On the sufficiency of finite support duals in semi-infinite linear programming
aa r X i v : . [ m a t h . O C ] A p r On the sufficiency of finite support duals in semi-infinitelinear programming
Amitabh Basu a , Kipp Martin b , Christopher Thomas Ryan b a The Johns Hopkins University b University of Chicago, Booth School of Business
Abstract
We consider semi-infinite linear programs with countably many constraintsindexed by the natural numbers. When the constraint space is the vectorspace of all real valued sequences, we show that the finite support (Haar)dual is equivalent to the algebraic Lagrangian dual of the linear program.This settles a question left open by Anderson and Nash [2]. This resultimplies that if there is a duality gap between the primal linear program andits finite support dual, then this duality gap cannot be closed by consideringthe larger space of dual variables that define the algebraic Lagrangian dual.However, if the constraint space corresponds to certain subspaces of all real-valued sequences, there may be a strictly positive duality gap with the finitesupport dual, but a zero duality gap with the algebraic Lagrangian dual.
Keywords: semi-infinite linear programs; finite support duals; duality gaps
1. Introduction
We begin with a brief review of notation and basic definitions for semi-infinite linear programs. Let Y be a vector space. The algebraic dual of Y isthe set of linear functionals with domain Y and is denoted by Y ′ . Let ψ ∈ Y ′ .The evaluation of ψ at y is denoted by h y, ψ i ; that is, h y, ψ i = ψ ( y ). Weemphasize that the theory presented here deals with algebraic dual spaces and not topological dual spaces. Discussion of how our work relates to topologicalduals appears in Remarks 2.2 and 2.3. Email addresses: [email protected] (Amitabh Basu), [email protected] (Kipp Martin), [email protected] (Christopher Thomas Ryan)
Preprint submitted to Operations Research Letters July 30, 2018 et P be a convex cone in Y . A convex cone P is pointed if and only if P ∩ − P = { } . In the rest of the paper all convex cones are assumed to bepointed. A pointed convex cone P in Y defines a vector space ordering (cid:23) P of Y , with y (cid:23) P y ′ if y − y ′ ∈ P . The algebraic dual cone of P is P ′ = { ψ ∈ Y ′ : h y, ψ i ≥ y ∈ P } . Elements of P ′ are called positive linear functionals on Y (see for instance,page 17 of [9]). Let A : X → Y be a linear mapping from vector space X to vector space Y . The algebraic adjoint A ′ : Y ′ → X ′ is a linear operatordefined by A ′ ( ψ ) = ψ ◦ A and satisfies h x, A ′ ( ψ ) i = h A ( x ) , ψ i where ψ ∈ Y ′ and x ∈ X . Using this notation, define the primal conic optimization probleminf x ∈ X h x, φ i s.t. A ( x ) (cid:23) P b (ConLP)where b ∈ Y and φ is a linear functional on X .Now define the standard algebraic Lagrangian dual for (ConLP). sup ψ ∈ P ′ inf x ∈ X {h x, φ i + h b − A ( x ) , ψ i} = sup ψ ∈ P ′ inf x ∈ X (cid:8) h x, φ i + h b, ψ i − h A ( x ) , ψ i (cid:9) = sup ψ ∈ P ′ (cid:8) h b, ψ i + inf x ∈ X {h x, φ i − h A ( x ) , ψ i} (cid:9) = sup ψ ∈ P ′ (cid:8) h b, ψ i + inf x ∈ X {h x, φ i − h x, A ′ ( ψ ) i} (cid:9) = sup ψ ∈ P ′ (cid:8) h b, ψ i + inf x ∈ X h x, φ − A ′ ( ψ ) i (cid:9) . Since x ∈ X is unrestricted, if φ − A ′ ( ψ ) is not the zero linear functional on X, then the inner minimization goes to negative infinity, so require φ − A ′ ( ψ ) = θ X , where θ X is the zero linear functional on X . Then the Lagrangian dualof (ConLP) is sup h b, ψ i s . t . A ′ ( ψ ) = φψ ∈ P ′ . (ConDLP)This problem is called the algebraic Lagrangian dual of (ConLP) since thelinear functionals ψ that define the dual problem are in Y ′ , which is thealgebraic dual of Y. emi-infinite linear programs. Consider the case where X = R n and Y = R I ,i.e., the vector space of real-valued functions with domain I where