Basis and Dimension of Exponential Vector Space
aa r X i v : . [ m a t h . AG ] J un Basis and Dimension of Exponential Vector Space
Jayeeta Saha a ∗ and Sandip Jana b a Department of Mathematics, Vivekananda College, Thakurpukur, Kolkata-700063, INDIA b Department of Pure Mathematics, University of Calcutta, Kolkata-700019, INDIA
Abstract
Exponential vector space [shortly evs ] is an algebraic order extension of vectorspace in the sense that every evs contains a vector space and conversely every vectorspace can be embedded into such a structure. This evs structure consists of a semi-group structure, a scalar multiplication and a partial order. In this paper we havedeveloped the concepts of basis and dimension of an evs by introducing the ideas oforderly independent set and generating set with the help of partial order and alge-braic operations. We have found that like vector space, an evs does not contain basisalways. We have established a necessary and sufficient condition for an evs to have abasis. It was shown that equality of dimension is an evs property but the converse isnot true. We have studied the dimension of subevs and found that every evs containsa subevs with all possible lower dimensions. Lastly we have computed basis and di-mension of some evs which help us to explore the theory of basis by creating counterexamples in different aspects.
AMS Classification: 08A99, 06F99, 15A99Key words : Vector space, exponential vector space, testing set, generator, orderlyindependent set, basis, dimension, feasible set.
Exponential vector space is an algebraic ordered extension of vector space. The word‘extension’ is used because of the fact that every exponential vector space contains a vectorspace and conversely every vector space can be embedded into such a structure. Thisstructure comprises a semigroup structure, a scalar multiplication and a compatible partialorder. We now start with the definition of evs. ∗ Corresponding Author, e-mail : [email protected] [email protected] efinition 1.1. [7] Let ( X, ≤ ) be a partially ordered set, ‘+’ be a binary operation on X [called addition ] and ‘ · ’: K × X −→ X be another composition [called scalar multiplication , K being a field]. If the operations and the partial order satisfy the following axioms then( X, + , · , ≤ ) is called an exponential vector space (in short evs ) over K [This structure wasinitiated with the name quasi-vector space or qvs by S. Ganguly et al. in [1]]. A : ( X, +) is a commutative semigroup with identity θA : x ≤ y ( x, y ∈ X ) ⇒ x + z ≤ y + z and α · x ≤ α · y, ∀ z ∈ X, ∀ α ∈ KA : (i) α · ( x + y ) = α · x + α · y (ii) α · ( β · x ) = ( αβ ) · x (iii) ( α + β ) · x ≤ α · x + β · x (iv) 1 · x = x, where ‘1’ is the multiplicative identity in K, ∀ x, y ∈ X, ∀ α, β ∈ KA : α · x = θ iff α = 0 or x = θA : x + ( − · x = θ iff x ∈ X := n z ∈ X : y z, ∀ y ∈ X r { z } o A : For each x ∈ X, ∃ p ∈ X such that p ≤ x. In the above definition, the axiom A (iii) indicates a rapid growth of the elements of X due to the fact x + x ≥ x and the axiom A gives some positive sense of each elements.These two facts express the exponential behaviour of the elements of an evs.In the axiom A , we can notice that X is precisely the set of all minimal elements of theevs X with respect to the partial order on X and forms the maximal vector space (within X ) over the same field as that of X ([1]). We call this vector space X as the ‘ primitivespace ’ or ‘ zero space ’ of X and the elements of X as ‘ primitive elements ’.Also given any vector space V over some field K , an evs X can be constructed (asshown below) such that V is isomorphic to X . In this sense, “exponential vector space”can be considered as an algebraic ordered extension of vector space. Example 1.2. [7] Let X := n ( r, a ) ∈ R × V : r ≥ , a ∈ V o , where V is a vector space oversome field K . Define operations and partial order on X as follows : for ( r, a ) , ( s, b ) ∈ X and α ∈ K ,(i) ( r, a ) + ( s, b ) := ( r + s, a + b );(ii) α ( r, a ) := ( r, αa ), if α = 0 and 0( r, a ) := (0 , θ ), θ being the identity in V ;(iii) ( r, a ) ≤ ( s, b ) iff r ≤ s and a = b .Then X becomes an exponential vector space over K with the primitive space { } × V which is evidently isomorphic to V . 2nitially the idea of this structure was given by S. Ganguly et al. with the name “quasi-vector space” in [1] and the following example of the hyperspace was the main motivationbehind this new stucture. Example 1.3. [1] Let C ( X ) be the topological hyperspace consisting of all non-emptycompact subsets of a Hausdörff topological vector space X over the field K of real orcomplex numbers. Then C ( X ) becomes an evs with respect to the operations and partialorder defined as follows. For A, B ∈ C ( X ) and α ∈ K ,(i) A + B := { a + b : a ∈ A, b ∈ B } (ii) αA := { αa : a ∈ A } (iii) The usual set-inclusion as the partial order.We now topologise an exponential vector space. For this we need the following concept. Definition 1.4. [2] Let ‘ ≤ ’ be a preorder in a topological space Z ; the preorder is said tobe closed if its graph G ≤ ( Z ) := n ( x, y ) ∈ Z × Z : x ≤ y o is closed in Z × Z (endowed withthe product topology). Theorem 1.5. [2]
A partial order ‘ ≤ ’ in a topological space Z will be a closed orderiff for any x, y ∈ Z with x y , ∃ open neighbourhoods U, V of x, y respectively in Z such that ( ↑ U ) ∩ ( ↓ V ) = ∅ , where ↑ U := { z ∈ Z : z ≥ u for some u ∈ U } and ↓ V := { z ∈ Z : z ≤ v for some v ∈ V } . Definition 1.6. [7] An exponential vector space X over the field K of real or complexnumbers is said to be a topological exponential vector space if there exists a topology on X with respect to which the addition and the scalar multiplication are continuous and thepartial order ‘ ≤ ’ is closed (Here K is equipped with the usual topology). Remark 1.7. If X is a topological exponential vector space then its primitive space X becomes a topological vector space, since restriction of a continuous function is continuous.Moreover, the closedness of the partial order ‘ ≤ ’ in a topological exponential vector space X readily implies (in view of Theorem 1.5) that X is Hausdörff and hence X becomes aHausdörff topological vector space. Example 1.8. [3] Let X := [0 , ∞ ) × V , where V is a vector space over the field K of real orcomplex numbers. Define operations and partial order on X as follows : for ( r, a ) , ( s, b ) ∈ X and α ∈ K ,(i) ( r, a ) + ( s, b ) := ( r + s, a + b ),(ii) α ( r, a ) := ( | α | r, αa ),(iii) ( r, a ) ≤ ( s, b ) iff r ≤ s and a = b .Then [0 , ∞ ) × V becomes an exponential vector space with the primitive space { } × V which is clearly isomorphic to V . 3n this example, if we consider V as a Hausdörff topological vector space then [0 , ∞ ) × V becomes a topological exponential vector space with respect to the product topology, where[0 , ∞ ) is equipped with the subspace topology inherited from the real line R .If instead of V we take the trivial vector space { θ } in this example then, the resultingtopological evs is [0 , ∞ ) × { θ } which can be clearly identified with the half ray [0 , ∞ ) ofthe real line.In this paper we have developed the concept of basis and dimension of evs. We knowthat basis of a vector space is a minimal part of it which generates the entire space. Butin evs it is impossible to express every element as a linear combination of some particularelements due to the exponential behaviour of its elements. In this paper, with the help ofpartial order we have developed the ideas of generating sets, orderly independent sets whichhelp us to define basis. It has been shown that basis of an evs is identified by a minimalgenerating set whereas maximal orderly independent set fails to form a basis [shown bya counter example], though every basis is a maximal orderly independent set. The maindifference between a vector space and an evs in this respect is that an evs may not havea basis always (like vector space). But for a topological evs, we have shown that, if it hasa basis then it contains uncountably many bases. We have found out a property of everyelement of a basis which helped us to give a necessary and sufficient condition for an evs tohave a basis. After that we have introduced the concept of dimension of an evs and shownthat equality of dimension is an evs property, though two non order-isomorphic evs mayhave same dimension.Lastly we have studied the dimension of subevs and shown that every evs containssubevs(s) with all possible lower dimensions. In the last section of this paper computationof basis and dimension of some evs are shown. In this section we have discussed some definitions, results and examples of exponentialvector space which are very much required to develop the main context. We now first startwith the definition of subevs.
