On M-decomposable sets

J. Math. Anal. Appl. 485 (2020) 123816

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

On M-decomposable sets Valeriu Soltan Department of Mathematical Sciences, George Mason University, USA

a r t i c l e

i n f o

Article history: Received 5 April 2019 Available online xxxx Submitted by A. Daniilidis Keywords: Asymptotic plane Exposed structure Extreme structure M-decomposable set M-polyhedral set

a b s t r a c t According to Goberna, González, Martínez-Legaz, and Todorov (2010), an Mdecomposable set in Rn is a closed convex set which is the sum of a compact convex set and a closed convex cone. Complementing the existing results on Mdecomposable sets, we study their extreme, exposed, and asymptotic properties. Also, we consider M-polyhedral sets which are sums of compact convex sets and polyhedral cones and establish some characteristic properties of such sets. © 2019 Elsevier Inc. All rights reserved.

1. Introduction The family of closed convex sets in Rn contains three well-known subfamilies, consisting, respectively, of compact convex sets, closed convex cones, and (convex) polyhedra. A new class of closed convex sets enclosing all these subfamilies was considered in [4–6]: it consists of M-decomposable subsets of Rn , each being the sum of a compact convex set and a closed convex cone. The motivation for this terminology came from a result of Motzkin [18] (Theorem E3 on p. 47), which states that a nonempty polyhedron P ⊂ Rn is expressible as the sum of a polytope and a polyhedral cone. (We recall that a polyhedron is the intersection of finitely many closed halfspaces, and a polytope is the convex hull of finitely many points.) Further development of this topic involves the study of M-predecomposable sets. Following [9], a convex set K ⊂ Rn is called M-predecomposable if it is the sum of a compact convex set and a convex cone, not necessarily closed. Two more related classes of sets, weakly M-predecomposable and OM-decomposable sets, are considered in [17] and [10]. Line-free closed convex sets K ⊂ Rn with bounded set ext K were independently considered in [3]. This paper complements the study of M-decomposable sets, given in [4–6]; it contains some new results on geometric properties of line-free M-decomposable sets, which include their extreme, exposed, and asymptotic properties. Also, we consider M-polyhedral sets, which are sums of compact convex sets and polyhedral cones, and establish some characteristic properties of such sets. The concept of M-polyhedral set is not new: it can E-mail address: [email protected]. https://doi.org/10.1016/j.jmaa.2019.123816 0022-247X/© 2019 Elsevier Inc. All rights reserved.

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be traced in various results on minimization of quadric functions over certain types of convex sets (see, e.g., [16] for various results and references on this matter). The important tools in the study of these properties are related to closedness of images under affine transformations (in particular, of orthogonal projections onto suitable planes) and nonexistence of asymptotic planes. Discussions on existing results on geometric properties of M-polyhedral sets are given below, prior to the proofs of Lemma 3 and Theorems 4 and 6. 2. Preliminaries This section contains necessary definitions, notation, and results on convex sets in the n-dimensional Euclidean space Rn (see, e.g., [19] and [20] for details). In what follows, o stands for the origin (the zero vector) of Rn . The elements of Rn are called vectors, or points (we make no difference between these two notions, and their use should support the reader’s intuition). Closed and open segments with endpoints u, v ∈ Rn are denoted [u, v] and (u, v), respectively. Also, u·v stands for the dot product of vectors u and v. By an r-dimensional plane L in Rn , where 0 ≤ r ≤ n, we mean a translate of a suitable r-dimensional subspace S of Rn : L = c + S, where c ∈ Rn . If the vector c is orthogonal to S, then the orthogonal projection pL on L is given by pL (x) = c + pS (x), where pS means the (linear) orthogonal projection on S. The orthogonal complement of a subspace S ⊂ Rn is denoted S ⊥ . For a set X ⊂ Rn , its closure, interior, affine span, and convex hull are denoted, respectively, by cl X, int X, aff X, and conv X. The dimension of X is denoted dim X. In what follows, K stands for a convex set in Rn . For simplicity of arguments, we consider only nonempty convex sets. A convex set K ⊂ Rn is called line-free if it contains no line. The relative interior and relative boundary of K are denoted rint K and rbd K, respectively. It is known (see, e.g., [20, Corollary 2.57]), that a convex set K ⊂ Rn has empty relative boundary if and only if K is a plane. As a consequence, a line-free convex set of positive dimension in Rn has nonempty relative boundary. A set D ⊂ Rn is called a cone with apex a if a + λ(x − a) ∈ D whenever λ ≥ 0 and x ∈ D. This definition implies (letting λ = 0) that D contains its apex a, although a stronger condition λ > 0 can be beneficial; see, e.g., [15]. A convex cone is a cone which is a convex set. The recession cone of a convex set K ⊂ Rn is defined by rec K = {e ∈ Rn : x + λe ∈ K whenever x ∈ K and λ ≥ 0}. It is known that rec K is a convex cone with apex o. If K is closed, then rec K is closed; furthermore, rec K = {o} if and only if K is unbounded. The lineality space of K is the subspace defined by lin K = rec K ∩ (−rec K). It is well known (see, e.g., [19, p. 65] and [20, Theorem 5.20]) that K = lin K + (K ∩ S), where S is the orthogonal complement of lin K and lin (K ∩ S) = {o} (in particular, K ∩ S is line-free provided K is closed). We recall that a convex subset F of a convex set K ⊂ Rn is an extreme face of K provided [u, v] ⊂ F whenever u, v ∈ K and (u, v) ∩ F = ∅. It is easy to verify that any extreme face of K is closed if K is closed. A subset G of K is called an exposed face of K if there exists a hyperplane H ⊂ Rn supporting K such that G = H ∩ K. It is well known that every exposed face of K is its extreme face. In what follows, ext K and extr K (respectively, exp K and expr K) denote the set of extreme points and the union of extreme halflines (respectively, the set of exposed points and the union of exposed halflines) of a line-free closed convex set K ⊂ Rn . It is easy to see that the endpoint of an extreme halfline (in particular, of an exposed halfline) of K is an extreme point of K. The assertions below hold for any line-free closed convex set K ⊂ Rn (see [12] for the proof of (1) and (3), and [11] and [8, p. 25] for the proof of (2)):

V. Soltan / J. Math. Anal. Appl. 485 (2020) 123816

∅ = exp K ⊂ ext K ⊂ cl (exp K),

expr K ⊂ extr K,

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(1)

K = conv (ext K ∪ extr K) = conv (ext K) + rec K,

(2)

K = cl (conv (exp K ∪ expr K)) = cl (conv (exp K)) + rec K.

