Coloring Intersection Graphs of ArcConnected Sets in the Plane
Micha Lason
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Piotr Micek
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Arkadiusz Pawlik
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Bartosz Walczak
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M. Lason Institute of Mathematics of the Polish Academy of Sciences
, Warsaw,
Poland
A family of sets in the plane is simple if the intersection of any subfamily is arcconnected, and it is pierced by a line L if the intersection of any member with L is a nonempty segment. It is proved that the intersection graphs of simple families of compact arcconnected sets in the plane pierced by a common line have chromatic number bounded by a function of their clique number. A proper coloring of a graph is an assignment of colors to the vertices of the graph such that no two adjacent ones are assigned the same color. The minimum number of colors sufficient to color a graph G properly is called the chromatic number of G and

denoted by (G). The maximum size of a clique (a set of pairwise adjacent vertices)
in a graph G is called the clique number of G and denoted by (G). It is clear that
(G) (G).
The chromatic and clique numbers of a graph can be arbitrarily far apart. There are
various constructions of graphs that are trianglefree (have clique number 2) and still
have arbitrarily large chromatic number. The first one was given in 1949 by Zykov [16],
and the one perhaps best known is due to Mycielski [11]. However, these classical
constructions require a lot of freedom in connecting vertices by edges, and many
important classes of graphs derived from specific (e.g. geometric) representations
have chromatic number bounded in terms of the clique number. A class of graphs is
called bounded if there is a function f : N N such that (G) f ((G)) holds
for any graph G from the class.
In this paper, we focus on the relation between the chromatic number and the clique
number for geometric intersection graphs. The intersection graph of a family of sets F
is the graph with vertex set F and edge set consisting of pairs of intersecting elements
of F . We consider finite families F of arcconnected compact sets in the plane which
are simple in the sense that the intersection of any subfamily of F is also arcconnected.
We usually identify the family F with its intersection graph and use such terms as
chromatic number, clique number or boundedness referring directly to F .
In the onedimensional case of subsets of R, the only arcconnected compact sets
are closed intervals. They define the class of interval graphs, which have chromatic
number equal to their clique number. The study of the chromatic number of intersection
graphs of geometric objects in higher dimensions was initiated in the seminal paper
of Asplund and Grnbaum [1], where they proved that the families of axisaligned
rectangles in R2 are bounded. On the other hand, Burling [2] showed that intersection
graphs of axisaligned boxes in R3 with clique number 2 can have arbitrarily large
chromatic number.
Since then, a lot of research focused on proving boundedness of the families
of geometric objects in the plane with various restrictions on the kind of objects
considered, their positions, or the way they can intersect. Gyrfs [5,6] proved that
the families of chords of a circle are bounded. This was generalized by Kostochka
and Kratochvl [8] to the families of convex polygons inscribed in a circle. Kim et
al. [7] showed that the families of homothetic (uniformly scaled) copies of a fixed
convex compact set in the plane are bounded. Fox and Pach [3] showed that the
intersection graphs of any arcconnected compact sets in the plane that do not contain
a fixed bipartite subgraph H have chromatic number bounded by a function of H .
This easily implies that the families of pseudodiscs, that is, closed disc homeomorphs
in the plane the boundaries of any two of which cross at most twice, are bounded.
Note that families of pseudodiscs are simple. The abovementioned results of [7] and
[3] are actually strongerthey state that the number of edges of the intersection graph
of a respective family F is bounded by f ((F ))F  for some function f .
A family of sets F is pierced by a line L if the intersection of any member of F
with L is a nonempty segment. McGuinness [9] proved that the families of Lshapes
(shapes consisting of a horizontal and a vertical segments of arbitrary lengths, forming
the letter L) pierced by a fixed vertical line are bounded. Later [10], he showed that
the trianglefree simple families of compact arcconnected sets in the plane pierced
by a common line have bounded chromatic number. Suk [15] proved boundedness
of the simple families of x monotone curves intersecting a fixed vertical line. In this
paper, we generalize the results of McGuinness, allowing any bound on the clique
number, and of Suk, removing the x monotonicity condition.
Theorem 1 The class of simple families of compact arcconnected sets in the plane
pierced by a common line is bounded.
By contrast, Pawlik et al. [12,13] proved that there are intersection graphs of
straightline segments (or geometric sets of many other kinds) with clique number
2 and arbitrarily large chromatic number. This justifies the assumption of Theorem
1 that the sets are pierced by a common line. The best known upper bound on the
chromatic number of simple families of curves in the plane with clique number is
lloogg n O(log ) due to Fox and Pach [4].
The bound on the chromatic number following from our proof of Theorem 1 is
double exponential in terms of the clique number.