Definition 2.1. [5] A subset Y of an exponential vector space X is said to be a subexponential vector space ( subevs in short) if Y itself is an exponential vector space withrespect to the compositions of X being restricted to Y . Note 2.2. [5] A subset Y of an exponential vector space X over a field K is a sub expo-nential vector space iff Y satisfies the following :(i) αx + y ∈ Y, ∀ α ∈ K , ∀ x, y ∈ Y .(ii) Y ⊆ X T Y , where Y := n z ∈ Y : y (cid:2) z, ∀ y ∈ Y r { z } o (iii) For any y ∈ Y , ∃ p ∈ Y such that p ≤ y .4f Y is a subevs of X then actually Y = X ∩ Y , since for any Y ⊆ X we have X ∩ Y ⊆ Y . [0 , ∞ ) × { θ } is clearly a subevs of the evs [0 , ∞ ) × V .We have used the following result to form a non-topological exponential vector space. Result 2.3. [4]
In a topological evs X if a = a + x for some a, x ∈ X then x = θ . To talk about an evs property of this space we have to know the idea of order-morphism.
Definition 2.4. [3] A mapping f : X −→ Y ( X, Y being two exponential vector spacesover the field K ) is called an order-morphism if(i) f ( x + y ) = f ( x ) + f ( y ), ∀ x, y ∈ X (ii) f ( αx ) = αf ( x ), ∀ α ∈ K , ∀ x ∈ X (iii) x ≤ y ( x, y ∈ X ) ⇒ f ( x ) ≤ f ( y )(iv) p ≤ q (cid:16) p, q ∈ f ( X ) (cid:17) ⇒ f − ( p ) ⊆↓ f − ( q ) and f − ( q ) ⊆↑ f − ( p ).A bijective (injective, surjective) order-morphism is called an order-isomorphism ( order-monomorphism , order-epimorphism respectively).If X, Y are two topological evs over K then an order-isomorphism f : X −→ Y is saidto be a topological order-isomorphism if f is a homeomorphism. Definition 2.5.
A property of an evs is called an evs property if it remains invariant underorder-isomorphism.The concept of order-isomorphism is competent enough to extract the structural beautyof an evs by judging the invariance of its various properties. Since the composition of twoorder-isomorphisms, the inverse of an order-isomorphism and the identity map are againorder-isomorphisms, the concept thereby produces a partition on the collection of all evsover some common field; this helps one to distinguish two evs belonging to two differentclasses under this partition.
Definition 2.6. [5] In an evs X the primitive of x ∈ X is defined as the set P x := { p ∈ X ◦ : p ≤ x } The axiom A in Definition 1.1 ensures that the primitive of each element of an evs isnonempty. Definition 2.7. [5] An evs X is said to be a single primitive evs if P x is a singleton set foreach x ∈ X. Also, in a single primitive evs X , P x + y = P x + P y and P αx = αP x , ∀ x, y ∈ X and for all scalar α .Single primitivity is an evs property [5]. Definition 2.8. [5] An evs X is said to be a comparable evs if ∀ x, y ∈ X , P x = P y ⇒ x and y are comparable with respect to the partial order of X .This is also an evs property [5]. 5e now give some examples of exponential vector space to build up some counterexamples of the main section. Example 2.9. [4] (Arbitrary product of exponential vector spaces)
Let { X i : i ∈ Λ } be an arbitrary family of exponential vector spaces over a common field K and X := Y i ∈ Λ X i be the Cartesian product. Then, X becomes an exponential vector space over K withrespect to the following operations and partial order :For x = ( x i ) i , y = ( y i ) i ∈ X and α ∈ K we define (i) x + y := ( x i + y i ) i , (ii) αx := ( αx i ) i ,(iii) x ≪ y if x i ≤ y i , ∀ i ∈ Λ.Here the notation x = ( x i ) i ∈ X means that the point x ∈ X is the map x : i x i ( i ∈ Λ),where x i ∈ X i , ∀ i ∈ Λ. The additive identity of X is given by θ = ( θ i ) i , θ i being theadditive identity in X i . Also the primitive space of X is given by X = Y i ∈ Λ [ X i ] .This product space X becomes a topological exponential vector space over the field K whenever each factor space X i is a topological evs over K and X is endowed with the producttopology, which is the weakest topology on X so that each projection map p i : X −→ X i given by p i : x x i is continuous.Thus for any cardinal number β , [0 , ∞ ) β becomes a topological evs. Example 2.10.
Let X be an evs over the field K (either R or C ) and V be a vector spaceover the same field K . We now give operations on X × V like [0 , ∞ ) × V , i.e.for ( x , e ) , ( x , e ) , ( x, e ) ∈ X × V and α ∈ K (i) ( x , e ) + ( x , e ) := ( x + x , e + e ).(ii) α ( x, e ) := ( αx, αe ).The partial order ‘ ≤ ’ is defined as : ( x , e ) ≤ ( x , e ) iff x ≤ x and e = e . Then X × V becomes an evs over the field K . Justification of this is straight forward. Example 2.11.
Let C θ ( X ) be the collection of all compact subsets of a Hausdörff topolog-ical vector space X containing θ (the identity in X ). So C θ ( X ) ⊆ C ( X ). If we take any twomembers A, B ∈ C θ ( X ) and any α ∈ K then αA + B ∈ C θ ( X ). Again [ C θ ( X )] = n { θ } o =[ C ( X )] ∩ C θ ( X ). For any A ∈ C θ ( X ), { θ } ⊆ A . This shows that C θ ( X ) is a subevs of C ( X ) [by note 2.2]. Example 2.12. [5] Let X be a vector space over the field K of real or complex numbers.Let L ( X ) be the set of all linear subspaces of X . We now define + , · , ≤ on L ( X ) as follows: For X , X ∈ L ( X ) and α ∈ K define(i) X + X := span( X ∪ X ), (ii) α · X := X , if α = 0 and α · X := { θ } , if α = 0 ( θ beingthe additive identity of X ), (iii) X ≤ X iff X ⊆ X .Then (cid:16) L ( X ) , + , · , ≤ (cid:17) is an exponential vector space over K .Since every element of L ( X ) is an idempotent h ∵ X + X = X , for all X ∈ L ( X ) i we can say that there is no topology with respect to which L ( X ) can be a topological evs6 Since a topological evs cannot contain any idempotent element, as follows from the Result2.3 i . Example 2.13. [6] Let us consider D ([0 , ∞ )) := [0 , ∞ ) × [0 , ∞ ). We define + , · , ≤ on D ([0 , ∞ )) as follows :For ( x , y ) , ( x , y ) ∈ D ([0 , ∞ )) and α ∈ C we define(i) ( x , y ) + ( x , y ) = ( x + x , y + y )(ii) α · ( x , y ) = ( | α | x , | α | y )(iii) ( x , y ) ≤ ( x , y ) ⇐⇒ either x < x or if x = x then y ≤ y [ dictionary order ]Then (cid:16) D ([0 , ∞ )) , + , · , ≤ (cid:17) becomes a non-topological exponential vector space over thecomplex field C . Note 2.14. [6] For a wel l-ordered set I and an evs X , if we consider D ( X : I ) := X I then also, like above example, it forms a non-topological evs with dictionary order. If I = { , , . . . , n } we shall usually denote the evs D ( X : I ) as D n ( X ). We can alsogeneralise this by taking different evs i.e. D ( X α : α ∈ I ) := Y α ∈ I X α , which also becomes anon-topological evs with dictionary order. In this section we have introduced the concepts of basis and dimension of an exponentialvector space. These concepts are different from those already in a vector space. Like vectorspace it is not true that every evs contains a basis, rather it behaves like a module in thisrespect. We have found a necessary as well as sufficient condition for an evs to have a basis.Finally, we have computed basis and dimension of some particular evs.
Definition 3.1.
