267
Internat. J. Math. & Math. Scl.
Vol. 8 No. 2 (1985) 267-273
A HELLY NUMBER FOR UNIONS OF TWO BOXES IN R2
MARILYN BREEN
Department of Mathematics
University of Oklahoma
Norman, Oklahoma 73019 U.S.A.
(Received February 13, 1985)
Let
ABSTRACT.
A
into sets
be a polygonal region in the plane with edges parallel to the
S
5
If every
coordinate axes.
B
and
S
or fewer boundary points of
so that
AU
cony
5
The number
convex sets, each a rectangle.
B c S,
conv
can be partitioned
S
then
is a union of two
is best possible.
S,
Without suitable hypothesis on edges of
Moreover, an
the theorem fails.
example reveals that there is no finite Helly number which characterizes arbitrary
unions of two convex sets, even for polygonal regions in the plane.
KEY WORDS AND PHRASES. Helly type theorems, un of convex ss.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODE. 52.A35, 52.A0, 52.AI0.
i.
INTRODUCTION.
We begin with some preliminary definitions,
x
point
S
in
N
is some neighborhood
of
q
S
(x
is visible from
S.
lies in
if for
I
Points
i
<
j
y
via
x l,...,xn
n,
x
S,
in
For points
S.
i
S
in
3
set of all such points
in
p
S
S.
If
then
q
is called a
point of local noncon-
and
y
in
S,
and ker
S
x
and
y,
Inc S
R(x,y)
y
sees
is the
xj
Finally,
p
S
Set
S.
via
S,
S
S
at least one of the corres-
S
sees each point of
via
S.
The
Conv
S,
bdry
S.
will be the set of inc points of
will represent the ray from
[9]
if and only
is called star-shaped if and only if
(convex) kernel of
The reader is referred to Valentine
via
Ix,y]
is said to be
will denote the convex hull, boundary, and kernel of set
respectively, while
points
fails to be
x
we say
The following terminology will be used throughout the paper:
S,
S
is convex.
points in
such that
A
if and only if there
are visually independent via
does not see
pondlng llne segments lies in
p
S
S) if and only if the corresponding segment
3-convex if and only if for every
there is some point
x
d.
R
be a subset of
N
S
such that
x
locally convex at some point
vexity (inc point) of
S
let
point of local convexity of
is called a
and to Imy
S.
S,
For distinct
x emanating through
[5]
y.
for a discussion of
these concepts.
An interesting theorem by Lawrence, Hare, and Kenelly
following characterization for unions of convex sets:
For
[4]
S
provides the
a subset of a linear
M. BREEN
268
is a union of
S
topological space,
convex sets if and only if each finite
k
F.I --c
k. A
i
S, I
with conv
k
such
characterize
to
number
Helly
of
finite
a
existence
the
related problem concerns
be
unions, and it is natural to ask when ’finite subset’ in the hypothesis may
S
of
F
subset
FI,...,F
has a k-partition
replaced by ’j-member subset’ for an appropriate j. (See [3] for similar results.)
Unfortunately, the Lawrence, Hare, Kenelly Theorem cannot be improved for
However,
(See Example 3.)
is a compact subset of the plane.
S
even when
2,
k
the problem of obtaining a finite Helly number to characterize certain unions of
Recent results suggest sets for which such a theorem
convex sets remains open.
Grnbaum [2],
In work by Danzer and
might hold.
a kind of Helly number (called a
[i]
Moreover, in
piercing number) is found for certain families of boxes.
another
kind of Helly number (a Krasnosel’skii number for unions of starshaped sets) is
established for certain polygonal regions whose edges are parallel to the coordinate
This is
Therefore, it seems reasonable to examine such polygonal regions.
axes.
the situation studied here, and the following result is obtained:
be a polygonal region in the plane with edges parallel to the coordi-
S
Let
A
sets
B
and
or fewer boundary points of
5
If every
nate axes.
so that
A U
cony
cony
The number
sets, each a rectangle.
restrictions on the edges of
5
B c S,
then
S
can be partitioned into
S
is a union of two convex
Without suitable
is best possible.
the result fails and in fact no finite Helly
S,
number exists.