(3)

We will need the following assertions about a closed convex set K ⊂ Rn (see [4–6,9] for their proofs). (P1) If K is M-decomposable (M-predecomposable) and is expressed as K = C + D, where C is a compact convex set and D is a convex cone with apex o, then D = rec K. (P2) If S is the orthogonal complement of lin K, then K is M-decomposable if and only if the set K ∩ S is M-decomposable. (P3) If K is closed and line-free, then K is M-decomposable if and only if the set ext K is bounded. (P4) If K is line-free and M-decomposable, then the set K  = cl (conv (ext K))

(4)

is the smallest among all compact convex sets E ⊂ Rn satisfying the condition K = E + rec K. (P5) If K is line-free, unbounded, and M-decomposable, then there is a hyperplane H with the following properties: (a) the set H ∩ M is compact, (b) an open halfspace determined by H, say W , contains K  , (c) K ∩ (Rn \ W ) is a union of closed halflines emanating from H ∩ K. Remark 1. Any hyperplane H ⊂ Rn satisfying (P5) cuts K (that is, K meets both open halfspaces determined by H). Consequently, rint (H ∩ K) = H ∩ rint K

and dim (H ∩ K) = dim K − 1,

(5)

as follows from [20], Corollary 2.33. Remark 2. Because cl (conv X) = conv (cl X) for any bounded set X ⊂ Rn (see [20], Theorem 3.17), the inclusions (1) imply that the set K  from (P4) can be described in any of the following ways: K  = conv (cl (ext K)) = cl (conv (exp K)) = conv (cl (exp K)).

(6)

Although the results of this paper hold for all n ≥ 1, we will be assuming that n ≥ 2 to avoid trivial cases which require separate arguments. Finally, a closed convex set K ⊂ Rn will be called M-indecomposable if it is not M-decomposable. 3. Extreme and exposed structures The main results of this section are given in Theorems 1 and 2 below. Lemma 1. For a line-free unbounded M-decomposable set K ⊂ Rn and a hyperplane H ⊂ Rn satisfying condition (P5), the following assertions hold. (a) A point u ∈ H ∩ K belongs to ext (H ∩ K) if and only if it is contained in an extreme halfline of K. (b) A point u ∈ H ∩ K belongs to exp (H ∩ K) if and only if it is contained in an exposed halfline of K.

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Proof. Put N = H ∩ K. By (P5), N is a compact convex set. Hence it has extreme and exposed points, as follows from (1). We may exclude the obvious case when K is a halfline. Indeed, in this case K is an extreme (and exposed) face of itself, and N is a singleton, which is an extreme (and exposed) point of itself. Due to this argument, we will assume that K is not a halfline. Then dim K ≥ 2 and dim N ≥ 1, according to (5). (a) Let u ∈ ext N . Then u ∈ H and thus u ∈ / K  due to (P5). By (P4), we can write u = x + e, where x ∈ K  and e is a nonzero vector in rec K. Therefore, u is a relatively interior point of the halfline h = {x + λe : λ ≥ 0} ⊂ x + rec K ⊂ K. Denote by F the extreme face of K generated by u (by the definition, F is the smallest extreme face of K containing u). Then h ⊂ F and u ∈ rint h ⊂ rint F (see [20, Theorem 7.9]). Since H cuts F (because h ⊂ H), we have dim F = dim (H ∩ F ) + 1, as follows from (5). Furthermore, Corollary 7.14 from [20] implies that the set G = H ∩ F is the smallest extreme face of N containing u. Clearly, G = {u} because u ∈ ext N . Consequently, 1 = dim h ≤ dim F = dim (H ∩ F ) + 1 = dim G + 1 = dim {u} + 1 = 1. Hence dim F = 1, which shows that F is a subset of a line. Since K is line-free, F cannot be a line. This argument and the inclusion h ⊂ F implies that F is an extreme halfline containing u. Conversely, suppose that a point u ∈ N belongs to an extreme halfline F of K. By the choice of H, we have H ∩ F = {u}. Theorem 7.13 from [20] shows that the singleton {u} = H ∩ F is an extreme face of N . Equivalently, u ∈ ext N . (b) Let u ∈ exp N . Then u ∈ rbd N due to dim N ≥ 1. Choose in H a plane L ⊂ H of dimension dim H − 1 (= n − 2) such that L ∩ N = {u}. Then L ∩ K = (L ∩ H) ∩ K = L ∩ (H ∩ K) = L ∩ N = {u}. Since u ∈ rbd K (otherwise u ∈ rint N by (5)), one has L ∩ rint K = ∅. By [20, Theorem 6.8], there is a hyperplane T ⊂ Rn that contains L and nontrivially supports K (that is, K ⊂ T ). Because u ∈ T ∩ ext N , an argument from the proof of assertion (a) shows that u is a relatively interior point of an extreme halfline F of K. Denote by V ⊂ Rn the closed halfspace determined by T and containing K. Then F ⊂ K ⊂ V , and the inclusion u ∈ T ∩ rint F implies that F ⊂ T (see [20, Corollary 1.32]). As in the proof of assertion (a), one has 1 = dim F ≤ dim (T ∩ K) = dim ((T ∩ K) ∩ H) + 1 = dim ((T ∩ H) ∩ (K ∩ H)) + 1 = dim (T ∩ N ) + 1 = dim{u} + 1 = 1. Hence dim (T ∩ K) = 1, and the inclusion F ⊂ T ∩ K immediately implies that F = T ∩ K. Summing up, F is an exposed halfline of K containing u. Conversely, suppose that a point u ∈ N belongs to an exposed halfline F of K. By Theorem 8.9 from [20], the singleton {u} = H ∩ F is an exposed face of N . Equivalently, u ∈ exp N .  Corollary 1. A line-free unbounded M-decomposable set K ⊂ Rn of dimension r (≥ 1) has at least r exposed halflines. Proof. Choose a hyperplane H ⊂ Rn satisfying condition (P5). By (5), the set N = H ∩ K is (r − 1)-dimensional. Then N has at least r exposed points, say u1 , . . . , ur (otherwise the dimension of N =

