On Unions and Intersections of Nested Families of Cones 1 ...

On Unions and Intersections of Nested Families of Cones Jim Lawrence and Valeriu Soltan Department of Mathematical Sciences, George Mason University Fairfax, Virginia 22030, USA e-mails: [email protected], [email protected]

Abstract. We prove that the union and the intersection of a nested family of cones in Rn (not necessarily convex or with a common apex) also are cones. Keywords: Apex, cone, convex, nested family, intersection, union. 2000 Mathematics Subject Classification: 52A20.

1

Introduction

We recall that a set C ⊂ Rn , possibly empty, is a cone with apex s ∈ Rn provided s + t(x − s) ∈ C whenever t > 0 and x ∈ C. In particular, C is a cone with apex o (the origin of Rn ) if and only if tC = C for all t > 0. In what follows, the set of all apices of a cone C ⊂ Rn , denoted ap C, is called the apex set of C. Clearly, an apex s of the cone C may be outside C. For instance, the zero point of the plane R2 is the only apex of the positive quadrangle {(x, y) : x, y > 0}, while the x-axis of R2 is the apex set of the open halfplane {(x, y) : y > 0}. The cone C is called convex cone if it is a convex set. There is a vast literature on convex cones and their applications to the duality theory, convex analysis, and optimization (see, for instance, the monographs [1], [2], and [3]). Our goal is to add one more result of general nature on convex cones in Rn . It is easy to see that the union and the intersection of any family of cones in Rn with a common apex s also are cones with this apex. Trivial examples show that this assertion does not hold for cones with distinct apices. Nevertheless, as shown in Theorem 1 below, this can be true in special cases. We recall that an indexed family F = {Cλ : λ ∈ Λ} of sets in Rn is called nested if either Cγ ⊂ Cµ or Cµ ⊂ Cγ whenever Cγ , Cµ ∈ F. The main result of this paper is given in the following theorem. Theorem 1. The union and the intersection of a nested family F = {Cλ : λ ∈ Λ} of cones in Rn (not necessarily convex or with a common apex) also are cones. While Theorem 1 looks like an assertion of affine geometry, its proof requires some topological arguments, combined with separation properties of convex sets (compare, for instance, with [4] and [5]). It would be interesting to know whether 1

Theorem 1 can be generalized within axiomatic convexity theory, as treated in [7] and [8]. The key points in the proof of Theorem 1 may be described as follows. 1. Instead of showing the existence of a particular apex of the set C = ∪ (Cλ : Cλ ∈ F), we are looking for the entire apex set of C. 2. To specify the properties of all possible apices of C, we show that the apex set of any cone in Rn is a plane (see Theorem 2). 3. We prove that the apex set of C contains the unique minimal plane in Rn which is an accumulating plane of the sequence of particular apices of the cones Cλ (see page 6). In a standard way, cl X, rint X, and rbd X stand, respectively, for the closure of a set X ⊂ Rn , the relative interior of X (that is, the interior of X with respect to its affine span aff X), and the relative boundary of X, defined by rbd X = cl X \ rint X. By a plane L in Rn we mean a translate of a subspace S ⊂ Rn : L = u + S for a suitable vector u ∈ Rn . The dimension of L, denoted dim L, is defined to be the dimension of S. It is easy to see that a nonempty set L ⊂ Rn is a plane if and only if (1 − t)x + ty ∈ L whenever t ∈ R and x, y ∈ L (see, e. g., [6, Theorem 1.2]). If M and N are planes in Rn such that M ⊂ N , dim M = m, and dim N = m + 1, then an open halfplane of N determined by M is one of the two components, say W1 and W2 , of the set N \ M . Their closures Vi = M ∪ Wi , i = 1, 2, are called the closed halfplanes of N determined by M . Clearly, the sets Wi and Vi are (m + 1)-dimensional (see, e. g., [9, Corollary 1.80]). Theorem 2. For a nonempty cone F ⊂ Rn , the following assertions hold. 1. The apex set of F is a plane, which lies in cl F . 2. Either ap F ⊂ F or F ∩ ap F = ∅. 3. If ap F is a plane of dimension m, 0 6 m 6 n, then F has one of the shapes below: (a) F = ap F , (b) F is a union of closed (m + 1)-dimensional halfplanes bounded by ap F , (c) F is a union of open (m+1)-dimensional halfplanes bounded by ap F . Replacing the condition t > 0 in the definition of cone with a weaker one, t > 0, we obtain a smaller class of cones (say, proper cones), which contain their apices and are suitable in dealing with various closedness conditions for convex sets. The example below shows that Theorem 1 does not hold for this class of cones.

