ON THE MINIMAL SPACE OF SURJECTIVITY QUESTION FOR LINEAR TRANSFORMATIONS ON VECTOR SPACES WITH APPLICATIONS TO SURJECTIVlTY OF DIFFERENTIAL OPERATORS ON LOCALLY CONVEX SPACES

We use transfinite induction to show that if L is an epimorphism of a vector space V and maps a vector subspace W of V into a proper subspace of itself, then there is a smallest subspace E of V containing W such that L(E) E (or a minimal space of surjectivity or solvability) and we give examples where there are infinitely many distinct minimal spaces of solvability. We produce an example showing that if L 1 and L 2 are two epimorphisms of a vector space V which are endomorphisms of a proper subspace W of V such that LI(W) n L2(W is a proper subspace of W, then there may not exist a smallest subspace E of V containing W such that LI(E) E L2(E). While no nonconstant linear partial differential operator maps the field of meromorphic functions onto itself, we construct a locally convex topological vector space of formal power series containing the meromorphic functions such that every linear partial differential operator with constant coefficients maps this space linearly and continuously onto itself. Furthermore, we show that algebraically there is for every linear partial differential operator P(D) with constant coefficients a smallest extension E of the meromorphic functions in n complex variables, where


INTRODUCTION.
A straightforward argument using transfinite induction and the axiom of choice will show that if L is a linear transformation of the vector space V over the field F onto itself, and W is a subspace of V such that L(W) is a proper subset of W, then there is a subspace E of V containing W such that L(E) E and such that if F is any proper subspace of E containing W, then L(F) F. That is to say there is a smallest extension E of W in which the inhomogeneous equation Lu f (I.I) has a solution u in E for every f in E.
In section two of this paper we show that there is a vector space V, a subspace W of V, a pair of linear transformations L 1 and L 2 of V onto itself such that LI(W n L2(W is a proper subspace of W and such that if E is any vector space containing W and contained in V such that LI(E) L2(E) E, then there is also a proper subspace F of E containing W such that LI(F) L2(F) F. Thus the minimal space of surjectivity question for families of mappings is not solvable.
If c is a category whose objects are sets, possibly equipped with some structure, and whose morphisms are mappings between the sets, which preserve the structure, then we can define the minimal space of surjectivity question as follows.Let C be such a category.Let V be an object in C and let W be a subobject of V, a subset of V which has the structure (if any) induced by that of V. Let L be a mapping that is an epimorphism of V in the sense that (e.g.Northcott  [1], chapter Ill) V is the unique object in the category of vector spaces and linear transformations such that IvL and LI V are defined where V is the identity mapping of V and L(V) V. Further assume that if U is any subobject of V, then the restriction of L to U defines a morphism of the category whose range can be any subobject of V containing U. Then we say E is a solution of the minimal space of surjectivity problem defined by the triple (V,W,L) satisfying the preceding conditions if L(E) E, L(W) is a proper subspace of W, and if F is any subspace of E containing W, then L(F) F. If there is a triple for which there is no solution to the minimal space of surjectivity problem we say that for the category the MSS question has a negative answer.If there is a triple (V,W,L) satisfying the above conditions for which the MSS problem does not have a unique answer, we say that there is nonuniqueness for the MSS question for the category.
In section 3 of this paper we show that the MSS question has a positive answer for the category of vector spaces and linear transformations, but in section 4 we show that we have nonuniqueness in this category.
In section 5 of this paper we show, for every nonzero linear partial differential operator with constant coefficients, the existence of a smallest extension of the meromorphic functions on which the operator is an epimorphism.We do this by exhibiting a locally convex topolgoical vector space containing the meromorphic functions on which every linear partial differential operator is an epimorphism and applying the results of the previous sections.