The ultimate goal of this quest is to understand the border line between the classes
of graphs (and classes of geometric objects) that are bounded and those that are not.
In a preliminary version of this paper, we proposed the following two problems.
Problem 1 Are the families (not necessarily simple) of x monotone curves in the
plane pierced by a common vertical line bounded?
Rok and Walczak [14] proved recently that the answers to both these questions are
positive. However, the bound on the chromatic number in terms of the clique number
resulting from their proof is enormous (greater than an exponential tower), which is
much worse than the double exponential bound of Theorem 1.
2 Topological Preliminaries
All the geometric sets that considered in this paper are assumed to be subsets of the
Euclidean plane R2 or, further in the paper, subsets of the closed upper halfplane
R [0, +). An arc between points x , y R2 is the image of a continuous injective
map : [0, 1] R2 such that (0) = x and (1) = y. A set X R2 is arcconnected
if any two points of X are connected by an arc in X . The union of two arcconnected
sets that have nonempty intersection is itself arcconnected. More generally, if X
is a family of arcconnected sets whose intersection graph is connected, then X
is arcconnected. For a set X R2, the relation {(x , y) X 2 : X contains an arc
between x and y} is an equivalence, whose equivalence classes are the arcconnected
components of X . Every arcconnected component of an open set is itself an open set.
All families of sets that we consider are finite. A family F of sets in R2 is simple
if the intersection of any subfamily of F is arcconnected (possibly empty). A set X
is simple with respect to a family Y if {X } Y is simple.
Lemma 2 Let X be a compact arcconnected set and Y be a family of compact
arcconnected sets such that X is simple with respect to Y and the intersections of the
members of Y with X are pairwise disjoint. Between any points x1, x2 X , there is
an arc A X that is simple with respect to Y.
Proof Let Y = {Y1, . . . , Yn}. For i {0, . . . , n}, we construct an arc Ai X between
x1 and x2 that is simple with respect to {Y1, . . . , Yi }. As X is arcconnected, we pick
A0 to be any arc between x1 and x2 within X . We construct Ai from Ai1 as follows.
If Ai1 Yi = , then we take Ai = Ai1. Otherwise, let y1 and y2 be respectively the
first and the last points on Ai1 that belong to Yi (which exist as Ai Yi is nonempty
and compact). To obtain Ai , replace the part of Ai1 between y1 and y2 by any arc
between y1 and y2 in X Yi (which exists because X Yi is arcconnected). Clearly,
Ai is simple with respect to Yi . Since X Yi is disjoint from each of Y1, . . . , Yi1, Ai
remains simple with respect to Y1, . . . , Yi1.
A Jordan curve is the image of a continuous map : [0, 1] R2 such that (0) =
(1) and is injective on [0, 1). The famous Jordan curve theorem states that if
C R2 is a Jordan curve, then R2 C has exactly two arcconnected components, one
bounded and one unbounded. An extension of this, called JordanSchnflies theorem,
adds that there is a homeomorphism of R2 that maps C to a unit circle, the bounded
arcconnected component of R2 C to the interior of this circle, and the unbounded
arcconnected component of R2 C to the exterior of the circle.
We will use a special case of the Jordan curve theorem for arcs in the closed upper
halfplane R [0, +). Namely, if x and y are two points on the horizontal axis
R {0} and A is an arc between x and y such that A {x , y} R (0, +), then
the set (R [0, +)) A has exactly two arcconnected components, one bounded
and one unbounded. This in particular implies that for any four points x1, x2, y1 and
y2 in this order on the horizontal axis, every arc in R [0, +) between x1 and y1
intersects every arc in R [0, +) between x2 and y2.
3 Grounded Families
In Theorem 1, a family F compact arcconnected sets in the plane is assumed to be
pierced by a common line. We assume without loss of generality that this piercing line
is the horizontal axis R {0} and call it the baseline. The base of a set X , denoted by
base(X ), is the intersection of X with the baseline.
We fix a positive integer k and assume (F ) k. The intersection graph of the
bases of the members of F is an interval graph, so it can be properly colored with k
colors. To find a proper coloring of F with a number of colors bounded in terms of k,
we can restrict our attention to one color class in the coloring of this interval graph.
Therefore, without loss of generality, we assume that no two members of F intersect
on the baseline and show that F can be colored properly with a bounded number of
colors. Moreover, it is clear that the families F + = {X (R [0, +)) : X F } and
F = {X (R (, 0]) : X F } are simple. It suffices to obtain proper colorings
+ and of F + and F , respectively, with bounded numbers of colors, since then
F may be colored by pairs (+, ). We only focus on coloring F +, as F can be
Fig. 1 A grounded family of sets
handled by symmetry. To simplify notation, we rename F + to F . Therefore, each set
X F is assumed to satisfy the following:
and F is assumed to be simple. Any set that satisfies the conditions above is called
grounded, and any simple family of grounded sets with pairwise disjoint bases is also
called grounded (see Fig. 1).