Let X be an evs over the field K and x ∈ X r X . Define L ( x ) := { z ∈ X : z ≥ αx + p, α ∈ K ∗ , p ∈ X } , where K ∗ ≡ K r { } We name these sets L ( x ) for different x ’s in X r X as testing sets of X .We discuss below some properties of L ( x ). First of all note that L ( x ) = ↑ ( K ∗ x + X ). Proposition 3.2. (i) ∀ x ∈ X r X , x ∈ L ( x ) and ↑ L ( x ) = L ( x ) . (ii) x ≤ y ( x, y ∈ X r X ) ⇒ L ( x ) ⊇ L ( y ) . (iii) If x = αy + p for some α ∈ K ∗ , p ∈ X and y ∈ X r X , then L ( x ) = L ( y ) . (iv) L ( x ) ∩ X = ∅ . (v) If a ∈ L ( b ) then L ( a ) ⊆ L ( b ) . (vi) For any x, y ∈ X r X , L ( x ) ∩ L ( y ) = ∅ .Proof. (i) Immediate from definition. (ii)
Let z ∈ L ( y ) ⇒ ∃ α ∈ K ∗ and p ∈ X such that αy + p ≤ z . Now x ≤ y ⇒ αx + p ≤ αy + p ≤ z ⇒ z ∈ L ( x ). 7 iii) As y ∈ X r X , so x ∈ X r X . Therefore we can talk about L ( x ). Let z ∈ L ( y ) ⇒ ∃ α z ∈ K ∗ and p z ∈ X such that α z y + p z ≤ z ⇒ α z α − ( x − p ) + p z ≤ z ⇒ α z α − x + ( p z − α z α − p ) ≤ z ⇒ z ∈ L ( x ). Therefore L ( y ) ⊆ L ( x ). Again x = αy + p ⇒ y = α − ( x − p ). So by above argument we also have L ( x ) ⊆ L ( y ). Thus L ( x ) = L ( y ). (iv) Let y ∈ L ( x ) ⇒ ∃ α ∈ K ∗ and p ∈ X such that αx + p ≤ y . If y ∈ X then αx + p ∈ X ⇒ x ∈ X . This contradiction proves that L ( x ) ∩ X = ∅ . (v) a ∈ L ( b ) ⇒ a ∈ X r X and ∃ α ∈ K ∗ , p ∈ X such that αb + p ≤ a ⇒ L ( a ) ⊆ L ( αb + p ) = L ( b ) [by (ii) and (iii) above]. (vi) For any p ∈ X with p ≤ y , x + p ≤ x + y ⇒ x + y ∈ L ( x ). Similarly we can saythat x + y ∈ L ( y ). So, x + y ∈ L ( x ) ∩ L ( y ) ⇒ L ( x ) ∩ L ( y ) = ∅ . Definition 3.3.
A subset B of X r X is said to be a generator of X r X if X r X = [ x ∈ B L ( x ) Note 3.4.
The set X r X always generates X r X . So generator always exists for X r X .It is clear that any superset of a generator of X r X is also a generator of X r X . Definition 3.5.
Two elements x, y ∈ X r X are said to be orderly dependent if either x ∈ L ( y ) or y ∈ L ( x ). Definition 3.6.
Two elements x, y ∈ X r X are said to be orderly independent if theyare not orderly dependent i.e neither x ∈ L ( y ) nor y ∈ L ( x ).A subset B of X r X is said to be orderly independent if any two members x, y ∈ B are orderly independent. Remark 3.7.
Let Y be a subevs of an evs X . Then any two orderly dependent elementsof Y r Y are also orderly dependent in X r X because of the fact Y ⊆ X . In otherwords, any two elements of Y r Y which are orderly independent in X r X are also orderlyindependent in Y r Y . But converse is not true in general i.e two orderly independentelements in Y r Y may not be orderly independent in X r X [in contrast to the case of linearindependence in vector space]. For example, { , , } and { , − , } are orderly independentin C θ ( R ) [see Example 2.11], since ∄ any α ∈ R ∗ such that α { , , } ⊆ { , − , } or α { , − , } ⊆ { , , } h Here [ C θ ( R )] = n { } o i . But these two elements are not orderlyindependent in C ( R ), as we can write { , − , } = { , , } + {− } , where {− } ∈ [ C ( R )] .In the above context it should thus be noted that while discussing the orderly indepen-dence of two elements of a subevs Y of an evs X , there are two types of orderly independence—— one with respect to Y and the other with respect to X ; while considering orderly in-dependence with respect to Y the testing sets should be of the form L Y ( y ) := { z ∈ Y : z ≥ αy + p, α ∈ K ∗ , p ∈ Y } for any y ∈ Y r Y ,and when considering orderly independence with respect to X the testing sets must be ofthe form 8 X ( y ) := { z ∈ X : z ≥ αy + p, α ∈ K ∗ , p ∈ X } for any y ∈ Y r Y .Since Y ⊆ X we have L Y ( y ) ⊆ L X ( y ), for any y ∈ Y r Y . Thus it follows that an orderlyindependent set in Y r Y need not be orderly independent in X r X . However a set B ( ⊆ Y r Y ) which is orderly independent in X r X must be so in Y r Y . Definition 3.8.
A subset B of X r X is said to be a basis of X r X if B is orderlyindependent and generates X r X . Note 3.9.
For each x ∈ X either x ∈ X or x ∈ X r X . If x ∈ X then it can be expressedas a finite linear combination of some basic vectors of some basis of X [as a vector space].If x ∈ X r X then there exists a member of some basis [if exists] of X r X which generates x . So we can say that to represent an evs X it is necessary to consider a basis of X r X together with a basis of X [in the sense of vector space]. Thus a basis of an evs X shouldbe composed of two components, one for X and the other for X r X . To express thisfact in an easiest way we shall represent a basis of an evs X as [ B : B ], where B is a basisof X r X and B is a basis of X [as a vector space]. If for an evs X , X = { θ } then weshall consider B = { θ } , since in that case X has no basis. Theorem 3.10.
For a topological evs X , X r X either has no basis or has uncountablymany bases.Proof. Let B be a basis of X r X . Then G α := { αx : x ∈ B } and H p := { x + p : x ∈ B } are also bases of X r X , for any α ∈ K ∗ and any p ∈ X . This holds because of theresult L ( x ) = L ( αx + p ) [proposition 3.2]. If G α = G β for any α, β ∈ K ∗ then αx = βx , ∀ x ∈ B [ ∵ αx = βy for any x, y ∈ B as B is orderly independent]. If we choose α, β ∈ K ∗ such that | α | < | β | then using continuity of the scalar multiplication of the topological evs X we must have x = θ [ ∵ αx = βx ⇒ ( αβ − ) n x = x , ∀ n ∈ N which implies by takinglimit n → ∞ that x = θ , as | αβ − | <
1] — a contradiction. Thus it follows that for any α, β ∈ K ∗ with | α | < | β | we must have G α = G β . This immediately justifies that there areuncountably many bases of X r X .If X contains more than one element then for p, q ∈ X we may consider H p , H q . If H p = H q then B being orderly independent we must have x + p = x + q , ∀ x ∈ B . Then byresult 2.3 it follows that p = q . Since X is a topological evs, X is a Hausdörff topologicalvector space. So if X = { θ } then it must be uncountable and hence ensures the existenceof uncountably many bases of X r X .For a non-topological evs it may so happen that G α = B for every α ∈ K ∗ [this will bediscussed in the next section]. However, an evs (topological or not) need not have a basis.We show in the next section that the evs D (cid:16) [0 , ∞ ) : N (cid:17) discussed in Note 2.14 cannot havea basis. The following result shows that having basis is an evs property.9 esult 3.11. Let φ : X −→ Y be an order-isomorphism. Then (1) for any generator B of X r X , φ ( B ) is a generator of Y r Y . (2) for any orderly independent subset B of X r X , φ ( B ) is also an orderly independentsubset of Y r Y .Thus, for a basis B of X r X , φ ( B ) becomes a basis of Y r Y .Proof. (1) B ⊆ X r X ⇒ φ ( B ) ⊆ Y r Y [As φ ( X ) = Y ]. Let y ∈ Y r Y ⇒ φ − ( y ) ∈ X r X ⇒ ∃ b ∈ B and α ∈ K ∗ , p ∈ X such that φ − ( y ) ≥ αb + p ⇒ y ≥ αφ ( b ) + φ ( p ) ⇒ y ∈ L ( φ ( b )) ⊆ [ b ∈ B L ( φ ( b )). Therefore Y r Y ⊆ [ b ∈ B L ( φ ( b )). Again by the proposition 3.2, L ( φ ( b )) ∩ Y = ∅ , ∀ b ∈ B . So Y r Y = [ b ∈ B L ( φ ( b )). Thus φ ( B ) is a generator of Y r Y .(2) We first show that for any two orderly dependent members y , y of Y r Y , φ − ( y ) , φ − ( y ) are orderly dependent in X r X . As y , y are orderly dependent sowithout loss of generality we can take y ∈ L ( y ) ⇒ ∃ α ∈ K ∗ and p ∈ Y such that αy + p ≤ y . Then φ − also being an order-isomorphism we have φ − ( αy + p ) ≤ φ − ( y ) ⇒ αφ − ( y ) + φ − ( p ) ≤ φ − ( y ) ⇒ φ − ( y ) ∈ L ( φ − ( y )) [as φ − ( p ) ∈ X ]. This justifiesour assertion. Then contra-positively, the result follows.The next theorem characterises a basis (if exists) of X r X , for any evs X . Theorem 3.12.