I is anarbitrary (potentially infinite) set. Let a , a , . . . , a n and b be functions in Y = R I . Let A : R n → R I be the linear mapping x ( a ( i ) x + a ( i ) x + . . . + a n ( i ) x n : i ∈ I ). Let R I + denote the pointed cone of u ∈ R I such that u ( i ) ≥ i ∈ I and let P = R I + . With this specification for the vectorspaces X and Y , the map A , right hand side b and cone P , problem (ConLP)reduces to the standard semi-infinite linear programinf x ∈ R n φ ⊤ x s.t. P nk =1 a k ( i ) x k ≥ b ( i ) for all i ∈ I. (SILP)There is a slight abuse of notation here. When X = R n , the algebraic dual X ′ is isomorphic to R n so each linear functional φ ∈ X ′ can be mapped toa vector in R n . Thus, the primal objective function h x, φ i in (ConLP), isreplaced by the inner product φ ⊤ x with φ now treated as a vector in R n . Next consider two alternative duals of (SILP): the algebraic Lagrangiandual and the finite support dual due to Haar [8]. Recall that ( R I + ) ′ denotesthe algebraic dual cone of P = R I + . The algebraic Lagrangian dual of (SILP)using (ConDLP) is sup h b, ψ i s . t . A ′ ( ψ ) = φψ ∈ ( R I + ) ′ . (DSILP)A second dual is derived as follows. Instead of considering every linearfunctional ψ ∈ ( R I + ) ′ as above, consider a subset of these linear functionals,called the finite support elements . For u ∈ R I , the support of u is the setsupp( u ) = { i : u ( i ) = 0 } . The subspace R ( I ) denotes those functions in R I with finite support. Let R ( I )+ denote the pointed cone of v ∈ R ( I ) such that v ( i ) ≥ i ∈ I . Under the natural embedding of R ( I ) into ( R I ) ′ for u ∈ R I and v ∈ R ( I ) , write h u, v i = P i ∈ I u ( i ) v ( i ). The latter sum is well-defined since v has finite support. Under this embedding, R ( I )+ is a subset of( R I + ) ′ . Moreover, under this embedding, A ′ : ( R I ) ′ → X ′ (= R n ) restricted to R ( I ) becomes the map A ′ ( v ) = ( P i ∈ I a k ( i ) v ( i )) nk =1 . The finite support dual is sup P i ∈ I b ( i ) v ( i )s . t . P i ∈ I a k ( i ) v ( i ) = φ k , k = 1 , . . . , nv ∈ R ( I )+ . (FDSILP)3he finite support dual (FDSILP) is restricted to the linear functionals ψ that can be mapped to v ∈ R ( I )+ under the standard embedding of R ( I ) into ( R I ) ′ . Therefore v (FDSILP) ≤ v (DSILP) where the optimal value ofoptimization problem ( ∗ ) is denoted by v ( ∗ ). This leads naturally to thefollowing question. Question 1.
Is it possible that v (SILP) = v (DSILP) and yet v (SILP) >v (FDSILP)? In other words, can there exist a duality gap between the pri-mal and its finite support dual that is closed by considering the algebraicLagrangian dual?This question is significant for the study of semi-infinite linear program-ming for at least two reasons. First, most duality theory has been developedfor the finite support dual [3, 5, 6, 10, 12, 14]. Moreover, the only other dualgiven significant attention in the literature is the “continuous dual” (see forinstance [4, 7]) and this dual shares many of the same duality properties asthe finite support dual. Indeed, as stated by Goberna in [4]: “all knownduality theorems guaranteeing the existence of a zero duality gap have thesame hypotheses for both dual problems [the finite support dual and thecontinuous dual]”. He even goes so far to say that the finite support dualand the continuous dual are “equivalent in practice.”Second, the algebraic Lagrangian dual is notoriously challenging to char-acterize and work with. Indeed, to the author’s knowledge, little has beensaid about the algebraic dual in the semi-infinite programming literature(only a few studies mention it, and they do not draw conclusions about itsconnection with the finite support dual [2, 13]).To the authors’ knowledge Question 1 has not been settled for I = N , i.e.,semi-infinite linear programs with countably many