ON MAXIMAL S-FREE CONVEX SETS∗
DIEGO A. MORÁN R.† AND SANTANU S. DEY‡
Abstract. Let S ⊆ Zn satisfy the property that conv(S) ∩ Zn = S. Then a convex set K is
called an S-free convex set if int(K) ∩ S = ∅. A maximal S-free convex set is an S-free convex set
that is not properly contained in any S-free convex set. We show that maximal S-free convex sets
are polyhedra. This result generalizes a result of Basu et al. [6] for the case where S is the set of
integer points in a rational polyhedron and a result of Lovász [18] and Basu et al. [5] for the case
where S is the set of integer points in some affine subspace of Rn .
Key words. Integer nonlinear programming, Cutting planes, Maximal lattice-free convex sets
AMS subject classifications. 90C11, 90C57
1. Introduction. A convex set is called lattice-free if it contains no integer
points in its interior. A maximal lattice-free convex set is a lattice-free convex set not
strictly contained in any lattice-free convex set. Lovász [18] and Basu et al. [5] proved
that maximal lattice-free convex sets are polyhedra. Given a convex set P ⊆ Rn , let
S be the set of integer points contained in P, that is S = P ∩ Zn . A set K is called
S-free convex set if int(K) ∩S = ∅. Hence the concept of S-free convex sets represents
a generalization of the concept of lattice-free convex sets. Maximal S-free convex sets
are defined analogously to maximal lattice-free convex sets. Dey and Wolsey [12] show
that maximal S-free convex sets are polyhedra under restrictive conditions. Fukasawa
and Günlük [14] show that maximal S-free convex sets are polyhedra where P is a
rational polyhedron in R2 . Basu et al. [6] prove that if P is any general rational
polyhedron, then maximal S-free convex sets are polyhedra. A natural question then
is to ask whether the polyhedrality of maximal S-free convex sets carries through to
the case where P is an arbitrary convex set, that is the case where S ⊆ Zn is assumed
only to satisfy the property that conv(S) ∩ Zn = S.
In this paper we verify that maximal S-free convex sets are polyhedra, where S is
the set of integer points in any convex set P ⊆ Rn . Since the proof in Basu et al. [6]
explicitly uses that fact that conv(S) is a rational polyhedron when S is the set of
integer points contained in a rational polyhedron, the result in this paper also provides
an alternative proof to the result in [6]. Finally, we establish the polyhedrality of
maximal S-free convex sets in some cases where S does not satisfy conv(S) ∩ Zn = S.
We now briefly describe one motivation for the study of maximal S-free convex
sets. A key algorithmic technique in solving mixed integer optimization problems is
to sequentially obtain tighter approximations of the convex hull of integer feasible
solutions. This is achieved by the addition of cutting planes, that is inequalities
that separate fractional point from the convex hull of integer feasible points. See for
example Marchand et al. [19] and Johnson et al. [17] for description of cutting plane
methods. A connection between S-free convex sets and cutting planes for mixed
integer linear programming was first discovered by Balas [3]. The main idea is the
∗ This
research was supported by NSF Grant CMMI 1030422.
Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology,
Atlanta, GA, USA. ([email protected]).
‡ H. Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology,
Atlanta, GA, USA. ([email protected]).
† H.
1
2
D. A. MORÁN R. AND S. S. DEY
following. Consider the mixed integer set
T := {(x, y) ∈ Zp × Rq : (x, y) ∈ C},
(1.1)
where C is a convex set. Now suppose that we are able to identify a set S ⊆ Zn such
that S ⊇ Projx (T ). If B ⊆ Rp is an S-free convex set, then by letting B̂ = B × Rq
we can construct a valid relaxation of conv(T ) as conv(C \ int(B̂)). Often by a good
choice of B, conv(C \ int(B̂)) is a much better approximation of the convex hull of T
than C. A classical example of this procedure is that of split disjunctions; see Cook
et al. [10]. Notice that if B 1 and B 2 are S-free convex sets such that B 1 ⊇ B 2 , then
conv(C \ int(Bˆ1 )) ⊆ conv(C \ int(Bˆ2 )). This motivates the search for maximal S-free
convex sets. Various families of cutting planes based on maximal S-free convex sets
(for different S) have been proposed. See for example Andersen et al. [1], Andersen et
al. [2], Basu et al. [6], Borozan and Cornuéjols [9], Cornuéjols and Margot [11], Dey
and Wolsey [12], Fukasawa and Günlük [14], Johnson [16], Zambelli [23]. Since we
verify that maximal S-free convex sets are polyhedra, one possible way to compute
conv(C \ int(B̂)) is by computing the convex hull of the following disjunction
t
[
(C ∩ {(x, y) : hai , xi ≥ bi }),
(1.2)
i=1
where B = {x ∈ Rp : hai , xi ≤ bi , i ∈ {1, ..., t}}. For many classes of convex
sets C, the convex hull of the union of sets in (1.2) can be described ‘conveniently’;
see for example the case of second order conic representable sets in Ben-Tal and
Nemirovski [8].
The rest of the paper is organized as follows. In Section 2, we review some
standard results from convex analysis and present a key result about maximal latticefree convex sets in affine subspaces due to Basu et al. [5]. In Section 3, we present
the characterization of maximal S-free convex sets. In Section 4, we discuss some
differences between the properties of maximal S-free convex sets in the general case
to the case where S is the set of integer points contained in a rational polyhedron and
point out some generalizations of the results presented in Section 3.
2. Preliminaries. Let W be an affine subspace of Rn . We use intW (A) to
denote the interior of the set A ⊆ W with respect to the topology induced by Rn on
W . Therefore rel.int(A) = intaff.hull(A) (A). We will call a set H a half-space (resp.
hyperplane) of W if H is the intersection of W with a half-space (resp. hyperplane)
of Rn and W * H. For u ∈ Rn and ε > 0, B(u, ε) is the open ball around u of radius
ε > 0, that is B(u, ε) := {x ∈ Rn : ||x − u|| < ε}.
We will frequently use the following basic result from convex analysis that we
prove for completeness.
Lemma 2.1. Let U ⊆ V ⊆ Rn be affine subspaces and let A ⊆ V be a convex set
such that intV (A) ∩ U 6= ∅. Then intV (A) ∩ U = intU (A ∩ U ).
Proof. The inclusion intV (A) ∩ U ⊆ intU (A ∩ U ) is straightforward. For the
other inclusion, assume that y ∈ intU (A ∩ U ). Then there exists ε > 0 such that
B(y, ε) ∩ U ⊆ A ∩ U . Since intV (A) ∩ U 6= ∅, there exists x ∈ intV (A) ∩ U . If x = y,
then the proof is complete. Otherwise, since U is a affine subspace, we obtain that
ε
(y − x) ∈ B(y, ε) ∩ U . It follows that y is a strict convex combination
z = y + 2kx−yk
of z ∈ A and x ∈ intV (A), so y ∈ intV (A), and then y ∈ intV (A) ∩ U .