2.
THE RESULTS.
The following lemma will be useful.
LEMMA 1
k
3.
A
and
If every
B
or fewer boundary points of
k
such that
A U conv B c S,
conv
PROOF.
S
Clearly
S
is closed.
S
If
is not
S
k
inc S #
S
locally convex set
is 3-convex.
connected, then it is easy to
Hence without loss of
assert that for every
in bdry S,
z
there would be a convex neighborhood
inc S c ker S.
S.
via
points
x,y
Inc S,
q
Since
N
in
S
bdry
S
N
[q,z]
S,
Ix,y]
with
S,
and
S
q
a
q
and
ker S,
<
[q,w]
b
<
w.
c S:
q
Let
[6].
inc S #
q
.
S
is closed,
N N S
seeing
Otherwise, since
with no point of
We
z
i
bdry S,
S.
However, then points
would
x,y,z
We
contradicting our hypothesis.
and the assertion is established.
Now it is easy to show that
[q,w]
of
is connected.
is not convex, and standard arguments produce
be visually independent points in
conclude that
[q,z]
N
S
for otherwise the closed, connected,
will be convex by a theorem of Tietze
We begin the proof by showing that
S
or fewer points of
generality we may restrict our attention to the case in which
Furthermore, we may assume that
be a fixed integer,
can be partitioned into sets
has exactly two components, each convex.
,
k
and let
S
then every
Moreover,
can be partitioned in this way as well.
show that
d
R
be a bounded subset of
S
Let
S
q
ker S:
If not, then for some
would contain an open interval
However,
the desired result.
[q,b]
c S.
(a,b)
w
with
in
a,b
S,
bdry
Again we have a contradiction, and
269
HELLY NUMBER FOR UNIONS OF TWO BOXES
.
x
Finally, to establish the lemma, let
S #
R(q,x i) meet bdry S
at
q
with
Yi
inc S,
q
Since
A
,xk}
{xj+I
conv
Similarly
S.
!
and
{Yl
({q}UA) ! S and
conv
>
k
inc
and let ray
Yj}’
conv {x
Since
k},
i
q
and select
k,
i
i
By our hypothesis, the points
B such that conv A U conv B ! S.
}"
B
{Yj+I
Yi"
A
Relabeling if necessary, assume that
i
{x i
q
< xi
can be partitioned into sets
yl,...,y k
S,
i
Without loss of generality, assume
Yk
({q} U A)! S.
xj} ! conv
I
it is easy to see that
3,
S
is
3-convex, and the lemma is proved.
Without the requirement that set
i
be bounded, Lemma
S
fails, as Example
i
illustrates.
EXAMPLE I.
A
and
2
which lle either in the first
(See Figure i.)
quadrant or on one of the coordinate axes.
partitioned into sets
R
be the set of points in
S
let
The set
bdry S
satisfying the hypothesis of the lemma.
B
the conclusion of the lemma fails for every
k
>
may be
However,
3.
We are ready to establish the following Helly-type theorem for unions of
rectangles in the plane.
THEOREM i.
let
the coordinate axes.
S
be a polygonal region in the plane with edges parallel to
If every
5
S
or fewer boundary points of
into sets A and B so that conv A U
convex sets, each a rectangle. The number
conv B c S,
5
S
then
can be partitioned
is a union of two
is best posole.
Figure i.
We begin with some preliminary observations.
PROOF.
Moreover, using [8, Theorem i]
Now in case S has O, i, or 2 inc points,
3-convex.
S
that
S
By Lemma i, set S is
is starshaped and inc S c ker S.
[8, Theorem 3]
we may use
to conclude
is a union of two or fewer closed convex sets, the desired result.
fore, throughout the argument it suffices to assume that
S
has at least
There-
3
inc
points.