All the geometric sets that we consider from now on are contained in R [0, +).
To prove Theorem 1, it suffices to show the following.
Proposition 3 For k
family F with (F )
The case k = 1 is trivial, and the case k = 2 with some additional assumptions
meant to avoid topological pathologies was settled by McGuinness [10].
We write X Y if base(X ) is entirely to the left of base(Y ). The relation is a
total order on a grounded family and naturally extends to its subfamilies (or any other
families of grounded sets with pairwise disjoint bases): for example, X Y denotes
that X Y for any Y Y. For grounded sets X1 and X2 such that X1 X2, we
define F (X1, X2) = {Y F : X1 Y X2}. For a grounded set X , we define
F (, X ) = {Y F : Y X } and F (X, +) = {Y F : X Y }.
The proof of Proposition 3 heavily depends on two decomposition lemmas, which
given a grounded family with large chromatic number find its subfamily with large
chromatic number and some special properties. The first one is a reformulation of
Lemma 2.1 in [9].
Lemma 4 Let F be a grounded family with (F ) > 2a(b + 1), where a, b 0.
There is a subfamily H of F that satisfies (H) > a and (F (H1, H2)) > b for any
intersecting H1, H2 H.
Proof We partition F into subfamilies F0 Fn so that (Fi ) = b + 1 for
0 i < n. This can be done by adding sets to F0 in the increasing order until we
get (F0) = b + 1, then following the same procedure with the remaining sets to form
F1, and so on. Let F 0 = i F2i and F 1 = i F2i+1. Since (F 0 F 1) > 2a(b +1),
Fig. 2 An externally supported family of sets
we have (F k ) > a(b + 1) for k = 0 or k = 1. We now color each F2i+k properly
using the same set of b + 1 colors. This coloring induces a partitioning of the entire
F k into subfamilies H0, . . . , Hb such that for 0 i n, 0 j b the family
Fi H j is independent. We set H = H j , where H j has the maximum chromatic
number among H0, . . . , Hb. Since (F k ) > a(b + 1), we have (H) > a. It remains
to show that (F (H1, H2)) > b for H1, H2 H with H1 H2 = . Indeed, such
sets H1 and H2 must lie in different families F2i1+k and F2i2+k , respectively, so
(F (H1, H2)) (F2i1+k+1) = b + 1 > b, as required.
For a set X , we define ext(X ) to be the only unbounded arcconnected component
of (R [0, +)) X . For a grounded family F , we define ext(F ) = ext( F ). A
subfamily G of a grounded family F is externally supported in F if for any X G
there exists Y F such that Y X = and Y ext(G) = (see Fig. 2). The
idea behind the following lemma is due to Gyrfs [5] and was subsequently used in
[9,10,15].
Lemma 5 Let F be a grounded family with (F ) > 2a, where a 1. There is a
subfamily G of F that is externally supported in F and satisfies (G) > a.
Proof For convenience, we restrict F to its connected component with maximum
chromatic number. Let X0 be the least member of F . For i 0, let Fi be the
family of members of F that are at distance i from X0 in the intersection graph of F .
It follows that F0 = {X0} and, for i j  > 1, each member of Fi is disjoint from
each member of F j . Clearly, ( i F2i ) > a or ( i F2i+1) > a, and therefore
there is d 1 with (Fd ) > a. We claim that Fd is externally supported in F . Fix
Xd Fd , and let X0, . . . , Xd be a shortest path from X0 to Xd in the intersection
graph of F . Since X0 ext(Fd ) = and X0, . . . , Xd2 are disjoint from Fd , we
have X0, . . . , Xd2 ext(Fd ). Thus Xd1 ext(Fd ) = and Xd1 Xd = .
4 Cliques and Brackets
Let F be a grounded family with (F ) k. A kclique in F is a family of k pairwise
intersecting members of F . For a kclique K, we denote by int(K) the only
arcconnected component of (R [0, +)) K containing the part of the baseline
Fig. 3 A kbracket B with kclique K and support S
between the two least members of K. A kbracket in F is a subfamily of F consisting
of a kclique K and a set S called the support such that S K or K S and
S int(K) = . For such a kbracket B, we denote by int(B) the only arcconnected
component of (R [0, +)) B containing the part of the baseline between S
and K (see Fig. 3).
