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K ′ slshv enotie rmboth from obtained been have esults ( hti eie from derived is that = c pso utn-lnshv enin- been have cutting-planes of ypes u [8 ] a h rtcutting- first the was 5]) ([18, cut ) e o ovn mxd nee pro- integer (mixed) solving for ues ′ lobe ieyipeetdinto implemented widely been also ) tl rtoa)polytope? (rational) a still ⌋} e e [9]. set vex T nfisaltecretyknown currently the all unifies . f r h tutrlpoete of properties structural the are ∈ F eal opoeta,for that, prove to able re opc ovxstada and set convex compact { fisrcsincn sa is cone recession its if x mr C)coueof closure (CG) omory ∈ R n | yifiieymany infinitely ly fx [email protected] c K sdfie as: defined is ⌊ ≤ ′ σ is K σ K finitely- denote ( f ) ⌋} . . As attempts to fully understand this question, a series of studies have been conducted for the polyhedrality of CG closure of various convex sets. In [12], Dey and Vielma show that, the CG closure of a bounded full-dimensional ellipsoid, described by rational data, is a rational polytope. In [8], Dadush, Dey and Vielma show the CG closure of a set obtained as an intersection of a strictly convex body and a rational polyhedron is a polyhedron. Along this line of work, in [9], the same group of authors extend the same result to compact convex sets, therefore giving affirmation answer to the long-standing open problem raised by Schrijver. Almost simultaneously this problem was also proved by Dunkel and Schulz [14] independently, where they specifically prove for the case of irrational polytope, instead of a more general compact . All their proofs are very much involved, a few years later Braun and Pokutta [4] give a short proof for the same result as [9]. However, no matter how different these proofs might seem, they all share some high- level similarities. For example, they all rely heavily on the homogeneity property of CG closure: F ′ = K′ F for any face F of K. By inductive hypothesis that F ′ is rational polyhedral and some ∩ additional argument, they will be able to obtain the polyhedrality of K′. As we will see later, in contrast to all these work in the literature, we are taking a completely different perspective and do not make use of the homogeneity property of CG closure. Key is here a characterization result for general cutting-plane closures from which a fundamental Theorem 1 is derived. We believe that the basic proof technique here lends itself to potentially many more classes of cutting-planes. Now we highlight the main results of this paper in the next section.

1.1 Main Results With respect to the CG closure of general closed convex set, we have the first main result: Theorem 1. Given a closed convex set K in Rn, then K′ is finitely-generated if and only if there exists a finite subset F Zn, such that x Rn fx σ (f) , f F K. ⊆ { ∈ | ≤ ⌊ K ⌋ ∀ ∈ }⊆ Based on this above theorem, we are able to prove the following result. Theorem 2. If K is a Motzkin-decomposable set, then the following statements are equivalent: 1. K′ is a finitely-generated. 2. K′ is a rational polyhedron. 3. K has rational polyhedral recession cone. Here a set K Rn is called Motzkin-decomposable (see, e.g., [17, 20]), if there exist a compact ⊆ convex set C and a closed D such that K = C + D. By Minkowski-Weyl theorem, a polyhedron is a Motzkin-decomposable set, thus the last theorem generalizes and unifies all the currently known results for rational polyhedron and compact convex set. Moreover, using the characterization result in [11] for the polyhedrality of integer hull, we can immediately obtain the next result as a corollary: Corollary 1. If K is a Motzkin-decomposable set in Rn and contains integer points in its interior, then K′ is a rational polyhedron if and only if conv(K Zn) is a polyhedron. ∩ Here conv(K Zn) is called the integer hull of K. As a footnote in [9], the authors wrote the ∩ following sentences to justify the reason why they focus on the case of a compact convex set: “ If the of integer points in a convex set is not polyhedral, then the CG closure cannot be expected to be polyhedral. Since we do not have a good understanding of when this holds for unbounded convex set, we restrict our attention here to the CG closure of compact convex sets. ”

2 Therefore, our Corollary 1 directly addresses their concern. This paper is organized as follows. In Section 2 we present some characterizations for the polyhedrality of general cutting-plane closure and some preliminary results that will be used later. In Section 3, we verify Theorem 1, and in Section 4, we verify Theorem 2 and Corollary 1.

Notations and assumptions. For any x Rn and a L Rn, we denote by ∈ ⊆ proj x the orthogonal projection of x onto L, and for any X Rn, proj X := proj x x X . L ⊆ L { L | ∈ } For any set S Rn, lin(S) := S ( S) denotes the lineality space of S, which is the largest ⊆ ∩ − linear subspace contained in S. cone(S) := k λ s k N, λ 0,s S i [k] denotes {Pi=1 i i | ∀ ∈ i ≥ i ∈ ∀ ∈ } the conical hull of S, and (S) := λs λ 0,s S denotes the cone that contains all non- + { | ∀ ≥ ∈ } negative multiplication of elements in S. For a linear subspace L, we denote by L⊥ the orthogonal complement of L. For a closed convex set K, rec(K) := r k + λr K, k K, λ 0 denotes { | ∈ ∀ ∈ ≥ } the recession cone of K, and ext(K) denotes the set of extreme points of K. O (x∗) = x Rn ǫ { ∈ | x x∗ ǫ denotes the ǫ-ball centered at x∗ in its ambient space. Throughout, all norm k − k ≤ } k · k refers to the Euclidean norm. For a matrix M, we denote by ker(M) the kernel of M.

2 Preliminary Results

The well-known Dickson’s lemma will be used in our later proof, and it also played an important role in some other relevant closure papers, see, e.g., [1, 10, 28]. Lemma 1 (Dickson’s lemma [13]). For any X Nn, the partially-ordered set (poset) (X, ) has ⊆ ≤ no infinite antichain. In order theory, an antichain (chain) is a subset of a poset such that any two distinct elements in the subset are incomparable (comparable). Now we define a new concept for the convergence of rays in a cone. i n ∗ n Definition 1. Given a sequence α i∈N R and α = 0 R , if there exists λi i∈N > 0 such { } ⊆ 6 ∈ c { } that lim λ αi = α∗, then we say αi conically converges to α∗, or αi α∗. i→∞ i { } −→ For the conical convergence, we have the following easy result. i n i c ∗ i c ∗ Lemma 2. Given a sequence α N R such that α α and α β when i , then { }i∈ ⊆ −→ −→ → ∞ there exists λ> 0, such that α∗ = λβ∗. i ∗ i ∗ Proof. By assumption, we know there exists γi , µi R+, such that γiα α ,µiα β . Let ∗ ∗ i i i ∗ γi i {∗ } { }⊆γi kα k ∗ →kα k ∗ → β := µ α . Then we have: β β , β α . Hence ∗ , and α = ∗ β . i → µi → µi → kβ k kβ k In a recent paper [28], the authors study the equivalent condition for a general cutting-plane closure to be polyhedral. In this section, we will follow the same notations and definitions as in [28], and exploit the characterization results therein to derive new results for CG closure. For the completeness of this paper, we will include the proofs for those results in the Appendix. Given a family of cutting-planes αx β for any (α, β) Ω, it is referred to as “a family of cuts ≤ ∈ given by Ω”. Then the corresponding (cutting-plane) closure is defined as:

n I (Ω) := \ x R αx β . (1) { ∈ | ≤ } (α,β)∈Ω Here without loss of generality (w.l.o.g.) we can assume that (0,..., 0, 1) Ω, since (0,..., 0, 1) ∈ corresponds to the trivial inequality 0 x 1. · ≤ For the valid inequality of the closure, we have the following result.