2

Example. Every closed halfplane Ck = {(x, y) : y > k1 } of R2 , k > 1, is a proper convex cone whose apices fulfil the line y = k1 . The union of these cones is the open convex cone C = {(x, y) : y > 0} of R2 , whose apex set, the x-axis, does not meet C. Similarly, the intersection of proper convex cones Dk = {(x, y) : y > 0} ∪ {(x, 0) : x > k},

k > 1,

is the open halfplane C = {(x, y) : y > 0}, which is not a proper cone. Since the union and the intersection of any nested family of convex sets in Rn also are convex sets, it is natural to ask for other families of sets which are closed with respect to nested unions and intersections. The example below shows that starshaped sets, which are similar to convex sets in many aspects, do not have this property. We recall that a set S ⊂ Rn is called starshaped provided there is a point s ∈ S such that the line segment [s, x] lies in S whenever x ∈ S. Example. It is easy to see that every set Sk = {(x, y) : |x| 6 k, y 6 x2 } ∪ {(x, y) : |x| > k, y 6 k 2 },

k > 1,

is starshaped with respect to every point s = (0, c), where c 6 −k 2 . The union S = S1 ∪ S2 . . . is the closed set {(x, y) : y 6 x2 }, which is not starshaped. Similarly, every set Tk = {(x, y) : x > 0, |y| > 1/x} ∪ {(x, y) : x > k},

k > 1,

is starshaped with respect to every point s = (c, 0), where c > 2k. The intersection T = T1 ∩ T2 . . . is the closed set {(x, y) : x > 0, |y| > 1/x} consisting of two components, and is not starshaped.

2

Proof of Theorem 2

1. The assertion is obvious if ap F is a singleton. So, let ap F contain more than one point. It suffices to show that for distinct points s1 , s2 ∈ ap F , the line l through s1 and s2 lies in ap F . Choose any point s ∈ l. Then s = (1 − t)s1 + ts2 for a suitable scalar t. Express t as t = a − b, where a, b > 0. Let F1 = F − s1 and F2 = F − s2 . Clearly, both F1 and F2 are cones with apex o. Also, F2 = F − s2 = (F1 + s1 ) − s2 = F1 + (s1 − s2 ). By the same argument, for any scalar c > 0, one has (s1 − s2 ) + F1 = F2 = cF2 = c((s1 − s2 ) + F1 ) = c(s1 − s2 ) + cF1 = c(s1 − s2 ) + F1 . Consequently, t(s1 − s2 ) + F1 = (a(s1 − s2 ) + F1 ) − b(s1 − s2 ) = ((s1 − s2 ) + F1 ) − b(s1 − s2 ) = (s1 − s2 ) + (F1 − b(s1 − s2 )) = (s1 − s2 ) + (F1 − (s1 − s2 )) = F1 . 3

From here it follows that F − s = F − (s1 + t(s2 − s1 )) = (F − s1 ) + t(s1 − s2 ) = F1 + t(s1 − s2 ) = F1 . Since F1 is a cone with apex o, it follows that F is a cone with apex s. Summing up, l ⊂ ap F , which shows that ap F is a plane. The inclusion ap F ⊂ cl F is obvious if ap F = F . Assume that ap F 6= F and choose any points s ∈ ap F and x ∈ F \ ap F . Then the whole open halfline h = (s, xi = {(1 − t)s + tx : t > 0} lies in F . Hence s ∈ cl F because s is an accumulation point of h. Thus ap F ⊂ cl F . 2. For the second assertion, assume the existence of an apex s ∈ F ∩ ap F . If ap F = {s}, then the inclusion ap F ⊂ F is obvious. Suppose the existence of another apex s0 ∈ ap F . Then the whole open halfline (s0 , si lies in F . If s00 is the midpoint of the open segment (s, s0 ), then s00 ∈ F , implying that s0 ∈ (s, s00 i ⊂ F . Hence ap F ⊂ F . 3. For the last assertion, suppose that F 6= ap F (the case F = ap F happens if and only if F is a plane). Obviously, F \ ap F 6= ∅. Choose any point x in F \ ap F , and denote by Nx the plane aff (ap F ∪ {x}). Clearly, dim Nx = m + 1. Let Wx be the open halfplane of Nx determined by ap F and containing x. Our goal is to prove the inclusion Wx ⊂ F . For this, choose any point y ∈ Wx . Further, let s1 , . . . , sm+1 be an affine basis for ap F . Then (see [9, Theorem 1.71]) y can be written as an affine combination y = t1 s1 + · · · + tm+1 sm+1 + tm+2 x, t1 + · · · + tm+2 = 1, tm+2 > 0. We consider the cases tm+2 6= 1 and tm+2 = 1 separately. 3a. Assume first that tm+2 6= 1. Since the point s=