constraints. Indeed, onpage 66 of Anderson and Nash’s seminal work [2] they write: “It seems to behard, if not impossible, to find examples of countable semi-infinite programswhich have a duality gap in this formulation [the finite support dual], buthave no duality gap when we take W to be a wider class of sequences”where W refers to the vector space of dual variables. In our notation, W =( R N ) ′ in (DSILP) and W = R ( N ) in (FDSILP). Semi-infinite linear programswith countably many constraints have been well-studied in the literature,particularly from the perspective of duality [3, 10, 11]. In fact, one caneven show that, theoretically, there is no loss in generality in consideringthe countable case. Theorem 2.3 in [11] shows that every semi-infinite linear4rogram with uncountable many constraints can be equivalently reposed overa countable subset of the original constraints.The main result of this paper (Theorem 2.4) proves that the answerto Question 1 is no for the case of I = N , settling Anderson and Nash’sopen question. We show that v (DSILP) = v (FDSILP) by establishing that(DSILP) and (FDSILP) are equivalent programs.However, there is a subtlety in Question 1 to keep in mind for semi-infinitelinear programs with countably many constraints. In the above discussion,a semi-infinite linear program with countably many constraints was cast asan instance of (ConLP) with X = R n , Y = R N , A : X → Y defined by A ( x ) = ( a ( i ) x + a ( i ) x + . . . + a n ( i ) x n : i ∈ I ), and P = R N + . Then (DSILP)was formed using (ConDLP). However, if the functions a , a , . . . , a n and b lie in a subspace V ⊆ R N , then we may use Y = V and P = V ∩ R N + to write the semi-infinite linear program as an instance of (ConLP). Thecorresponding (ConDLP) issup h b, ψ i s . t . A ′ ( ψ ) = φψ ∈ ( V ∩ R N + ) ′ (DSILP( V ))where ( V ∩ R N + ) ′ ⊆ V ′ is the dual cone of P = V ∩ R N + , which lies in thealgebraic dual of V .It is quite possible that a positive linear functional defined on ( V ∩ R N + ) ′ cannot be extended to ( R N + ) ′ . This implies (DSILP) (with I = N ) may havea smaller value than (DSILP( V )), i.e., v (DSILP) < v (DSILP( V )). In thiscontext, the following question is a natural extension of Question 1. Question 2.
Is it possible that v (SILP) = v (DSILP( V )) and v (SILP) >v (FDSILP) = v (DSILP) when a , . . . , a n , b ∈ V for some subspace V ⊆ R N ?In other words, when the constraint space V lies in a subspace of R N , can thereexist a duality gap between the primal and its finite support dual (FDSILP),that is closed by considering the algebraic Lagrangian dual defined accordingto that subspace?We show in Section 3 that this can happen. More concretely, in Exam-ple 3.5 in Section 3, there is a duality gap between (SILP) and the finitesupport dual (FDSILP). However, if a , . . . , a n , b are considered as elementsof the space of convergent real sequences c , then (SILP) is a special caseof (ConLP) with X = R n , V = Y = c , A : X → Y and P = c + (the cone of5onvergent sequences with nonnegative entries), and there is no duality gapwith its algebraic Lagrangian dual (DSILP( c ))sup h b, ψ i s . t . A ′ ( ψ ) = φψ ∈ ( c + ) ′ (DSILP( c ))where ( c + ) ′ is the algebraic dual cone of c + . The same result also holds when Y is the subspace of bounded real sequences ℓ ∞ .
2. Main resultLemma 2.1.
For every ψ ∈ ( R N + ) ′ , there exists u ∈ R ( N )+ such that h y, ψ i = P i ∈ N y ( i ) u ( i ) for every y ∈ R N . In other words, every positive linear func-tional on R N can be represented by a positive finite support dual vector. Proof.
Consider any ψ ∈ ( R N + ) ′ , i.e., h y, ψ i ≥ y ≥
0. We show that ψ can be represented by a finite support linear functional. Claim 1.
There exists M ∈ N such that for all v ∈ R N + whose first M components are zero, h v, ψ i = 0. Proof.