3 Proposition 1 (Proposition 1 [28]). Given Ω Rn+1 containing (0,..., 0, 1), such that I (Ω) = . ⊆ 6 ∅ Then αx β is a valid inequality to I (Ω) if and only if (α, β) cl cone(Ω). ≤ ∈ For any set S, we use cl(S) to denote the smallest closed set containing S, which is also called closure in topology. To avoid confusion, we will only use cl(S) to refer the topological closure. This above proposition immediately implies the following consequence. Corollary 2. Given Ω Rn+1 containing (0,..., 0, 1) with I (Ω) = . Then I (Ω) is finitely- ⊆ 6 ∅ generated if and only if there exists a finite subset Ω¯ Ω such that cone(Ω)¯ = cl cone(Ω). ⊆ The proofs for both Proposition 1 and Corollary 2 can be found in Appendix A. From this above Corollary 2, we know that in order to show I (Ω) is finitely-generated, it suffices to show cl cone(Ω) is a polyhedral cone and can be generated by finitely many elements from Ω. The next easy lemma is helpful for characterizing cl cone(Ω). Here the denotes the direct sum. ⊕ n Lemma 3. For any Ω R , let L = lin(cl cone(Ω)). Then cl cone(Ω) = cl cone(proj ⊥ Ω) L, ⊆ L ⊕ where cl cone(projL⊥ Ω) is a pointed, closed convex cone. Recall that a cone is called pointed if its lineality space is the origin. In order to prove such result, we will also require the following lemma. Lemma 4 (fact 9 [26]). Given a non-empty closed convex cone K, K lin(K)⊥ is a pointed cone ∩ and K = (K lin(K)⊥) lin(K). ∩ ⊕ ⊥ Proof of Lemma 3. By Lemma 4, it suffices to show: cl cone(projL⊥ Ω) = cl cone(Ω) L . First, we ⊥ ∩ ⊥ want to show cl cone(projL⊥ Ω) cl cone(Ω) L . The relation cl cone(projL⊥ Ω) cl cone(L )= ⊥ ⊆ ∩ ⊆ L is obvious. Moreover, for any ω Ω, proj ⊥ ω = ω r, for some r L. Hence proj ⊥ Ω ∈ L − ∈ L ⊆ Ω+ L cl cone(Ω) + cl cone(Ω) = cl cone(Ω). Therefore, cl cone(proj ⊥ Ω) cl cone(Ω), which ⊆ L ⊆ completes the proof of this direction. ⊆ ⊥ ∗ Then, we want to show that cl cone(projL⊥ Ω) cl cone(Ω) L . Arbitrarily pick x ⊥ ∗ ∗ ⊇ ∗ ∩ ∈ cl cone(Ω) L . If x cone(Ω), then x = projL⊥ x projL⊥ cone(Ω) = cone(projL⊥ Ω). If i ∗ ∩ ∈ i i ∈ ∗ i x x for a sequence of x cone(Ω), then projL⊥ x x where projL⊥ x projL⊥ cone(Ω) = → ∗{ }⊆ → ∈ cone(proj ⊥ Ω). Hence x cl cone(proj ⊥ Ω). This completes the proof. L ∈ L It is well-known that, a pointed, closed convex cone is a polyhedral cone, if and only if it has finitely many different extreme rays. For a pointed cl cone(Ω), its extreme rays can be exactly characterized by elements in Ω, as stated by the next lemma. We include its proof in Appendix B. Lemma 5 (Corollary 1, Lemma 3 [28]). Given Ω Rn+1 with (0,..., 0, 1) Ω and 0 / Ω. If ⊆ ∈ ∈ cl cone(Ω) is pointed, then for any extreme ray r of cl cone(Ω), either r (Ω)+, or there exist c ∈ different ri Ω such that ri r. { }⊆ −→ Henceforth, when we mention a ray r of a cone, we will make no distinction between r and its positive scalar multiplication. In other words, we say two rays r1 and r2 are different, if and only if there does not exist λ> 0, such that r1 = λr2.

3 Chv´atal-Gomory Closure of Closed Convex Set.

We will prove Theorem 1 in this section. For a given closed convex set K, we denote the family of CG cuts of K to be:

Ω := (0,..., 0, 1) (c, σ (c) ), c Zn . (2) CG { } ∪ { ⌊ K ⌋ ∀ ∈ }

4 ′ Then by definition of CG closure, there is K = I (ΩCG). For ease of notation, when it is clear from the context, we will not specify what is the corresponding closed convex set K of ΩCG. Throughout, a CG cut cx σ (c) will sometimes also be referred to as a vector (c, σ (c) ). ≤ ⌊ K ⌋ ⌊ K ⌋ Before presenting the proof for the main Theorem 1, we will need the following lemmas. Lemma 6 (Gordan’s lemma). Given a lattice L Zn and a rational polyhedral cone C Rn. ⊆ ⊆ Then there exists a finite set of lattice points g1,...,gm C L such that every point x C L { }⊂ ∩ ∈ ∩ is an integer conical combination of these points: x = m λ gj, λ N for all j [m]. Pj=1 j j ∈ ∈ Here the finite generator g1,...,gm of C L in the above lemma is usually referred to as the { } ∩ Hilbert basis of C (see, e.g., [7]). From Gordan’s lemma we obtain the next result. n i n Lemma 7. Given a rational polyhedral cone C R , a sequence of integer vectors v N C Z , ⊆ { }i∈ ⊆ ∩ and a rational vector q∗ Qn. Then there must exist an infinite set I N and i∗ N, such that ∗ ∈ ∗ ∗ ⊆ ∈ for any i I, vi vi C, and viq∗ viq∗ = vi q∗ vi q∗ . ∈ − ∈ − ⌊ ⌋ − ⌊ ⌋ Proof. By Lemma 6, we know there exist g1,...,gm C Zn, such that x C Zn if and only if ∈ ∩ ∈ ∩ x can be written as the integer conical combination of these points. Therefore, for each vi, there exists λi Nm such that vi = m λi gj. ∈ Pj=1 j Folklore An infinite poset contains either an infinite chain or an infinite antichain. i m Within the infinite poset Λ := λ N N ordered by component-wise order , from the { }i∈ ⊆ ≤ Dickson’s Lemma 1 and this above folklore, we know there must exist an infinite index set I′ N, ⊆ such that λi ′ is an infinite chain within Nm. { }i∈I Since q∗ is a rational vector, we can write it as q∗ := 1 z, where z Zn, and D is the least D ∈ common multiple of the denominators of each q ,...,q . So for any vi Zn, viq∗ can be written 1 n ∈ as 1 viz , where ciz Z. Therefore, viq∗ viq∗ i I′ 0, 1 ,..., D−1 , which is a finite D · ∈ { − ⌊ ⌋ | ∈ } ⊆ { D D } set. Here I′ is an infinite index set, by the pigeonhole principle, there also exists another infinite index set I I′, such that viq∗ viq∗ = vjq∗ vjq∗ for any i, j I. ⊆ − ⌊ ⌋ − ⌊ ⌋ ∈ i m i j i So far we have obtained an infinite index set I, such that for any i I, v = Pj=1 λjg , λ i∈I m i ∗ i ∗ j ∗ j ∗ ∈ { i} is an infinite chain within N , and v q v q = v q v q for any i, j I. Since λ i∈I − ⌊ ⌋ − ⌊ ⌋ ∗ ∈ { } is an infinite chain within Nm, then there must exist i∗ I such that λi is the least element i ∈ i ∗ i ∗ i∗ ∗ i∗ ∗ (a.k.a. minimum element) of λ i∈I . Therefore, for any i I, v q v q = v q v q and ∗ { } ∗ ∗ ∈ − ⌊ ⌋ − ⌊ ⌋ λi λi , which implies that vi vi = m (λi λi )gj C. ≥ − Pj=1 j − j ∈ The next lemma states that, within any infinite sequence of CG cuts of K, there must exist a conically convergent subsequence which converges to a valid inequality of K. i i Lemma 8. Given a closed convex set K and a sequence (r , σK (r ) ) i∈N ΩCG. Then there { ⌊ ⌋ } ⊆ c exists a subsequence (ri, σ (ri) ) , such that when i I, i , (ri, σ (ri) ) (r∗, r∗) for { ⌊ K ⌋ }i∈I ∈ →∞ ⌊ K ⌋ −→ 0 some valid inequality r∗x r∗ of K. ≤ 0 1 i i Proof. Picking γ := i i for all i N. Then γ (r , σ (r ) ) O (0) which is a compact i k(r ,⌊σK (r )⌋)k ∈ i ⌊ K ⌋ ∈ 1 i i set. By Bolzano-Weierstrass theorem, we can find a convergent subsequence γi(r , σK (r ) ) i∈I . i i ∗ ∗ i n { ⌊ ⌋ } Denote γi(r , σK (r ) ) (r , r0). Since each r Z , we know there exists an infinite subsequence i ⌊ ⌋ i→ ∈ i i ∗ of r i∈I such that r . W.l.o.g. we assume r when i , since γir r , then { } k k→∞ i k∗ k→∞ i →∞ i → i there is γi 0. Furthermore, because γi σK (r ) r0 and γi(σK (r ) 1) < γi σK (r ) γiσK(r ), → i ∗ ⌊ ⌋→ ∗ ∗ − ⌊ ⌋≤ we know that γiσK(r ) r0. Now we want to argue that r x r0 is valid to K. If not, then there ∗ →∗ ∗ ∗ ∗ ∗ ∗ ≤ i i ∗ ∗ exists x K such that r x > r0. Denote ǫ := r x r0 > 0. Since γi(r ,σK (r )) (r , r0), then ∈ ǫ − → there exist N N and κ := ∗ , such that when i N : 0 ∈ 2(1+kx k) ≥ 0 r∗ γ ri κ, r∗ γ σ (ri) κ. k − i k≤ | 0 − i K |≤