Using Lemma 2.1, we restate a version of separation theorem for convex sets that
we use later.
On Maximal S-free Convex Sets
3
Theorem 2.2 (Separation Theorem). Let W ⊆ Rn be an affine subspace. If
A, B ⊆ W , A and B are convex sets, and rel.int(A) ∩ rel.int(B) = ∅, then there exists
half-space H of W such that A ⊆ H and intW (H) ∩ B = ∅.
Proof. By Theorem 4.14 in Hiriart-Urrut and Lemaréchal [15], there exists s ∈ Rn
such that
supy∈A hs, yi ≤ infy∈B hs, yi
infy∈A hs, yi < supy∈B hs, yi.
(2.1)
(2.2)
Let H̃ = {x ∈ Rn : hs, xi ≤ supy∈A hs, yi} and H = H̃ ∩ W . By (2.2), s is not
orthogonal to W . Therefore H is a halfspace of W . Finally, by Lemma 2.1 we obtain
intRn (H̃) ∩ W = intW (H), which completes the proof.
Listed below are some properties of interiors and relative interiors of convex sets
that are also used frequently. See Hiriart-Urrut and Lemaréchal [15] and Rockafellar [20] for proofs.
Proposition 2.3. Let W be an affine subspace and let A, B ⊆ W be convex sets.
Then,
1. intW (A) ∩ intW (B) = intW (A ∩ B).
2. rel.int(A) ∩ rel.int(B) ⊆ rel.int(A ∩ B).
3. If rel.int(A) ∩ rel.int(B) 6= ∅, then rel.int(A) ∩ rel.int(B) = rel.int(A ∩ B).
4. A ⊆ rel.int(A), where C represent the closure of C.
5. rel.int(A) + rel.int(B) = rel.int(A + B).
Definition 2.4 (S-free and Lattice-free Convex sets). Let W ⊆ Rn be an affine
subspace of dimension dim(W ) ≥ 1, P ⊆ W be a convex set and S = P ∩ Zn . A set
K is called S-free (resp. lattice-free) convex set of W if K ⊆ W , K is convex and
intW (K) ∩ S = ∅ (resp. intW (K) ∩ Zn = ∅). A convex set K ⊆ W is called maximal
S-free (resp. lattice-free) convex set of W if K is S-free (resp. lattice-free) convex set
and there does not exists a set K 0 ⊆ W such that K 0 6= K, K 0 ⊇ K and K 0 is an
S-free (resp. lattice-free) convex set of W .
Basu et al. [5] proved that maximal lattice-free convex sets of affine subspaces
are polyhedra. This is a crucial ingredient in the proof presented in this paper. We
present this result next.
Theorem 2.5 (Basu et al. [5]). Let W ⊆ Rn be an affine subspace containing
an integral point and V be the affine hull of W ∩ Zn . A set K ⊆ W is a maximal
lattice-free convex set of W if and only if one of the following holds:
1. K is a polyhedron in W whose dimension equals dim(W ), K ∩V is a maximal
lattice-free convex set of V whose dimension equals dim(V ), and for every
facet F of K, F ∩ V is a facet of K ∩ V ,
2. K is an affine hyperplane of W such that K ∩ V is an irrational hyperplane
of V ,
3. K is a half-space of W that contains V on its boundary.
The proof of Theorem 2.5 in Basu et al. [5] involves showing that if K is a latticefree convex set of W , then K must be contained in maximal lattice-free convex set of
W . We will therefore use the following simplified version of Theorem 2.5.
Theorem 2.6 (Basu et al. [5]). Let W ⊆ Rn be an affine subspace containing an
integral point. If K ⊆ W is a lattice-free convex set of W , then it is contained in a
maximal lattice-free convex set B of W , where B is a polyhedron.
3. Maximal S-free Convex Sets. As observed in Basu et al. [5], the existence
of maximal S-free convex sets is a consequence of Zorn’s Lemma.
4
D. A. MORÁN R. AND S. S. DEY
We will verify the following results in this section.
Theorem 3.1. Let W ⊆ Rn be an affine subspace of dimension dim(W ) ≥ 1, V
the affine hull of W ∩ Zn , P ⊆ W a convex set, S = P ∩ Zn and P = conv(S). If
K ⊆ W is a maximal S-free convex set, then one of the following holds:
1. dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6= ∅: Then K is a polyhedron
in W , K ∩ V is a maximal S-free convex set of V whose dimension equals
dim(V ), and for every facet F of K, F ∩ V is a facet of K ∩ V ,
2. dim(K) < dim(W ): Then K is an hyperplane of W ,
3. dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) = ∅: Then K is a halfspace of
W,
4. S = ∅ and K = W .
If S is the set of integer points contained in an rational polyhedron, then every
facet of maximal S-free polyhedron contains a point of S in its relative interior. A
slightly weaker result holds in the general case. Before we present this result, we
require some additional notation. Let K be a polyhedron of W , that is let K = {x ∈
W : hai , xi ≤ bi ∀i ∈ {1, ..., N }}. Then we denote the ith facet of K as Fi (K). Also
for all ε > 0, let Fiε (K) = {x ∈ W : haj , xi < bj ∀ j 6= i and bi < hai , xi < bi + ε}.
Theorem 3.2. Let W ⊆ Rn be an affine subspace of dimension dim(W ) ≥ 1,
P ⊆ W a convex set and S = P ∩ Zn . Let K be a S-free convex set such that
dim(K) = dim(W ). If K is a maximal S-free convex set, then K is a polyhedron with
N facets such that
1. (rel.int(Fi ) ∪ Fiε (K)) ∩ S 6= ∅
∀ ε > 0 ∀ i ∈ {1, . . . , N } and
2. N ≤ 2dim(conv(S)) .
The rest of the section is organized as follows. In Section 3.1, we prove that
maximal S-free convex sets are polyhedra. In Section 3.2, we verify properties of facets
of maximal S-free convex sets. In particular, if dim(K) = dim(W ) and rel.int(K) ∩
rel.int(P ) 6= ∅, then we show that K ∩ V is a maximal S-free convex set of V whose
dimension equals dim(V ), and for every facet F of K, F ∩V is a facet of K ∩V (where
V = aff.hull(W ∩Zn )). We also verify part (1.) of Theorem 3.2 in this section. Finally
in Section 3.3, we obtain an upper bound on the number of facets of maximal S-free
convex sets, completing the proof of Theorem 3.2. This upper bound is obtained
using the upper bound result for maximal lattice-free sets (Doignon [13], Bell [7],
Scarf [21]) but involves a little more work as facets of maximal S-free convex sets do
not in general contain points of S in their relative interior.
