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The ample cone consists of real numerical divisor classes represented by ample real divisors. It is an open convex cone, and Kleiman's criterion identifies it with the interior of the nef cone.

The closed cone of curves is the closure in $N_1(X{)}_{\mathbb{R}}$ of the convex cone generated by classes of integral curves. A nef divisor pairs nonnegatively with this entire closed cone. Its boundary can contain limiting classes which are not represented by one curve.

A convex cone that is a closed set in its ambient topology. In a finite-dimensional normed vector space, this means it contains every limit of its convergent sequences. For example, the positive semidefinite cone is closed because each test $X\mapsto x^{T}Xx$ is continuous.

The convex cone of completely positive matrices:It is a closed convex cone and the dual cone of the copositive cone under the Frobenius inner product. The finite-sum definition imposes no closure by fiat; closedness of the completely positive cone supplies that fact.

A cone in toric geometry is a convex cone generated by finitely many lattice vectors and containing no nonzero linear subspace. Such cones are the pieces of a fan in toric geometry.

A cone in toric geometry is the nonnegative real span of finitely many lattice vectors. It is rational when those generators lie in a lattice and strictly convex when it contains no nonzero linear subspace.

The set of all finite conic combinations of elements of $S$, including zero. It is the smallest convex cone containing $S$. It need not be closed even when $S$ is closed: the closed set $S=\{(t,1):t\ge 1\}$ generates vectors converging to $(1,0)$ via $(t,1)/t$, but no nonzero generated vector has second coordinate zero.

The real symmetric matrices that are copositive matrices form a closed convex cone:Each fixed $x$ gives a closed linear inequality in $A$. Their intersection is therefore closed and convex, and it is preserved by nonnegative scaling.

Two nonempty disjoint convex sets in a finite-dimensional real vector space admit a separating nonzero linear functional: for an appropriate $h\ne 0$ and $a$, $h\cdot x\le a\le h\cdot y$ on the respective sets. Strict separation requires additional assumptions, such as one set being compact and the other closed. For an open convex cone $C$ excluding zero, separation from $\{0\}$ gives $h\cdot c\ge 0$ for all $c\in C$; scaling cone elements forces the separating level to be zero. This finite-dimensional result is related to the supporting hyperplane theorem.

For a centrally symmetric convex body in John position, there are contact points $u_j\in \partial K\cap S^{d-1}$ and positive weights $c_j$ such that ${\sum }_j{c}_j{u}_j\otimes u_j=I$. Taking the trace gives ${\sum }_j{c}_j=d$. A supporting hyperplane at $u_j$ shows $|⟨u_j,x⟩|\le \Vert x{\Vert }_K$.

Here is the maximum-volume argument for the decomposition. If $I$ were outside the convex cone generated by the contact matrices, a separating hyperplane would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract a small positive multiple of $I$ so these inequalities become strict while the trace remains positive. The support function of $(I+tA)B_2^d$ is $|(I+tA)v|$. It decreases below $1$ near all contact directions for small positive $t$; away from them, the strict gap $h_K(v)>1$ preserves containment. Its volume increases, since $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$, a contradiction. The cone is closed because its generators have trace $1$ and form a compact set; finite-dimensional convexity gives a finite decomposition.

The nef cone is the closed convex cone of nef divisor classes. It is dual to the closed cone of curves. Its interior is the ample cone on a projective scheme.

Normalize the riskless financial asset to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain random vector is $Y=S_1/b-S_0$. A zero-cost portfolio holding $h\in {\mathbb{R}}^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $bh\cdot Y$. Thus an arbitrage meansAn equivalent martingale measure is a probability measure $Q$ with the same null sets as $P$, under which $Y$ is integrable and ${\mathbb{E}}_{Q}Y=0$. The discounted risky prices are then martingales over the single trading period.

If such $Q$ exists, a nonnegative terminal gain has expectation zero under $Q$. A nonnegative random variable with zero expectation vanishes almost surely: on $\{X\ge 1/n\}$ its expectation is at least $\mathbb{P}(X\ge 1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding arbitrage.