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cl (conv (exp N )) would be at most r − 1). By Lemma 1, every point ui is contained in an exposed halfline Fi of K, 1 ≤ i ≤ r. Since H ∩ Fi = {ui } for all 1 ≤ i ≤ r, the halflines F1 , . . . , Fr are pairwise distinct.  Lemma 2. If K ⊂ Rn is an M-decomposable set, then an (n − 1)-dimensional subspace S ⊂ Rn supports rec K if and only if there is a translate of S which supports K. Proof. We can write S = {x ∈ Rn : x ·e = 0}, where e is any nonzero vector orthogonal to S. Assume first that S supports rec K. Denote by V a closed halfspace determined by S and containing rec K. Without loss of generality, we may assume that V = {x ∈ Rn : x ·e ≤ 0}. Let α = max {x ·e : x ∈ C}, where C is the compact convex set for which K = C + rec K, according to (P1). Choose a point w ∈ C for which w ·e = α. Obviously, w ∈ K. We state that the hyperplane H = {x ∈ Rn : x ·e = α} supports K. Indeed, any point x ∈ K = C + rec K can be written as x = y + z, with y ∈ C and z ∈ rec K. Consequently, x·e = (y + z)·e = y·e + z·e ≤ α + 0 = α. Hence K lies in the closed halfspace V  = {x ∈ Rn : x ·e ≤ α} determined by H. Since w ∈ H ∩ K, the hyperplane H supports K. Conversely, assume that a hyperplane H, which is a translate of S, supports K at a point u. Denote by P a closed halfspace determined by H and containing K. Without loss of generality, we may write P = {x ∈ Rn : x · c ≤ γ} for a suitable nonzero vector c ∈ Rn and a scalar γ. Then u · c = γ and u + rec K ⊂ K ⊂ P . Consequently, rec K lies in the closed halfspace P − u = {(x − u) ∈ Rn : (x − u)·c ≤ γ − γ} = {v ∈ Rn : v·c ≤ 0}. Since o ∈ rec K and o ·c = 0, the subspace S = H − u supports rec K.  We observe that Lemma 2 does not hold for M-indecomposable convex sets. Indeed, if K is the closed convex set in R2 given by K = {(x, y) : y ≥ x2 }, then rec K is the halfline h = {(0, y) : y ≥ 0}. The y-axis of R2 supports rec K, while every vertical line cuts K. Theorem 1. For a line-free unbounded M-decomposable set K ⊂ Rn , the following assertions hold. (a) Every extreme halfline of K is a translate of a suitable extreme halfline of rec K. (b) Every extreme halfline of rec K is a translate of a suitable extreme halfline of K. (c) Every exposed halfline of K is a translate of a suitable exposed halfline of rec K. Proof. (a) Let F be an extreme halfline of K. Since K = K  + rec K, the set F can be expressed as F = E + G, where E is an extreme face of K  and G is an extreme face of rec K (see [20, Theorem 7.15]). Because F is 1-dimensional, both sets E and G have dimension at most 1. Clearly, E is either a singleton, say {u}, or a line segment, say [u, v], parallel to F . Similarly, G is a closed halfline with apex o parallel to F (if G were a singleton of a segment, then the sum F = E + G would not be a halfline). If u = v, then we assume that the vector v − u has the same direction as F . In either case, F = u + G, which shows that F is a translate of G. (b) First, we consider the trivial case when rec K is a halfline, say G. Let H be a hyperplane satisfying (P5). From (2) we conclude that H meets an extreme halfline, say E, of K. According to assertion (a), E is a translate of an extreme halfline of rec K, which is G (= rec K) in our case. Consequently, G is a translate of E, as desired.

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We are going to prove assertion (b) by induction on r = dim K (≥ 1). The case r = 1 is obvious: K is a closed halfline. Then rec K is a halfline, and the assertion follows from the above argument. Suppose that the induction hypothesis holds for all s ≤ r − 1, where r ≥ 2, and let K be a line-free unbounded M-decomposable set of dimension r. By the above argument, we may suppose that dim (rec K) ≥ 2. Choose an extreme halfline F of rec K and a hyperplane H ⊂ Rn containing F and nontrivially supporting rec K (so that rec K ⊂ H). The existence of H follows from [20, Corollary 6.9]. By Lemma 2, there is a translate H  of H which supports K. Clearly, K ⊂ H  (otherwise rec K ⊂ H). Then the set L = K ∩ H  has dimension less than r. Furthermore, rec L = H ∩ rec K (see [20, Corollary 5.8]). The inclusions F ⊂ rec L ⊂ rec K immediately imply that F is an extreme halfline of rec L. By the inductive assumption, L has an extreme halfline F  which is a translate of F . Since L is an extreme face of K, the halfline F  is an extreme halfline of K (see [20, Theorem 7.3]). Summing up, F is a translate of the extreme halfline F  , as desired. (c) Let F be an exposed halfline of K. Since K = K  + rec K, the set F can be expressed as F = E + G, where E is an exposed face of K  and G is an exposed face of rec K (see [20, Theorem 8.10]). Since all three sets F, E, and G also are extreme faces of the respective sets, an argument of the proof of assertion (a) implies that F is a translate of G.  There may be no one-to-one correspondence between the extreme (respectively, exposed) halflines of K and those of rec K. Indeed, let K be a one-way infinite circular cylinder, given by K = {(x, y, z) : x2 + y 2 ≤ 1, z ≥ 0}. Then K has continuum many extreme (exposed) halflines, all being translates of the halfline rec K = {(0, 0, z) : z ≥ 0}. The next example shows that a similar to part (b) of Theorem 1 assertion does not hold for the case of exposed halflines (compare with Theorem 3 below).  Example 1. Let C = {(x, y, z) : z ≥ x2 + y 2 } be the solid circular cone in R3 . Put K = conv (C ∪ (e + C)), where e = (1, 0, 0). Clearly, K is the M-decomposable line-free convex set of the form K = [o, e] + C and rec K = C. The halfline h = {(0, y, y) : y ≥ 0} is an exposed halfline of rec K, and there is a unique plane H ⊂ R3 supporting rec K along h, given by H = {(x, y, y) : y ∈ R}. Any translate of h that lies in the boundary of K belongs to the set K ∩ H, which is the convex hull of the union of h and e + h. Since neither h nor e + h is an exposed halfline of K, we conclude that no translate of h is an exposed halfline of K. Following [1], we denote by ext1 K (respectively, by exp1 K) the union of all extreme faces of K (respectively, the union of all exposed faces of K), each of dimension at most one. Clearly, ext1 K = ext K ∪ exts K ∪ extr K,