3.1. Polyhedrality of Maximal S-free Convex Sets. To show that a maximal S-free convex set is a polyhedron, it is sufficient to show that every S-free convex
set is contained in a polyhedral S-free convex set. This is verified next.
Proposition 3.3. Let W ⊆ Rn be an affine subspace of dimension dim(W ) ≥ 1,
P ⊆ W be a convex set, S = P ∩ Zn and P = conv(S). Let K ⊆ W be an S-free
convex set of W . Then one of the following holds:
1. dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6= ∅: K is contained in an
S-free convex set B ⊆ W such that B is a polyhedron,
2. dim(K) < dim(W ): Then K is contained in an S-free hyperplane of W ,
3. rel.int(K) ∩ rel.int(P ) = ∅: Then K is contained in an S-free halfspace of W ,
4. S = ∅ and W is an S-free convex set (K is contained in W ).
Since the proof of Proposition 3.3 is technical, we summarize the main steps here.
The proof is by induction on the dimension of W . We first prove cases (2.), (3.) and
(4.). The proof of these three cases does not require the use of the induction hypothesis. To prove case (1.), we carefully separate K from S, i.e., determine half-spaces of
On Maximal S-free Convex Sets
5
W such that the polyhedron defined by the intersection of these half-spaces contains
K and does not contain any point of S in its interior (wrt W ). To achieve this goal, we
first observe that K ∩ P is a lattice-free convex set and thus there exists a maximal
lattice-free polyhedron Q ⊆ W containing K ∩ P . Then we examine each facet of
Q. Corresponding to each facet F of Q, we separate K from the subset of S that
the inequality defining F separates from K ∩ P , using a finite number of separating
half-spaces. In most cases, this is achieved via a straightforward application of the
Separation Theorem. However for facets of Q where the defining hyperplane, say H = ,
has a non-empty intersection with S and also with the interior of K, these separating
half-spaces can be found by first determining a maximal (S ∩ H = )-free polyhedron
contained in (W ∩ H = ). The existence of this polyhedron is by the induction hypothesis. In the last step of the proof, it is verified that the polyhedron constructed as the
intersection of the above separating half-spaces (corresponding to all the facets of Q)
does not contain any point of S in its interior (wrt W ). Since Q has a finite number
of facets, the intersection of all these separating half-spaces is a S-free polyhedron
containing K.
Proof. The proof of (1.), (2.), (3.) and (4.) is by induction on the dimension of
W . Consider first the case where dim(W ) = 1. Then since K ⊆ W is a convex set
and dim(K) ≤ 1, we conclude that K is an S-free polyhedron. Therefore the cases
(1.), (2.), (3.) and (4.) are easily verifiable when dim(W ) = 1.
Assume now that we have (1.), (2.), (3.) and (4.) for every affine subspace of
dimension less than dim(W ). We first prove cases (2.), (3.) and (4.) and then we
prove the more complicated case (1.)
If S = ∅, then W is an S-free convex set and this completes the proof. Hence, we
can assume S 6= ∅. Observe that since P is convex, S = P ∩ Zn .
If dim(K) < dim(W ), then K is contained in a hyperplane H of W . Notice that
we must have intW (H) = ∅, for otherwise dim(H) = dim(W ) and therefore H = W ,
a contradiction with W ( H. Since intW (H) ∩ S = ∅, the proof is complete.
If rel.int(K) ∩ rel.int(P ) = ∅, then by separation theorem there exists a half-space
H ≤ of W such that K ⊆ H ≤ and P ∩intW (H ≤ ) = ∅. Thus, K is contained in an S-free
half-space. Therefore we can assume dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6=
∅.
Since K is a S-free convex set, we obtain that intW (K ∩ P ) ∩ Zn = intW (K) ∩
intW (P ) ∩ Zn ⊆ intW (K) ∩ (P ∩ Zn ) = intW (K) ∩ S = ∅. So we conclude K ∩ P is
a lattice-free convex set of W . Also since S 6= ∅, we obtain that W ∩ Zn 6= ∅. Hence,
by Theorem 2.6, there exists a maximal lattice-free convex set of W , Q ⊆ W such
that K ∩ P ⊆ Q and Q is a polyhedron. Observe also that since S 6= ∅, Q ( W . We
therefore obtain that
Q=
m
\
Hi≤ ,
i=1
where each Hi≤ , i = 1, . . . , m is a half-space of W defining a facet of Q. For
i ∈ I = {1, . . . , m}, denote Hi≥ = W \ (intW (Hi≤ )) and Hi= = Hi≤ ∩ Hi≥ .
We partition the set I into four disjoints subsets.
1. I1 := {i ∈ I : rel.int(P ) ∩ rel.int(Hi≥ ) 6= ∅}. If i ∈ I1 , then rel.int(P ∩ Hi≥ ) =
6
D. A. MORÁN R. AND S. S. DEY
rel.int(P ) ∩ rel.int(Hi≥ ). Therefore
rel.int(K) ∩ rel.int(P ∩ Hi≥ ) = rel.int(K) ∩ rel.int(P ) ∩ rel.int(Hi≥ )
= rel.int(K ∩ P ) ∩ rel.int(Hi≥ )
⊆ Q ∩ rel.int(Hi≥ ) = ∅.
The second equality holds because rel.int(K)∩rel.int(P ) 6= ∅. Since rel.int(K)∩
rel.int(P ∩ Hi≥ ) = ∅, we obtain that there exists Gi , a half-space of W , such
that
intW (Gi ) ∩ P ∩
K ⊆ Gi ,
(3.1)
Hi≥
(3.2)
= ∅.
2. Consider I2 := {i ∈ I \ I1 : intW (K) ∩ Hi= = ∅, S ∩ Hi= 6= ∅}.
In this case, we verify that K ⊆ Hi≤ . Since K ⊆ rel.int(K) = intW (K)
and Hi≤ is a closed convex set, it is sufficient to verify that intW (K) ⊆ Hi≤ .