For the converse we prove the needed positive-density separation proof of the one-period asset-pricing theorem. First make the gain integrable without changing its null sets. Define an equivalent probability measure $P_0$ byLet $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^{\perp }$ be its orthogonal complement. Taking a finite basis of $N$ shows $Y\in L$ almost surely. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the positive-weight expectation conePositive combinations of weights show that $C$ is a convex cone. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it toThese weights remain bounded and strictly positive, whileFor every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive probability; otherwise $v\in N\cap L$. ConsequentlyThe matrix is finite because $\Vert Y{\Vert }^2/(1+\Vert Y\Vert )\le \Vert Y\Vert$. The kernel of $A_Z$ on $L$ is zero, so it defines an invertible matrix on the finite-dimensional vector space $L$. For any sufficiently small $\delta >0$, every $y$ with $\Vert y\Vert <\delta /\Vert A_Z^{-1}\Vert$ is the image of a vector in the radius-$\delta$ ball, since $\Vert A_Z^{-1}y\Vert <\delta$. Hence the perturbations give a neighborhood of zero, proving relative openness of $C$.

If $0\notin C$, the hyperplane separation theorem gives a nonzero $h\in L$ with $h\cdot c\ge 0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=1_A+\epsilon$. Taking the limit as $\epsilon \downarrow 0$ yields ${\mathbb{E}}_{P_0}[(h\cdot Y)1_A]\ge 0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an arbitrage, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies ${\mathbb{E}}_{P_0}[ZY]=0$. Normalize it by $dQ/d{P}_0=Z/{\mathbb{E}}_{P_0}Z$. This gives an equivalent martingale measure with integrable gains, establishing the fundamental theorem of asset pricing here without any further existence theorem.

Now retain only the $m$ positive-probability states of the finite probability space. Let $V\subseteq {\mathbb{R}}^m$ be the linear span of the discounted terminal asset payoffs; it contains the constant vector $1$. A complete market means $V={\mathbb{R}}^m$. An equivalent martingale measure is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V={\mathbb{R}}^m$, the indicator functions of individual states are attainable, so these prices determine every $q_j$ uniquely.

If $V$ is proper, choose $0\ne v\in V^{\perp }$. Since $1\in V$, ${\sum }_j{v}_j=0$. For sufficiently small $\epsilon >0$, both $q+\epsilon v$ and $q-\epsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct equivalent martingale measures. Therefore market completeness is equivalent to uniqueness of the equivalent martingale measure.

We prove exhaustiveness, including existence of the positive density when neither form of arbitrage occurs. Use the subspace $L$ from part (a)(ii), and considerThis is a nonempty convex cone closed under positive scaling: ${\nu }_0=(1+\Vert A_1\Vert {)}^{-1}$ supplies one element and guarantees weighted integrability even when $A_1$ itself has no first moment. Crucially, $\mathcal{K}$ is relatively open in $L$; it need not be closed.

To prove openness at a point $z=\mathbb{E}[\nu A_1]$, put $Z=\nu A_1$. The vectors $\mathbb{E}[hZ]$, for bounded real measurable $h$, span $L$. Indeed, a vector orthogonal to all of them would satisfy $\mathbb{E}[h{u}^{\top }Z]=0$ for every such $h$, forcing $u^{\top }Z=0$ almost surely. Since $\nu >0$, this says $u\in L^{\perp }$. Choose finitely many bounded $h_j$ for which $\mathbb{E}[h_{j}Z]$ form a basis of $L$. For coefficients $t_j$ sufficiently small,This gives a relative neighbourhood of $z$. It also proves the openness step behind the strictly positive barycentre cone lemma without assuming a closed moment cone.

If $A_0\notin L$, the second arbitrage form was already constructed. If $A_0\in L\setminus \mathcal{K}$, the separation of a point and an open convex set, applied inside the finite-dimensional space $L$, gives a nonzero $x\in L$ with $x^{\top }{A}_0\le {\inf }_{z\in \mathcal{K}}{x}^{\top }z$. Positive scaling in the cone forcesWe claim $x^{\top }{A}_1\ge 0$ almost surely. Otherwise $E=\{x^{\top }{A}_1<0\}$ has positive probability, and the admissible positive weights ${\nu }_m={\nu }_0(1+m{1}_E)$ givecontradicting the separation inequality. Since $x\in L$ is nonzero, $x^{\top }{A}_1$ cannot vanish almost surely; thus it is strictly positive on an event of positive probability. This is the first arbitrage form.