Lemma 6 Let F be a grounded family, X, Y, Z F , and X Y = . Let C1 and
C2 be any two distinct arcconnected components of (R [0, +)) (X Y ). Let
z, z Z C1. If every arc between z and z within Z intersects C2, then every such
arc intersects both X and Y .
Proof Suppose there is an arc A Z between z and z such that A X = or
A Y = . If A X = and A Y = , then A C1, so A C2 = . Now, suppose
A X = and A Y = . Let y and y be respectively the first and last points of A
in Y . Since Z Y is arcconnected, there is an arc A in Z that is simple with respect
to Y and goes along A from z to y, then to y inside Y , and finally along A to z . It
follows that A C1 Y , so A C2 = . The case that A X = and A Y =
is symmetric.
Corollary 7 Let F be a grounded family, K be a kclique in F , and X F . If
x , y X int(K) (or x , y X ext(K)) and every arc between x and y within X
intersects ext(K) (int(K), respectively), then X intersects every member of K.
Proof Let K = {K1, . . . , Kk } and K1 int(K) K2 Kk . The statement
follows directly from Lemma 6 and the fact that int(K) and ext(K) belong to distinct
arcconnected components of (R [0, +)) (K1 Ki ) for 2 i k.
Corollary 8 Let F be a grounded family and B be a bracket in F with clique K and
support S. Let X F . If X int(B) = and X ext(B) = , then X intersects S
or every member of K.
Proof Let x X int(B) and x X ext(B), and suppose X S = . By the Jordan
curve theorem, every arc between x and x within X must intersect S int(K) and
thus int(K). Since x , x X ext(K), it follows from Corollary 7 that X intersects
every member of K.
5 Proof of Proposition 3
The proof goes by induction on k. Proposition 3 holds trivially for k = 1 with 1 = 1.
Therefore, we assume that k 2 and that the statement of the proposition holds for
k 1. This context of the induction step is maintained throughout the entire remaining
part of the paper. A typical application of the induction hypothesis looks as follows: if
F is a grounded family with (F ) k, G F , and there is X F G intersecting
all members of G, then (G) k 1 and thus (G) k1.
Define k = 8kk21, k,k = 0, k, j = k + 2k, j+1 + 2k1(kk1 + k + 2) + 2
for k 1 j 0, and finally k = 2k+2(k,0 + 2k1 + 1).
We say that a grounded set X (a grounded family X ) is surrounded by a set S
if X (every member of X , respectively) is disjoint from S ext(S). For a set S
and a grounded set R such that base(R) is surrounded by S, let cut(R, S) denote
the closure of the unique arcconnected component of R S containing base(R).
For a set S and a grounded family R of sets whose bases are surrounded by S, let
cut(R, S) = {cut(R, S) : R R}.
First, we present a technical lemma, which generalizes similar statements from
[10] (Lemma 3.2) and [15] (Lemma 4.1), and which we will prove in Sect. 6. Loosely
speaking, it says that one can color properly, with the number of colors bounded in
terms of k, all the members of F surrounded by a set S which intersect cut(R, S) for
any set R F intersecting S.
Suppose for the sake of contradiction that there is a grounded family F with
(F ) k and (F ) > k = 2k+2(k,0 + 2k1 + 1). A repeated application of
Lemma 5 yields a sequence of families F = Fk+1 Fk F0 such that Fi
is externally supported in Fi+1 and (Fi ) > 2i+1(k,0 + 2k1 + 1), for 0 i k.
The following claim is the core of the proof.
Claim 10 For 0 j
the following properties:
k, there are families S, G F j and sets S1, . . . , S j S with
(i) G is surrounded by S,
(ii) (G) > k, j ,
(iii) the sets S1, . . . , S j pairwise intersect,
(iv) every member of F intersecting ext(S) and some member of G also intersects
each of S1, . . . , S j .
Proof The proof goes by induction on j . First, let j = 0. Apply Lemma 4 to find
H F0 such that (H) > 1 and for any intersecting H1, H2 H we have
(F0(H1, H2)) > k,0 + 2k1. Since (H) > 1, such two intersecting H1, H2 H
exist. Let S = {H1, H2} and G be the family of those members of F0(H1, H2) that
are disjoint from H1 H2. It is clear that (i) holds. Since the members of F0(H1, H2)
intersecting H1 H2 have chromatic number at most 2k1, we have (G) > k,0, so
(ii) holds. The conditions (iii) and (iv) are satisfied vacuously.
Now, assume that j 1 and the claim holds for j 1, that is, there are families
S , G F j1 and sets S1, . . . , S j1 S satisfying (i)(iv). Let
R =
D =
R F : base(R) is surrounded by
D G : D
S ) = .