tm+1 t1 s1 + · · · + sm+1 1 − tm+2 1 − tm+2

is an affine combination of points s1 , . . . , sm+1 from ap F , and since ap F is a plane, it follows that s ∈ ap F (see [6, § 1]). Consequently, y = (1 − tm+2 )s + tm+2 x = s + tm+2 (x − s) ∈ (s, xi ⊂ F. 3b. Let tm+2 = 1. Then t1 + · · · + tm+1 = 0. Put s = (1 − t1 )s1 − t2 s2 − · · · − tm+1 sm+1 ,

z = s1 + 21 (x − s1 ).

Similarly to the above, s, as an affine combination of the points s1 , . . . , sm+1 , belongs to ap F . Because s1 ∈ ap F and x ∈ F , the point z also belongs to F . The equalities y = t1 s1 + · · · + tm+1 sm+1 + x = (s1 + x) − ((1 − t)s1 − t2 s2 − · · · − tm+1 sm+1 ) = 2z − s = s + 2(z − s) 4

show that y ∈ F . Summing up, Wx ⊂ ap F . From the above argument it follows that the set F \ ap F is the union of (m+1)-dimensional open halfplanes of the form Wx , x ∈ F \ap F . If F ∩ap F = ∅, then F = ∪ (Wx : x ∈ F \ ap F ). If ap F ⊂ F , then every set Vx = ap F ∪ Wx , x ∈ F \ ap F , is a closed (m + 1)dimensional halfplane determined by ap F , and F = ap F ∪ (F \ ap F ) = ∪ (ap F ∪ Wx : x ∈ F \ ap F ) = ∪ (Vx : x ∈ F \ ap F ).

3

Auxiliary Lemmas

Given a set Y ⊂ Rn and a closed ball Bρ ⊂ Rn of radius ρ > 0 centered at the origin o, the set Bρ (Y ) = Y + Bρ is called a ρ-neighborhood of Y . Obviously, Bρ (Y ) = ∅ if Y = ∅. Lemma 1. Let L1 and L2 be planes in Rn . For any scalar ρ > 0, there is a scalar σ > 0 such that Bσ (L1 ) ∩ Bσ (L2 ) ⊂ Bρ (L1 ∩ L2 ).

(1)

Proof. First, we consider the case L1 ∩ L2 = ∅. Then the inf -distance δ(L1 , L2 ) = inf{kx1 − x2 k : x1 ∈ L1 , x2 ∈ L2 } is positive (see, e. g., [9, Theorem 1.98]). If a scalar σ > 0 satisfies the inequality σ < δ(L1 , L2 )/2, then Bσ (L1 ) ∩ Bσ (L2 ) = ∅, and the inclusion (1) trivially holds. Suppose that L1 ∩ L2 6= ∅. Choose a vector u ∈ L1 ∩ L2 , and let Si = Li − u, i = 1, 2. Then both sets S1 and S2 are subspaces, and S1 ∩ S2 = (L1 − u) ∩ (L2 − u) = L1 ∩ L2 − u. Furthermore, for any plane L ⊂ Rn , expressed as a translate L = v + S of a subspace S, and for any scalar γ > 0, one has Bγ (S) = Bγ (L − v) = (L − v) + Bγ = (L + Bγ ) − v = Bγ (L) − v. This argument shows that (1) is equivalent to the following inclusion: Bσ (S1 ) ∩ Bσ (S2 ) ⊂ Bρ (S1 ∩ S2 ). To prove the inclusion (2), let p = dim S1 ,

q = dim S2 ,

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r = dim (S1 ∩ S2 ).