Assume by contradiction, that there exists x ∈ intW (K) ∩ intW (Hi≥ ). Since
∅ 6= rel.int(K) ∩ rel.int(P ) ⊆ K ∩ P ⊆ Hi≤ , we obtain that rel.int(K) ∩
Hi≤ 6= ∅. However, note that since the affine hull of rel.int(K) and Hi≤ is
W , rel.int(K) is an open set (wrt W ) and Hi≤ is a half space of W , we
obtain that rel.int(K) ∩ rel.int(Hi≤ ) 6= ∅. (Choose ỹ ∈ rel.int(K) ∩ Hi≤ . If
ỹ ∈ rel.int(H ≤ ), then we are done. Otherwise, since ỹ ∈ rel.int(K) and affine
hull of rel.int(K) is W , it is possible to choose a neighborhood B(ỹ, ε) ∩ W
of ỹ contained in rel.int(K). However, since ỹ ∈
/ rel.int(Hi≤ ), we obtain that
≤
ỹ ∈ Hi= . Therefore, B(ỹ, ε) ∩ W ∩ rel.int(Hi ) 6= ∅ and thus there exists y ∈
rel.int(K)∩rel.int(Hi≤ )). Let y ∈ rel.int(K)∩rel.int(Hi≤ ). Hence every convex
combination of x and y belongs to intW (K) since x, y ∈ intW (K). Moreover,
since x ∈ intW (Hi≥ ) and y ∈ intW (Hi≤ ), there exists a convex combination of
x and y that belongs to intW (K) ∩ Hi= . Therefore intW (K) ∩ Hi= 6= ∅, which
contradicts the fact that i ∈ I2 . If i ∈ I2 , then define Gi = Hi≤ . Therefore
from the above, we obtain
K ⊆ Gi ,
Hi= ∩ intW (Gi ) = ∅.
(3.3)
(3.4)
3. Consider I3 := {i ∈ I \ I1 : intW (K) ∩ Hi= 6= ∅, S ∩ Hi= 6= ∅}. If i ∈ I3 ,
then by Lemma 2.1 we obtain that intHi= (K ∩ Hi= ) = intW (K) ∩ Hi= .
Since K is an S-free convex set and intHi= (K∩Hi= ) = intW (K)∩Hi= we obtain
that intHi= (K ∩ Hi= ) ∩ S = ∅. Hence, K ∩ Hi= is a (S ∩ Hi= )-free convex set
of Hi= . Note that since W ( Hi≤ , we obtain that dim(Hi= ) = dim(W ) − 1,
i.e., 1 ≤ dim(Hi= ) < dim(W ). By the induction hypothesis, there exists a
polyhedron Ti ⊆ Hi= such that K ∩ Hi= ⊆ Ti , Ti is a (S ∩ Hi= )-free convex set
of Hi= . Note that since (S ∩ Hi= ) 6= ∅, we obtain that Ti 6= Hi= . Therefore
Tm i ≤
T = j=1
Fij , where Fij≤ ⊆ Hi= , j = 1, . . . mi are the half-spaces of Hi=
defining the facets of Ti . Denote Fij≥ = Hi= \ (intHi= (Fij≤ )). We have:
7
On Maximal S-free Convex Sets
Case I1=
H
S
Case I2
Case I3
Gi1
Gi2
G
S
Q
S
Q
Q
H=
H=
K
K
K
Fig. 3.1. Illustration of case I1 , I2 , and I3 .
rel.int(Fij≥ ) ∩ rel.int(K) = rel.int(Fij≥ ) ∩ Hi= ∩ intW (K)
⊆ Fij≥ ∩ intHi= (K ∩ Hi= )
⊆ Fij≥ ∩ intHi= (Ti ) = ∅.
Therefore there exists Gij , half-spaces of W , such that:
Fij≥
K ⊆ Gij ,
(3.5)
∩ intW (Gij ) = ∅.
(3.6)
Tm i
Gij .
In this case, define Gi = j=1
We verify a property that we require later. Note that since intW (K) ⊆
intW (Gij ) and intW (K)∩Hi= 6= ∅, we obtain that intW (Gij )∩Hi= 6= ∅. Therefore using Lemma 2.1 we obtain the equality intHi= (Gij ∩ Hi= ) = intW (Gij ) ∩
Hi= . Therefore, (3.6) is equivalent to
Fij≥ ∩ intHi= (Gij ∩ Hi= ) = ∅.
(3.7)
4. Finally define I4 := {i ∈ I \ I1 : S ∩ Hi= = ∅}. Note that I = I1 ∪ I2 ∪ I3 ∪ I4 .
Before the final step in the proof, we verify that if rel.int(P ) ∩ rel.int(Hi≥ ) = ∅,
then
P ∩ rel.int(Hi≥ ) = ∅.
(3.8)
As rel.int(P ) ∩ rel.int(Hi≥ ) = ∅, we obtain rel.int(P ) ⊆ Hi≤ . Since Hi≤ is a closed
set, we obtain rel.int(P ) ⊆ Hi≤ . Finally, note that since P is a convex set, we obtain
≥
P ⊆ rel.int(P ) ⊆ Hi≤ (see Proposition 2.3). Thus P ∩ rel.int(H
T i ) = ∅.
To complete the proof we will verify that the set B = i∈I\I4 Gi ⊆ W is an Sfree convex set of W containing K. By (3.1), (3.3) and (3.5), we obtain that K ⊆ B.
We need to prove that B is an S-free convex set to complete the proof. Assume by
contradiction that there exists y ∈ S ∩ intW (B), that is y ∈ S and y ∈ intW (Gi )
∀i ∈ I \ I4 .
8
D. A. MORÁN R. AND S. S. DEY
We have
P =W
" ∩P
#
[ ≥
=
Hi ∪ intW (Q) ∩ P
"
⊆
"
=
i∈I
[
#
(Hi≥
∩ P ) ∪ intW (Q)
i∈I
[
(Hi≥
∩ P) ∪
i∈I1
[
(Hi= ∩ P ) ∪
i∈I2
[
(Hi= ∩ P ) ∪
i∈I3
[
#
(Hi= ∩ P )
i∈I4
∪ intW (Q).
(3.9)
The last equality is a consequence of (3.8), since for i ∈ I2 ∪ I3 ∪ I4 we have that
(Hi≥ ∩ P ) = (rel.int(Hi≥ ) ∩ P ) ∪ (Hi= ∩ P ) = (Hi= ∩ P ).
Since y ∈ S = P ∩ Zn and S ∩ Hi= = ∅ for i ∈ I4 , we obtain that y ∈
/ Hi= ∩ P for
i ∈ I4 . Moreover Q is lattice-free. Therefore, using (3.9) there are three cases:
1. For i ∈ I1 , if y ∈ Hi≥ ∩ P and y ∈ intW (Gi ), then we obtain a contradiction
to (3.2).
2. For i ∈ I2 , if y ∈ Hi= and y ∈ intW (Gi ), then we obtain a contradiction to
(3.4).
Tm i
Tm i
intHi= (Gij ∩
intW (Gij )∩Hi= = j=1
3. For i ∈ I3 , if y ∈ Hi= , then since y ∈ j=1
Tm i
Tm i
intHi= (Fij ≤ ) (the last inclusion is a conse(Hi= \ Fij ≥ ) = j=1
Hi= ) ⊆ j=1
quence of (3.7)), we obtain that y ∈ intHi= (Ti ) which is in contradiction with
the fact that Ti is an S ∩ Hi= -free convex set of Hi= .