Suppose no such M exists. Then for every n ∈ N , there exists v n ∈ R N + such that h v n , ψ i > n components of v n are zero. Consider the sequence of vectorsˆ v n = v n h v n ,ψ i for n ∈ N . Observe that h ˆ v n , ψ i = 1 for all n ∈ N . Now, considervector a ∈ R N + where a ( i ) = P n ∈ N ˆ v n ( i ) = P in =1 ˆ v n ( i ) + P n>i ˆ v n ( i ) for all i ∈ N . Since ˆ v n ( i ) = 0 for all n > i , P n>i ˆ v n ( i ) = 0. This implies P n ∈ N ˆ v n ( i )is a finite sum and therefore well-defined. Given any N ∈ N , it follows fromthe definition of a ( i ), that a ( i ) − P Nn =1 ˆ v n ( i ) = P n ∈ N ˆ v n ( i ) − P Nn =1 ˆ v n ( i ) = P n>N ˆ v n ( i ). We have P n>N ˆ v n ( i ) ≥ v n are in R N + . This implies a − P Nn =1 ˆ v n ≥
0. Thus, h a − P Nn =1 ˆ v n , ψ i ≥ N ∈ N . By linearityof ψ , this implies that h a, ψ i ≥ P Nn =1 h ˆ v n , ψ i = N for every N ∈ N . But thismeans h a, ψ i cannot be a finite number, which is a contradiction of the factthat ψ , being a linear functional, is real valued.Let M be the natural number from Claim 1. Now let e i denote the elementof R N with 1 in the i -th coordinate and 0 everywhere else. Let u ∈ R ( N ) bea finite support element given as follows u ( i ) = (cid:26) h e i , ψ i , i ≤ M , i > M. laim 2. h y, ψ i = P i ∈ N u ( i ) y ( i ) for every y ∈ R N + . Proof.
Observe that any y ∈ R N + can be represented as y = P Mi =1 y ( i ) e i + v where v ∈ R N + has zeros in its first M components. By Claim 1, h v, ψ i = 0.Hence h y, ψ i = P Mi =1 y ( i ) h e i , ψ i + h v, ψ i = P i ∈ N u ( i ) y ( i ).Claim 2 only applies to y ≥
0, i.e., the nonnegative elements in R N . Tocomplete the proof, we need to show that for arbitrary y ∈ R N , h y, ψ i = P i ∈ N u ( i ) y ( i ). Define y + , y − ∈ R N + as follows: for each i ∈ N , y + ( i ) =max { y ( i ) , } and y − ( i ) = max {− y ( i ) , } . Thus, we can write y ∈ R N as y = y + − y − . Therefore, h y, ψ i = h y + , ψ i − h y − , ψ i = P i ∈ N u ( i ) y + ( i ) − P i ∈ N u ( i ) y − ( i )= P i ∈ N u ( i )( y + ( i ) − y − ( i ))= P i ∈ N u ( i ) y ( i )where the second equality follows from Claim 2.Recall every u ∈ R ( N )+ maps to a positive linear functional ψ over R N via ψ ( x ) = P ∞ i =1 u ( i ) x ( i ). Thus, R ( N )+ can be embedded into a subset of ( R N + ) ′ .Combined with Lemma 2.1, this implies ( R N + ) ′ ∼ = R ( N )+ . In other words, thealgebraic dual cone of R N + is isomorphic to the positive cone R ( N )+ in R ( N ) . Remark 2.2.
The fact that ( R N + ) ′ is isomorphic to R ( N )+ does not contradictthe well-known fact that the full algebraic dual ( R N ) ′ of R N is difficult tocharacterize. Indeed, the full algebraic dual ( R N ) ′ is of uncountable dimension(see page 195 of [1]) whereas the algebraic dual cone ( R N + ) ′ is isomorphic toa subset of R ( N ) , which has countable dimension. One intuitive justificationis that R N + is not a “full dimensional” subset of R N . This is because R N + has empty interior in every linear topology. For a justification of this factsee Appendix A. Remark 2.3.
Our Lemma 2.1 shares some similarities with Theorem 16.3 of[1]. We emphasize that Lemma 2.1 does not contradict, nor depend on, The-orem 16.3 of [1]. The latter result states that, under the usual product topol-ogy of R N , the topological dual ( R N ) ∗ of R N is isomorphic to R ( N ) . Lemma 2.1differs from this result in two ways. First, Lemma 2.1 demonstrates an equiv-alence between the algebraic dual cone of R N + and the positive cone of R ( N ) .7n other words, Theorem 16.3 of [1] concerns all continuous linear functionalson R N whereas our result concerns only positive linear functionals. Second,our result is a statement about algebraic duality whereas Theorem 16.3 of [1]concerns topological duality. Our proof of Lemma 2.1 is direct and entirelyalgebraic. It does require Theorem 16.3 of [1] or any topological concepts.Using Lemma 2.1, Question 1 is answered for I = N . Theorem 2.4.
When I = N , v (SILP) = v (DSILP) if and only if v (SILP) = v (FDSILP). Proof.