exp1 K = exp K ∪ exps K ∪ expr K,

where exts K (respectively, exps K) denotes the union of all extreme (respectively, of all exposed) segments of K. Clearly, exp1 K ⊂ ext1 K. The following inclusion, initially proved in [1] for the cases of convex bodies, was extended in [20, Theorem 8.39] to the case of line-free closed convex sets in Rn : ext1 K ⊂ cl (exp1 K).

(7)

The example below shows that a stronger form of (7), ext K ∪ extr K ⊂ cl (exp K ∪ expr K), does not hold even for the case of M-decomposable sets.

(8)

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Example 2. Let X = {o} ∪ C ⊂ R3 , where C is a circle in the plane z = 1, given by C = {(x, y, 1) : x2 + (y − 1)2 = 1}. Obviously, the set L = conv X is compact and ext L = X. Let P = L + h, where h is the nonnegative halfline of the z-axis: h = {(0, 0, z) : z ≥ 0}. Then P is a line-free M-decomposable set, with rec P = h. Now, put K = conv (P ∪ (e + P )), where e = (1, 0, 0). Then K is a line-free M-decomposable set (as the sum of the compact convex set L + [o, e] and the halfline h). Clearly, rec K = h and the exposed points of K are either o and e, or belong to the plane z = 1. Also, every exposed halfline of K is a vertical halfline originated at an exposed point of K which belongs to the plane z = 1. There are precisely two halflines, h and e + h, which are extreme but not exposed for K. Consider the open segment I = (o, v), where v = (0, 0, 1). This segment lies in h, which is a part of extr K. Furthermore, any point x = (0, 0, t) ∈ I, 0 < t < 1, does not belong to the set cl (exp K ∪ expr K). (We observe that x is the limit of a sequence of points from exps K, which is the union of exposed segments of K, with their endpoints in o or in e and the plane z = 1.) So, I ⊂ extr K \ cl (exp K ∪ expr K), and the inclusion (8) does not hold. If the inclusion (8) were true, then an immediate consequence of (1) and (2) would give us the following representation of a line-free M-decomposable set K ⊂ Rn : K = conv (cl (exp K ∪ expr K)). Nevertheless, despite the fact that (8) is generally false, the above equality still holds, as proved in the next theorem. Theorem 2. If K ⊂ Rn is a line-free M-decomposable set, then K = conv (cl (exp K ∪ expr K)).

(9)

Proof. First, we are going to show that conv (cl (exp K)) = conv (cl (exp K ∪ exps K)).

(10)

Indeed, since the inclusion conv (cl (exp K)) ⊂ conv (cl (exp K ∪ exps K)) is obvious, it remains to prove the opposite one. The endpoints of any exposed segment of K are extreme points of K. Hence exps K ⊂ conv (ext K). Because both sets ext K and exp K are bounded, we have cl (conv (ext K)) = conv (cl (ext K)),

cl (conv (exp K)) = conv (cl (exp K)).

This argument and (1) give cl (exps K) ⊂ cl (conv (ext K)) = conv (cl (ext K)) ⊂ conv (cl (cl (exp K))) = conv (cl (exp K)) = cl (conv (exp K)). Therefore, conv (cl (exp K ∪ exps K)) = conv (cl (exp K) ∪ cl (exps K)) ⊂ conv (cl (exp K) ∪ cl (conv (exp K))) = conv (cl (conv (exp K))) = cl (conv (conv (exp K))) = cl (conv (exp K)) = conv (cl (exp K)).

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Summing up, the equality (10) holds. For the equality (9), we observe that a combination of (2) and (7) gives K = conv (cl (exp K ∪ exps K ∪ expr K)).

(11)

Since conv (X ∪ Y ) = conv (conv X ∪ conv Y )

whenever X, Y ⊂ Rn ,

the equalities (10) and (11) imply that K = conv (cl (exp K ∪ exps K ∪ expr K)) = conv (cl (exp K ∪ exps K) ∪ cl (expr K)) = conv (conv (cl (exp K ∪ exps K)) ∪ conv (cl (expr K))) = conv (conv (cl (exp K)) ∪ conv (cl (expr K))) = conv (cl (exp K) ∪ cl (expr K)) = conv (cl (exp K ∪ expr K)).