5 Therefore, when i N : ≥ 0 γ rix∗ γ σ (ri) r∗x∗ κ x∗ r∗ κ i − i K ≥ − k k− 0 − = ǫ κ(1 + x∗ ) − k k 1 = ǫ> 0. 2 This gives the contradiction since rix σ (ri) is valid to K. ≤ K Next we present the most crucial result for establishing the proof of the main theorem.

Proposition 2. Given a closed convex set K Rn and a rational polyhedron P K such that n ⊆ n i i ⊆ P = x R fx σ (f) , f F for some finite set F Z . If (c , σ (c ) ) N ΩCG is { ∈ | ≤ ⌊ K ⌋ ∀ ∈ } ⊆ { ⌊ K ⌋ }i∈ ⊆ a sequence of vectors with σ (ci) > σ (ci) for any i N, then there exist a finite set Λ ΩCG P ⌊ K ⌋ ∈ ⊆ and an infinite index set I N, such that (ci, σ (ci) ) cone(Λ) for any i I. ⊆ ⌊ K ⌋ ∈ ∈ i i Proof. For each i N, since P K, there is σP (c ) σK (c ) < , we know there exists extreme i ∈ i i ⊆ i ≤ ∞ point p of P , such that c p = σP (c ). So from the condition of this proposition, we have the following inequalities: σ (ci) cipi = σ (ci) > σ (ci) . K ≥ P ⌊ K ⌋ Hence cipi > σ (ci) = cipi , for all i N. This can be visualized in Fig. 1. Since the number of ⌊ K ⌋ ⌊ ⌋ ∈ extreme points of polyhedron P is finite, by the pigeonhole principle, we know there exist a single extreme point p∗ of P and an infinite subset I N, such that pi = p∗ for any i I . Note that 1 ⊆ ∈ 1 for a rational polyhedron P and an extreme point p∗ P , cip∗ = σ (ci) if and only if ∈ P ci C := x Rn (p∗ p)x 0 p ext(P ),rx 0 r rec(P ) , ∈ { ∈ | − ≥ ∀ ∈ ≤ ∀ ∈ } where C is a rational polyhedral cone. For the rational vector p∗ and rational polyhedral cone C, ∗ by Lemma 7, we know there exist another infinite subset I I1 and i I1, such that for any ∗ ∗ ∗ ⊆ ∈ i I, ci ci C, and cip∗ cip∗ = ci p∗ ci p∗ . Now we denote ∈ − ∈ − ⌊ ⌋ − ⌊ ⌋ ∗ ∗ Λ := (f, σ (f) ) f F, (ci , σ (ci ) ), (0,..., 0, 1) . { ⌊ K ⌋ ∀ ∈ ⌊ K ⌋ } Here we have Λ Ω and Λ is a finite set. ⊆ CG Lastly, we want to show that the above constructed I and Λ satisfy the condition of this i i ∗ proposition, namely, for any i I, there is (c , σK (c ) ) cone(Λ). By condition of I and i , we ∗ ∈ ∗ ⌊ ∗ ⌋ ∈ know ci ci C, which means (ci ci )x (ci ci )p∗ is valid to P . By definition of P , so this − ∈i i∗ i ∗ i∗ ∗ − ≤ − implies that (c c , c p c p ) cone( (f, σK (f) ) f F, (0,..., 0, 1) ). Note that for any − ∗ − ∗ ∈ { ⌊ ⌋ ∀ ∈ } i I, cip∗ cip∗ = ci p∗ ci p∗ , therefore: ∈ − ⌊ ⌋ − ⌊ ⌋ i i i i ∗ (c , σK (c ) ) = (c , c p ) ⌊ ⌋ ∗ ⌊ ∗ ⌋ ∗ ∗ = (ci , ci p∗ ) + (ci ci , cip∗ ci p∗) ∗ ⌊ ⌋∗ − ∗ − ∗ = (ci , σ (ci ) ) + (ci ci , cip∗ ci p∗) ⌊ K ⌋ − − cone(Λ). ∈ Hence we complete the proof.

The following is the last piece of result we will need to prove Theorem 1.

6 p∗ P K

Figure 1: The red solid lines represent the CG cuts of K that cut off extreme point p∗, and the dashed red and blue lines represent the corresponding valid inequalities of K and P . Then these two CG cuts of K are also the CG cuts of P .

Proposition 3. Given a closed convex set K in Rn, and there exists a finite subset F Zn, such ⊆ that x Rn fx σ (f) , f F K. Then for any v cl cone(ΩCG), there exists a finite { ∈ | ≤ ⌊ K ⌋ ∀ ∈ } ⊆ ∈ set Λ ΩCG, such that v cone(Λ ). v ⊆ ∈ v Proof. If v cone(Ω ), then by Carath´eodory’s theorem, there exists a finite subset Λ Ω ∈ CG ⊆ CG with at most dim(Ω ) elements, such that v cone(Λ). CG ∈ If v / cone(ΩCG), since v L cl cone(ΩCG), we can find a sequence in cone(ΩCG) converging ∈ d i,j ∈ ⊆ i,j to v. Assume Pj=1 λi,jv v when i , here d = dim(ΩCG) and v ΩCG, λi,j 0 → →n ∞ ∈ ≥ for all i N, j [d]. Denote P = x R fx σK (f) , f F , which is contained in K. ∈ ∈ i,1 { ∈ | ≤ ⌊ ⌋ ∀ ∈ } First, within the set v N Ω , there must exist an infinite index subset I N such { }i∈ ⊆ CG 1 ⊆ that CG cuts within vi,1 are either all valid to P , or all invalid to P . If vi,1 all { }i∈I1 { }i∈I1 correspond to valid inequalities of P , then vi,1 cone( (f, σ (f) ) f F, (0,..., 0, 1) ), { }i∈I1 ⊆ { ⌊ K ⌋ ∀ ∈ } where (f, σ (f) ) f F, (0,..., 0, 1) Ω . If they all correspond to invalid inequalities of { ⌊ K ⌋ ∀ ∈ } ⊆ CG P , then by Proposition 2, there exists another infinite index set I¯ I and finite set Λ Ω , 1 ⊆ 1 1 ⊆ CG such that vi,1 cone(Λ ) for all i I¯ . In other words, no matter whether CG cuts within vi,1 ∈ 1 ∈ 1 { }i∈I1 are all valid to P or not, we can always find an infinite index set I¯1 I1 and finite set Λ1 ΩCG, i,1 i,2 ⊆ ⊆ such that v cone(Λ ) for all i I¯ . Now, within the set v ¯ Ω , we can do the above ∈ 1 ∈ 1 { }i∈I1 ⊆ CG argument one more time, and obtain another infinite index subset I¯ I¯ and another finite set 2 ⊆ 1 Λ Ω , such that vi,2 cone(Λ ) for all i I¯ . In fact, since I¯ I¯ , we also have vi,1 cone(Λ ) 2 ⊆ CG ∈ 2 ∈ 2 2 ⊆ 1 ∈ 1 for all i I¯ . After doing such argument for d times, eventually, we will obtain an infinite index ∈ 2 set I¯ N and d finite sets Λ ,..., Λ Ω , such that vi,j cone(Λ ) for any i I¯ , j [d]. d ⊆ 1 d ⊆ CG ∈ j ∈ d ∈ Note that d λ vi,j v when i I¯ , i , and d λ vi,j cone(Λ . . . Λ ), for any Pj=1 i,j → ∈ d → ∞ Pj=1 i,j ∈ 1 ∪ ∪ d i I¯ . Therefore, we obtain v cone(Λ . . . Λ ). Here Λ . . . Λ Ω is a finite set, by ∈ d ∈ 1 ∪ ∪ d 1 ∪ ∪ d ⊆ CG picking Λ := Λ . . . Λ we conclude the proof. v 1 ∪ ∪ d Now we are ready to verify the main theorem in this section.