A subset of X r X is a basis of X r X iff it is a minimal generatingsubset of X r X . h Here minimal generating subset B of X r X means there does notexist any proper subset of B which can generate X r X . i Proof.
Let us suppose B be a basis of X r X . Then B generates X r X . Now B beingan orderly independent subset of X r X , if we take an element x ∈ B then ∀ y ∈ B r { x } , x and y are orderly independent. Therefore x / ∈ L ( y ), ∀ y ∈ B r { x } . This shows that B r { x } cannot generate X r X and this holds for any x ∈ B . Therefore B is a minimalgenerator of X r X .Conversely, suppose B be a minimal generator of X r X . For any two members x, y ∈ B if x ∈ L ( y ) then by proposition 3.2, L ( x ) ⊆ L ( y ) = ⇒ B r { x } also generates X r X ,which contradicts that B is a minimal generator of X r X . Again, if y ∈ L ( x ) we getsimilar contradiction. So neither x ∈ L ( y ) nor y ∈ L ( x ) = ⇒ x, y are orderly independent.Arbitrariness of x, y shows that B is an orderly independent subset of X r X . Consequently, B is a basis of X r X . Result 3.13.
Every basis of X r X is a maximal orderly independent subset of X r X . h Here maximal orderly independent subset B of X r X means there does not exist anyorderly independent subset of X r X containing B . i Proof.
Let B be a basis of X r X . Then for any x ∈ X r ( B ∪ X ), ∃ b ∈ B such that x ∈ L ( b ). This shows that B ∪ { x } is not orderly independent. Thus B is maximal orderlyindependent in X r X . 10onverse of above result is not true in general i.e maximal orderly independent subsetof X r X may not be a basis of X r X . For example, in the evs C θ ( X ) [discussed in2.11]let us consider the collection G := n A ∈ C θ ( X ) : A consists of three distinct elements of X o Then G ⊂ C θ ( X ) r n { θ } o . Now we define a relation ‘ ∼ ’ on G by “ A ∼ B iff A = αB for some α ∈ K ∗ ”. Then this relation becomes an equivalence relation on G . Let usconsider the subcollection H of G taking exactly one member from each equivalence classproduced by the equivalence relation ‘ ∼ ’. Then H becomes an orderly independent subsetof C θ ( X ) r n { θ } o , because any two elements A, B ∈ G are orderly dependent iff A = αB for some α ∈ K ∗ and hence belong to the same equivalence class. For any member C ∈ C θ ( X ) \ ( H ∪ [ C θ ( X )] ) if card( C ) ≥ A C ∈ H and α ∈ K ∗ such that αA C ⊆ C . If card( C ) = 2 then also there exists β ∈ K ∗ and A C ∈ H suchthat C ⊆ βA C . This shows that H ∪ { C } is orderly dependent. So we can say that H forms a maximal orderly independent set in C θ ( X ) r n { θ } o . But it does not generate C θ ( X ) r n { θ } o . In fact, for any D ∈ C θ ( X ) with card( D ) = 2 there does not exist any A ∈ H such that D ∈ L ( A ), since each member of L ( A ) contains three or more elementsof X . Hence H cannot be a basis of C θ ( X ) r [ C θ ( X )] although it is maximal orderlyindependent h here note that [ C θ ( X )] = n { θ } oi . Remark 3.14. If A is an orderly independent set in X r X then for any a , a ∈ A with a = a neither a ∈ L ( a ) nor a ∈ L ( a ). In other words, if a ∈ L ( a ) for some a , a ∈ A then a = a . Moreover any two elements of an orderly independent set A must be incomparable with respect to the partial order ‘ ≤ ’ of the evs X ; in fact, x ≤ y ⇒ y ∈ L ( x ). Lemma 3.15.
Let A and B be two bases of X r X . Then for any a ∈ A , there exists oneand only one b a ∈ B such that L ( a ) = L ( b a ) .Proof. As B is a basis of X r X , so for the member a ∈ A , there must exist some b ∈ B such that a ∈ L ( b ). Let us suppose, ∃ b , b ∈ B such that a ∈ L ( b ) ∩ L ( b ) ⇒ L ( a ) ⊆ L ( b ) ∩ L ( b ) [by proposition 3.2] —— ( ∗ ). Again a ∈ L ( b ) ⇒ ∃ α ∈ K ∗ and p ∈ X such that αb + p ≤ a —— ( ∗∗ ). Now since A is a basis, so for b , b ∈ B , ∃ a , a ∈ A such that b ∈ L ( a ) and b ∈ L ( a ) ⇒ L ( b ) ⊆ L ( a ) and L ( b ) ⊆ L ( a ) [by proposition3.2]. By ( ∗ ), L ( a ) ⊆ L ( a ) and L ( a ) ⊆ L ( a ) ⇒ a ∈ L ( a ) and a ∈ L ( a ). As a, a , a aremembers of the basis A , so we can say that a = a = a [by above remark 3.14]. Therefore b , b ∈ L ( a ) ⇒ ∃ α , α ∈ K ∗ and p , p ∈ X such that α a + p ≤ b and α a + p ≤ b ——( ∗ ∗ ∗ ). From ( ∗∗ ) and ( ∗ ∗ ∗ ), we get b ≥ α a + p ≥ α ( αb + p ) + p = α αb + ( α p + p ). ∴ b ∈ L ( b ) [As α α ∈ K ∗ and α p + p ∈ X ]. Since b , b are members of the basis B so b = b [by above remark 3.14]. Thus there exists one and only one member (say) b a ∈ B such that a ∈ L ( b a ) and also b a ∈ L ( a ) ⇒ L ( a ) ⊆ L ( b a ) and L ( b a ) ⊆ L ( a ) ⇒ L ( a ) = L ( b a ). 11 heorem 3.16. If A and B are two bases of X r X then card ( A ) = card ( B ) .Proof. From the proof of the above lemma 3.15 we can say that for each a ∈ A , ∃ unique b ∈ B such that L ( b ) = L ( a ). This property creates a one to one correspondence between A and B . Hence the theorem.This theorem motivates us to introduce the concept of dimension of an evs. Definition 3.17.
For an evs X we define dimension of X r X asdim( X r X ) := card( B ), where B is a basis of X r X .Then we shall represent dimension of the evs X as dim X := [dim( X r X ) : dim X ]. If X = { θ } , dimension of X will be taken as 0, since then X has no basis [as vector space]. Note 3.18.
Theorem 3.16 makes the above definition well-defined. From result 3.11 wecan say that if X and Y are order-isomorphic evs then dim X = dim Y . Here by “dim X =dim Y ” we mean dim( X r X ) = dim( Y r Y ) as well as dim X = dim Y . However,converse of this is not true in general i.e there are evs X, Y such that dim X = dim Y but X, Y are not order-isomorphic. This will be clear in the next section, when we shallcompute the dimension of some particular evs.
Result 3.19.