The following example shows that the equality (9) does not generally hold for the case of Mindecomposable sets. Example 3. Consider in R3 the line-free convex set P = conv (Q ∪ {a1 , a2 }), where Q = {(x, 0, z) : z = x2 } is the parabola in the xz-plane of R3 and a1 = (1, 1, 2) and a2 = (−1, 1, 2). The set P is not closed, and its closure, K = cl P , is the disjoint union F ∪ P , where F is a one-way infinite slab of the plane y = 1, given by F = {(x, 1, z) : −1 ≤ x ≤ 1, z > 2}. It is easy to see that ext K = exp K = Q ∪ {a1 , a2 }, while extr K is the union of two closed halflines h1 = {(1, 1, z) : z  2} and h2 = {(−1, 1, z) : z  2}. There is a unique plane through h1 (or h2 ) supporting K, and this plane contains F . This argument and the inclusion expr K ⊂ extr K give expr K = ∅. Summing up, conv (cl (exp K ∪ expr K)) = conv (Q ∪ {a1 , a2 }) = P = K. 4. M-polyhedral sets Definition 1. An M-decomposable set K ⊂ Rn is called M-polyhedral provided its recession cone rec K is polyhedral. In view of [4], a convex set K ⊂ Rn is M-polyhedral provided it can be expressed as K = C + D, where C is a compact convex set and D is a convex polyhedral cone with apex o. For the next theorem, we need some notation and terminology. Given a nonempty set X ⊂ Rn , the set X ◦ = {e ∈ Rn : x·e ≤ 0 ∀ x ∈ X}

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is a closed convex cone with apex o, called the (negative) polar cone of X. If K ⊂ Rn is a closed convex set and p ∈ K, then the polar cone Np (K) = (K − p)◦ = {e ∈ Rn : (x − p)·e ≤ 0 ∀ x ∈ K} is called the normal cone of K at p. The following result complements Theorem 1. Theorem 3. If K ⊂ Rn is a line-free unbounded M-polyhedral set, then every exposed halfline of rec K is a translate of a suitable exposed halfline of K. Proof. Let H ⊂ Rn be a hyperplane satisfying (P5). Choosing a vector v ∈ K  , where K  is given by (4), and translating both sets K and H on −v, we may assume that o ∈ K  . Consequently, rec K ⊂ K and the nonempty set P = H ∩ rec K is compact (as a closed subset of the compact set H ∩ K). By the assumption on rec K, the set P is a polytope. Choose an exposed halfline F or rec K. Then H ∩ F consists of a single point, say u, which is an exposed point of P (see Lemma 1). Hence u is a vertex of P . Denote by S the (n − 1)-dimensional subspace H − u. Select in H an (n − 2)-dimensional plane L supporting P such that L ∩ P = {u}. Clearly, L can be described as Le = {x ∈ H : x ·e = γe }, where e is a suitable unit vector in S and γe is a scalar. Without loss of generality, we may assume that P lies in the closed halfplane Qe = {x ∈ H : x ·e ≤ γe } of H (the case of the opposite halfplane is similar). Since u is a vertex of P , the closed convex cone Mu (P ) = {z ∈ H : (x − u)·(z − u) ≤ 0 ∀ x ∈ P } with apex u is full-dimensional in H; that is dim Mu (P ) = n − 1. Consequently, the closed convex cone Ru (P ) = Mu (P ) − u = {z ∈ S : (x − u)·z ≤ 0 ∀ x ∈ P } is (n − 1)-dimensional and has apex o. Clearly, Ru (P ) = Nu (P ) ∩ S, where Nu (P ) is the normal cone of P at u. The condition Le ∩ P = {u} is equivalent to the inclusion e ∈ rint Ru (P ) (see [20], Corollary 5.49). Hence there is a scalar ε > 0 such that c ∈ rint Ru (P ) for any unit vector c ∈ S satisfying the condition

e − c < ε. Let U = {c ∈ S : c = 1 and e − c < ε}. By the same corollary, for any vector c ∈ U there is a scalar γc such that the closed halfplane Qc = {x ∈ H : x·c ≤ γc } of H supports P and its boundary plane Lc = {x ∈ H : x·c = γc } has the property Lc ∩ P = {u}. Since the set H ∩ K is compact, for any vector c ∈ U , there is a closed halfplane Qc = {x ∈ H : x·c ≤ μc } which supports H ∩ K. The set of all vectors c ∈ U such that the boundary plane of Qc meets H ∩ K at a single point is dense in U (see [20, Theorem 8.23]). So, there is a vector b ∈ U such that the closed halfplane Qb supports H ∩ K at an exposed point, say w (equivalently, the boundary plane Lb of Qb satisfies the condition Lb ∩ (H ∩ K) = {w}). As in the proof of Lemma 1, we can find a hyperplane T  ⊂ Rn containing Lb and supporting K along an exposed halfline, say G . By an argument of proof of Theorem 1, the subspace T = T  − w supports rec K along an exposed halfline, say G, which is a translate of G . Since the planes Lb and Lb are translates of each other, the choice of b implies that the plane H ∩ T supports P at u only, implying that G = F . Summing up, F is a translate of an exposed halfline G of K, as desired.  We recall that a mapping f : Rn → Rm is called affine transformation if it can be expressed as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. It is well known that an unbounded polyhedral cone C ⊂ Rn with apex a can be described as