Proof of Theorem 1. It suffices for us to show the “if” direction: if there exists a finite subset F Zn such that x Rn fx σ (f) , f F K, then K′ is finitely-generated. Denote ⊆ { ∈ | ≤ ⌊ K ⌋ ∀ ∈ } ⊆ P = x Rn fx σ (f) , f F , and L = lin(cl cone(Ω )). { ∈ | ≤ ⌊ K ⌋ ∀ ∈ } CG By Lemma 3, cl cone(Ω ) = cl cone(proj ⊥ Ω ) L and cl cone(proj ⊥ Ω ) is a pointed, CG L CG ⊕ L CG closed convex cone. We start our argument by analyzing the extreme rays of cl cone(projL⊥ ΩCG). By Lemma 5, any extreme ray of cl cone(projL⊥ ΩCG) is either in (projL⊥ ΩCG)+, or can be conically ↑ converged by elements from projL⊥ ΩCG. DefineΩ as the set of extreme rays of cl cone(projL⊥ ΩCG)

7 ∗ that can be conically converged by elements from projL⊥ ΩCG, and define Ω as the set of extreme rays of cl cone(projL⊥ ΩCG) that are in (projL⊥ ΩCG)+. ↑ ∗ ∗ ↑ ↑ First we consider the set Ω . Let (r , r0) Ω . By assumption of vectors in Ω , we know there i i ∈ exists a sequence (r , σ (r ) ) N Ω and γ N R , such that { ⌊ K ⌋ }i∈ ⊆ CG { i}i∈ ⊆ + i i ∗ ∗ γ proj ⊥ (r , σ (r ) ) (r , r ) when i . (3) i L ⌊ K ⌋ → 0 →∞ i i i i By Lemma 8, within this sequence (r , σK (r ) ) i∈N, there exists a subsequence (r , σK (r ) ) i∈I , { ⌊ c ⌋ } { ⌊ ⌋ } such that when i I, i , (ri, σ (ri) ) (v∗, v∗) for some valid inequality v∗x v∗ of K. Let ∈ →∞ ⌊ K ⌋ −→ 0 ≤ 0 λ (ri, σ (ri) ) (v∗, v∗) when i I, i . (4) i ⌊ K ⌋ → 0 ∈ →∞

Here each λi > 0. We can rewrite (4) as follows:

i i i i ∗ ∗ λ proj ⊥ (r , σ (r ) )+ λ proj (r , σ (r ) ) (v , v ) when i I, i . (5) i L ⌊ K ⌋ i L ⌊ K ⌋ → 0 ∈ →∞ If (v∗, v∗) L, then for each i I, take the inner product of both sides of (5) with the vec- 2 0 γi ∈ ∈ ⊥ i i 2 ⊥ i i 2 tor λi projL (r , σK (r ) ), this gives us γi projL (r , σK (r ) ) 0. Together with (3) we ⌊ ⌋ ∗ ∗ k ∗ ⌊∗ ⌋ k → ∗ ∗ get the contradiction, since (r , r0) = 0. Hence (v , v0) / L, and projL⊥ (v , v0 ) = 0. From i i ∗ ∗ 6 ∈ i 6 i c λi(r , σK (r ) ) (v , v0) for i I, i , we simply obtain that projL⊥ (r , σK (r ) ) ⌊ ∗ ∗ ⌋ → ∈ → ∞ i i c ∗ ∗ ⌊ ⌋ −→ projL⊥ (v , v0), for i I, i . Since there is also projL⊥ (r , σK (r ) ) (r , r0), by Lemma 2, ∗ ∗ ∈ → ∞∗ ∗ ∗ ⌊ ∗ ⌋ −→ we know (r , r )= λ proj ⊥ (v , v ), for some λ> 0. Since v x v is valid to K, which contains 0 L 0 ≤ 0 P , so (v∗, v∗) cone(Ω¯ ↑), where 0 ∈ Ω¯ ↑ := (f, σ (f) ) f F, (0,..., 0, 1) Ω . { ⌊ K ⌋ ∀ ∈ }⊆ CG ∗ ∗ ∗ ∗ ∗ ∗ ¯ ↑ ∗ ∗ ↑ From (r , r0) = λ projL⊥ (v , v0), we also have (r , r0) cone(projL⊥ Ω ). Since here (r , r0) Ω ↑ ∈ ↑ ↑ ∈ is arbitrary, in the end, we have shown Ω cone(proj ⊥ Ω¯ ), for some finite subset Ω¯ of Ω . ⊆ L CG Now we consider the other set Ω∗. Assuming Ω∗ contains infinitely many different extreme i i ∗ rays of cl cone(projL⊥ ΩCG): let (r , r0) i∈N Ω be one sequence of different extreme rays of i i { } i⊆ i i i i n cl cone(projL⊥ ΩCG), where (r , r0) = projL⊥ (c , σK (c ) ) for CG cut c x σK (c ) , c Z . Since i i i i ⌊ ↑⌋ ≤ ⌊i ⌋i ∈ ↑ (r , r ) is an extreme ray, then (r , r ) / cone(proj ⊥ Ω¯ ), so there is also (c , σ (c ) ) / cone(Ω¯ ). 0 0 ∈ L ⌊ K ⌋ ∈ By Proposition 1 and definition of Ω¯ ↑, this implies that inequality cix σ (ci) is not valid to ≤ ⌊ K ⌋ P , for any i N. By Proposition 2, we know there exists a finite set Λ ΩCG and an infinite set ′ ∈ i i ′ i i ⊆ i i I N, such that (c , σ (c ) ) cone(Λ) for any i I . Hence (r , r ) = proj ⊥ (c , σ (c ) ) ⊆ ⌊ K ⌋ ∈ ∈ 0 L ⌊ K ⌋ ∈ projL⊥ cone(Λ) = cone(projL⊥ Λ) cone(projL⊥ ΩCG). However, by our above assumption, for any ′ i i ∗ ⊆ ∗ i I , (r , r0) Ω is extreme ray of cl cone(projL⊥ ΩCG), we get the contradiction. So Ω can only ∈ ∈ ∗ contain finitely many different extreme rays of cl cone(projL⊥ ΩCG). By assumption of Ω , here we ¯ ∗ ∗ ¯ ∗ can find a finite subset Ω ΩCG, such that Ω cone(projL⊥ Ω ). ⊆ ⊆ ↑ ∗ ↑ ∗ So far, we have shown that, there exists finite subsets Ω¯ and Ω¯ of ΩCG, such that Ω Ω ↑ ∗ ↑ ∗ ∪ ⊆ cone proj ⊥ (Ω¯ Ω¯ ) . Since Ω Ω contains all the extreme rays of cl cone(proj ⊥ Ω ), we L ∪  ∪ L CG have: ↑ ∗ cl cone(proj ⊥ Ω ) = cone proj ⊥ (Ω¯ Ω¯ ) . L CG L ∪  For the lineality space L, which is a subset of cl cone(Ω), by Proposition 3, we can find a finite subset Ω¯ Ω , such that L cone(Ω¯ ). Hence: L ⊆ CG ⊆ L ↑ ∗ cl cone(Ω ) = clcone(proj ⊥ Ω ) L cone Ω¯ Ω¯ Ω¯ . CG L CG ⊕ ⊆ ∪ ∪ L Therefore, we obtain that: cl cone(Ω ) = cone Ω¯ ↑ Ω¯ ∗ Ω¯ , where Ω¯ ↑, Ω¯ ∗ and Ω¯ are finite CG ∪ ∪ L L subsets of ΩCG. By Corollary 2 we conclude the proof.