Let X be an evs and B be a basis of X r X . Then ↓ x r X ⊆ L ( x ) , foreach x ∈ B .Proof. Let x ∈ B and y ∈↓ x r X . Since B is a basis of X r X , ∃ x ∈ B such that y ∈ L ( x ) ⇒ ∃ α ∈ K ∗ and p ∈ X such that α x + p ≤ y ⇒ α x + p ≤ x [ ∵ y ≤ x ] ⇒ x ∈ L ( x ). Since B is orderly independent and both x, x ∈ B , we can say by remark3.14 that x = x . Therefore y ∈ L ( x ). Thus ↓ x r X ⊆ L ( x ), for each x ∈ B .This result reveals an important property of each member of a basis of X r X whichhelps us to set up a precise domain of basic elements of X r X . The collection of all x in X r X satisfying the property stated in the above result 3.19 makes our task of finding abasis of X r X easier. To make this assertion precise let us consider the following :For an evs X , let Q ( X ) := n x ∈ X r X : ( ↓ x r X ) ⊆ L ( x ) o From result 3.19 we can say that B ⊆ Q ( X ), for any basis B of X r X . It is thus enoughto find any basis of X r X within Q ( X ). We call this set Q ( X ) as feasible set of X . At thispoint it is important to note that Q ( X ) may be empty; in fact, if for an evs X , Q ( X ) = ∅ then such evs X cannot have any basis (as we have claimed earlier). We shall encountersuch evs later. If for an evs X , Q ( X ) = ∅ then also X may not have a basis. In fact, weshall prove shortly a theorem which will characterise, in terms of Q ( X ), when X will havea basis. We now prove a lemma which will be useful in the sequel. Lemma 3.20.
For an evs X , if x ∈ Q ( X ) then for each y ∈↓ x r X , L ( x ) = L ( y ) . roof. y ∈↓ x r X ⇒ y ≤ x . So by proposition 3.2 we have L ( x ) ⊆ L ( y ). Again byconstruction of Q ( X ), x ∈ Q ( X ) ⇒ ↓ x r X ⊆ L ( x ) ⇒ y ∈ L ( x ) ⇒ L ( y ) ⊆ L ( x ) [byproposition 3.2]. Thus L ( x ) = L ( y ).The following theorem may be compared with the so-called ‘ Replacement theorem ’ inthe context of basis of a vector space.
Theorem 3.21.
For an evs X , let B be a basis of X r X and x ∈ B . Then for any y ∈↓ x r X , (cid:16) B r { x } (cid:17) ∪ { y } is also a basis of X r X .Proof. Let A = (cid:16) B r { x } (cid:17) ∪ { y } . As y ∈↓ x r X and x ∈ B ⊆ Q ( X ) so by lemma 3.20, L ( x ) = L ( y ). Therefore X r X = [ z ∈ B L ( z ) = [ z ∈ A L ( z ) ⇒ A generates X r X . To showthat A is orderly independent it is sufficient to show that for any z ∈ B r { x } , z, y areorderly independent. If not, then for some z ∈ B r { x } either y ∈ L ( z ) or z ∈ L ( y ).Now if y ∈ L ( z ) then by proposition 3.2, L ( y ) ⊆ L ( z ) ⇒ x ∈ L ( x ) = L ( y ) ⊆ L ( z )which contradicts that x, z are two members of the basis B . Again if z ∈ L ( y ) then z ∈ L ( y ) = L ( x ) which again contradicts that x, z are orderly independent. Thus itfollows that A is orderly independent.The above theorem makes it convenient to construct new basis from old one. Thefollowing theorem is the key to ensure the existence of a basis of an evs. Theorem 3.22.
An evs X has a basis iff Q ( X ) is a generator of X r X .Proof. Let us suppose X has a basis [ B : B ], where B and B are bases of X r X and X respectively. Then by the result 3.19, B ⊆ Q ( X ). As B is a generator of X r X so Q ( X ) is also a generator of X r X .Conversely, suppose Q ( X ) is a generator of X r X ⇒ Q ( X ) = ∅ . We now give arelation ∼ in Q ( X ) as follows : For x, y ∈ Q ( X ), we say x ∼ y ⇔ L ( x ) = L ( y ). Thenobviously this becomes an equivalence relation on Q ( X ). Let us consider a collection takingexactly one representative from each equivalence class and denote this collection as B . Then B ⊆ Q ( X ) ⊆ X r X . Also x, y ∈ B with x = y ⇔ x, y ∈ Q ( X ) and L ( x ) = L ( y ). Weclaim that B is a basis of X r X . Let z ∈ X r X ⇒ ∃ x z ∈ Q ( X ) [ ∵ Q ( X ) is a generator]such that z ∈ L ( x z ) ⇒ ∃ an element x ′ z ∈ B such that L ( x z ) = L ( x ′ z ) and hence z ∈ L ( x ′ z ) ⇒ B generates X r X . Now we have to show that B is orderly independent. Supposenot, then ∃ two distinct elements x , x ∈ B such that they are orderly dependent. Sowithout loss of generality we can think that x ∈ L ( x ). Now x ∈ L ( x ) ⇒ ∃ α ∈ K ∗ and p ∈ X such that αx + p ≤ x . Since x ∈ Q ( X ) [as B ⊆ Q ( X )] and αx + p ∈↓ x r X ,using lemma 3.20 we can say that L ( αx + p ) = L ( x ) and then by proposition 3.2 we have L ( x ) = L ( αx + p ) = L ( x ) —— which contradicts that x , x are two distinct elementsof B . Thus B becomes a basis of X r X . Let us take any basis B of X . Then [ B : B ]becomes a basis of X . 13rom the proof of the above theorem it is clear that the hypothesis of Q ( X ) beinga generator of X r X is only used to justify that B , constructed using the equivalencerelation ∼ within Q ( X ), is a generator of X r X ; to ensure the orderly independence of B , the structure of Q ( X ) is enough. Thus we can conclude that for any evs X , Q ( X )(if nonempty) always contains an orderly independent set like B (as constructed in theproof of the above theorem 3.22). This orderly independent set is also a maximal orderlyindependent set in Q ( X ). In fact, if D be another orderly independent set in Q ( X ) suchthat B ⊂ D then for any x ∈ D , ∃ z ∈ B such that L ( x ) = L ( z ) ⇒ x = z [ ∵ x, z ∈ D and D is orderly independent] and hence x ∈ B . Thus B = D . Summarising all these facts weget the following theorem. Theorem 3.23.
For an evs X , if Q ( X ) = ∅ then it contains a maximal orderly independentset. The next theorem is useful in finding a basis of X r X , for any evs X . Theorem 3.24.
For an evs X , every maximal orderly independent set of Q ( X ) is a basisof X r X , provided Q ( X ) generates X r X .Proof. Let B be a maximal orderly independent set in Q ( X ). Since Q ( X ) generates X r X ,for any x ∈ X r X , ∃ z ∈ Q ( X ) such that x ∈ L ( z ). If z ∈ B we are done. If z / ∈ B then B being a maximal orderly independent set in Q ( X ), B ∪{ z } is orderly dependent. So ∃ b ∈ B such that either z ∈ L ( b ) or b ∈ L ( z ). If z ∈ L ( b ) then by proposition 3.2, x ∈ L ( z ) ⊆ L ( b ).If b ∈ L ( z ), ∃ α ∈ K ∗ and p ∈ X such that b ≥ αz + p . Then b ∈ B ⊆ Q ( X ) ⇒ L ( z ) = L ( b )[by lemma 3.20] ⇒ x ∈ L ( b ). Thus B generates X r X . Consequently B is a basis of X r X .The above theorem 3.24 shows the converse of the result 3.13 to some extent; as wehave explained, just after the result 3.13, through the example of C θ ( X ) that every maximalorderly independent subset of X r X need not be a basis of X r X , the above theorem 3.24shows that every maximal orderly independent subset of Q ( X ) [but not only of X r X ]becomes a basis of X r X , provided of course Q ( X ) generates X r X [note that thenecessity of Q ( X ) being a generator of X r X is the principal key for an evs X to have abasis]. From remark 3.14 we may recall one more point that