(2)

Choose a basis e1 , . . . , er for S1 ∩ S2 and complete it into bases e1 , . . . , er , er+1 , . . . , ep

and e1 , . . . , er , ep+1 , . . . , eq

for S1 and S2 , respectively. From linear algebra we know that the combined list e1 , . . . , eq is a basis for S1 + S2 . Expand this basis to a basis e1 , . . . , en for Rn , and consider the family of parallelotopes P (ε) of the form P (ε) = {x1 e1 + · · · + xn en : |xi | 6 ε for all 1 6 i 6 n},

ε > 0.

We assert that (S1 + P (ε)) ∩ (S2 + P (ε)) = S1 ∩ S2 + P (ε),

ε > 0.

(3)

Indeed, the equality (3) immediately follows from the descriptions below: (a) S1 + P (ε) consists of all points (x1 + y1 , . . . , xp + yp , xp+1 , . . . , xn ) such that |xi | 6 ε, 1 6 i 6 n, and yi ∈ R, 1 6 i 6 p. (b) S2 + P (ε) consists of all points (x1 + y1 , . . . , xr + yr , xr+1 , . . . , xp , xp+1 + yp+1 , . . . , xq + yq , xq+1 , . . . , xn ) such that |xi | 6 ε, 1 6 i 6 n, and yi ∈ R, 1 6 i 6 r and p + 1 6 i 6 q. (c) S1 ∩ S2 + P (ε) consists of all points (x1 + y1 , . . . , xr + yr , xr+1 , . . . , xn ) such that |xi | 6 ε, 1 6 i 6 n, and yi ∈ R, 1 6 i 6 r. Since every parallelotope P (ε), ε > 0, is a convex body centered at o, a continuity argument shows the existence of positive scalars σ and ε0 such that Bσ ⊂ P (ε0 ) ⊂ Bρ . Consequently, Bσ (S1 ) ∩ Bσ (S2 ) = (S1 + Bσ ) ∩ (S2 + Bσ ) ⊂ (S1 + P (ε0 )) ∩ (S2 + P (ε0 )) = S1 ∩ S2 + P (ε0 ) ⊂ S1 ∩ S2 + Bρ = Bρ (S1 ∩ S2 ). Let α be a limit ordinal number, and let W (α) denote the set of all ordinal numbers λ < α. Choose any net (i. e, a transfinite sequence) X = {sλ : λ ∈ W (α)} of points in Rn . We will say that a set Y ⊂ Rn is an accumulating set of X provided for any positive scalar ρ > 0 the ordinal set Λ(Y, ρ) = {λ ∈ W (α) : sλ ∈ X \ Bρ (Y )} is not cofinal with W (α); that is, if sup {µ : µ ∈ Λ(Y, ρ)} < α. 6

Lemma 2. With the above notation, there is the smallest accumulating plane L ⊂ Rn of the net X. Proof. First, we will show that the intersection of any two accumulating planes of X also is an accumulating plane. Indeed, let L1 and L2 be accumulating planes of X. Choose any scalar ρ > 0. By Lemma 1, there is a scalar σ > 0 such that the inclusion (1) holds. Consequently, X \ Bρ (L1 ∩ L2 ) = X ∩ (Rn \ Bρ (L1 ∩ L2 )) ⊂ X ∩ (Rn \ (Bσ (L1 ) ∩ Bσ (L2 ))) = (X ∩ (Rn \ Bσ (L1 ))) ∪ (X ∩ (Rn \ Bσ (L2 ))) = (X \ Bσ (L1 )) ∪ (X \ Bσ (L2 )). Hence the set Λ(L1 ∩ L2 , ρ) is not cofinal with W (α) as a subset of the union of two non-cofinal with W (α) sets Λ(L1 , σ) and Λ(L2 , σ). Thus the plane L1 ∩L2 is an accumulating set of X. Since any accumulating set of X must be nonempty, one has L1 ∩ L2 6= ∅, implying that L1 ∩ L2 is a plane. Next, denote by P the family of all accumulating planes of X. Then P = 6 ∅ due to Rn ∈ P. A dimension argument shows the existence of finitely many planes L1 , . . . , Lk ∈ P such that ∩ (Lβ : Lβ ∈ P) = L1 ∩ · · · ∩ Lk . By the above proved, the induction on i = 1, . . . , k implies that every intersection L1 ∩ · · · ∩ Li is an accumulating plane of X. Clearly, L is the smallest accumulating plane of X.

