing

e~ , Then

O

U G)

We conclude Moreover,

, aff(p

that

since

cI(U{F

: F

Recall

that

B n P cU ~

in

N N S

N' n S

is not

: F

lies

in a finite

{xn}

: F

in

s EU

{F : F

collection

of closed

in F }

is establisbed.

sets,

cl(B n P)

c

N

is selected

in F}. (j+l)-flat

for every union

P

where

neighborhood

of flats

in F } .

N n S

and hence

in F } , and tbe a s s e r t i o n

in the

Moreover,

N' N S ~ U { F

sequence

{F : F

is a finite

F }) = U{F

is minimal~

Hence

(p,s]

EF

EF

O

s E aff(p

j+l

j-flat

cl(P n S)

in F } . aff E

Moreover, E F

E

aff(p U E)

and

F } .

a

o

, s EU {F : F

p

is either

in F } .

p ~ aff E ~ then

e

c

F

Let

meets

at most

to

, then there are two p o s s i b i l i t i e s

G

b)

E

aff(p U G)

in F } .

, (p,s)

has d i m e n s i o n

s E L(p,s)

j

Hence

aff(p U G)

{F : F N S)

s E aff(p U E) = aff E

and a)

Assign

B N P !

and since

dimension

(j-l)-flat.

in

P

to

x

of

having

By a standard

converging

N'

x

dimension

argument,

so that no

with

so that N' c

at most

we may select

x

N j a

is in U {F : F D

in F } ~

Then

S c p , so

{xn} i S

~

(U{F

: F

B N N ~ B N N n S !

in

F }) !

B N P , and

(S ~ cl(B N N)) = N N ( S ~ cl B) .

S ~cl(B

n P)

{x } c N n

We have a sequence

~

However,

(S ~ cl(B {x }

in

N N

N p)) c

N

N

S ~ cl B

n

converging

to

x , and

x

E cl(S ~ cl B)

, the desired

result.

Tbis

finishes

Case i, CASE 2,

Suppose

that for some n e i g h b o r h o o d

lies in a finite union j .

Choose

a sequence

and only one of these M

n

with

minimal

M and

n

n S

in a

kn K j

.

of flats {yn } flats.

to

Thus each

Yn

By o u r

P

of

x

P , each of which

converging

k -flat n

in

N'

n

where

induction

x

with

N' c

has dimension

so that each

YD

has an associated M

~

is

hypothesis,

E cl(S

, N'n S at most

lies in one

neighborhood

chosen so that Yn

N

~

cl

k B)

n

is for

114

each

Breen

n.

Hence there is a sequence

{z

} nm

Moreover,

we may assume that

dist(yn,Znm)

arguments reveal that the sequence cl B) , again the desired result.

{Znn }

in S ~ cl B I < --m for each converges to

This completes

converging

Standard

x ~ and

x ~ cl(S

Case 2.

k , 1 ~ k ~ j < d ~ it is true for

By induction,

k , 1 < k ~ d .

it is true for all

for all

x E S , so

the two sets are equal.

Thus

Yn ~

m .

Thus when the result is true for

cl B)

to

We conclude that

j§ x

.

~ cl(S

S ! cl(S ~ el B) , cl $ c cl(S ~ cl B) ~ and B

is nowhere dense in

S , and the proof of

Theorem 1 is established. Observe that when In conclusion,

d = 2 , the number

[i, Example 3]

finite Krasnosel'skii

f(d)

in the theorem is best possible~

shows that the theorem fails and in fact no

number is possible without

the requirement

that

~

S

have finitely many components.

REFERENCES [i]

BREEN, Mari!yn: 'A Krasnosel~skii-type theorem for nonclosed the plane,' Journal of Geometry 18 (1982), 28-42.

[2]

BREEN, Marilyn: 'An improved Krasnosel'skii theorem for nonclosed in the plane,' Journal of Geometry 21 (1983), 97-100.

[3]

BREEN, Marilyn: 'The dimension of the kernel of a planar set, ~ :Pacific J. Math. 82 (1979), 15-21.

[4]

KRASNOSEL'SKII, M.A.: 'Sur un critere pour Math. Sb. (61) 19 (1946), 309-310.

[5]

LAY, Steven R.: York, 1982.

[6]

VALENTINE,

qu' u n

Convex Sets and Their Applications,

F.A.:

Convex Sets, McGraw-Hill,

(Eimgegangen

am

29.

Nai

q98@)

sets

domain soit etoi!~, ~ John Wiley, New

New York, 1964.

University of Oklahoma Norman, Ok 73019 U.s.A.

sets in