while finding a basis of X r X ,we have to gather only suitable incomparable elements from Q ( X ). In this context it shouldalso be noted that any two elements of Q ( X ) need not be orderly independent. In fact, forany x ∈ Q ( X ) if y ∈↓ x r X then also y ∈ Q ( X ) [see result 3.26(ii)]. Clearly this x, y areorderly dependent, since L ( x ) = L ( y ). Result 3.25. If X and Y are order-isomorphic then Q ( X ) and Q ( Y ) are in a one-to-onecorrespondence. roof. Let φ : X −→ Y be an order-isomorphism. We now show that φ ( Q ( X )) = Q ( Y ).Let x ∈ Q ( X ) ⇒ ↓ x r X ⊆ L ( x ). Also let y ∈↓ φ ( x ) r Y ⇒ y ≤ φ ( x ) and y / ∈ Y ⇒ φ − ( y ) ≤ x and φ − ( y ) / ∈ X ⇒ φ − ( y ) ∈↓ x r X ⊆ L ( x ) ⇒ ∃ α ∈ K ∗ and p ∈ X such that αx + p ≤ φ − ( y ) ⇒ αφ ( x ) + φ ( p ) ≤ y ⇒ y ∈ L ( φ ( x )) [ ∵ φ ( p ) ∈ Y ]. Therefore ↓ φ ( x ) r Y ⊆ L ( φ ( x )) ⇒ φ ( Q ( X )) ⊆ Q ( Y ). Similarly we can say that φ − ( Q ( Y )) ⊆ Q ( X )[ ∵ φ − is an order-isomorphism from Y onto X ] ⇒ Q ( Y ) ⊆ φ ( Q ( X )). ∴ φ ( Q ( X )) = Q ( Y ). Thus Q ( X ) and Q ( Y ) are in a one-to-one correspondence. Result 3.26. (i) If x ∈ Q ( X ) then for any α ∈ K ∗ and p ∈ X , αx + p ∈ Q ( X ) i.e Q ( X ) is closed under dilation and translation by primitive elements. (ii) If x ∈ Q ( X ) then ↓ x r X ⊆ Q ( X ) i.e ↓ Q ( X ) r X ⊆ Q ( X ) .Proof. (i) x ∈ Q ( X ) ⇒ ↓ x r X ⊆ L ( x ). We now show that ↓ ( αx + p ) r X ⊆ L ( αx + p ).Let y ∈↓ ( αx + p ) r X ⇒ y ≤ αx + p and y / ∈ X ⇒ α − ( y − p ) ≤ x and y / ∈ X ⇒ α − ( y − p ) ∈↓ x r X ⇒ α − ( y − p ) ∈ L ( x ) ⇒ ∃ β ∈ K ∗ and q ∈ X such that βx + q ≤ α − ( y − p ) ⇒ α ( βx + q ) + p ≤ y ⇒ αβx + αq + p ≤ y ⇒ y ∈ L ( x ) [ ∵ αq + p ∈ X ] ⇒ y ∈ L ( αx + p ) [ ∵ L ( αx + p ) = L ( x ), by proposition 3.2]. Therefore αx + p ∈ Q ( X ).(ii) Let y ∈↓ x r X . Then by lemma 3.20, L ( x ) = L ( y ). Now for each z ∈↓ y r X wehave z ≤ y ≤ x with z / ∈ X ⇒ L ( z ) = L ( x ) [by lemma 3.20] ⇒ z ∈ L ( x ) = L ( y ). Thus ↓ y r X ⊆ L ( y ). Consequently, y ∈ Q ( X ) and hence ↓ x r X ⊆ Q ( X ).As we have explained in remark 3.7 regarding orderly independence in a subevs of anevs, the theory of basis of a subevs does not behave nicely like the theory of basis of asubspace of a vector space. However we have the following theorems and examples whichreveal some technical aspects of dimension theory of evs. Theorem 3.27.
Every evs contains a subevs of dimension [1 : 0] .Proof.
Let X be an evs over K and B ( x ) := ( n X i =1 α i x : α i ∈ K , n ∈ N ) , where x ∈↑ θ r { θ } .Then for any α, β ∈ K and any n X i =1 α i x , m X j =1 β j x ∈ B ( x ) we have α n X i =1 α i x + β m X j =1 β j x = n X i =1 αα i x + m X j =1 ββ j x ∈ B ( x ). Also [ B ( x )] = { θ } = B ( x ) ∩ X and for any y ∈ B ( x ), θ ≤ y [ ∵ θ ≤ x ]. So B ( x ) forms a subevs of X for any x ∈↑ θ r { θ } . In this case { x } forms a basis of B ( x ) r [ B ( x )] . In fact, for any n X i =1 α i x ∈ B ( x ), α j x + θ ≤ n X i =1 α i x for some j ∈ { , , . . . , n } for which α j = 0. Again any singleton set consisting of a non-zero elementis always orderly independent. So we can say that dim B ( x ) = [1 : 0].The following example shows that corresponding to any cardinal α , there exists an evsof dimension [ α : 0]. 15 xample 3.28. For any cardinal number α , let us consider the evs [0 , ∞ ) α , discussed inExample 2.9. Let us take a set I such that card( I ) = α . We now show that B := { e i : i ∈ I } is a basis of [0 , ∞ ) α , where e i = ( δ ij ) j ∈ I and δ ij = , when i = j , when i = j For any x ∈ [0 , ∞ ) α r h [0 , ∞ ) α i (cid:20) here h [0 , ∞ ) α i = { θ } and θ = ( z j ) j ∈ I , where z j = 0, ∀ j ∈ I (cid:21) with representation x = ( x j ) j ∈ I , ∃ p ∈ I such that x p = 0 ⇒ x p e p ≤ x [ ∵ x j ≥ ∀ j ∈ I ] ⇒ x p e p + θ ≤ x ⇒ x ∈ L ( e p ) ⇒ B generates [0 , ∞ ) α r h [0 , ∞ ) α i . Now clearly anytwo members of B are orderly independent in [0 , ∞ ) α r h [0 , ∞ ) α i . This shows that B is abasis of [0 , ∞ ) α r h [0 , ∞ ) α i . Therefore dim[0 , ∞ ) α = [ α : 0], since card( B ) = card( I ) = α and dim h [0 , ∞ ) α i = dim { θ } = 0.Thus for any two cardinal numbers α, β with α = β , [0 , ∞ ) α and [0 , ∞ ) β cannot beorder-isomorphic, since they are of different dimension. We now show that for any twocardinal numbers α, β , there exists an evs of dimension [ α : β ]. For this we need thefollowing theorem first. Theorem 3.29.
For an evs X and a vector space V , both being over the common field K ,the evs Y := X × V has a basis iff the evs X has a basis [The evs X × V is discussed inexample 2.10] . Also dim( X × V ) = [dim( X r X ) : dim X + dim V ] .Proof. Let X has a basis. We first show that A := { ( b, θ V ) : b ∈ B } is a basis of Y r Y ,where B is a basis of X r X and θ V is the identity of V . As B is orderly independent in X r X we can say that any two members of A are orderly independent ⇒ A is an orderlyindependent set in Y r Y . Let ( x, v ) ∈ Y r Y ⇒ x ∈ X r X [ ∵ Y = X × V ]. Since B generates X r X , for this x , ∃ b ∈ B such that x ∈ L ( b ) ⇒ αb + p ≤ x for some α ∈ K ∗ and p ∈ X ⇒ α ( b, θ V ) + ( p, v ) = ( αb + p, v ) ≤ ( x, v ) and ( p, v ) ∈ [ X × V ] ⇒ ( x, v ) ∈ L (cid:16) ( b, θ V ) (cid:17) ⇒ A is a generator of Y r Y . So A becomes a basis of Y r Y . Consequently Y has a basis.Now dim( Y r Y ) = card( A ) = card( B ) = dim( X r X ) and dim[ X × V ] = dim( X × V ) =dim X + dim V . Therefore dim( X × V ) = [dim( X r X ) : dim X + dim V ].Conversely, suppose Y := X × V has a basis. Let B be a basis of Y r Y . Now consider B ′ := { x : ( x, v x ) ∈ B for some v x ∈ V } . Then x ∈ B ′ ⇒ x / ∈ X . Therefore B ′ ⊆ X r X .We now show that B ′ forms a basis of X r X . For any z ∈ X r X , ( z, θ V ) ∈ Y r Y . As B is a basis of Y r Y , ∃ ( x, v x ) ∈ B such that α ( x, v x ) + ( p, v ) ≤ ( z, θ V ) for some α ∈ K ∗ and( p, v ) ∈ [ X × V ] = X × V ⇒ ( αx + p, αv x + v ) ≤ ( z, θ V ) ⇒ αx + p ≤ z ⇒ z ∈ L ( x ). So B ′ generates X r X . If two members of B ′ say x ′ , z ′ are orderly dependent then withoutloss of generality we can take x ′ ∈ L ( z ′ ) ⇒ ∃ α ∈ K ∗ and p ∈ X such that αz ′ + p ≤ x ′ ⇒ α ( z ′ , v z ′ ) + ( p, v x ′ − αv z ′ ) ≤ ( x ′ , v x ′ ) ⇒ ( x ′ , v x ′ ) and ( z ′ , v z ′ ) are orderly dependent in Y r Y .Therefore we can say that B ′ is orderly independent in X r X as B is orderly independentin Y r Y . So B ′ becomes a basis of X r X . Consequently, X has a basis.16 xample 3.30. For any two cardinal numbers α, β there exists an evs X such that dim X =[ α : β ]. For example, if we consider the evs X := Y × E , where Y is an evs whose dimensionis [ α : 0] (existence of such evs has been established in example 3.28) and E is a vectorspace with dimension β , then by above theorem 3.29 dim X = [ α : β ]. Theorem 3.31.