4

Proof of Theorem 1

Put C = ∪ (Cλ : Cλ ∈ F). Since the cases when C is empty or is a singleton are obvious, we may assume that dim C > 1. We also may suppose that all cones Cλ are nonempty. Using the transfinite induction, one can choose in F a well-ordered subfamily F 0 such that C = ∪ (Cµ : Cµ ∈ F 0 ). Consequently, we may assume that the family F is well-ordered itself. Denote by α the ordinal number of F. If α is a successor ordinal, then C = Cα , implying that C is a cone. Hence, it remains to consider the case when α is a limit ordinal. For any λ < α, choose an apex sλ of Cλ , and put X = {sλ : λ < α}. By Lemma 2, there is the smallest accumulating plane L of X. Our goal is to show that every point of L is an apex of C, thus concluding that C is a cone. This assertion will be proved by induction on r = dim C, r > 1. The case r = 1 is obvious: either C is a halfline and L is the endpoint of C, or C is a line such that C = L; in either case L ⊂ ap C and C is a cone. Suppose that the assertion holds for all r 6 m−1, m > 2, and let dim C = m. Assume, for contradiction, that a certain point z ∈ L is not an apex of C. Then there is a point u ∈ C \{z} and a scalar t > 0 such that the point v = z +t(u−z) does not belong to C. The inclusion u ∈ C means that u ∈ Cβ for a suitable 7

ordinal number β < α. Put Dβ = {u}, and define inductively the sequence of cones Dλ , β < λ < α, as follows. 1. If λ is a successor ordinal, then put Dλ = {sλ + t(x − sλ ) : t > 0, x ∈ Dλ−1 }. 2. If λ is a limit ordinal, then put Dλ = {sλ + t(x − sλ ) : t > 0, x ∈ ∪(Dµ : µ < λ)}. Obviously, the family {Dλ : β < λ < α} is nested. Using the transfinite induction, we assert that every cone Dλ is convex. Indeed, this is trivial for λ = β. Let λ > β. If λ is a successor ordinal, then Dλ is convex as the conic hull of the convex set Dλ−1 (see [9, Theorem 4.30]). If λ is a limit ordinal, then the set ∪ (Dµ : µ < λ) is convex as the union of a nested family of convex sets Dµ (see [9, Theorem 2.8]), again implying the convexity of Dλ . Since (i) the initial set Dβ is relatively open, (ii) the conic hull of a relatively open convex set is relatively open (see [9, Theorem 4.41]), and (iii) the union of a nested family of relatively open convex sets is relatively open (see [9, Theorem 2.28]), we conclude, applying the above argument, that all cones Dλ , β < λ < α, are relatively open and their union D = ∪ (Dλ : β < λ < α) is a relatively open convex set. Due to the obvious inclusions Dλ ⊂ Cλ for all β < λ < α, one has D ⊂ C. We assert now that D is not a plane. Indeed, assume for a moment that D is a plane. Then sλ ∈ D for all β < λ < α. Hence the plane D is an accumulating plane of X, implying the inclusion L ⊂ D (see Lemma 2). Because D ⊂ C ⊂ aff C and C 6= aff C (otherwise C = aff C would be a plane, and whence a cone, contrary to the above assumption), D is a proper subplane of aff C. Therefore, dim D < dim (aff C) = dim C = m. For every β < λ < α, the set Cλ0 = D ∩ Cλ is a cone with apex sλ . Clearly, ∪ (Cλ0 : β < λ < α) = D ∩ (∪ (Cλ : β < λ < α)) = D ∩ C. Hence the union C 0 of the nested family F 0 = {Cλ0 : β < λ < α} of cones is a set of dimension 6 m − 1, and, by the inductive assumption, is a cone such that every point of L is an apex of C 0 . Next, we observe that u ∈ / D (otherwise u ∈ D ∩ C, implying the inclusion v ∈ (z, ui ⊂ D ∩ C ⊂ C, which contradicts the choice of v ∈ / C). On the other hand, {u} = Dβ ⊂ D. The obtained contradiction shows that D cannot be a plane. The assumption v ∈ / C implies that v ∈ / D. Therefore, there is a hyperplane H properly separating D and v (that is, D and v belong, respectively, to the opposite closed halfspaces V and V 0 of Rn determined by H such that {v}∪D 6⊂ H). Our goal is to show that H may be chosen to satisfy the following two conditions: (a) D ∩ H = ∅, (b) H either supports cl D or is asymptotic to cl D. 8