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C = conv (h1 ∪ · · · ∪ hm ), where h1 , . . . , hm are finitely many closed halflines, all with endpoint a (see, e.g., [20, Theorem 9.19]). The connection between the polyhedrality of convex cones and closedness of their planar projections was emphasized in [13, Theorem 4.11]. The theorem below has a partial overlap with Theorem 1 from [16], which deals with the case of (possibly nonconvex) M-decomposable sets and arbitrary planes (without specifying their dimension). Theorem 4. For a M-predecomposable set K ⊂ Rn , the following conditions are equivalent. (a) K is M-polyhedral. (b) For any affine transformation f : Rn → Rm , the set f (K) is closed. (c) Given a positive integer r, where 2 ≤ r ≤ n − 1, all orthogonal projections of K on r-dimensional planes of Rn are closed. Proof. (a) ⇒ (b). Let K be M-polyhedral. We can write K = C + rec K, such that C is a compact convex set and rec K = conv (h1 ∪ · · · ∪ hm ), where h1 , . . . , hm are closed halflines with common endpoint o. Express the affine transformation f : Rn → Rm as f (x) = a + g(x), where a ∈ Rm and g : Rn → Rm is a linear transformation. Then f (K) = a + g(K) = a + g(C) + g(rec K) = a + g(C) + g(conv (h1 ∪ · · · ∪ hm )) = a + g(C) + conv (g(h1 ) ∪ · · · ∪ g(hm )). Since g is a continuous mapping, the set f (C) is compact. Every set g(hi ), 1 ≤ i ≤ m, is either {o} or a closed halfline with endpoint o. Hence the set g(rec K) = conv (g(h1 ) ∪ · · · ∪ g(hm )) is a polyhedral cone. Summing up, the set f (K) = a + g(C) + g(rec K) is M-polyhedral. The implication (b) ⇒ (c) is obvious. So, it remains to prove that (c) ⇒ (a). By (P1), K can be expressed as K = C +rec K, where C is a compact convex set. First, we state that, under assumption (c), all orthogonal projections of rec K on r-dimensional planes of Rn are closed. Indeed, let L ⊂ Rn be an r-dimensional plane, and let pL denote the orthogonal projection on L. Then pL (K) = pL (C + rec K) = pL (C) + pL (rec K). Since the set pL (C) is compact and pL (rec K) is a convex cone, we conclude that pL (K) is an Mpredecomposable set. By the assumption, pL (K) is closed. Then Corollary 9 from [9] shows that pL (rec K) must be closed. As follows from [13, Theorem 4.11], the closedness of all orthogonal projections of the cone rec K on r-dimensional planes of Rn implies that rec K is polyhedral. Consequently, the set K = C + rec K is M-polyhedral.  Remark 3. The restriction r ≥ 2 in condition (c) of Theorem 4 is essential (that is, one cannot let r = 1). Indeed, consider the convex cone K = (0, 0) ∪ {(x, y) : x, y > 0} in R2 . Obviously, K is not closed and whence is not polygonal, while every orthogonal projection of K on a line of R2 is closed.

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For a convex set K ⊂ Rn and a point p ∈ Rn , the set Cp (K) = {p + λ(x − p) : x ∈ K}, is called the cone with apex p generated by K; it is the smallest convex cone with proper apex p containing K (see, e.g., Corollary 2.6.3 from [19]). Closedness of generated cones Cp (K) provides characteristic properties of polyhedral cones and boundedly polyhedral sets. Namely, a closed convex cone D ⊂ Rn distinct from a plane is polyhedral if and only if all generated cones Cp (D), p ∈ rbd D, are closed (see [2]). Similarly, Proposition 5.8 from [13] shows that a closed convex set K ⊂ Rn distinct from a plane is boundedly polyhedral if and only if all generated cones Cp (K), p ∈ rbd K, are closed. (A convex set K ⊂ Rn is called boundedly polyhedral if all intersections of K with polytopes are polytopes.) The next theorem provides an assertion of a similar nature. As mentioned in Section 2, a nonempty convex set K ⊂ Rn has nonempty relative boundary if and only if K is not a plane. Theorem 5. For a non-planar M-decomposable set K ⊂ Rn , the following conditions are equivalent. (a) K is a polyhedron. (b) All generated cones Cp (K), p ∈ rbd K, are closed. Proof. (a) ⇒ (b). Let K be a polyhedron and p be a point in rbd K. Denote by F1 , . . . , Fr all facets of K which contain p. It is well known that these facets are exposed faces of K (see, e.g., [20, Section 9.2]). Therefore, there are hyperplanes H1 , . . . , Hr ⊂ Rn satisfying the conditions Hi ∩ K = Fi , 1 ≤ i ≤ r. Denote by Vi a closed halfspace determined by Hi and containing K, 1 ≤ i ≤ r. Then Cp (K) = aff K ∩ V1 ∩ · · · ∩ Vr (see, e.g., [20], Theorem 9.25). Since the plane aff K can be expressed as the intersection of finitely many closed halfspaces, the cone Cp (K) is polyhedral, and thus is a closed set. (b) ⇒ (a). By [13, Proposition 5.8], the closed convex set K satisfying condition (b) is boundedly polyhedral. Express K as K = lin K + (K ∩ S), where S is the orthogonal complement of S and K ∩ S is line-free. By (P4), K ∩ S = (K ∩ S) + rec (K ∩ S). Choose an n-dimensional convex polytope P ⊂ Rn such that (K ∩ S) ⊂ int P (this is possible because the set (K ∩ S) is compact). Since K is boundedly polyhedral, the set Q = (P ∩ K) ∩ S = P ∩ (K ∩ S) is a polytope. Because every extreme point of K ∩ S is a vertex of Q, we conclude that the set ext (K ∩ S) is finite. Furthermore, for every extreme halfline h of K ∩ S, the intersection h ∩ Q is an edge of Q. So, the number of extreme halflines of K ∩ S is finite. Consequently, the line-free closed convex set K ∩ S = conv (ext (K ∩ S) ∪ extr (K ∩ S)) is a polyhedron. Summing up, the set K = lin K + (K ∩ S) is polyhedral.  We recall (see, e.g., [14] and [21]) that a plane L ⊂ Rn is called asymptotic to a nonempty set X ⊂ Rn if L ∩ cl X = ∅ and δ(L, X) = inf{ x − y : x ∈ L, y ∈ X} = 0.