Let X be an evs whose dimension is [ α : β ] . Also let γ and δ be two cardinalnumbers such that γ ≤ α and δ ≤ β . Then ∃ a subevs Y of X such that dim Y = [ γ : δ ] .Proof. Let B be a basis of X r X . Then card( B ) = α . Since γ ≤ α , there exists C ⊆ B such that card( C ) = γ . For each c ∈ C we choose one element p c ∈ P c and fix it. Case 1 : If δ < γ then ∃ E $ C such that card( E ) = δ . Consider the set D := E ∪ { c − p c : c ∈ C r E } Since C is orderly independent it follows that card( D ) = card( C ) = γ . As L ( c − p c ) = L ( c )it follows that D is an orderly independent set in X r X . Also consider for any d ∈ D , q d = p d if d ∈ E otherwise q d = θ . Then there exists a subspace W of the vector space X such that q d ∈ W , ∀ d ∈ D and dim W = δ . Case 2 : If γ ≤ δ then consider D = C and q d = p d , ∀ d ∈ D . Then also there exists asubspace W of X such that q d ∈ W , ∀ d ∈ D and dim W = δ .Thus for both cases we get(i) an orderly independent set D in X r X whose cardinality is γ .(ii) a subspace W of X such that q d ∈ W where q d < d , ∀ d ∈ D and dim W = δ .Now we consider the set G ( D ) := ( n X i =1 α i d i + p : α i ∈ K , d i ∈ D, p ∈ W, n ∈ N ) Step 1 :
In this step we will prove that G ( D ) becomes a subevs of X with D ⊆ G ( D ) and[ G ( D )] = W .For any d ∈ D , d = 1 .d + θ ∈ G ( D ) ⇒ D ⊆ G ( D ). Also for any p ∈ W , 0 .d + p ∈ G ( D ) ⇒ W ⊆ G ( D ). For any two elements x = m X i =1 α i d i + p , y = n X j =1 β j d j + q in G ( D ) and anytwo scalars α, β , αx + βy = m X i =1 αα i d i + n X j =1 ββ j d j + ( αp + βq ) ∈ G ( D ) [as W is a subspace].Let y ∈ [ G ( D )] . Then y is a minimal element of G ( D ). As y ∈ G ( D ), y can be written as y = n X i =1 α i d i + p . Our claim is that all α i = 0. If not, there exists j ∈ { , , . . . , n } such that α j = 0. Then there exists q d j ∈ W such that q d j < d j ⇒ n X i =1 α i q d i + p < y which contradictsthat y ∈ [ G ( D )] , as n X i =1 α i q d i + p ∈ W ⊆ G ( D ). So all α i = 0. Therefore y = p ∈ W ⇒ [ G ( D )] ⊆ W ⊆ G ( D ) ∩ X . Therefore [ G ( D )] = G ( D ) ∩ X = W [by Note 2.2]. Also for Here the notation ‘ q d < d ’ is used to mean that q d ≤ d but q d = d . x = n X i =1 α i d i + p ∈ G ( D ), n X i =1 α i q d i + p ∈ W = [ G ( D )] such that x ≥ n X i =1 α i q d i + p . Thusit follows that G ( D ) is a subevs of X . Step 2 :
In this step we shall show that D is a basis of G ( D ) r [ G ( D )] . Since D is an orderly independent subset of X r X and G ( D ) is a subevs of X containing D ,by remark 3.7 we can say that D is orderly independent in G ( D ) r [ G ( D )] . Now let y ∈ G ( D ) r [ G ( D )] . Then y can be written as y = n X i =1 α i d i + p , where not all α i = 0. Let α j = 0. Then α j d j + n X i =1 i = j α i q d i + p ≤ y . As n X i =1 i = j α i q d i + p ∈ W = [ G ( D )] , so y ∈ L ( d j )in G ( D ) r [ G ( D )] . Thus D becomes a basis of G ( D ) r [ G ( D )] .Therefore dim G ( D ) = [card( D ) : dim W ] = [ γ : δ ]. In this section we shall discuss the existence of basis of some particular evs and therebycompute their dimensions. We show that there are evs which do not have basis.
Theorem 4.1.
Let X be a single-primitive comparable topological evs. Then X has a basisand dim X = [1 : dim X ] .Proof. Since X is single-primitive, for each z ∈ X let us write P z = { p z } . Let x ∈↑ θ with x = θ . Then P x = { p x } = { θ } . Now for y ∈ X r X , y − p y ∈↑ θ . Then X being comparableevs, x and y − p y are comparable as P x = P y − p y = { θ } . If x ≤ y − p y then x + p y ≤ y ⇒ y ∈ L ( x ). If x > y − p y our claim is that there exists α ∈ K ∗ such that αx ≤ y − p y with | α | <
1. For, otherwise we can choose a sequence { α n } in K ∗ such that y − p y < α n x ∀ n ∈ N and α n → n → ∞ . Since X is a topological evs we then have y − p y ≤ θ [taking limit n → ∞ ] —— a contradiction. So there must exist one α ∈ K ∗ such that αx ≤ y − p y ⇒ y ∈ L ( x ). Thus L ( x ) = X r X . Clearly { x } is orderly independent. Therefore { x } is abasis of X r X . Consequently X has a basis and dim X = [1 : dim X ].As the evs [0 , ∞ ) × V is a single primitive comparable evs by above theorem we cansay that dim([0 , ∞ ) × V ) = [1 : dim V ], for any Hausdörff topological vector space V . Soin particular, if V = { θ } then the resulting evs is order-isomorphic to [0 , ∞ ) and hencedim[0 , ∞ ) = [1 : 0]. We have shown in the previous section that dim[0 , ∞ ) α = [ α : 0], forany cardinal α . This can also be justified from the following more general example. Example 4.2.
Let { X i : i ∈ I } be an arbitrary collection of exponential vector spaces,over the common field K , each having a basis. Let B i be a basis of X i r [ X i ] , ∀ i ∈ I .Consider the product evs X := Y i ∈ I X i [see Example 2.9]. Then X = Y i ∈ I [ X i ] . For any18 ∈ I consider the set D j := Y i ∈ I C i , where C i := { θ X i } , when i = jB j , when i = j . Here θ X i is theidentity in X i . Then D j ⊆ X r X , ∀ j ∈ I . Let D := [ j ∈ I D j . Then D ⊆ X r X . Nowtwo different members in different D i are orderly independent. As each B i is a basis of X i , so two different members of one D i are orderly independent. Thus any two differentmembers of D are orderly independent and hence D is orderly independent in X r X . Wenow show that D is a basis of X r X . For any x = ( x i ) i ∈ I ∈ X r X , ∃ some k ∈ I suchthat x k ∈ X k r [ X k ] ⇒ ∃ b k ∈ B k , α k ∈ K ∗ and p k ∈ [ X k ] such that α k b k + p k ≤ x k .Now for i = k , ∃ p i ∈ [ X i ] such that p i ≤ x i . Let b = ( b i ) i ∈ I , where b i = θ X i for i = k and p = ( p i ) i ∈ I ∈ X . Then α k b + p = ( α k b i + p i ) i ∈ I ≤ ( x i ) i ∈ I = x and b ∈ D k ⊂ D ⇒ x ∈ L ( b ).This shows that D generates X r X and hence is a basis of X r X . Consequently, X hasa basis and dim X = [card( D ) : dim X ].If I be finite then card( D ) = X i ∈ I card( D i ) = X i ∈ I card( B i ) = X i ∈ I dim (cid:16) X i r [ X i ] (cid:17) anddim X = X i ∈ I dim[ X i ] . For any four cardinal number α, β, γ, δ if we use the notation[ α + γ : β + δ ] = [ α : β ] + [ γ : δ ] then we can writedim Y i ∈ I X i = "X i ∈ I dim ( X i r [ X i ] ) : X i ∈ I dim[ X i ] = X i ∈ I h dim ( X i r [ X i ] ) : dim[ X i ] i If I be infinite then also we get the similar expression as above, provided the sums (over I ) be properly defined.If all X i ’s are same, say X i = Y, ∀ i ∈ I and card( I ) = α then we havedim( Y α ) = [ α · dim( Y r Y ) : α · dim Y ]Thus it follows that for any cardinal α , dim[0 , ∞ ) α = [ α : 0], since dim[0 , ∞ ) = [1 : 0]. Theorem 4.3.