It suffices to show that v (DSILP) = v (FDSILP). Lemma 2.1 implies( R N + ) ′ ∼ = R ( N )+ and thus the feasible regions of (DSILP) and (FDSILP) areequivalent under the standard embedding of R ( N ) into ( R N ) ′ . The objectivesare also equivalent under that embedding and thus v (DSILP) = v (FDSILP).
3. Duality gaps in proper subspaces of R N By Theorem 2.4, the optimal value of the finite support dual is equalto the optimal value of the algebraic Lagrangian dual for semi-infinite lin-ear programs with countably many constraints, when we model them as aninstance of (ConLP) using Y = R N . However, this is not necessarily truefor problems with countably many constraints when they are modeled as(ConLP) with Y as a proper subspace of R N . In this section, we give anaffirmative answer to Question 2 via Example 3.5.Two examples of proper subspaces of R N are ℓ ∞ , the space of all boundedreal sequences, and c, the space of all convergent sequences. Clearly, c ⊂ ℓ ∞ ⊂ R N . We extend the notion of positive linear functionals to these sub-spaces. A linear functional ψ ∈ X ′ on any subspace X ⊆ R N , is called apositive linear functional on X if h v, ψ i ≥ v ∈ R N + ∩ X .Restricting to a subspace of R N allows for more positive linear functionals.Define the limit functional ψ on c by h v, ψ i = lim i →∞ v ( i ) . (3.1)Clearly, ψ is a positive linear functional over c . The next result shows that ψ cannot be extended to a positive linear functional over all of R N .8 emma 3.1. The limit functional ψ defined in (3.1) cannot be extended toa positive linear functional on R N , i.e., it cannot be extended to an elementof ( R N + ) ′ ⊆ ( R N ) ′ . Proof.
Assume ψ is an extension of ψ and that ψ is a positive linear functionalon R N . We shall derive a contradiction. Let ∈ R N + ∩ c be the all onessequence. Then h , ψ i = h , ψ i = 1. Let v = ( n ) n ∈ N be the sequence(1 , , , . . . ). Since v is in the nonnegative orthant h v, ψ i = α ≥
0. Let M = ⌈ α ⌉ + 1. Consider the sequence m = (0 , , , ..., M, M, M, . . . ) wherethe first M − v − m is in the nonnegative orthant,but h v − m, ψ i = h v, ψ i − h m, ψ i = α − h m, ψ i = α − M <
0, and thisis a contradiction to the assumption that ψ is a positive linear functional.Therefore ψ cannot be extended to a positive linear functional on R N .Although positive linear functionals on the space c cannot be extendedto ( R N + ) ′ , they can be extended to positive linear functionals on ℓ ∞ as shownin Lemma 3.3 below.First, recall the notion of a core point . Given a vector space X and asubset A ⊆ X , a point a ∈ A is called a core point of A if for every x ∈ X ,there exists ǫ > a + λx ∈ A for all 0 ≤ λ ≤ ǫ . Some authors callsuch a point an internal point (see for instance Definition 5.58 in [1]). Thefollowing is a useful result for extending positive linear functionals. Theorem 3.2 (Krein-Rutman theorem, see Holmes [9] p. 20) . Let X bea vector space ordered by (cid:23) P where P is a pointed, convex cone in X .Furthermore, let M be a linear subspace of X ordered by (cid:23) P ∩ M . If P ∩ M contains a core point (with respect to X ) of P , then any positive linearfunctional on M admits a positive linear extension to all of X . In otherwords, if ψ : M → R satisfies h m, ψ i ≥ m ∈ M ∩ P , then there existsa ψ : X → R with h x, ψ i ≥ x ∈ P and h x, ψ i = h x, ψ i for all x ∈ M . Lemma 3.3.
Every positive linear functional on c can be extended to apositive linear functional on ℓ ∞ . Proof.
Let P = R N + ∩ ℓ ∞ be the nonnegative cone in ℓ ∞ . Take the convergentsequence = (1 , , . . . ). This convergent sequence is an element of P ∩ c .Also, is a core point of P with respect to ℓ ∞ . To see that is a core pointof P , take any sequence ( a n ) n ∈ N ∈ ℓ ∞ . Since ( a n ) is in ℓ ∞ , sup n | a n | < ∞ and + λ ( a n ) ∈ P for all λ ∈ (0 , / sup n | a n | ) . Since P has a core point with9espect to ℓ ∞ , apply the Krein-Rutman theorem to extend positive linearfunctionals defined on c to positive linear functionals defined on ℓ ∞ . Corollary 3.4.