S and R
S = ,
It follows from Lemma 9 that (D) k and thus (G D) > 2k, j +2k1(kk1 +
k + 2) + 2. Since the chromatic number of a graph is the maximum chromatic number
of its connected component, there is G G D such that the intersection graph of
G is connected and (G ) > 2k, j + 2k1(kk1 + k + 2) + 2. Partition G into three
subfamilies X , Y, Z so that X Y Z and (X ) = (Z) = k, j +(k +1)k1 +1.
It follows that (Y) > 2k1(kk1 + 1). Apply Lemma 4 to find H Y such that
(H) > k1 and for any intersecting H1, H2 H we have (Y(H1, H2)) > kk1.
Since (H) > k1, there is a kclique K H. The members of Y intersecting K
have chromatic number at most kk1, so there is P Y that is contained in int(K).
Since F j1 is externally supported in F j , there is S j F j such that S j P =
and S j ext(S ) S j ext(F j1) = . Therefore, since S and G satisfy (iv), S j
intersects each of S1, . . . , S j1 and thus (iii) holds for S1, . . . , S j .
We show that S j G or G S j . Suppose that neither of these holds. It follows
that base(S j ) is surrounded by S , which yields S j R. Moreover, base(S j ) is
surrounded by G , as the intersection graph of G is connected. Therefore, we have
cut(S j , S ) G = , so there is X G such that X cut(S j , S ) = . This
means that X D, which contradicts the definition of G .
Now, we have S j int(K) S j P = and S j K or K S j , so the kclique
K and the support S j form a kbracket. Let S = S K {S j }. If S j G , then
S j X K. In this case, let G be the family of those members of X that are disjoint
from K S j . It is clear that (i) holds. Since (X ) > k, j + (k + 1)k1 and the
members of X intersecting K S j have chromatic number at most (k + 1)k1,
we have (G) > k, j , so (ii) holds. Since ext(S) ext(K {S j }), it follows from
Corollary 8 that every member of F intersecting ext(S) and some member of G
intersects S j . Hence (iv) holds. If G S j , then let G be the family of those members
of Z that are disjoint from K S j . An analogous argument shows that (i), (ii), and
(iv) are satisfied.
Let S, G, and S1, . . . , Sk be as guaranteed by Claim 10 for j = k. By (ii), we have
(G) > 0, so there is P G. Since Fk is externally supported in F , there is Sk+1 F
such that Sk+1 P = and Sk+1 ext(S) = . By (iii) and (iv), we conclude that
S1, . . . , Sk+1 pairwise intersect. This contradicts the assumption that (F ) k, thus
completing the proof of Proposition 3.
Fig. 4 The setting of the proof of Lemma 9: the set S with a dashed arc S and three sets from R
6 Proof of Lemma 9
The proof of Lemma 9 goes along similar lines to the proof of Lemma 4.1 in [15]. Since
D is surrounded by S, there is an arc S S such that D is surrounded by S . We can
assume without loss of generality that base(S ) = { p, q} for some points p and q on
the baseline such that p D q. For every R R we have cut(R, S) cut(R, S ).
Hence every member of D intersects cut(R, S ) (see Fig. 4).
Proof Consider a relation < on cut(R, S ) defined as follows: R1 < R2 if and only
if R1 R2 and R1 R2 = . It is clear that < is irreflexive and antisymmetric. It is
also transitive, which follows from the fact that if R1, R2, R3 R, R1 R2 R3,
and R1 R3 = , then R2 (R1 R3) = . Therefore, < is a strict partial order.
The intersection graph of cut(R, S ) is the incomparability graph of <, so it is perfect,
which implies (cut(R, S )) = (cut(R, S )) k.
By Claim 11, there is a coloring of R with k colors such that for any R1, R2 R
with (R1) = (R2), we have cut(R1, S ) cut(R2, S ) = . For a color c, let
Rc = {R R : (R) = c} and Dc = {D D : D cut(Rc, S ) = }. We are
going to show that (Dc) 8k21. Once this is obtained, we will have (D)
c (Dc) 8kk21 = k .
Since the sets cut(R, S ) for R Rc are pairwise disjoint, the curve S and the
families cut(Rc, S ) and Dc satisfy the assumptions of Lemma 9. To simplify the
notation, we assume for the remainder of the proof that S = S , R = cut(R, S )
for every R R , R = cut(Rc, S ), and D = Dc. By JordanSchnflies theorem,
c
the segment pq and the arc S form a Jordan curve which is the boundary of a set J
homeomorphic to a closed disc. In this new setting, S is an arc and RD is a grounded
family with the following properties:
We enumerate the members of R as R1, . . . , Rm in the order, that is, so that R1
Rm . We are going to show that (D) 8k21.