Therefore B is S-free polyhedron containing K.
Cases (2.), (3.) and (4.) of Theorem 3.1 follow from maximality of K and cases
(2.), (3.) and (4.) of Proposition 3.3 respectively. Case (1.) of Proposition 3.3
shows that maximal S-free convex sets are polyhedra when dim(K) = dim(W ) and
rel.int(K) ∩ rel.int(P ) 6= ∅. To complete the proof of Theorem 3.1, we need to show
that K ∩ V is a maximal S-free convex set of V whose dimension equals dim(V ), and
for every facet F of K, F ∩ V is a facet of K ∩ V (where V = aff.hull(W ∩ Zn )). This
is verified in the next section.
3.2. Structure of Facets of Maximal S-free Convex Sets. For convenience
we repeat the definition of some notation. Let K be a polyhedron of W , that is let
K = {x ∈ W : hai , xi ≤ bi ∀i ∈ {1, ..., N }}. Then we denote the ith facet of K as
Fi (K). Also for all ε > 0, let Fiε (K) = {x ∈ W : haj , xi < bj ∀j 6= i and bi <
hai , xi < bi + ε}.
Before completing the proof of Theorem 3.1, we prove part (1.) of Theorem 3.2.
Proposition 3.4. Let K = {x ∈ W : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }} ⊆ Rn be
a maximal S-free convex set, such that dim(K) = dim(W ). Then
(rel.int(Fi (K)) ∪ Fiε (K)) ∩ S 6= ∅
∀ ε > 0 ∀ i ∈ {1, . . . , N }.
Proof. For ε > 0 and i ∈ {1, ..., N } consider the set
Kiε = {x ∈ W : haj , xi ≤ bj ∀j 6= i and hai , xi ≤ bi + ε}.
Since K ( Kiε we obtain that Kiε is not an S-free convex set, that is, intW (Kiε )∩S 6= ∅.
Hence, the set intW (Kiε ) \ intW (K) must contain a point of S. Observe that:
On Maximal S-free Convex Sets
9
intW (Kiε ) \ intW (K) = {x ∈ W : haj , xi < bj ∀ j 6= i and bi ≤ hai , xi < bi + ε}
= Fiε (K) ∪ rel.int(Fi (K)).
Therefore ∅ 6= intW (Kiε ) \ intW (K) ∩ S = (Fiε (K) ∪ rel.int(Fi (K))) ∩ S.
Note that Proposition 3.4 highlights an important difference between maximal
S-free convex sets for general S and for the case where S is the set of integer points
contained in a rational polyhedron. In the case of general S, there is no guarantee
that every facet of a full-dimensional (wrt W ) maximal S-free convex set contains
points belonging to S in its relative interior.√This is illustrated in the next example.
2
Example 3.5.
√ Let S = {x ∈ Z : 2x1 + x2 ≤ 0, x1 ≥ 1}. Then the set
2
K = {x ∈ R : 2x1√+ x2 ≥ 0} is a maximal S-free convex set, but the facet of K
defined by {x ∈ R2 : 2x1 + x2 = 0} contains no point belonging to S.
Now we complete the proof of Theorem 3.1. The proof of the first part of the
next proposition is similar to the proof of a similar property for maximal lattice-free
convex sets, appearing in Basu et al. [5].
Proposition 3.6. Let K = {x ∈ W : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }} ⊆ W be a
maximal S-free convex set such that dim(K) = dim(W ) and intW (K) ∩ rel.int(P) 6= ∅.
Let V be the affine hull of W ∩ Zn . Then K ∩ V is a maximal S-free convex set of
V such that dim(K ∩ V ) = dim(V ) and F ⊆ V is a facet of K ∩ V if and only if
F = Fi (K) ∩ V for some i ∈ {1, . . . , N }.
Proof. Since intW (K)∩rel.int(P) 6= ∅, we obtain that intW (K)∩V 6= ∅. Therefore
Lemma 2.1 implies that intW (K) ∩ V = intV (K ∩ V ).
We first verify that K ∩ V is a maximal S-free convex set of V . Since K is an
S-free convex set and intW (K) ∩ S = intW (K) ∩ S ∩ V = intV (K ∩ V ) ∩ S, we obtain
that K ∩ V is an S-free convex set of V . If K ∩ V is not maximal, then there exist
B ⊆ V , an S-free convex set, such that B ) K ∩ V . Consider K 0 = conv(K ∪ B).
Then K 0 ∩ V = B. We have intW (K 0 ) ∩ S = intW (K 0 ) ∩ S ∩ V = intV (B) ∩ S = ∅.
Therefore K 0 is an S-free convex set of W . Since K 0 ⊇ K and B ) K ∩ V , we obtain
K 0 ) K which is in contradiction with the fact that K is maximal S-free convex set.
Therefore, K ∩ V is a maximal S-free convex set of V .
Since intV (K ∩ V ) 6= ∅, we obtain dim(K ∩ V ) = dim(V ). Finally, we verify that
F ⊆ V is a facet of K ∩ V if and only if F = Fi (K) ∩ V for some i ∈ {1, . . . , N }.
(⇒) If F is a facet of K ∩ V = {x ∈ V : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }}, then
F = Fi (K) ∩ V for some i ∈ {1, . . . , N }.
(⇐) Given ε > 0, by the use of Proposition 3.4 we obtain that for all i = 1, . . . , N ,
∅ 6= (relint(Fi (K)) ∪ Fiε (K)) ∩ S = (rel.int(Fi (K)) ∪ Fiε (K)) ∩ V ∩ S ⊆ intW ({x ∈ W :
hak , xi ≤ bk , ∀ k ∈ {1, . . . , N } \ {i}}) ∩ V ∩ S = intV ({x ∈ V : hak , xi ≤ bk , ∀ k ∈
{1, . . . , N } \ {i}}) ∩ S. Note that the last equality is obtained as a consequence of
Lemma 2.1, {x ∈ W : hak , xi ≤ bk , ∀ k ∈ {1, . . . , N }\{i}} ⊇ K and intW (K)∩V 6= ∅.
As K ∩ V is an S-free convex set of V and dim(K ∩ V ) = dim(V ), we conclude that
hai , xi ≤ bi must define a facet of K ∩ V , that is Fi (K) ∩ V is a facet of K ∩ V for all
i ∈ {1, ..., N }.
3.3. Upper Bound on the Number of Facets of Maximal S-free Convex
Sets. When S is the set of integer points contained in a general convex set, fulldimensional (wrt W ) maximal S-free convex sets do not need to have points of S in
the relative interior of each facet. Therefore, proving upper bound on the number of
facets is slightly more involved than the case of maximal lattice-free convex sets.
10
D. A. MORÁN R. AND S. S. DEY
We first begin with a Lemma that states what is known about maximal S-free
convex sets in the easy case when the set S is defined by a polytope.