8 4 Chv´atal-Gomory Closure of Motzkin-Decomposable Set.

In this section, we will prove that, the CG closure of a Motzkin-decomposable set is a rational polyhedron if and only if it has rational polyhedral recession cone. Before presenting the proof for such main result, we first develop some intuition by examining the following examples. As we shall see in a moment, in some sense, the assumptions of Motzkin-decomposable is necessary.

Example 1. Consider closed convex set

K = x R2 x x 2 , 1 { ∈ + | 1 · 2 ≥ } 2 see fig. 2a. Note that rec(K1)= R+ a rational polyhedral cone, but K1 is not Motzkin-decomposable 2 since there does not exist a compact convex set C such that K1 = C + R+. Moreover, K1 has integer hull

conv(K Z2)= x R2 x + x 3,x 1,x 1 , 1 ∩ { ∈ | 1 2 ≥ 1 ≥ 2 ≥ } while K′ is not finitely-generated. To observe this, realize that conv(K Z2) K′ K where 1 1 ∩ ⊆ 1 ⊆ 1 rec(K ) = rec(conv(K Z2)) = R2 , so rec(K′ )= R2 . If K′ is finitely-generated, then K′ will have 1 1 ∩ + 1 + 1 1 facet-defining inequalities x α ,x α for some α , α 0, and inequalities x α ,x α 1 ≥ 1 2 ≥ 2 1 2 ≥ 1 ≥ 1 2 ≥ 2 are both CG cuts of K1. Clearly α1 = α2 = 1. However, there does not exist any fractional β1, β2 (0, 1), such that x1 β1 and x2 β2 are valid to K1, so x1 1,x2 1 cannot be CG cuts ∈ ′ ≥ ≥ ≥ ≥ of K1, which means K1 is not finitely-generated. Example 2. Consider another closed convex set

K = x R2 (x 0.2) (x 0.2) 2,x > 0.2,x > 0.2 , 2 { ∈ | 1 − · 2 − ≥ 1 2 } see fig. 2b. Note that K2 here can be obtained by translating the closed convex set K1 in Example 1. Here K2 is also not Motzkin-decomposable, and

conv(K Z2)= x R2 x + x 4,x 1,x 1 . 2 ∩ { ∈ | 1 2 ≥ 1 ≥ 2 ≥ } However, x 0.2,x 0.2,x + x 0.4 + 2√2 are all valid inequalities of K , so x 1,x 1 ≥ 2 ≥ 1 2 ≥ 2 1 ≥ 2 ≥ 1,x + x 4 are CG cuts of K . Therefore, K′ is finitely-generated. 1 2 ≥ 2 2 Henceforth, we will only consider exposed faces of closed convex sets, and for the sake of brevity we refer to them as faces. In other words, a face F of a closed convex set K is a subset of the form F = x K πx = σ (π) for some supporting hyperplane πx = σ (π). We will call the face F { ∈ | K } K the π-face of K. Detailed definitions and properties of faces can be found in [25] and [2]. For any vector π Rn, we can associate an unique rational linear subspace V with it. ∈ π Definition 2. Given π Rn, define V := x Rn αT x = 0 for any α Qn such that αT π Q . ∈ π { ∈ | ∈ ∈ } Lemma 9. Given a rational linear subspace L Rn, and π L. Then V L. ⊆ ∈ π ⊆ Proof. Let L := x Rn Bx = 0 , where B Qk×n. Then BT π = 0 Q, here B is the ℓ-th row { ∈ | } ∈ ℓ ∈ ℓ of B. By Definition 2, we know V x Rn BT x = 0 ℓ [k] = L. π ⊆ { ∈ | ℓ ∀ ∈ } For the rational linear subspace V associated with any vector π Rn, we have the following π ∈ well-known simultaneous diophantine approximation theorem which is due to Kronecker [22]. Note that the version we used here is similar to the one used by [4]. We include its proof in Appendix C.

9 x2 x2

x1 0.2 x1 0.2 (a) The blue region is K1, whose integer hull (b) The blue region denotes K2, red region is marked in red and it is rational polyhe- 2 ′ denotes conv(K2 Z ), and dashed lines dral, while K1 is not. 1 ∩ 1 x1 = 2 and x2 = 2 are two asymptotes of K2. Red line represents the CG cut of K2, derived from the blue dashed line.

Figure 2: Figures (a) and (b) demonstrate two congruent closed convex sets, whose integer hulls are both rational polyhedral, while their CG closures have completely different properties.

Lemma 10 ([22, 27, 4]). Let n, N N and π Rn with π = 0. Then Zn πZ contains a 0 ∈ ∈ 6 − >N0 dense subset of Vπ. We will also require the following classic result about the sensitivity of Linear Programming.

n ∗ n Lemma 11 (Sticky face lemma [23]). If P is a polyhedron in R , x0 is a point of R and F is the set of maximizers of x∗, on P (a face of P ). Then for any x∗ close enough to x∗, the maximizers h 0 ·i 0 of x∗, on P are just its maximizers on F . h ·i 4.1 Sufficient Condition In this section, we want to establish the sufficient condition in Theorem 2, for finitely-generated property of the CG closure. Using the main Theorem 1, it suffices for us to show the following result.

Proposition 4. If K is a Motzkin-decomposable set in Rn with rational polyhedral recession cone, then there exists a finite subset F Zn, such that x Rn fx σ (f) , f F K. ⊆ { ∈ | ≤ ⌊ K ⌋ ∀ ∈ }⊆ Before presenting the proof of Proposition 4, we will need the following auxiliary results. The first lemma can be viewed as an extension of the continuity Lemma 1 in [4] and sticky face lemma 11. Note that unlike Proposition 4, here we do not assume rational polyhedral recession cone.

Lemma 12. Let K be a Motzkin-decomposable set with polyhedral recession cone, and F is a π-face of K. For any δ > 0, let F := x K x′ F s.t. x x′ δ . Then there exists ǫ> 0, such δ { ∈ |∃ ∈ k − k≤ } that for any π′ with π′ π < ǫ,σ (π′)= σ (π′). k − k K Fδ Proof. By assumption, since K is Motzkin-decomposable with polyhedral recession cone, then we write K = C + cone(R) for a compact convex set C and a finite set of extreme rays R. Let F = x K πx = π be the π-face of K and πx = π is a supporting hyperplane (π < ), { ∈ | 0} 0 0 ∞ we know that πr 0 for any r R, and R := r R πr = 0 is the set of extreme rays of F . ≤ ∈ 0 { ∈ | } Clearly R0 is also the set of extreme rays of Fδ. We prove the statement of this lemma by contradiction: there exists a convergent sequence πi π and σ (πi) >σ (πi). Note that here σ (πi) < , which implies πir 0 for any r R . → K Fδ Fδ ∞ ≤ ∈ 0

10 By definition of R , we know that for any r R R , there is πr < 0. Here R R is a finite set, 0 ∈ \ 0 \ 0 hence for any πi close enough to π, we also have πir< 0 for any r R R . Therefore, for any πi ∈ \ 0 close enough to π, there is πir 0 for any r R. W.l.o.g., we can assume that for our sequence ≤ ∈ πi ,πir 0 for any r R and i 1. From σ (πi) > σ (πi) for any i 1, we know there { }i≥1 ≤ ∈ ≥ K Fδ ≥ must exist xi K F , such that πixi > σ (πi) = max πix. From our above assumption ∈ \ δ Fδ x∈Fδ∩C that πir 0 for any r R, here we can further assume that xi C F . Since xi C which ≤ ∈ ∈ \ δ ∈ is a compact set, by Bolzano-Weierstrass theorem, there is a convergent subsequence of xi . { }i≥1 W.l.o.g. we still assume the convergent subsequence of xi is itself, and xi x∗ C. Note { }i≥1 → ∈ that xi / F , so we have x∗ / F . Therefore, from πixi > max πix, we have: ∈ δ ∈ x∈Fδ∩C ∗ i i i πx = lim π x lim max π x = max πx = π0. i→∞ ≥ i→∞ x∈Fδ∩C x∈Fδ∩C Since πx = π is a supporting hyperplane of K, we obtain that πx∗ = π and x∗ F , which gives 0 0 ∈ the contradiction.