Claim 12 For 1
Ri+1, . . . , R j1.
m, any arc in J between Ri and R j intersects all
Proof Let A be an arc in J between points xi Ri and x j R j . For any R
{Ri+1, . . . , R j1}, base(R) is surrounded by Ri A R j and we have R (Ri A
R j ) = , as R S = . Since R is disjoint from Ri and R j , we have R A = .
A point x J is a neighbor of Ri if there is an arc in J between x and Ri disjoint
from all R1, . . . , Rm except Ri . It follows from Claim 12 that each point in J is a
neighbor of at most two consecutive sets of R1, . . . , Rm . For 1 i < m, let Ii denote
the set of points in J that are neighbors of Ri and Ri+1.
Claim 13 Any arcconnected subset of J intersects an interval of sets in the sequence
R1, I1, R2, . . . , Im1, Rm .
Proof Let X be an arcconnected subset of J . First, we show that if X intersects Ri
and Ri+1, then it also intersects Ii . This is guaranteed by the compactness of Ri and
Ri+1. Indeed, take a minimal arc in J between Ri and Ri+1. By Claim 12, the
interior of this arc is disjoint from all R1, . . . , Rm and therefore must lie in Ii .
Now, let i be the least index such that X (Ri Ii ) = and j be the greatest index
such that X (I j1 R j ) = . Let A be an arc in X between points xi X (Ri Ii )
and x j X (I j1 R j ). Since xi is a neighbor of Ri , there is an arc Ai J between
xi and Ri disjoint from all R1, . . . , Rm except Ri . Similarly, since x j is a neighbor of
R j , there is an arc A j J between x j and R j disjoint from all R1, . . . , Rm except
R j . There is an arc A Ai A A j between Ri and R j . By Claim 12, A intersects
all Ri+1, . . . , R j1. But Ri+1, . . . , R j1 are disjoint from Ai and A j , hence they
intersect A.
For convenience, define I0 = Im = . For D D, define leftclip(D) = D Ii and
rightclip(D) = D I j , where i and j are chosen so that Ri+1 is the first and R j is the last
member of R intersecting D (see Fig. 5). This definition extends to families M D:
leftclip(M) = {leftclip(M ) : M M} and rightclip(M) = {rightclip(M ) : M M}.
Proof We present the proof only for leftclip(D), as for rightclip(D) it is analogous.
The sets I1, . . . , Im1 are open in J , as they are arcconnected components of the set
J R, which is open in J . Each member of leftclip(D) is a difference of a compact
set in D and one of I1, . . . , Im1 and thus is compact as well.
To prove that leftclip(D) is simple and consists of arcconnected sets, we need
to show that leftclip(M) is arcconnected for any M D. Let x , y
leftclip(M) M. Since M is arcconnected, Lemma 2 provides us with
an arc A M between x and y that is simple with respect to R. It
suffices to show A leftclip(M ) for each M M. To this end, fix M M
and let Ri be the least member of R intersecting M . Suppose there is a point
z A (M leftclip(M )) Ii1. Since x , y leftclip(M ) mj=i ( R j I j ), it
follows from Claim 13 that the parts of A from x to z and from z to y intersect Ri .
This and z / Ri contradict the simplicity of A with respect to Ri .
Claim 15 leftclip(D) and rightclip(D) have clique number at most k 1.
Proof Again, we present the proof only for leftclip(D). Let K be a clique in leftclip(D).
By Claim 14, each member of K is arcconnected. Therefore, by Claim 13, each
member of K intersects an interval of sets in the sequence R1, I1, . . . , Rm1, Im1, Rm ,
the first set in this interval being of the form R j . Since the members of K pairwise
intersect, these intervals also pairwise intersect, which implies that they all contain a
common R j . Thus K { R j } is a clique, and (R D) k yields K k 1.
L = { X D : base(leftclip( X )) = },
R = { X D : base(rightclip( X )) = }.
D
D
STihnecreefeoarceh, mit eismebneoruogfhDtoinstehroswectthsaatt lea(Dst Lo)ne of4Rk12, 1. .a.n,dRm(,DweR )have4Dk2L1.DWRe=onDly.
present the proof of (D R ) 4k21, as the proof of the other inequality is analogous.
By Claims 14 and 15, rightclip(D R ) is a grounded family with clique number at
most k 1. This and the induction hypothesis yield (rightclip(D R )) k1. We fix a
coloring R of D R with k1 colors so that R ( X ) = R (Y ) for any X, Y D R with
rightclip( X ) rightclip(Y ) = . Let M D R be a family of sets having the same
color in R . In particular, we have rightclip( X ) rightclip(Y ) = for any X, Y M.