Consequently absence of both forms forces $A_0\in \mathcal{K}$, which supplies the required $\nu$. Conversely either form excludes every such density by part (a). We have proved exactly one of arbitrage or a strictly positive pricing density holds. If $L=\{0\}$, the same argument reduces to the elementary cases $A_0=0$ and $A_0\ne 0$, so no nondegeneracy assumption has been hidden.

We prove the positive state-price density alternative using finite-dimensional convex cones. Put $w=(1+\Vert A_1\Vert {)}^{-1}$ and defineAll displayed expectations are finite, since $w\Vert A_1\Vert \le 1$. The set $D$ is a convex cone, so $K$ is a closed convex cone. A vector $x$ is nonnegative on $K$ exactly when $x^{\top }{A}_1\ge 0$ almost surely. Indeed, one direction follows by taking every nonnegative bounded test weight $f$; for the other direction, if $x^{\top }{A}_1<0$ on a set of positive probability, take its indicator as $f$ to obtain a strictly negative expectation. This also identifies $K$ as the closed cone generated by the support of a probability distribution of $A_1$.

The three cases $A_0\notin K$, $A_0\in K\setminus \mathrm{ri}K$, and $A_0\in \mathrm{ri}K$ are treated below. They give respectively a negative-cost separating portfolio, a supporting zero-cost portfolio with a strictly positive payoff on an event of positive probability, or the required positive pricing weight. Thus the alternatives are exhaustive.

They are mutually exclusive. If $A_0=\mathbb{E}[\nu A_1]$ with $\nu >0$ and a portfolio has nonnegative payoff, then $x^{\top }{A}_0=\mathbb{E}[\nu x^{\top }{A}_1]\ge 0$. A negative initial cost is impossible. If the payoff is strictly positive with positive probability, this expectation is strictly positive, so nonpositive initial cost is impossible too. The given weighted integrability permits taking the expectation of this scalar product.

In a one-period market, $x^{\top }{A}_0$ is the portfolio's initial cost and $x^{\top }{A}_1$ its terminal value. The first two cases are investment-consumption arbitrage: nonnegative wealth can be taken out at both dates, with a strict benefit at one of them. If a strictly positive bank account is traded, any negative initial cost can be invested in it, converting the opportunity into a zero-cost arbitrage with nonnegative terminal payoff and positive payoff with positive probability.

In the third case, $\nu$ is a strictly positive state-price density. If a traded numéraire has price $B_0>0$ and terminal value $B_1>0$, its pricing identity is $\mathbb{E}[\nu B_1]=B_0$. ConsequentlyThe density has mean one and is strictly positive, so $Q$ is equivalent to the physical measure. Thus discounted asset prices are martingales in the one-period sense. This is the fundamental theorem of asset pricing formulation of absence of arbitrage; normalization to a probability measure uses the positive traded numéraire, not an unsupported assumption that $\mathbb{E}\nu =1$.

Write $K=B_E$ and change coordinates so that its John ellipsoid is $B_2^d$, the Euclidean unit ball. Then the standard symmetric-body inequalities areWe give the contact argument underlying the construction. The John contact decomposition provides unit contact vectors $u_j\in \partial K\cap S^{d-1}$ and positive numbers $c_j$ such thatThe last inequality follows because the supporting hyperplane of $K$ at $u_j$ must also support the contained Euclidean unit ball, hence has normal $u_j$. For completeness, the decomposition follows from maximal volume as follows. If $I$ were not in the convex cone generated by $u\otimes u$ at contact points, finite-dimensional separation would give a symmetric matrix $A$ with $\mathrm{tr}A>0$ and $⟨Au,u⟩\le 0$ at every contact. Subtract $\delta I$ for small $\delta >0$, retaining positive trace and obtaining strict negative values. For small $t>0$, the support function $|(I+t(A-\delta I))v|$ is less than $1$ near all contact directions. On the compact complement of those neighbourhoods, $h_K(v)-1$ has a positive minimum, so the perturbed ellipsoid remains inside $K$ there too. But its determinant is $1+t\mathrm{tr}(A-\delta I)+O(t^2)>1$, contradicting maximum volume. The cone is closed: its generating contact matrices form a compact set of trace $1$, so bounded traces bound their nonnegative coefficient sums. Finite-dimensional convexity then supplies finitely many contacts in the decomposition. Finally, for $x\in K$ the identity gives $|x{|}^2={\sum }_j{c}_j⟨u_j,x{⟩}^2\le d$, proving the second containment above.