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The question whether a given plane L ⊂ Rn is an asymptote of a set X ⊂ Rn can be reformulated in terms of a suitable orthogonal projection of X, as shown in the lemma below (this approach can be traced in the papers [14,16,21]). We provide the proof of this lemma for the sake of completeness. Lemma 3. Let a plane L ⊂ Rn be expressed as L = c + S, where S ⊂ Rn is a subspace and c ∈ Rn is a nonzero vector orthogonal to S. Let T ⊂ Rn be the orthogonal complement of S. If pT denotes the orthogonal projection on T , then L is an asymptote of a closed set X if and only if c ∈ cl pT (X) \ pT (cl X). Proof. Assume first that L is an asymptote of X. Since pT (L) = L ∩ T = {c}, one has c ∈ / pT (cl X). The equality δ(L, X) = 0 shows the existence of sequences u1 , u2 , . . . ∈ L and v1 , v2 , . . . ∈ X such that limi→∞ ui − vi = 0. Then pT (ui ) = c and pT (vi ) ∈ pT (X) for all i ≥ 1. Because lim c − pT (vi ) = lim pT (ui ) − pT (vi ) ≤ lim ui − vi = 0,

i→∞

i→∞

i→∞

we obtain the inclusion c ∈ cl pT (X). Conversely, let c ∈ cl pT (X) \ pT (cl X). As above, the condition pT (L) = {c} ⊂ pT (cl X) show that L ∩ cl X = ∅. Next, let w1 , w2 , . . . be a sequence of points in pT (X) such that c = limi→∞ wi . Choose in X points v1 , v2 , . . . satisfying the conditions pT (vi ) = wi , i ≥ 1. Now, let ui = c + (vi − wi ), i ≥ 1. Then ui ∈ c + S = L because vi − wi ∈ T ⊥ = S. Furthermore, δ(L, X)  lim ui − vi = lim c + (vi − wi ) − vi = lim c − wi = 0. i→∞

i→∞

i→∞

Summing up, L is an asymptote of X.  The theorem below is proved in [16] for the case of (possibly nonconvex) M-decomposable sets and arbitrary planes (without specifying their dimension). Theorem 6. For an M-decomposable set K ⊂ Rn , the following conditions are equivalent. (a) K is M-polyhedral. (b) Given an integer r, where 1 ≤ r ≤ n − 2, no r-dimensional plane is asymptotic to K. Proof. (a) ⇒ (b). Choose any plane L ⊂ Rn of dimension r, where 1 ≤ r ≤ n − 2, which is disjoint from K and express it as L = c + S, where S is a suitable r-dimensional subspace of Rn and c ∈ Rn is a vector orthogonal to S. Denote by T the orthogonal complement of S. Then 2 ≤ dim T ≤ n − 1 and L ∩ T = {c}. If pT denotes the orthogonal projection on T , then pT (L) = {c}. The set pT (K) is closed, as follows from Theorem 4. Hence the set cl pT (K) \ pT (K) is empty, and Lemma 3 shows that L cannot be asymptotic to K. (b) ⇒ (a). Choose any plane N ⊂ Rn of dimension n − r, where 1 ≤ r ≤ n − 2, and express it as N = a + S, where S is a suitable (n − r)-dimensional subspace of Rn and a is orthogonal to S. Clearly, 2 ≤ n − r ≤ n − 1. Let T = S ⊥ . Then dim T = r. By assumption (b), no translate of T is asymptotic to K. Therefore, Lemma 3 implies that the orthogonal projection pN (K) = a + pS (K) is a closed set. Finally, Theorem 4 shows that the set K is M-polyhedral.  Following [19, p. 95], we will say that nonempty sets X1 and X2 in Rn are strongly separated by a hyperplane if there is an ε > 0 such that the open ε-neighborhoods Uε (X1 ) and Uε (X2 ) lie in the opposite open halfspaces determined by H. As proved in [18, Theorem G5 on p. 63], any two disjoint polyhedra are strongly separated by a hyperplane. The next corollary generalizes this assertion to the case of M-polyhedral sets.

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Corollary 2. Any two disjoint M-polyhedral sets K1 and K2 in Rn are strongly separated by a hyperplane. Proof. By Theorem 6, neither K1 nor K2 admits an asymptotic plane. A combination of Theorems 6.32 and 6.37 from [20] implies that K1 and K2 are strongly separated by a suitable hyperplane.  5. Asymptotic planes This section contains some results on asymptotic planes of M-decomposable sets. In view of Theorem 6, these sets cannot be M-polyhedral. The author thanks the referee for the following method of proof of Theorem 7. Theorem 7. No M-decomposable set K ⊂ Rn has asymptotic hyperplanes. Proof. Assume, for contradiction, the existence of a hyperplane H ⊂ Rn which is asymptotic to K. Let H = {x ∈ Rn : x ·e = γ}. Since H is disjoint from K, one of the open halfspaces of Rn determined by H contains K. Without loss of generality, we may suppose that namely the halfspace W = {x ∈ Rn : x·e > γ} contains K. Then the linear functional ϕ(x) = x ·e is bounded below on K but does not attain a minimum over K. The latter is impossible, as proved in [4, Proposition 13]. Summing up, H cannot be asymptotic to K.  Corollary 3. Let K ⊂ Rn be an M-decomposable set. If a plane L ⊂ Rn is asymptotic to K, then there is a hyperplane H ⊂ Rn which contains L and properly supports K. Proof. Since L ∩ rint K = ∅, there is a hyperplane H ⊂ Rn such that L ⊂ H and H ∩ rint K = ∅ (see [20, Theorem 6.2]). Because 0 ≤ δ(H, K) ≤ δ(L, K) = 0, one has δ(H, K) = 0. By Theorem 7, H cannot be asymptotic to K. Hence H ∩ K = ∅, implying that H nontrivially supports K.  Given an n-dimensional closed convex set K ⊂ Rn , let α(K) denote the set of integers j between 1 and m − 1 such that K admits an j-dimensional asymptotic plane. As shown in [14] and [7], for any set J ⊂ {1, . . . , m − 1} (possibly, J = ∅) there is an n-dimensional closed convex set K ⊂ Rn such that α(K) = J. The following corollary complements this assertion. Remark 4. No M-decomposable set K ⊂ Rn of dimension one or two has asymptotic planes. Indeed, if K = C + rec K, where C is a compact convex set, such that dim K ≤ 2, then dim (rec K) ≤ 2. Therefore, rec K is either a singleton, a closed halfline, a line, or a closed 2-dimensional cone. In either case, rec K is a polyhedral set, and Theorem 6 implies that K has no asymptotic planes. Consequently, if Rn contains an M-decomposable set which is not M-polyhedral, then n ≥ 3. Corollary 4. If K ⊂ Rn , n ≥ 3, is an M-decomposable set which is not M-polyhedral, then for every integer r = 1, . . . , n − 2 there is a plane L ⊂ Rn of dimension r which is asymptotic to K.  Proof. Let t = n − r. Then 2 ≤ t ≤ n − 1. By Theorem 4, there is a t-dimensional plane L ⊂ Rn such that the orthogonal projection pL (K) is not closed. Hence there is a point u ∈ cl pL (K) \ pL (K). Denote by S the t-dimensional subspace parallel to L, and let N = u + S ⊥ . Then N is a plane of dimension n − t = r satisfying the conditions: K ∩ N = ∅ and δ(K, N ) = 0 (see Lemma 3). Thus N is asymptotic to K. 