It remains to prove that (M) 4k1.
We show how to construct a coloring L of M with k1 colors such that
leftclip( X ) leftclip(Y ) = for any X, Y M with L ( X ) = L (Y ). We exploit the
fact that members of M have pairwise disjoint intersections with each Ri to simplify
the topology of M and R1, . . . , Rm . Recall that S is an arc with base(S) = { p, q}.
For 1 i m, by Lemma 2, there is an arc Qi Ri between base( Ri ) and
Ri S that is simple with respect to M. We assume without loss of generality that
base(Qi ) = {ui } and Qi S = {vi }. The points v1, . . . , vm occur in this order on
S as moving from p to q. Moreover, the arcs Q1, . . . , Qm partition J into m + 1
sets J0, . . . , Jm , each homeomorphic to a closed disc, so that Ji1 Ji = Qi for
1 i m. It is clear that each arcconnected subset of J intersects an interval of sets
in the sequence J0, Q1, J1, . . . , Qm , Jm . Since Q1, . . . , Qm are simple with respect
to M, so are J0, . . . , Jm .
Since the sets J0, . . . , Jm are homeomorphic to a closed disc and so are rectangles
with bottom sides base( J0), . . . , base( Jm ), there are homeomorphisms 0, . . . , m
such that
i is constant on base( Ji ) and maps Ji onto a rectangle with bottom side base( Ji )
for 0 i m,
i1 and i agree on Qi for 1 i m.
Thus 0 m is a homeomorphism between J and a rectangle with bottom side
base( J ), and it extends to a homeomorphism of R [0, +) whose restriction
to each Ji is i . Let 0, . . . , m be horizontal translations such that 0(u)
m (u) for a point u on the baseline. Let x1, . . . , xm be the x coordinates of the points
u1, . . . , um , respectively, so that ui = (xi , 0) for 1 i m. Define
Note that Ji Ji and Qi Q i . For a set X , define
where [Y, Z ] denotes the rectangle with left side Y and right side Z . It is clear that the
map X X preserves compactness and arcconnectedness and is compatible with
unions and intersections. In particular, M = {X : X M} is a grounded family
with intersection graph isomorphic to that of M (see Fig. 6).
In the remainder of the proof, we will deal with M and R1, . . . , Rm , but for
simplicity we relabel them to M and R1, . . . , Rm , respectively. We also relabel I0 , . . . , Im
to I0, . . . , Im , Q1, . . . , Qm to Q1, . . . , Qm , S to S, J to J , and 0( J0), . . . , m ( Jm )
to J0, . . . , Jm . The following properties clearly follow:
J0, . . . , Jm are pairwise disjoint rectangles,
Qi is a rectangle whose left side is the right side of Ji1 and whose right side is the
left side of Ji , for 1 i m,
every arcconnected subset of J intersects an interval of sets in the sequence
J0, Q1, J1, . . . , Qm , Jm ,
Qi Ri for 1 i m,
the intersection of any member of M with any Ji is arcconnected,
the intersection of any member of M with any Qi is a rectangle or horizontal
segment spanning the entire width of Qi .
Fig. 6 The transformation X X ; top: a family M before transformation; bottom: the families M
(including the marked regions) and M+ (excluding the marked regions), and connections of the sets in
M to the baseline
with intersection graph isomorphic to that
Proof Let Mi = { X M : X Ri = } for 1 i m. It follows that leftclip( X ) =
X Ii for every X Mi+1 Mi and 1 i < m. We can assume without loss of generality that each member of Mi+1 Mi intersects Qi+1 for 1 i < m, as those that do not are isolated vertices in the intersection graph of leftclip(M) and thus do not influence the existence of M .
For X Mi+1 Mi and 1 i < m, let X + denote the part of X to the right of Qi+1
including the right side of Qi+1, that is, X + = X ( Ji+1 Qi+2 Ji+2 Qm Jm ).
It follows that X + leftclip( X ) X + Ri+1 and X + is compact and arcconnected.
For X M1, let X + = X = leftclip( X ). Let M+ = { X + : X M}. Since the
intersections of the members of leftclip(M) with each Ri+1 are pairwise disjoint, the
intersection graphs of M+ and leftclip(M) are isomorphic. We show how to extend
the sets X + for which base( X +) = base(leftclip( X )) = to connect them to the
baseline without creating any new intersections (see Fig. 6).
Let 1 i < m, X Mi+1 Mi , and base( X +) = . Thus base( X ) base( Ji ).