The limit functional defined in (3.1) can be extended to apositive linear functional over ℓ ∞ .Example 3.5 below provides an affirmative answer to Question 2. Thisexample gives an (SILP) with I = N , where v (SILP) = v (DSILP( V )) and v (SILP) > v (FDSILP) = v (DSILP), with the subspace V = c and V = ℓ ∞ .The equality v (FDSILP) = v (DSILP) follows from Theorem 2.4. Example 3.5.
The (SILP) isinf x x + i x ≥ i , i ∈ N . (3.2)We show that ( x , x ) = ( δ , δ ) is feasible to (3.2) for all δ >
0. For every i ∈ N , ( i − δ ) ≥ ⇒ i + δ − δi ≥ ⇒ i + δ ≥ δi ⇒ i + δ δi ≥ δiδi ⇒ δ + δi ≥ i . Letting δ → ∞ gives feasible solutions whose objective values converge to anobjective value of 0.Next observe that if (¯ x , ¯ x ) is any feasible solution, then x ≥ . Indeed,taking i → ∞ in (3.2) leaves x ≥
0. Therefore the optimal primal objectivevalue is equal to 0.The finite support dual (FDSILP) for this semi-infinite linear programis infeasible. The objective coefficient of x is 0 and the coefficient of x is strictly positive in the constraints, and so the only possible dual elementsatisfying the dual constraint corresponding to x is u = (0 , , . . . , , . . . );however, the objective coefficient of x is 1 and the dual constraint corre-sponding to x is not satisfied for u = 0. This shows that using the finitesupport dual leads to an infinite duality gap.In this example, a ( i ) = 1 for all i ∈ N so this sequence converges to 1.Also, a ( i ) = i for all i ∈ N and this sequence converges to 0. Likewise, b ( i ) = i for all i ∈ N so this sequence converges to 0. Therefore, this semi-infinite linear program is an instance of (ConLP) with Y = c (or ℓ ∞ ) ⊂ R N . V )) defined on the spaceof positive linear functionals on c (or ℓ ∞ ). The limit functional ψ definedin (3.1) is h v, ψ i = lim i →∞ v ( i ), for all v ∈ R N + ∩ c . This is a positive lin-ear functional on c (as is its extension to ℓ ∞ by Corollary 3.4) since anyconvergent sequence of nonnegative numbers has a nonnegative limit. Asobserved earlier, a ( i ) converges to 1 which is the coefficient of x in theobjective, and a ( i ) converges to 0 which is the coefficient of x in the objec-tive. Therefore ψ is a feasible dual solution. The dual objective value h b, ψ i is lim i ∈ N b ( i ) = lim i ∈ N i = 0. This is the optimal value of the primal andtherefore the duality gap is zero.This example illustrates Lemma 3.1. If ψ could be extended to a positivelinear functional ψ on R N , then ψ would be a feasible solution to (DSILP)with objective function value zero implying a zero duality gap between (3.2)and (DSILP). Since there is an infinite duality gap between (3.2) and itsfinite support dual (FDSILP), this would contradict Theorem 2.4. ⊳ Acknowledgements
The authors are very grateful to Greg Kuperberg, University of CaliforniaDavis, for discussions that led to the proof of Lemma 2.1. We are alsothankful for the useful suggestions of our anonymous reviewer.
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Appendix A. Details on Remark 2.2
This short appendix contains ancillary material pertaining to Remark 2.2and the unsubstantiated claim that R N has no interior points in any lineartopology. This claim follows from Corollary 9.41 in [1] and the fact R N has noorder unit. An order unit e in a Riesz space X is a positive element where forevery vector x ∈ X there exists a λ > | x | ≤ λe . See page 322 of [1]for more details. To see that R N has no order unit, consider candidate vector e = ( e , e , . . . ). There are two cases to consider. Case 1: the components e are bounded. Then there exists an M > e i ≤ M for all i . Take12 = ( M, M , . . . , M n , . . . ). Then e is not an order unit since, for sufficientlylarge n , there does not exist λ > M n ≤ λe i for all i . Case2: the components of e are unbounded. Then e contains a subsequence e i k where e i k → ∞ . Now consider x where x i = e i for i = 1 , , . . . . Setting λ ≥ e i k /e i k = e i k for all k is impossible since e i k → ∞ . Thus, e is not anorder unit since e i k → ∞→ ∞