It is well known (see, e. g., [6, Theorem 1.3]) that H can be expressed as H = {x ∈ Rn : x·c = γ} for a suitable nonzero vector c ∈ Rn and a scalar γ ∈ R. Without loss of generality, we assume that V = {x ∈ Rn : x·c 6 γ}. (a). Suppose that D ∩ H 6= ∅. Due to rint D = D, any point x ∈ D ∩ H is relatively interior in D, which implies the inclusion D ⊂ H (see, e. g., [9, Corollary 8.3]). Consequently, v ∈ / H. Since D is not a plane, there is a point w ∈ rbd D (see [9, Corollary 2.57]). Choose a hyperplane H0 properly supporting cl D at w (so that H0 ∩ D = ∅). The intersection H ∩H0 is a plane of dimension n−2. Slightly rotating H about H ∩ H0 , we obtain a new position of H which still properly separates D and v such that D ∩ H = ∅. (b). Suppose H satisfies condition (a) but not condition (b). Then γ 0 := sup{x·c : x ∈ D} < γ. Since the hyperplane H 0 = {x ∈ Rn : x·c = γ 0 } still properly separates D and v and satisfies both conditions (a) and (b), we may put H = H 0 . It is easy to see that conditions (a) and (b) can be rewritten, respectively, as x·c < γ for all x ∈ D,

sup{x·c : x ∈ D} = γ.

(4)

Now, we assert that H is an accumulating plane of X. Clearly, it suffices to show that H is an accumulating plane of the set X 0 = {sλ ∈ X : β < λ < α}. Assume, for contradiction, that H is not an accumulating plane of X 0 . Then there is a scalar ρ > 0 such that the set Ω = {λ : sλ ∈ X 0 \ Bρ (H)} is cofinal with W (α). Since Bρ (H) can be expressed as Bρ (H) = {x ∈ Rn : γ − τ 6 x·c 6 γ + τ } for a suitable scalar τ > 0, conditions (4) show the existence of a point y ∈ D such that γ − τ 6 y·c < γ. Choose an ordinal number λ < α with the property y ∈ Dλ . Since Ω is cofinal with W (α), there is an ordinal number µ > λ. Therefore, y ∈ Dµ , and the open halfline h = (sµ , yi lies in Dµ , and whence lies in D. On the other hand, sµ · c < γ − τ 6 y · c < γ by the choice of sµ and y. Consequently, 0 < y·c − sµ ·c < γ − sµ ·c. Choosing a positive scalar t, with the condition t>

γ − sµ ·c , y·c − sµ ·c

we see that the point p = sµ + t(y − sµ ) = (1 − t)sµ + ty ∈ (sµ , yi ∈ D

9

satisfies the inequality p·c = sµ ·c + t(y·c − sµ ·c) > sµ ·c + (γ − sµ ·c) = γ. Therefore, p ∈ / V , contrary to the inclusion D ⊂ V . The obtained contradiction shows that the hyperplane H is an accumulating plane of X 0 , and is an accumulating plane of X. Next, we observe that L 6⊂ H. Indeed, if L ⊂ H, then z ∈ L ⊂ H, and the open halfline (z, ui would lie in the open halfspace int V , contrary to the choice of v ∈ (z, ui ∩ V 0 . By Lemma 2, the planes L and H meet. Then the plane L ∩ H (of dimension dim L − 1) is smaller than L, in contradiction with the choice of L. Summing up, z is an apex of C. The second assertion of the theorem immediately follows from the first one. Indeed, if Cλ is a cone with apex sλ , then Rn \ Cλ also is a cone with apex sλ . By the above proved, the right-hand side of the equality Rn \ (∩ (Cλ : Cλ ∈ F)) = ∪ (Rn \ Cλ : Cλ ∈ F)

(5)

is a cone. Consequently, the left-hand side of (5) is a cone with an apex s, say. Hence the intersection ∩ (Cλ : Cλ ∈ F) also is a cone with apex s.

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