For every Y Mi , it is an immediate consequnce of the Jordan curve theorem and
arcconnectedness of X Ji and (Y Ri ) Ji that Y Qi+1 is empty or lies above
X Qi+1. Therefore, we can connect X + to base(Qi+1) by an arc inside Qi+1 that
is disjoint from any other Y + M+. Moreover, all these arcs for X Mi+1 Mi
with base( X +) = can be drawn so that they are pairwise disjoint. Doing so for all
i with 1 i < m, we transform M+ into a grounded family M with intersection
graph isomorphic to the intersection graph of M+ and hence of leftclip(M).
Claim 16 allows us to use the induction hypothesis on leftclip(M) to obtain a
coloring L of M
with L w(Xith) =k1Lc(oYl)o.rsLesutcNh that leftclip( X ) leftclip(Y ) = for any
X, Y M M be a family of sets having the same
color in L . In particular, we have leftclip( X ) leftclip(Y ) = and rightclip( X )
rightclip(Y ) = for any X, Y N . The following claim completes the proof of
Lemma 9. The planarity argument used in the proof applies the ideas of McGuinness
[10].
NNNiLLR === {{{XXX NNN L:: :XXX lreilgfethfcttlccipllii(ppX((XX) ))==}J,i}}, for 1 i < m,
NiR = { X N R : X rightclip( X ) Ji } for 1 i < m,
Ni = NiL NiR ,
C XL to be the arcconnected component of ( Ni ) Ji containing X leftclip( X ),
for X NiL and 1 i < m,
C XR to be the arcconnected component of ( Ni ) Ji containing X rightclip( X ),
for X NiR and 1 i < m,
C to be the family of arcconnected components of ( Ni ) Ji for 1 i < m,
that is, C = {C XL : X N L } {C XR : X N R }.
For X Ni and 1 i < m, the set X Ji is arcconnected and thus entirely contained
in one member of C. It is clear that C XL = C XR for X N L N R . Consider the graph
G with vertex set C and edges connecting C XL and C XR for all X N L N R . We are
to show that G is planar.
We construct sets V E N with the following properties:
(i) V is a finite subset of C,
(ii) E is a finite union of arcs with endpoints in V , pairwise disjoint outside of V ,
(iii) E C is arcconnected for every component C C,
(iv) every X N L N R contains an arc in E between C XL and C XR .
The construction proceeds by induction on the members of N . We start with V
containing one (arbitrary) point in each member of C, and with E = V . They clearly
satisfy (i)(iii). Then, for each X N L N R , we enlarge V and E to satisfy the
condition (iv) for X , as follows. Since C L
X X C XR is arcconnected, there is an arc
A C L X
X X C XR between E C XL and E C R . We can furthermore assume that
A E C XL = {v L } and A E C XR = {v R }. This implies that A E = {v L , v R },
aasndX A t(oCEXL. ACfteXRr)pirsodciessjosiinngt farlolmXanyNmLembNer Rof, tNhe re{sXul}t.inWgeseatdsdVv LanadndEv sRattoisfVy
(i)(iv).
We have thus obtained a planar representation of a graph H with vertex set V and
edge set consisting of maximal arcs in E internally disjoint from V . It follows from
(iii) that the subgraph of H induced on V C is connected for every C C. Consider
the graph obtained from H by contracting V C for every C C. Its vertices represent
members of C, and by (iv), its edges connect the vertices representing C XL and C XR for
all X N L N R . Hence this graph is isomorphic to G. This shows that G is planar,
as a contraction of a planar graph is planar.
Since G is planar, there is a proper coloring of the vertices of G with four colors
{1, 2, 3, 4}. Choose a coloring : N {1, 2, 3, 4} so that
Such a coloring exists, because (C XL ) = (C XR ) for any X N L N R . To see that
is a proper coloring of N , consider some X, Y N such that X Y and X Y = .
Since L ( X ) = L (Y ) and R ( X ) = R (Y ) we have leftclip( X ) leftclip(Y ) =
and rightclip( X ) rightclip(Y ) = . Therefore,
rightclip( X )) (Y
leftclip(Y )) = X Y = .
In particular, we have X N R and Y N L . The set X Y is arcconnected, so
C R
X = C L . This yields ( X ) = (C R ) = (CYL ) = (Y ).
Y X
Acknowledgments M. Lason, P. Micek, and A. Pawlik were supported and B. Walczak was partially
supported by Ministry of Science and Higher Education of Poland Grant 884/NESFEuroGIGA/10/2011/0
within ESF EuroGIGA project GraDR. B. Walczak was partially supported by Swiss National Science
Foundation Grant 200020144531.
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