As in
[7]
and
[i],
it is helpful to order
This in turn induces a natural order on the edges of
edge as
from
’right’,
bdry S.
ql,q2,...,qn,
n
3
’left’,
’up’,
or
’down’
bdry S
S,
in a clockwise direction.
and we may classify each
according to the order it inherits
For convenience of notation, we label the inc points of
again according to the clockwise order on
by a previous assumption.
concerning these inc points.
bdry S.
S
by
Recall that
The proof of the theorem will require some results
270
M. BREEN
It is easy to see that for each inc point of
’up’
edge and followed by an
’down’
by an
’right’
edge and followed by a
edge and followed by a ’right’ edge, or preceded
’left’
edge and followed by a
S,
cation together with the 3-convexlty of
ql
exactly one of the following
’left’
edge, preceded by a
’down’
edge, preceded by a
’up’
S,
The Inc point Is either preceded by a
must occur relative to our order:
Moreover, using this classifi-
edge.
it is not hard to show that no three
ql
q2
q2
Case I.
Case 2.
Case 3.
ql
q2
Case 4.
Case 6.
Case 5.
ql
ql
Case 7.
Case 8.
Case 9.
Figure 2.
consecutive Inc points of
is a
(in our order)
S
S,
appropriate orientation of
’down’ edge and whose successor is an
S,
independent points of
can be colllnear:
’right’
this would yield a
’up’
Otherwise, for an
edge whose predecessor
edge, producing three visually
impossible.
Furthermore, we assert that every two consecutive inc points of
a segment parallel to one of the coordinate axes:
condition fails for
ql
oriented in the plane so
and
q2"
that ql
possible classifications for
ql
ql
’down’
cannot be classified as
’right-up’.
determine
Suppose on the contrary that the
Without loss of generality, assume that
is to the left of
and
q2"
edge.
q2
and above
Using the fact that
our clockwise order, it is not hard to see that
edge and followed by a
S
ql
q2
q2"
S
is
We examine
follows
cannot be preceded by a
ql
For convenience of terminology, we say that
’left-down’.
Similarly,
q2
in
’left’
cannot be classified as
271
HELLY NUMBER FOR UNIONS OF TWO BOXES
Examine the remaining 9 possibilities.
(See Figure 2, Cases 1-9, for corres-
ponding illustrations.)
i.
2.
3.
4.
5.
6.
7.
8.
9.
’left-down’
’down-rlght’
is ’up-left’
’right-up’
’down-rlght’ and q2 is ’left-down’
’down-right’ and q2 is ’down-right’
’down-rlght’ and q2 is ’up-left’
’up-left’ and q2 is ’left-down’
’up-left’ and q2 is ’down-right’
’up-left’ and q2 is ’up-left’
’right-up’
is
ql
ql
is
ql
ql
is
is
is
ql
ql
ql
ql
ql
is
is
is
is
and
is
and
is
q2
q2
and q2
’right-up’
It is not hard to prove that each of Cases 2 through 9 above violates the condition
that
so none of these can occur.
ker S,
ql,q 2
Thus Case 1 is the only
possibility.
Moreover, by a similar argument involving the classification of Inc point
q3
must be
wise order.
that
ql
’up-left’:
Classifications
ker S
and
ker S, respectively.
q2
However, this forces
S
to have
5
ql’
P2
above
ql
(See Figure 3, Cases 1-2.)
boundary points
there is no partition satisfying our hypothesis:
preceding
and right of
q2’
Select
P3
Pi’
near
Pl
near
1
q2
i
ql
Case I.
Case 2.
Figure 3.
q2
P3
Figure 4.
5,
Figure 5.
for which
on the edge
on the edge
q2
P4
q3’
’down-right’ is impossible in our clockl)’right-up’ and 2)’left-down’ violate the facts
Classification
272
M. BREEN
following
q2’
near
P4
edge following
on the edge preceding
q3
(see Figure 4.)
q3"
S
false, and every two consecutive inc points of
to one of the coordinate axes.