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Theorem 8. If an M-decomposable set K ⊂ Rn has an asymptotic plane L ⊂ Rn , then there is a translate of L which is asymptotic to rec K. Proof. By the definition of L, we have K ∩ L = ∅, and there are sequences of points u1 , u2 , . . . ∈ K and v1 , v2 , . . . ∈ L such that ui − vi → 0 as i → ∞. Clearly, limi→∞ ui = limi→∞ vi = ∞. Indeed, assuming that at least one of the sets {u1 , u2 , . . . } and {v1 , v2 , . . . } is bounded, we would be able to choose converging subsequences of these sequences which have the same limit point in K ∩ L, contrary to the assumption K ∩ L = ∅. Consequently, K is unbounded, which shows that rec K = {o}. Let C be a compact convex set satisfying the condition K = C + rec K. Since limi→∞ ui = ∞, we may assume that all points u1 , u2 , . . . belong to K \ C. Then we can write ui = zi + λi ei , where zi ∈ C, λi > 0, and ei is a unit vector in rec K, i ≥ 1. By the compactness of C, the sequence z1 , z2 , . . . contains a subsequence which converges to a point z ∈ C. Without loss of generality, we may suppose that namely zi → z as i → ∞. Put ui = z + λi ei , i ≥ 1. Clearly, ui ∈ z + rec K. Choose any scalar ε > 0 and an index i0 ≥ 1 such that

zi − z < ε/2 and ui − vi < ε/2 for all i ≥ i0 . Then

ui − vi ≤ ui − ui + ui − vi

= z − zi + ui − vi <

ε ε + = ε, i ≥ i0 . 2 2

Therefore, δ(z + rec K, L) = 0. Furthermore, (z + rec K) ∩ L ⊂ K ∩ L = ∅. Summing up, L is an asymptotic plane of the cone u + rec K. Equivalently, the plane L − z, which is a translate of L, is asymptotic to rec K.  Remark 5. It is an open question whether the converse to Theorem 8 assertion holds. Acknowledgment The author thanks the anonymous referee for valuable comments on an earlier draft of this paper and for the information on the manuscript [16]. References [1] E. Asplund, A k-extreme point is the limit of k-exposed points, Israel J. Math. 1 (1963) 161–162. [2] A. Bastiani, Polyèdres convexes de dimension quelconque, C. R. Acad. Sci. Paris 247 (1958) 1943–1946. [3] R.G. Batson, Necessary and sufficient condition for boundedness of extreme points of unbounded convex sets, J. Math. Anal. Appl. 130 (1988) 365–374. [4] M.A. Goberna, E. González, J.E. Martínez-Legaz, M.I. Todorov, Motzkin decomposition of closed convex sets, J. Math. Anal. Appl. 364 (2010) 209–221. [5] M.A. Goberna, J.E. Martínez-Legaz, M.I. Todorov, On Motzkin decomposable sets and functions, J. Math. Anal. Appl. 372 (2010) 525–537. [6] M.A. Goberna, A.N. Iusem, J.E. Martínez-Legaz, M.I. Todorov, Motzkin decomposition of closed convex sets via truncation, J. Math. Anal. Appl. 400 (2013) 35–47. [7] P. Goossens, Hyperbolic sets and asymptotes, J. Math. Anal. Appl. 116 (1986) 604–618. [8] B. Grünbaum, Convex Polytopes, Interscience Publishers, New York, 1967. [9] A.N. Iusem, J.E. Martínez-Legaz, M.I. Todorov, Motzkin predecomposable sets, J. Global Optim. 60 (2014) 635–647. [10] A.N. Iusem, M.I. Todorov, On OM-decomposable sets, Comput. Appl. Math. 37 (2018) 2837–2844. [11] V.L. Klee, Extremal structure of convex sets, Arch. Math. 8 (1957) 234–240. [12] V.L. Klee, Extremal structure of convex sets. II, Math. Z. 69 (1958) 90–104. [13] V.L. Klee, Some characterizations of convex polyhedra, Acta Math. 102 (1959) 79–107. [14] V.L. Klee, Asymptotes of convex bodies, Math. Scand. 20 (1967) 89–90.

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[15] J. Lawrence, V. Soltan, On unions and intersections of nested families of cones, Beitr. Algebra Geom. 57 (2016) 655–665. [16] J.E. Martínez-Legaz, D. Noll, W. Sosa, Non-polyhedral extensions of the Frank-and-Wolfe theorem, arXiv:1805.03451, 2018. [17] J.E. Martínez-Legaz, M.I. Todorov, Weakly Motzkin predecomposable sets, Set-Valued Var. Anal. 25 (2017) 507–516. [18] T. Motzkin, Beiträge zur Theorie der linearen Ungleichungen, Azriel, Jerusalem, 1936, English translation in: Theodore S. Motzkin: Selected Papers, Birkhäuser, Boston, 1983, pp. 1–80. [19] R.T. Rockafellar, Convex Analysis, Princeton University Press, Princeton, NJ, 1970. [20] V. Soltan, Lectures on Convex Sets, World Scientific, Hackensack, NJ, 2015. [21] V. Soltan, Asymptotic planes and closedness conditions for linear images and vector sums of sets, J. Convex Anal. 25 (2018) 1183–1196.