Next we present the key lemma for establishing the proof of Proposition 4.

Lemma 13. Let K be a Motzkin-decomposable set with rational polyhedral recession cone. For any ′ π-face F of K, if F is finitely-generated, then there exists a rational polyhedron Pπ obtained from finitely many CG cuts of K and ǫ > 0, such that for any π′ with π′ π < ǫ ,π′x σ (π′) is π k − k π ≤ K valid to Pπ. Proof. Let K = C + cone(R), where C is a compact convex set and R is a finite set of rational extreme rays of K. Denote u := max x < . Here we can find a multiplier κ> 0, such that x∈C k k ∞ (α, α ) := κ(π,σ (π)) with α Z. In the following discussion, we simply denote the supporting 0 K 0 ∈ hyperplane πx = σK(π) of K as αx = α0. By assumption that F ′ is finitely-generated, we can denote F ′ = x Rn gx σ (g) , g { ∈ | ≤ ⌊ F ⌋ ∀ ∈ G , with g Zn, g G. Here w.l.o.g. we assume that 0 G, since 0x σ (0) trivially holds. } ∈ ∀ ∈ ∈ ≤ ⌊ F ⌋ Pick a small positive number

1+ σ (g) σ (g) δ < min ⌊ F ⌋− F , g∈G 2 and choose a neighborhood of F : δ F¯ := x K x′ F s.t. x x′ . { ∈ |∃ ∈ k − k≤ max g } g∈G k k ′ ′ By Lemma 12, we know there exists a positive number ǫ0 > 0, such that for any α with α α <ǫ0, ′ ′ k − k there is σK(α ) = σF¯(α ). Furthermore, there is a large enough integer number N0, such that for any positive integer m N and any vector c with c mα δ , we have c+g α ǫ0 ≥ 0 k − k ≤ u k kc+gk − kαk k ≤ kαk for any g G, which is (c + g) kαk α ǫ . ∈ k kc+gk − k≤ 0 By Lemma 10, Zn αZ contains a dense subset of V , so we can find some ci m α, λ [0, 1] − >N0 α − i i ∈ for i [k] with k λ = 1, such that ∈ Pi=1 i

i i δ i n i X λi(c miα) = 0, c miα , c Z ,mi N>N0 , c miα Vα. (6) − k − k≤ u ∈ ∈ − ∈ i∈[k]

Claim 1. For any x F¯ and v V ,vx v u. ∈ ∈ α ≤ k k

11 Proof of claim. Let R := r R αr = 0 denote the set of extreme rays of the face F . Then 0 { ∈ | } obviously R is also the set of extreme rays of the set F¯. Since α ker(R ), where ker(R ) is 0 ∈ 0 0 a rational linear subspace because R is assumed to be a finite set of rational vectors, hence from Lemma 9, we know V ker(R ). This implies that, for any v V and x = y + r C + R , there α ⊆ 0 ∈ α ∈ 0 is vx = vy v u. ≤ k k ⋄ Therefore, for any x F¯ , i [k] and g G, we have ∈ ∈ ∈ i i (c + g)x = gx + miαx + (c miα)x − (7) σ (g)+ δ + m α + δ. ≤ F i 0 i Here for any x F,gx¯ σF (g)+ δ is from the definition of F¯, and (c miα)x δ is from the ∈ ≤ i i δ − ≤ last claim and the fact that c miα Vα and c miα u . According to our construction of −i ∈kαk k −i k≤kαk N and ǫ , we know that σ ((c + g) i )= σ ¯ ((c + g) i ), for any i [k]. Hence: 0 0 K kc +gk F kc +gk ∈ i i σ (c + g)= σ ¯(c + g) σ (g)+ m α + 2δ. K F ≤ F i 0 Here the last inequality is from (7). This implies that

(ci + g)x σ (g)+ m α + 2δ = σ (g) + m α ≤ ⌊ F i 0 ⌋ ⌊ F ⌋ i 0 is a CG cut of K, for any i [k] and g G. Here the last equality is because δ is assumed to be ∈ ∈ less than min 1+⌊σF (g)⌋−σF (g) , and m , α Z. Now we denote g∈G 2 i 0 ∈ P := x Rn (ci + g)x σ (g) + m α , i [k], g G . (8) π { ∈ | ≤ ⌊ F ⌋ i 0 ∀ ∈ ∈ }

Here Pπ is a rational polyhedron which is obtained from finitely-many CG cuts of K. Claim 2. αx α is valid to P , and x P αx = α F . ≤ 0 π { ∈ π | 0}⊆ Proof of claim. For any g G, by definition of P , we know that inequality ( λ ci + g)x ∈ π Pi∈[k] i ≤ σ (g) + λ m α is valid to P . By (6), such inequality is just (γα + g)x σ (g) + γα , ⌊ F ⌋ Pi∈[k] i i 0 π ≤ ⌊ F ⌋ 0 where we denote γ := λ m . By assumption that 0 G, we know αx α is valid to P . Pi∈[k] i i ∈ ≤ 0 π Moreover, there is

x P αx = α x Rn αx = α , (γα + g)x σ (g) + γα , g G { ∈ π | 0} ⊆ { ∈ | 0 ≤ ⌊ F ⌋ 0 ∀ ∈ } x Rn gx σ (g) , g G ⊆ { ∈ | ≤ ⌊ F ⌋ ∀ ∈ } = F ′ F. ⊆ Hence this claim holds. ⋄ Lastly, we want to show that, for α′ close enough to α, α′x σ (α′) will always be valid to P . ≤ K π Denote F to be the α-face of P , and let σ := αx for any arbitrary x F . Here F = x P π π ∈ π π { ∈ π | αx = σ . By the sticky face lemma 11, for polyhedron Pπ, we know there exists ǫ1 > 0, such that } ′ ′ ′ ′ when α α < ǫ1, σPπ (α ) = σFπ (α ). It suffices for us to show, for α close enough to α there k ′ − k ′ ′ ′ ′ ′ is σ π (α ) σ (α ), because this will imply that σ π (α ) σ (α ), meaning α x σ (α ) is also F ≤ K P ≤ K ≤ K valid to P . Note that Claim 2 tells that αx α is valid to P , we have σ α . Next we argue π ≤ 0 π ≤ 0 by two cases:

1. Case σ = α0: In this case, Fπ = x Pπ αx = α0 . By Claim 2, there is Fπ F ′ { ∈′ | } ′ ⊆ ⊆ K, which implies that σFπ (α ) σK(α ). Hence in this case, when α is close enough to ′ ′ ′ ≤ α, σ π (α )= σ π (α ) σ (α ), completing the proof. P F ≤ K