For every Hausdörff topological vector space X , C ( X ) [discussed in 1.3]has a basis.Proof. Let us consider the relation ‘ ∼ ’ on X r { θ } , defined as x ∼ y ⇔ ∃ α ∈ K ∗ such that x = αy Then ‘ ∼ ’ becomes an equivalence relation on X r { θ } . Let us construct a set X ′ takingexactly one representative from each equivalence class relative to ‘ ∼ ’ and consider the set N := n { θ, x } : x ∈ X ′ o We now show that N becomes a basis of C ( X ) r [ C ( X )] . If A ∈ C ( X ) r [ C ( X )] , thenthere must exist two elements x, y of X with { x, y } ⊆ A and x = y . Then { θ, x − y } + { y } = { x, y } ⊆ A . Now x − y ∈ X r { θ } ⇒ ∃ z ∈ X ′ and α ∈ K ∗ such that x − y = αz . So we canwrite α { θ, z } + { y } ⊆ A ⇒ A ∈ L ( { θ, z } ). Therefore N generates C ( X ) r [ C ( X )] . We nowshow that N is an orderly independent set in C ( X ) r [ C ( X )] . For any two elements { θ, x } and { θ, y } in N , if { θ, x } ∈ L ( { θ, y } ) then ∃ α ∈ K ∗ such that α { θ, y } + { z } ⊆ { θ, x } for19ome z ∈ X ⇒ { z, αy + z } = { θ, x } [ ∵ z = αy + z ] ⇒ either z = θ or z = x . If z = θ then αy = x which means that x, y belong to the same equivalence class relative to ‘ ∼ ’ and hence { θ, x } , { θ, y } cannot be two distinct elements of N —— which is not the case. If z = x then αy + x = θ ⇒ x = − αy which again leads to the same contradiction. This proves thatany two elements of N are orderly independent. Therefore N is orderly independent in C ( X ) r [ C ( X )] and hence becomes a basis of C ( X ) r [ C ( X )] . Consequently, C ( X ) has abasis and dim C ( X ) = [card( N ) : dim X ]. Remark 4.4.
We have shown in the above theorem 4.3 that N forms a basis of C ( X ) r [ C ( X )] . We now show that this basis depends on a basis (as vector space) of X .(i) If X be a Hausdörff topological vector space of dimension 1, then any non-zeroelement of X is a scalar multiple of a single basic vector of X and hence N containsexactly one element. So dim C ( X ) = [1 : 1]. For that reason dimension of C ( R ) over R is[1 : 1] and dimension of C ( C ) over C is [1 : 1].(ii) Let X be a Hausdörff topological vector space of dimension 2 and B = { a, b } be abasis of X . We first show that X ′ = { a + βb : β ∈ K } ∪ { b } , where X ′ is as defined in theproof of the theorem 4.3. Any two distinct elements a + β b, a + β b ∈ X r { θ } must lie intwo different equivalence classes relative to ‘ ∼ ’, since for any α ∈ K ∗ if α ( a + β b ) = a + β b then α = 1 and hence β = β [as { a, b } is a linearly independent subset of X ] — thiscontradicts that a + β b = a + β b . Also linear independence of a, b implies that a + βb and b must lie in two different equivalence classes relative to ‘ ∼ ’, for any β ∈ K . Now for anynon-zero element x ∈ X , ∃ α, β ∈ K (not both zero) such that x = αa + βb [since { a, b } is a basis of X ]. If α = 0 then x = α ( a + βα − b ) ⇒ x lies in the class (relative to ‘ ∼ ’)whose representative is ( a + βα − b ). If α = 0 then x lies in the equivalence class (relativeto ‘ ∼ ’) whose representative is b . Therefore X ′ = { a + βb : β ∈ K } ∪ { b } . Now the map α a + αb creates a bijection between K and X ′ r { b } . So we can say that cardinality of X ′ and hence cardinality of N is c , the cardinality of the set of real numbers R . Thereforedim C ( X ) = [ c : 2]. For that reason dimension of C ( C ) over R is [ c : 2].(iii) In a similar manner as above we can show that for a well-ordered basis B of aHausdörff topological vector space X , X ′ = ( e + < B > ) ∪ ( e + < B > ) ∪ · · · ∪ ( e n + < B n > ) ∪ · · · where B = { e , e , . . . , e n , . . . } , B = B r { e } , B n = B n − r { e n } , ∀ n ≥ < B i > denotes the linear span of B i in X , ∀ i . Theorem 4.5.
For every vector space X , the evs L ( X ) has a basis. [The evs L ( X ) isdiscussed in Example 2.12] Proof.
Let T be the collection of all one dimensional subspaces of X . We now show that T forms a basis of L ( X ) r [ L ( X )] . For any non-trivial subspace Y of X , there existsa non-zero element x ∈ Y such that < x > ⊆ Y h here < x > denotes the linear span of x X i . So Y ∈ L ( < x > ). Also < x > ∈ T . Thus T generates L ( X ) r [ L ( X )] . Forany two distinct elements < x >, < y > ∈ T , if α < x > ⊆ < y > for some α ∈ K ∗ then < x > = α < x > ⊆ < y > ⇒ < x > = < y > which contradicts that < x > and < y > are distinct. So we can say that T is an orderly independent subset of L ( X ) r [ L ( X )] .Therefore T forms a basis of L ( X ) r [ L ( X )] . Consequently L ( X ) has a basis anddim L ( X ) = [card( T ) : 0], since [ L ( X )] = n { θ } o .From above theorem we can immediately get the following result. Also T is the only basis of L ( X ) r [ L ( X )] . Result 4.6. dim L ( X ) = [1 : 0] , when dim X = 1 and dim L ( X ) = [ c : 0] , when dim X =2 , c being the cardinality of the set of all reals R . Note 4.7.
From the previous result we can say that dim L ( R ) = [1 : 0] which is same withthe dim[0 , ∞ ). But L ( R ) and [0 , ∞ ) are not order-isomorphic as first one is non-topologicalevs whereas second one is a topological evs and being topological is an evs property. Thisexample shows that converse part of the statement that equality of dimension is an evsproperty which we have discussed in 3.18 is not true. Theorem 4.8.
For any n ∈ N , D n [0 , ∞ ) has a basis and dim D n [0 , ∞ ) = [1 : 0] .Proof. We first show that (0 , , . . . , ,
1) generates D n [0 , ∞ ) r [ D n [0 , ∞ )] . Let x = ( x , . . . , x n ) ∈ D n [0 , ∞ ) r [ D n [0 , ∞ )] . Since [ D n [0 , ∞ )] = n (0 , . . . , o , there exists i ∈ { , , . . . , n } such that x i = 0 and x j = 0, for all j < i . If i < n then obviously(0 , , . . . , , ≤ x . If i = n then x i (0 , , . . . , , ≤ x . In any case x ∈ L (cid:16) (0 , , . . . , (cid:17) .Since n (0 , . . . , , o is orderly independent it follows that n (0 , . . . , , o is a basis of D n [0 , ∞ ) r [ D n [0 , ∞ )] and hence dim D n [0 , ∞ ) = [1 : 0].Following exampale shows that there exist an evs which has no basis. Theorem 4.9. X := D (cid:16) [0 , ∞ ) : N (cid:17) has no basis.Proof. Let x = ( x i ) i ∈ N ∈ X r X . Since here X = n (0 , , . . . ) o , there must exist a leastpositive integer p such that x p = 0. If we consider y = ( y i ) i ∈ N , where y i = x i , ∀ i = p, p + 1and y p = 0, y p +1 = 1 then y ≤ x and y / ∈ X ; but there does not exist any α ∈ K ∗ suchthat αx ≤ y —— which means that y / ∈ L ( x ). This shows that x / ∈ Q ( X ) and this holdsfor any non-zero element x of X . Therefore Q ( X ) = ∅ . So D (cid:16) [0 , ∞ ) : N (cid:17) has no basis.Looking at the proof of the above theorems we can get the following generalised theorem. Theorem 4.10.
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