Construct an orthonormal basis recursively. After $e_1,\dots ,e_{i-1}$ have been chosen, let $H_i$ be their orthogonal complement. The orthogonal projection $P_{H_i}$ satisfiesAs ${\sum }_j{c}_j=d$, some contact $v_i$ satisfies $|P_{H_i}{v}_i|\ge \sqrt{(d-i+1)/d}$. Choose $e_i=P_{H_i}{v}_i/|P_{H_i}{v}_i|$. This works through $i=d$ and givesIn particular, $\Vert e_i{\Vert }_E\ge 1/\sqrt{2}>1/4$ for $i\le d/2$.

Keep the contact vectors $v_i$ used for this basis. They lie in $\mathrm{span}(e_1,\dots ,e_i)$. Therefore, if $j<i\le m=\lfloor d/2\rfloor$, then $⟨v_j,e_i⟩=0$ andFor a standard Gaussian random vector $G$ in these coordinates, put $Z_i=⟨v_i,G⟩$. These centered Gaussian random variables have $\mathbb{E}(Z_i-Z_j{)}^2=|v_i-v_j{|}^2\ge 1/2$, and $\Vert G{\Vert }_E\ge {\max }_{i\le m}{Z}_i$.

Here is a proof of the Sudakov-Fernique inequality needed for this finite family. Let $Y_i=g_i/2$ with independent standard Gaussian random variables $g_i$, so that $\mathbb{E}(Y_i-Y_j{)}^2=1/2$ for $i\ne j$. Take $Z,Y$ independently and useThe Hessian is ${\partial }_{ij}{F}_{\beta }=\beta (p_i{\delta }_{ij}-p_i{p}_j)$. Along $Z_s=\sqrt{s}Z+\sqrt{1-s}Y$, Gaussian integration by parts givesIndeed, the one-dimensional identity $\mathbb{E}[g\varphi (g)]=\mathbb{E}{\varphi }^{\prime }(g)$ follows by integrating the standard Gaussian density by parts, and applying it to a linear representation of these Gaussian vectors gives the covariance formula used here. Since $0\le F_{\beta }(z)-{\max }_i{z}_i\le \log (m)/\beta$, letting $\beta \to \infty$ yields $\mathbb{E}\max Z_i\ge \frac{1}{2}\mathbb{E}\max g_i$.

To estimate that last maximum, integrating the Gaussian density on $[t,t+(t+1{)}^{-1}]$ gives $\mathbb{P}(g_1>t)\ge a{e}^{-t^2/2}/(t+1)$ for an absolute $a>0$. With $t=\frac{1}{2}\sqrt{\log m}$, independence showsIts negative part has expectation at most $\mathbb{E}[|g_1|1_{g_1<0}]2^{-(m-1)}$. Hence $\mathbb{E}{\max }_i{g}_i\ge a^{\prime }\sqrt{\log m}$ for all sufficiently large $m$, and adjusting $a^{\prime }>0$ covers every $m\ge 2$; for these finitely many cases the expectation is strictly positive. For $d\ge 4$, $\log \lfloor d/2\rfloor \ge \frac{1}{3}\log d$. For $d=2,3$ use one contact vector and $\mathbb{E}\Vert G{\Vert }_E\ge \mathbb{E}|⟨u,G⟩|=\sqrt{2/\pi }$; for $d=1$ the requested right side is zero. Thusfor an absolute $c>0$, independent of both dimension and the norm.

Fix $\epsilon >0$ and a finite epsilon-net $x_1,\dots ,x_N$ in $X$. Choose a Borel map $r:X\to \{x_1,\dots ,x_N\}$ with $d(x,r(x))\le \epsilon$, for example the nearest point with ties broken by index. Its cells $A_i$ give discrete probability measures with masses $p_i=P(A_i)$ and $q_i=Q(A_i)$. LetThe feasible set is nonempty and compact, so the minimum exists. Pushing any coupling forward by $(r,r)$ changes its cost by at most $2\epsilon$, hence $W_{\epsilon }\le W+2\epsilon$. Conversely, lift a discrete coupling by the measureTerms with zero row or column masses vanish. This lift has marginals $P,Q$, and its cost differs by at most $2\epsilon$. Therefore