Lemma 3.7. Let K 0 ⊆ W , P 0 a polytope, S 0 = P 0 ∩ Zn , S 0 6= ∅ and dim(K 0 ) =
dim(W ). Then K 0 is a maximal S 0 -free convex set of W if and only if K 0 is a S-free
polyhedron with a point of S 0 in the relative interior of each of its facets.
The next Lemma is a standard result. See Schrijver [22] for a proof.
Lemma 3.8. If L ⊆ Rn is a rational affine subspace (i.e. aff.hull(L ∩ Zn ) = L),
then there exists an affine transformation T : Rdim(L) 7−→ L such that T is invertible
and T (Zdim(L) ) = L ∩ Zn .
Using Lemma 3.7 and Lemma 3.8 and a proof similar to that in Doignon [13],
Bell [7], Scarf [21] we obtain the following result.
Lemma 3.9. Let K 0 ⊆ W , P 0 a polytope, S 0 = P 0 ∩ Zn , S 0 6= ∅ and dim(K 0 ) =
dim(W ). If K 0 is a maximal S 0 -free convex set of W , then K 0 is a polyhedron with at
0
most 2dim(P ) facets.
We now have all the tools needed to verify the upper bound on the number of
facets of maximal S-free convex sets.
Proposition 3.10. If K = {x ∈ W : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }} ⊆ W is a
maximal S-free convex set such that dim(K) = dim(W ), then N ≤ 2dim(conv(S)) .
Proof. Let ε > 0. Consider the sets I1 and I2 defined as:
1. i ∈ I1 , if S ∩ rel.int(Fi (K)) 6= ∅.
2. i ∈ I2 , if S ∩ rel.int(Fi (K)) = ∅ and, Fiε (K) ∩ S 6= ∅.
By Proposition 3.4, these sets are well defined and I1 ∪ I2 = {1, ..., N }. If I1 6= ∅,
then for each i ∈ I1 take a point xi ∈ rel.int(Fi (K)) ∩ S and if I2 6= ∅, then for each
i ∈ I2 , take a point xi ∈ Fiε (K) ∩ S.
Define P 0 = conv({x1 , . . . , xN }) and S 0 = P 0 ∩ Zn . Since P 0 ⊆ P, we have that
0
S ⊆ S. This implies that intW (K) ∩ S 0 ⊆ intW (K) ∩ S = ∅, so K is an S 0 -free convex
set.
Note that P 0 is a rational polytope, so by Corollary 3.7 and Lemma 3.9 every
0
maximal S 0 -free convex set K 0 is a polyhedron that has at most 2dim(P ) facets and
contains an integer point of S 0 in the relative interior of each of its facets. Observe
that dim(P 0 ) ≤ dim(conv(S)).
0
If I2 = ∅, then K is a maximal S 0 -free convex set. Therefore N ≤ 2dim(P ) ≤
2dim(conv(S)) .
If I2 6= ∅, then consider i ∈ I2 . We will construct a polyhedron K1 with N facets
such that:
1. K1 is an S 0 -free convex set.
2. Fjε (K) ⊆ Fjε (K1 ), ∀ j ∈ I2 \ {i}.
3. K1 has N facets that are in one-to-one correspondence with the facets of K.
4. K1 has at least |I1 | + 1 facets with a point of S 0 in the relative interior.
We construct K1 from K by changing the right hand side of the ith inequality.
Since P 0 is bounded and S 0 ⊆ Zn , we obtain that |S 0 | is finite. So S 0 ∩ Fiε (K) is also
finite. Moreover, since xi ∈ S 0 ∩ Fiε (K), we have S 0 ∩ Fiε (K) 6= ∅. Hence, there exists
zi ∈ S 0 ∩ Fiε (K) such that
min{hai , xi : x ∈ S 0 ∩ Fiε (K)} = hai , zi i.
Denote b0i = hai , zi i. Consider the polyhedron
K1 = {x ∈ W : haj , xi ≤ bj , ∀ j ∈ I1 ∪ I2 \ {i}, hai , xi ≤ b0i }
We verify that K1 satisfies (1.), (2.), (3.), and (4.):
On Maximal S-free Convex Sets
11
1. Since K is S 0 -free, we only need to show that int(K1 )\int(K) does not contain
points of S 0 . Since b0i ≤ bi + ε we have:
int(K1 ) \ int(K) = {x ∈ W : haj , xi < bj , ∀ j ∈ I1 ∪ I2 \ {i}, bi ≤ hai , xi < b0i , }
⊆ {x ∈ W : haj , xi < bj ∀ j 6= i and bi ≤ hai , xi < bi + ε}
= Fiε (K) ∪ rel.int(Fi (K))
Since i ∈ I2 , S ∩ rel.int(Fi (K)) = ∅ so (int(K1 ) \ int(K)) ∩ S 0 ⊆ Fiε (K) ∩
S 0 . On the other hand, since ∀ x ∈ S 0 ∩ Fiε (K), ha, xi ≥ b0i , we conclude
(int(K1 ) \ int(K)) ∩ S 0 = ∅. Therefore, K1 is a S 0 -free convex set.
2. Using the fact that bi ≤ b0i and j ∈ I2 \ {i} we have,
Fjε (K) = {x ∈ W : hak , xi < bk ∀k 6= j and bj < haj , xi < bj + ε}
⊆ {x ∈ W : hak , xi < bk ∀k =
6 i, j, hai , xi < b0i and bj < haj , xi < bj + ε}
= Fjε (K1 ).
3. By definition, the point zi satisfies hai , zi i = b0i and ∀ j 6= i, haj , zi i < bj .
Therefore the inequality hai , xi ≤ b0i is a facet defining inequality for K1 .
Observe also that the inequalities haj , xi ≤ bj for j 6= i are facet-defining
inequality for K1 . Therefore K1 has N facets.
4. As verified earlier, zi belongs to the relative interior of the facet of K1 defined
by {x ∈ K1 : hai , xi = b0i }. Also for j ∈ I1 , since bi ≤ b0i , we have:
rel.int(Fj (K)) = {x ∈ W : hak , xi < bk ∀k 6= j and haj , xi = bj }
⊆ {x ∈ W : hak , xi < bk ∀k =
6 i, j, hai , xi < b0i and haj , xi = bj }.
Therefore, since xj ∈ rel.int(Fj ) ∩ S 0 , the facet of K1 defined by {x ∈ K1 :
haj , xi = bj } has a point of S 0 in its relative interior. In conclusion, K1 has
at least |I1 | + 1 facets with a point of S 0 in its relative interior.