12 2. Case σ < α0: We decompose the face Fπ as: Fπ = conv(Eπ) + cone(Rπ), where Eπ and Rπ denote the set of extreme points and extreme rays of F respectively. Arbitrarily pick r R , π ∈ π it is also an extreme ray of polyhedron P . By the definition (8) of P , we know (ci + g)r 0 π π ≤ for any i [k] and g G. Since r is an extreme ray of face Fπ = x Pπ αx = σ , we also ∈ ∈ { ∈ |′ } have αr = 0. By (6), we know there must exist some i′ [k], such that ci r 0. Combined ′ ∈ ≥ with the fact that (ci + g)r 0 for any g G, we have: gr 0, g G. Recall that ′ n ≤ ∈ ≤′ ∀ ∈ F = x R gx σF (g) , g G , hence r is also a ray of F , which is contained in K. { ∈ | ≤ ⌊ ′⌋ ∀ ∈ }′ ′ This implies that, for any α , if σFπ (α )= , then σK(α )= . Therefore, we only have to ′ ∞ ′ ′∞ ′ show, for any α close enough to α and σFπ (α ) < , then α x σK (α ) is valid to Pπ. For ′ ′ ∞ ≤ any α close enough to α with σ π (α ) < , there is F ∞ ′ ′ α0 σ σ + α0 σFπ (α ) = max α x max αx + − = . x∈Eπ ≤ x∈Eπ 2 2 ′ ′ Here the second inequality is because, σEπ (α ) σEπ (α) as α α. Moreover, arbitrarily ∗ ′ α0−σ → → pick a point x F , when α α < ∗ , there is ∈ k − k 2kx k σ + α α′x∗ = αx∗ + (α′ α)x∗ α α′ α x∗ > 0 . − ≥ 0 − k − k · k k 2

′ σ+α0 ′ ′ Hence, σ (α ) > σ π (α ) when α is sufficiently close to α. This concludes the proof K 2 ≥ F for this case.

′ Therefore, we have shown that, there exists a small constant ǫα > 0, such that for any α with ′ ′ ′ ′ ′ α α <ǫα, α x σK(α ) is always valid to Pπ. Note that α = κπ and σK(κπ )= κσK (π ), by k − k ǫα ≤ picking ǫπ := κ , we conclude the proof. Now we have all the tools needed to verify Proposition 4.

Proof of Proposition 4. Denote K = C + cone(R), where C is a compact convex set and R is a finite set of rational extreme rays of K. The proof proceeds via induction on the dimension of K. By inductive hypothesis, this proposition holds for proper faces of K. Further from Theorem 1, we know that the CG closure of any proper face of K is finitely-generated. Let

Π := π Rn π = 1, πr 0 r R . { ∈ | k k ≤ ∀ ∈ } Since a closed convex set can be exactly given by intersecting all of its supporting half-spaces, there is K = x Rn πx σ (π), π Π . For any π Π, since the CG closure of any { ∈ | ≤ K ∀ ∈ } ∈ proper face of K is finitely-generated, by Lemma 13, we know there exists a rational polyhedron ′ Pπ obtained from finitely many CG cuts of K and a positive number ǫπ, such that for any π with ′ ′ ′ π π < ǫ ,π x σ (π ) will be valid to P . Hence we obtain an open cover O π (π) k − k π ≤ K π { ǫ }π∈Π for Π. Because Π is a compact set, as a consequence, there exists a finite subset Π¯ Π with ⊆ Π O π (π) ¯ . Consider polyhedron ⊆ { ǫ }π∈Π

P := \ Pπ. (9) π∈Π¯ Here we know that P is also obtained from finitely many CG cuts of K. Moreover, for any ′ ′ ′ supporting half-space π x σK(π ) of K, since π Π Oǫπ (π) π∈Π¯ , we know there exists ′′ ′ ≤ ′′ ′ ′ ∈ ⊆ { } π Π¯ such that π O ′′ (π ). Hence π x σ (π ) is valid to P ′′ , which contains P . In other ∈ ∈ ǫπ ≤ K π words, we have shown that, for any supporting half-space of K, this half-space also contains P . Therefore, our constructed P in (9) is contained in K, and we complete the proof.

13 4.2 Proof of Theorem 2 To show the necessary condition for the rational polyhedrality of K′ in Theorem 2, we will need the following easy lemma. Lemma 14. For a closed convex set K, rec(K) = rec(K′). Proof. Since K′ K, it suffices to show: rec(K) rec(K′). Arbitrarily pick r rec(K), denote ⊆ ⊆ ∈ C := c Zn c r 0 . For any c Zn C , since c r > 0, we know σ (c) = . Therefore, r { ∈ | · ≤ } ∈ \ r · K ∞ K′ = x Rn cx σ (c) . Since c r 0 for any c C , we obtain r rec(K′). By the Tc∈Cr { ∈ | ≤ ⌊ K ⌋} · ≤ ∈ r ∈ arbitrariness of r rec(K), we conclude the proof of rec(K) rec(K′). ∈ ⊆ Now we are ready to prove Theorem 2.

Proof of Theorem 2. Let K be a Motzkin-decomposable set. First, assume rec(K) is a rational polyhedral cone. Then by Proposition 4 and Theorem 1, we know that K′ is finitely-generated. Now assume that K′ is a finitely-generated. Then K′ is also a rational polyhedron, which has rational polyhedral recession cone. By Lemma 14, we obtain that K also has rational polyhedral recession cone.

When K is further assumed to contain integer points in its interior, we have the following necessary condition for conv(K Zn) to be a rational polyhedron. ∩ Proposition 5 (Theorem 6 [11]). Let K be a closed convex set in Rn. If int(K) Zn = and ∩ 6 ∅ conv(K Zn) is a polyhedron, then rec(K) is a rational polyhedral cone. ∩ From the last proposition and Theorem 2, we obtain Corollary 1 as an immediate corollary.

Proof of Corollary 1. Let K be a Motzkin-decomposable set which contains integer points in its interior. First, assume conv(K Zn) is a polyhedron. Then by Proposition 5, we know that K has ∩ rational polyhedral recession cone. From Theorem 2, we obtain that K′ is a rational polyhedron. Now, assuming K′ is a rational polyhedron. By the fact that K′ Zn = K Zn, we know ∩ ∩ conv(K Zn) = conv(K′ Zn), which is a rational polyhedron. ∩ ∩ For closed convex sets which are not Motzkin-decomposable, as we have seen from Example 1 ′ and Example 2, the integer hull of K1 and K2 are both polyhedral, while K1 is non-polyhedral, ′ and K2 is polyhedral. The fact that K1 is congruent with K2 suggests that for more general closed convex set, the relationship between its integer hull and its CG closure is more subtle. Moreover, we should further remark that, the additional condition that int(K) Zn = is not artificial. ∩ 6 ∅ Example 3. Let K = x R2 √2x x = 0 , which is a straight line with irrational slope. { ∈ | 1 − 2 } Then K Z2 = 0 , and its integer hull is a singleton (also a polyhedron). However, K′ = K which ∩ { } is an irrational polyhedron.

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Appendices

A Proof of Proposition 1 and Corollaries We will require the next extended Farkas’ lemma for the proof of Proposition 1.

Lemma 15 (Extended Farkas’ lemma, Corollary 3.1.2 [16]). The inequality ax b is a conse- ≥ quence of the consistent system a x b ,t T if and only if (a, b) cl cone( (a , b ) t { t ≥ t ∈ } ∈ { t t ∀ ∈ T, (0,..., 0, 1) ). − } Proof of Proposition 1. Consider the linear system ω (x, 1) ω Ω . Since I (Ω) is essentially { · − ∀ ∈ } the feasible region given by this linear system which is also non-empty, we know that linear system ω (x, 1) ω Ω is consistent. By extended Farkas’ lemma 15 and the assumption that { · − ∀ ∈ } (0,..., 0, 1) Ω, we obtain αx β is valid to I (Ω) if and only if (α, β) cl cone(Ω). ∈ ≤ ∈ Proof of Corollary 2. By definition, I (Ω) is finitely-generated if there exists a finite subset Ω¯ Ω ⊆ with I (Ω)¯ = I (Ω). It suffices to show: for any finite subset Ω¯ Ω, I (Ω)¯ = I (Ω) if and only if ⊆ cone(Ω)¯ = cl cone(Ω). Note that I (Ω)¯ = I (Ω) is equivalent of saying: any inequality αx β is ≤ valid to I (Ω)¯ if and only if it is also valid to I (Ω). By Proposition 1, that is further equivalent of saying: (α, β) cone(Ω)¯ if and only if (α, β) cl cone(Ω). Thus we complete the proof. ∈ ∈ B Proof of Lemma 5 First, we present some well-known results in convex geometry that will be needed.