Here is the finite linear programming duality proof. Write $d_{ij}=d(x_i,x_j)$ and consider the convex coneIt is closed and finitely generated. One elementary justification of closedness is to reduce each nonnegative representation to linearly independent generators, eliminating a coefficient along any linear dependence; after choosing a subsequence using the same subset of generators, their independent coordinates have bounded, convergent coefficients whenever the represented vectors converge. For $\eta >0$, the point $(p,q,W_{\epsilon }-\eta )$ lies outside $C$. Separation from this closed convex cone gives coefficients $(a,b,\tau )$ withIf $\tau =0$, summing the middle inequalities against any feasible $\pi$ contradicts the last one. Thus $\tau >0$, and $f_i=-a_i/\tau$, $g_j=-b_j/\tau$ obey $f_i+g_j\le d_{ij}$ with dual objective greater than $W_{\epsilon }-\eta$. The weak inequality already proved gives the reverse bound. Letting $\eta \downarrow 0$ proves finite strong duality.

Reduce this discrete dual to a single Lipschitz function. For a feasible pair put ${\varphi }_i={\min }_j(d_{ij}-g_j)$. The triangle inequality gives $|{\varphi }_i-{\varphi }_l|\le d_{il}$, while $f_i\le {\varphi }_i$ and $g_i\le -{\varphi }_i$. ConsequentlyConversely, any $1$-Lipschitz vector $\varphi$ gives a feasible pair $(\varphi ,-\varphi )$. Adding a constant leaves its objective unchanged, so impose ${\varphi }_1=0$. The resulting feasible set is compact, and its maximum is $W_{\epsilon }$.

For a maximizing vector, the McShane extension theorem has the following explicit proof in this situation. DefineThe triangle inequality proves that this extension is $1$-Lipschitz, and the discrete Lipschitz condition makes $\varphi (x_i)={\varphi }_i$. Thus $(\varphi ,-\varphi )$ is a continuous feasible pair. Because $|\varphi (x)-\varphi (r(x))|\le \epsilon$,Letting $\epsilon \downarrow 0$, together with part i, proves $m_d(P,Q)=W(P,Q)$. The infimum defining $W$ is also attained: the set of couplings is a closed subset of the compact space of probability measures on $X\times X$, and integrating the continuous cost is continuous for the weak-star topology.

Let the finite primal supremum be $v$. First it is attained. The setis a finitely generated cone, hence closed by the argument in the root Solution. For a sequence of primal feasible points whose values approach $v$, the pairs $(b,c^{T}x)$ belong to $K$. Their limit $(b,v)$ therefore belongs to $K$. It gives a feasible point $x_{\ast }$ with $c^T{x}_{\ast }\ge v$, so necessarily $c^T{x}_{\ast }=v$.

Let $I$ index the tight primal inequalities at $x_{\ast }$ and let $Z$ index its zero coordinates. Consider the closed convex cone generated by $A_i^T$ for $i\in I$ and $-e_j$ for $j\in Z$. If $c$ were outside this cone, separation from a closed convex cone would give a direction $d$ withFor sufficiently small positive $\delta$, $x_{\ast }+\delta d$ remains a feasible point: the active constraints have the required signs, and every inactive constraint retains its positive slack. Its larger objective contradicts optimality. Therefore there are $y_i\ge 0$, supported on $I$, and ${\mu }_j\ge 0$, supported on $Z$, such thatThis makes $y$ a dual feasible point. Moreover, the support restrictions giveTogether with weak duality, this proves strong duality and dual attainment:

The dual linear program is the minimization problemIndeed, for $y\ge 0$ the optimization Lagrangian $c^{T}x+y^T(b-Ax)$ bounds the primal objective above on its feasible points. Its supremum over $x\ge 0$ is finite precisely when $A^{T}y\ge c$, and then equals $b^{T}y$. The two labelled parts below establish the relevant weak duality and strong duality conclusions.

For the subsequent infeasibility argument, introduce nonnegative slack variables $s$ and the convex coneThis is generated by the columns of $A$ and the coordinate vectors. Here is why this finitely generated cone is closed. Any representation with a minimal number of positive coefficients uses linearly independent generators: otherwise subtract a suitable multiple of a dependence until one positive coefficient vanishes, preserving nonnegativity. For each fixed independent selection of generators, its nonnegative combinations form a closed set, because the inverse coordinate map on its span is continuous. There are finitely many selections, so their union $C$ is closed. This is the closedness of finitely generated cones used in Farkas' lemma.