4
has at most
inc points.
[8, Theorem 3],
S
S
are collinear, it is not hard to
3,
n
3
cannot have exactly
Hence
4.
n
[8, Theorem 3],
it is easy to
may be expressed as a union of two rectangles, and the proof of
S
i
Theorem
S
we conclude that
is a union of two closed convex sets.
Finally, using decomposition techniques from
show that
on the
q3
must determine a segment parallel
Similarly,
inc points, and since we are assuming that
by
near
Using the assertion above together with the
fact that no three consecutive inc points of
S
P5
Thus the assertion is established.
The rest of the argument is easy.
show that
and
q3’
We have a contradiction, our supposition is
is complete.
5
To see that the number
in Theorem
is best possible, consider the
i
following example.
EXAMPLE 2.
Let
S
be the set in Figure 5.
partitioned into sets
A
and
B
Every
4
conv A U conv B
such that
S
points of
!
S.
may be
S
However,
is
not a union of fewer than three convex sets.
In conclusion, it is interesting to observe that the result in Theorem 1
fails without the restriction on edges of S. In fact, there is no finite Helly
number which characterizes unions of two convex sets, even for polygonal regions in
the plane, and the Lawrence, Hare, Kenelly Theorem is best possible in this case.
This is illustrated in Example
3.
a
a
2
b
5
e
b!
3
a
2
e
a
EXAMPLE 3.
b
For
n
4
>
e
2,
let
P(n)
modulo
e
i
2n + i,
such that
for
s
i
i
meets
i
<
si_ 2
2n +I,
at
a5
be a regular
i
and
s
[al,b i]
i
s
i
2
meets
b3
(2n + l)-gon
el,e2,...,e2n+l.
Let
a
b
3
Fgure 6.
are labeled in a clockwise direction by
4
s
whose edges
Adjusting our subscripts
be a segment containing
i+2
at
b
i"
Finally
let
HELLY NUMBER FOR UNIONS OF TWO BOXES
273
(The
U{s
i < i < 2n +l}.
be the simply connected region determined by
i
be
may
S(n)
of
points
2n
2 is illustrated in Figure 6.) Every
case for n
+ i
2n
the
partitioned into sets A and B so that conv A U cony B c__ S. However,
have no such partition, and S(n) is not a union of
i < 2n + 1
i
a
points
S(n)
<
i
fewer than three cot:vex sets.
REFERENCES
i.
BREEN, M.
A Krasnosel’skii-type theorem for unions of two starshaped
(to appear).
sets in the plane, Pacific J. Math.
Intersection properties of boxes in Rd
and GR0"NBAUM, B
Combinatorica 2 (3) (1982), 237-246.
2
DANZER, L
3.
DANZER, L. GRUNBAUM, B. and KLEE, V. Helly’s theorem and its relatives, Proc.
Symposia in Pure Math., Vol. VII (Convexity) (1963), 101-180.
4.
LAWRENCE, J. F., HARE, W. R. Jr. and KENELLY, J. W. Finite unions of convex
sets, Proc. Amer. Math. Soc. 34 (1972), 225-228.
5.
LAY, S. R. Convex Sets and Their Application, John Wiley, New York, 1982.
6.
TIETZE, H. Uber Konvexheit im kleinen und im grossen und 6er gewisse den
Pukten einer Menge zugeordnete Dimensionzahlen, Math. Z. 28 (1928),
697-707.
7.
TOSSAINT, G. T. and EL-GINDY, H.
Traditional galleries are starshaped if every
two paintings are visible from some common point,
Amer. Math. Monthly
(to appear).
8.
VALENTINE, F. A.
1227-1235.
A three point convexity property, Pacific J. Math. 7 (1957),
9.
VALENTINE, F. A.
Convex Sets, McGraw Hill, New York, 1964.