So we have a procedure to construct, from K, an S 0 -free polyhedron K1 with N
facets that has at least one more facet that contains a point of S 0 in its relative interior
than K has. If K1 is not a maximal S 0 -free convex set, then we can choose a facet
of K1 without a point of S 0 in its relative interior and by properties (1.) and (2.),
repeat the above procedure. We can repeat this a finite number of times, obtaining a
sequence K1 , K2 , . . . , KT of polyhedra with the same number of facets as K and such
that KT does not have any facet without a point of S 0 in its relative interior. So KT
0
is a maximal S 0 -free convex set. Thus, N ≤ 2dim(P ) ≤ 2dim(conv(S)) .
4. Notes.
4.1. Differences Between General Maximal S-free Convex Sets and the
Case Where S is the Set of Integer Points Contained in a Rational Polyhedron. As discussed in Section 3.2, in the case of general S, full-dimensional (wrt
W ) maximal S-free convex sets do not necessarily have integer points in the relative
interior of each facet.
There is another difference between maximal S-free convex sets in the general case
and the case when S is the set of integer points contained in a rational polyhedron.
The result of Lovász [18] states that maximal lattice-free convex sets are in the form of
a polytope plus a cylinder, that is, if r is a recession direction of maximal lattice-free
convex set, then so is −r. Similarly, Basu et al. [6] prove the following result.
12
D. A. MORÁN R. AND S. S. DEY
Proposition 4.1 (Basu et al. [6]). If S is a nonempty set of integer points
contained in a rational polyhedron P, K is a maximal S-free convex set such that
K ∩ conv(S) has nonempty interior, and r belongs to the recession cone of P and K,
then −r belongs to the recession cone of K.
Proposition 4.1 is an important property and is useful in verifying many results.
See for example Basu et al. [4]. The following example shows that this result is not
true in general.
√
√
Example 4.2. Let P = {u ∈ R3 : u3 ≥ u2 − 2u1 , u3 ≥ 2u1 − u2 }.
√ Then
K = {v ∈ R3 : v3 ≤ 1, v2 ≥ 0} is a maximal S-free convex set. Also r = (1, 2, 0) ∈
rec(K ∩ P). However, −r ∈
/ rec(K).
We present some sufficient conditions on P for a property like the one presented
in Proposition 4.1 to hold. For simplicity we restrict the discussion to the case where
W = Rn .
Given a closed convex set P ⊆ Zn , and a non-zero vector r ∈ rec(P), let Z(P, r) =
{x ∈ Zn : ∃λ ≥ 0, s.t. x + λr ∈ P}.
Definition 4.3 (Convex Set with Dirichlet Property). We say a closed convex
set P ⊆ Rn satisfies Dirichlet Property if P satisfies the following conditions: For
all r ∈ rec(P) \ {0} and for all x ∈ Z(P, r), given any ² > 0 and γ ≥ 0 there
exists x̄ ∈ P ∩ Zn such that the distance between the integer point x̄ and the half line
{x + λr : λ ≥ γ} is less than ².
We first show that the Dirichlet property indeed implies a property similar to
that presented in Proposition 4.1.
Proposition 4.4. Let P ⊆ Rn be a closed convex set satisfying Dirichlet Property, let S = P ∩ Zn , and let K be a full-dimensional maximal S-free convex set. If
r belongs to the recession cone of P and K, then −r belongs to the recession cone of
K.
Proof. By part (5.) of Proposition 2.3, we obtain int(K + {λr | λ ∈ R}) =
int(K) + rel.int({λr | λ ∈ R}) = int(K) + {λr | λ ∈ R}.
To prove the result of this proposition we need to verify that int(K + {λr | λ ∈
R}) ∩ S = ∅. Assume by contradiction that x ∈ int(K + {λr | λ ∈ R}) ∩ S. By the
previous claim, there exists λ̄ ∈ R+ such that x + λ̄r ∈ int(K). Therefore B(0, δ) +
{x + λr : λ ≥ λ̄} is contained in the interior of K for some suitable small and positive
δ.
Since r is a recession direction of P, we obtain that x ∈ Z(P, r). As P satisfies
Dirichlet property, there exists an integer point z belonging to P in the interior of the
set B(0, δ) + {x + λr : λ ≥ λ̄}. However, this set lies in the interior of K. Therefore
z ∈ P and lies in interior of K, a contradiction.
We note here that if conv(S) satisfies Dirichlet Property, then the statement of
Proposition 4.4 holds with P replaced by conv(S), i.e., if r belongs to the recession
cone of conv(S) and K, then −r belongs to the recession cone of K. This observation
is useful in conjunction with the result of Proposition 4.7 that is presented in the next
Section.
Our motivation for the name ‘Dirichlet property’ is due to the fact that we often
use the following consequence of Dirichlet Theorem to prove the ‘Dirichlet property’.
Lemma 4.5 (Basu et al. [5]). If x ∈ Zn and r ∈ Rn , then for all ² > 0 and γ ≥ 0,
there exists a point of Zn at a distance less than ² from the half line {x+λr : λ ≥ γ}.
4.1.1. Some Convex Sets Satisfying Dirichlet Property. Next we give
examples of convex sets with Dirichlet Property.
On Maximal S-free Convex Sets
13
Proposition 4.6. Every full dimensional closed convex set in R2 is a convex set
satisfying Dirichlet Property.
Proof. Let P be a full-dimensional, closed convex set in R2 . Consider any ² > 0
and γ ≥ 0. Let r be a non zero vector in the recession cone of P and x ∈ Z(P, r).
Denote η = min{λ ∈ R+ : x + λr ∈ P}.
1. Since P is a closed set, we obtain that if r is a rational vector, then there exists
x̄ ∈ P ∩Z2 such that the distance between x̄ and the half line {x+λr : λ ≥ γ}
is 0.
2. Suppose now that r is not a rational vector (i.e. λr ∈
/ Z2 ∀λ > 0). Assume by
2
contradiction that there exists no x̄ ∈ P ∩ Z such that the distance between
x̄ and the half line {x + λr : λ ≥ γ} is less than ². Since P is full dimensional
and L := {x + λr : λ ≥ η} ⊆ P, the set M := {y ∈ P : distance(y, L) ≤ ²}
is a full-dimensional convex subset of P. Note now that M is lattice-free
and r belongs to the recession cone of M . Therefore M is contained in
a maximal lattice-free convex set. Since the only class of unbounded fulldimensional maximal lattice-free convex set in R2 is the split set (a set of
form {y ∈ R2 : b ≤ ha, yi ≤ b + 1} where a ∈ Z2 and b ∈ Z), we obtain that
r is a rational vector which is a contradiction.
We next verify that all the examples which do not satisfy the property similar to
that presented in Proposition 4.1, also satisfy the condition that conv(S) is not closed.
Note that Proposition 4.6 can be used to construct examples where conv(S) is not
closed and yet the property similar to that presented in Proposition 4.1 is satisfied.