When the primal linear program is infeasible, $b\notin C$. Let $p$ be the nearest point of $C$ to $b$, and set $h=p-b$. The minimizing property gives $h^T(z-p)\ge 0$ for every $z\in C$. Taking $z=0,2p$ gives $h^{T}p=0$. ConsequentlyIn particular $h\ge 0$ and $A^{T}h\ge 0$. Rescaling $h$ by the positive number $-h^{T}b$ produces the requested Farkas' lemma certificate:Now a feasible point of the relaxed inequalities would satisfySince ${\lambda }^{T}b=-1$, the sum of the nonnegative ${\lambda }_i$ is positive. Thus robust infeasibility under uniform constraint relaxation givesThe ellipsoid method makes progress by reducing the volume of a containing ellipsoid. A nonempty linear polyhedron may have empty interior, so small volume alone cannot certify that the original system is infeasible. Relaxing the inequalities remedies this difficulty, but must not make an infeasible system appear feasible. The displayed positive gap proves exactly that sufficiently small relaxations preserve infeasibility. For rational input, determinant bounds give a Farkas' lemma certificate of polynomial encoding length, so a sufficiently small rational relaxation can be chosen from the input encoding length. Together with a bounding box and bounds on the volume of a relaxed feasible system, this lets the ellipsoid method distinguish infeasibility from a feasible system of very small or zero original volume in polynomial time. The certificate argument alone does not assert a polynomial precision bound for arbitrary unencoded real coefficients.

Choose $0<ϵ\ll 1$ as in (c). Thenis the sum of a nef divisor and an ample real divisor. The nef-plus-ample ampleness lemma givesFor clarity, this last lemma follows from Kleiman's criterion and the convex cone property: if $x$ is an interior point of the nef cone and $y$ lies in that cone, translating a small neighbourhood of $x$ by $y$ stays in the cone. Thus $x+y$ remains in its interior, which is the ample cone on a projective scheme.

The complete argument proves the real Nakai–Moishezon criterion rather than assuming it: curve positivity gives nefness, rational approximation and a proved section-count inequality give bigness, induction and the finite-support argument give a uniform ample subtraction, and the nef-plus-ample lemma concludes ampleness. The zero-dimensional case is automatic, and ampleness on reduced components handles reducibility and nilpotents.

For a convex cone $D$ in an inner product space, use the nonnegative-pairing conventionHere the pairing is the Frobenius inner product on real symmetric matrices. Let $C_0$ be the conic hull of the nonnegative rank-one matrices $x{x}^T$, and let $C=\overline{C_0}$. For $A\in K$, $⟨A,x{x}^T{⟩}_F=x^{T}Ax\ge 0$. This extends to conic combinations, and by continuity to their limits. Hence $C\subseteq K^{\ast }$.

For the converse, if $B\notin C$, separation from a closed convex cone supplies a symmetric $H$ withIn particular $x^{T}Hx\ge 0$ for every $x\ge 0$, so $H\in K$. The negative pairing then excludes $B$ from $K^{\ast }$. ThereforeThe same generator test gives $C^{\ast }=K$. This is the duality of copositive and completely positive cones; the next argument removes the closure.

The space of real numerical divisor classes is $N^1(X{)}_{\mathbb{R}}$. Write any big class as $[D]=[A]+[E]$, with $A$ an ample real divisor and $E$ effective. For a sufficiently small perturbation $\delta \in N^1(X{)}_{\mathbb{R}}$, the class $[A]+\delta$ remains in the open ample cone, so $[D]+\delta$ remains big by part (v). Thus $\mathrm{Big}(X)$ is open. Positive scaling and addition preserve the ample-plus-effective expression, so it is a convex cone.

For $n\ge 1$, fix a very ample divisor $H$. The linear functional $\ell ([D])=D\cdot H^{n-1}$ is strictly positive on every big class: the ample part contributes positively and the effective part nonnegatively. No nonzero linear subspace can be contained in this cone, since it would contain both $v$ and $-v$. In particular the zero class is not big in positive dimension. When $n=0$, $N^1(X{)}_{\mathbb{R}}=0$, so there is no positive-dimensional subspace to consider. Therefore