Lemma 16 (Supporting Hyperplane Theorem for pointed cone). Let K Rn be a closed convex ⊆ pointed cone. Then there is h Rn such that if x K and x = 0, then hT x> 0. ∈ ∈ 6 Proof. Since K is pointed, we know the polar cone K◦ is full-dimensional. So we can find an interior point x∗ K◦, which has x∗ x< 0 for all x K. By picking h = x∗ we complete the proof. ∈ · ∈ − Lemma 17 (Lemma 2.4 in [19], Theorem 3.5 [21]). Let S be a non-empty closed set in Rn. Then, every extreme point of cl conv(S) belongs to S.

16 Now we are ready to verify Lemma 5.

Proof of Lemma 5. From Lemma 16, we can find a supporting hyperplane hx = 0 such that hω > 0 for all ω = 0 cl cone(Ω). Denote the normalized version of Ω: Ω′ = ω ω Ω , which is well- 6 ∈ { h·ω | ∈ } defined since 0 / Ω, and for all ω = 0 Ω there is h ω > 0. ∈ 6 ∈ · Claim 3. x Rn hx = 1 cl cone(Ω) = cl conv(Ω′). { ∈ | }∩ Proof of claim. First, we show x Rn hx = 1 cl cone(Ω) cl conv(Ω′). Arbitrarily pick i ∗ ∗ { ∈ |i }∩ ⊆ i ∗ i α α such that hα = 1, and there exists α cone(Ω) such that α α . Denote β := i . { } ⊆ → hα Since αi α∗, hα∗ = 1, we know hαi 1. Hence we also have βi α∗, and here βi x → → → ∈ { ∈ Rn hx = 1 cone(Ω). In the following, we show: x Rn hx = 1 cone(Ω) conv(Ω′), | }∩ { ∈ | }∩ ⊆ which will imply that α∗ cl conv(Ω′) since βi α∗ and βi x Rn hx = 1 cone(Ω). ∈ → ∈ { ∈ | }∩ According to the arbitrariness of α∗ x Rn hx = 1 cl cone(Ω), this will complete the proof ∈ { ∈ | }∩ of x Rn hx = 1 cl cone(Ω) cl conv(Ω′). { ∈ | R}∩n ⊆ k i i Pick β x hx = 1 cone(Ω), we can write it as: β = Pi=1 λib for some λi > 0, b ∈ { ∈ | }∩ n k i ∈ Ω, i [k], k N. Here because β x R hx = 1 , we know i=1 λihb = 1. Therefore, we can i i P ∈ ∈ k i ∈ {b ∈ | b ′} k i ′ also write β as: β = (λ hb ) i , here i Ω , λ hb = 1. We get β conv(Ω ), which Pi=1 i · hb hb ∈ Pi=1 i ∈ concludes x Rn hx = 1 cone(Ω) conv(Ω′). { ∈ | }∩ ⊆ Lastly, we show the other direction x Rn hx = 1 cl cone(Ω) cl conv(Ω′). By definition, { ∈ | }∩ ⊇ Ω′ x Rn hx = 1 , which implies cl conv(Ω′) x Rn hx = 1 . On the other hand, clearly ⊆ { ∈ | } ⊆ { ∈ | } Ω′ cone(Ω), so cl conv(Ω′) cl cone(Ω), and we complete the proof for this claim. ⊆ ⊆ ⋄ Given an extreme ray r cl cone(Ω), w.l.o.g. we assume hr = 1. Then r Ω iff r Ω′. From ∈ ∈ ∈ the above claim, we also know r cl conv(Ω′). Lastly, we want to show that r is an extreme ∈ point of cl conv(Ω′). Assume r = k λ ai for λ > 0, k λ = 1 and r = ai cl conv(Ω′). Pi=1 i i Pi=1 i 6 ∈ From the definition of Ω′, we also have hai = 1, ai cl cone(Ω). According to the extreme ray ∈ assumption of r, while it can be written as the conical combination ( is also i conical combination) of other points in cl cone(Ω), we know there exists γi > 0 such that a = γir. i i Since hr = ha = 1, we have γi = 1, meaning a = r, which contradict to the assumption that r = ai. So for any extreme ray r cl cone(Ω) with hr = 1, r is an extreme point of cl conv(Ω′). 6 ∈ Since cl conv(Ω′) = cl conv(cl(Ω′)), so r is an extreme point of cl conv(cl(Ω′)). By Lemma 17, we ′ ′ obtain that r cl(Ω ). By definition of Ω , it implies that either r (Ω)+, or there exists different ∈ c ∈ ri Ω such that ri r. { }⊆ −→ C Proof of Lemma 10

The next lemma says, the rational linear subspace Vπ defined in Definition 2 can be characterized by any linear basis of 1,π ,...,π over Q. Let e1,...,en denote the canonical basis of Zn. { 1 n} Lemma 18. Let 1,π for i I be a linear basis of 1,π ,...,π over Q, with π = q + { i ∈ } { 1 n} j j,0 q π j / I, here q Q i 0 I,j / I. Then V = x Rn x = q x j / I . Pi∈I j,i i ∀ ∈ j,i ∈ ∀ ∈ { }∪ ∈ π { ∈ | j Pi∈I j,i i ∀ ∈ } Proof. Denote L := x Rn x = q x , j / I . First, we show that V L. For any { ∈ | j Pi∈I j,i i ∈ } π ⊇ α Qn such that αT π Q, since π = q + q π for any j / I, we have: αT π = α π + ∈ ∈ j j,0 Pi∈I j,i i ∈ Pi∈I i i α (q + q π ) = α q + (α + α q )π Q. Since α Qn,q Q, Pj∈ /I j j,0 Pi∈I j,i i Pj∈ /I j j,0 Pi∈I i Pj∈ /I j j,i i ∈ ∈ j,i ∈ and 1,π for i I are linearly independent over Q, hence we obtain that α + α q = 0 { i ∈ } i Pj∈ /I j j,i for any i I. For any x L, by definition of L, we have αT x = (α + α q )x , which ∈ ∈ Pi∈I i Pj∈ /I j j,i i is simply 0. Therefore, we have shown V L. Next, we show L V . For any j / I, define π ⊇ ⊇ π ∈

17 αj := ej q ei. Then easy to verify that, (αj)T π Q. So for any x V and j / I, there is − Pi∈I j,i ∈ ∈ π ∈ (αj)T x = 0. This is simply saying, for any j / I,x = q x , which implies that L V . ∈ j Pi∈I j,i i ⊇ π Proof of Lemma 10. W.l.o.g. we assume that a linear basis of 1,π ,...,π over Q is 1,π ,...,π . { 1 n} { 1 k} Let π = q + k q π for any j > k, here q Q i 0 [k],j > k. By Lemma 18, we j j,0 Pi=1 j,i i j,i ∈ ∀ ∈ { }∪ know V = x Rn x = k q x ,j >k . When k = n then V = Rn, and the statement of π { ∈ | j Pi=1 j,i i } π this lemma is a special case of Weyl’s criterion. We reduce the general case to this one. The following elements lie in x Rn x = k q x ,j>k , which is V : { ∈ | j Pi=1 j,i i } π n n i i j j e˜ := e + X qj,ie i k, π˜ = π X qj,0e . ∀ ≤ − j=k+1 j=k+1

k k By Weyl’s criterion, Z + (π1,...,πk)Z>N0 is dense in R . We reformulate this for V via the projection to the first k coordinates, which is an isomorphism between V and Rk: a dense subset k Z i Z Zn Z of Vπ is Pi=1 e˜ +π ˜ >N0 , which is a subset of + π >N0 . This completes the proof.

18