Proposition 4.7. If S ⊆ Zn and conv(S) is full-dimensional and closed, then
conv(S) satisfies the Dirichlet property.
Proof. Let P := conv(S). If n = 1, then the result is straightforward to verify.
The proof is now by induction on n. Consider any ² > 0 and γ ≥ 0. Let r be
a non zero vector in the recession cone of P and x ∈ Z(P, r). Denote η = min{λ ∈
R+ : x + λr ∈ P}.
1. Since P is a closed set, we obtain that if r is a rational vector, then there exists
x̄ ∈ P ∩Zn such that the distance between x̄ and the half line {x+λr : λ ≥ γ}
is 0.
2. Suppose now that r is not a rational vector. There are two cases:
(a) Suppose ∃ δ > 0 and ζ > 0 such that B(x + ζr, δ) ⊆ P. Let ²0 =
min{δ, ²}. Now as a consequence of the Dirichlet Theorem there exists
an integer point x̄ at a distance less that ²0 from the half line {x + λr :
λ ≥ max{ζ, γ}}. Then x̄ belongs to P, completing the proof.
(b) The half-line {x + λr : λ ≥ η} lies on the boundary of conv(S). Let
F be a proper face of conv(S) containing this half-line. (Recall that a
face F of a closed convex set P is a closed convex subset such that if
x ∈ F and x can be written as convex combination of x1 , x2 ∈ P, then
x1 , x2 ∈ F ). Then we claim:
i. F = conv(S ∩ F ): Since S ∩ F ⊆ F and F is a convex set, we
obtain that conv(S ∩ F ) ⊆ F . For the other inclusion, observe that
if x ∈ F , then x can be written as a convex combination of a finite
number of vectors in S (since F ⊆ conv(S)). However by definition
of a face, each of these vectors belong to F . Thus x ∈ conv(S ∩ F ),
or equivalently F ⊆ conv(S ∩ F ).
ii. aff.hull(F ) is a rational affine half-space containing an integer point:
14
D. A. MORÁN R. AND S. S. DEY
By the above claim S ∩ F 6= ∅ and aff.hull(F ) can be generated
by vectors in S, which are integral vectors. Thus, aff.hull(F ) is a
rational affine half-space.
Now, we apply the invertible affine transformation A : aff.hull(F ) 7−→
Rdim(aff.hull(F )) to F , where A is the function described in Lemma 3.8.
Observe now that A(F ) is a closed convex set, A(F ) = conv(A(S ∩
F )), Ar is a recession direction of A(F ), Ax ∈ Z(A(F ), Ar) and 1 ≤
dim(aff.hull(F )) < n. Therefore by the induction hypothesis, for all
δ > 0, there exists an integer point x̂ belonging to A(conv(S ∩ F )) that
lies within a distance of δ from the half-line {u : u = Ax + λAr, λ ≥
max{η, γ}}. However, this implies that there exists a point x̄ ∈ S such
that x̄ lies at a distance less than ² to the half line {u : u = x + λr, λ ≥
max{η, γ}}.
We remark here that by the use of Lemma 3.8, the result of Proposition 4.7
can be extended to the case where aff.hull(conv(S)) is not full-dimensional since
aff.hull(conv(S)) is a rational affine subspace.
Next we show an interesting class of non-polyhedral convex sets that satisfy
Dirichlet Property.
Definition 4.8. A set P is called a strictly convex set if it satisfies the following
property: If u ∈ P such that u is not an extreme point, then u belongs to the relative
interior of P.
Proposition 4.9. Every full dimensional, closed, strictly convex set satisfies
Dirichlet Property.
Proof. Let P be a full dimensional, closed, strictly convex set. Consider any ² > 0
and γ ≥ 0. Let r be a non zero vector in the recession cone of P and x ∈ Z(P, r).
Denote η = min{λ ∈ R+ : x + λr ∈ P}.
1. Since P is a closed set, we obtain that if r is a rational vector, then there exists
x̄ ∈ P ∩Zn such that the distance between x̄ and the half line {x+λr : λ ≥ γ}
is 0.
2. Suppose now that r is not a rational vector. Let γ̄ := max{γ, η + 1}. The
point x+ γ̄r belongs to P and is not an extreme point of P. Since P is strictly
convex, the point x + γ̄r lies in the interior of P. Therefore, there exists a ball
of radius δ > 0 around the point x + γ̄r that lies in P. Set ²0 := min{², 2δ }.
Now as a consequence of the Dirichlet Theorem there exists an integer point
x̄ at a distance less that ²0 from the half line {x + λr : λ ≥ γ̄}. Since x̄
lies at a distance less than δ from the half line {x + λr : λ ≥ γ̄} and the
set B(0, δ) + {x + λr : λ ≥ γ̄} belongs to P, we obtain that x̄ belongs to
P. Moreover, the point x̄ lies at a distance less that ² from the half line
{x + λr : λ ≥ γ}.
4.2. Other Extensions.
1. Instead of defining S ⊆ Zn , we can define S as a subset of points belonging
to a general lattice. All the results in Section 3 carry through.
2. The condition that S ⊆ Zn such that conv(S) ∩ Zn = S is not necessary for
the polyhedrality of maximal S-free convex sets. For example, the following
corollary of Theorem 3.1 can be proven.
Corollary 4.10. Let Si ⊆ Zn ∩ W , i = 1, . . . , N such that Si = conv(Si ) ∩ Zn .
SN
Denote S = i=1 Si . Let K ⊆ W be an S-free convex set of W . Then there exists an
On Maximal S-free Convex Sets
15
S-free polyhedron B ⊆ W such that K ⊆ B.
Proof. For all i = 1, . . . , N , intW (K) ∩ Si ⊆ intW (K) ∩ S = ∅. Therefore K is an
Si -free convex set of W . By Theorem 3.1,Tthere exists an Si -free polyhedron Bi ⊆ W
N
such that K ⊆ Bi . The polyhedron
B = i=1 Bi satisfies
i KS⊆ B and intW (B) ∩ S =
TN
SN
SN h
TN
N
i=1 intW (Bi ) ∩
j=1 Sj =
j=1 Sj ∩
i=1 intW (Bi ) ⊆
j=1 [Sj ∩ intW (Bj )] = ∅.
Thus B is an S-free polyhedron containing K.
An analysis similar to that in Section 3.1 and 3.2 can be carried out for this
more general class of S. The upper bound on the number of facets of maximal Sfree convex sets presented in Section 3.3 is not valid for this class of more general S.
For example, if S = {(0, 0), (1, 0), (−1, 1), (2, 1), (0, 2), (1, 2)}, then the hexagon with
vertices {(0.5, −0.25), (2, 0.5), (2, 1.5), (0.5, 2.25), (−1, 1.5), (−1, 0.5)} is a maximal Sfree convex set.
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