2.
NONEXISTENCE OF A MSS FOR SOME FAMILIES OF EPIMORPHISMS.It can be shown that if V is a vector space, W is a subspace of V, and L is an epimorphism of V such that L(W) is a proper subset of W, then there is a subspace E of V containing W which is minimal with respect to surjectivity in the sense that L(E) E, but if U is any proper subspace of E containing W, then L(U) does not contain U.It seems natural to ask the same question for families of linear transformations.
MSS Question for Families of Mappings.Le___t V be a vector space.Let W be a subspace of V. Let F be a family of epimorphisms of V such that L(W)C W for all L in F and L(W) # W for at.least one L inF.Does there exist a subspace E o__fV containing W such that L(E) E for every L in F having the additional property that if U is any subspace of E containinq W, then L(U) # U for some L in F i__f U E.
The following theorem shows that the MSS question for families of linear transformations fails in general to have a positive answer even if F contains only two mappings.
Theorem 2.1.There exists a vector space V, two epimorphisms L and L 2 o__f V, subspace W o__f V such that Lk(W) c W for k 1, 2, such that LI(W) n L2(W) is a proper subset of W, and having the additional property that if E is any subspace of V containing W such that Lk(E) E for k I, 2 then there is a proper subspace U o_f E containinq W such that Lk(U) U for k 1, 2.
Proof of Theorem 2.1.Let N denote the set of nonnegative integers.Let Q denote the set of all nonzero integer powers of the prime q.Let F denote an arbitrary field.For convenience we introduce the following.
Definition 2.1.If S is a set without a topology and v" S F is a mapping from the set S into a field F, then the support of v is defined by the rule, supp(V) {j S:(j) # O} We let V 0 denote the vector space of mappings from {0} into the field F. For every positive integer k we let V k denote the vector space of mappings from Q into F whose support is a finite subset of Q.Let P V k V k denote the projector onto the space of functions whose supports are subsets of Nq {q-n n 1,2,3 (2.1)Let T P(V k) P(V k) be a linear transformation defined by the rule T(Pk)(q'2n pk(q -n) T(Pk) (q for all positive integers n and all functions k in V k. -n -vt o Lemma 2.1.If ' is a function in P(V k) whose, support is precisely {q q , -n -2n -2n 2 -2n o r}, then the support of Tv is exactly {q q q r}. -2n Proof of Lemma 2.1.Suppose q were in the support of T. Then by definition n must be one of {n 1, n 2 nr}.
As a corollary of Lenna 2.1 we observe that Ker(T) is trivial.

Let
V k V k denote the projector defined by the rule (q2n) (q2n) (2.4) for all positive integers n, and )v(qm) 0 (2.., =) if m is an integer that is not equal to 2n for some positive integer n.
Let B VI VO be defined by the rule, BY (2.6) where .(0): V(q2n+l) Si(q2n-1) i(q-n) (2.9) and SV(q "n) 0 (2.10) for all positive integers n.Let V be the vector space over F defined to be the set of all (0' 1' Vk where Vk is a member of V k for all nonnegative integers k and k is identically zero for all but a finite number of nonnegative integers k, and let w denote the space of w (t 0, 1' O, ...) (2.11) where 0 is a member of V 0 and I is a member of (V I) n Ker(B) where p _1 (2.12)   Define a mapping L V V by the rule, + + L1(0'I 'n "'') (B(TI)'TPI + ]I + 2 TPn n n+l (2.13) Define a mapping L2"V V by the rule Sn n+l Lemma 2.2.Lk'V V is an epimorphism for k 1,2, and L2-V V is an s omorph sm.Proof.That each L k is an epimorphism is obvious since V 0 is one dimensional and n+l covers the part of the nth coordinate space not covered by TP n + ]n or S n, + is identically zero this implies that each v respectively.Also if each Sv n n+l n is identically zero, if Vn+l O, since S'V n (I-P)V n is an isomorphism.Now let us construct a space of surjectivity, E, for the operators L and L 2 which contains W. Let (0,0,0 denote a nonzero member of W. it in E and L lv LI.Now suppose that n > 1.Then the nth entry of LI(O, 1,...,vn,O,O is TP n + n" Since T:PV n PV n is a monomorphism by Lemma 2.1 we conclude that Pn and n are identically zero.Now we know that Pn-1 + n-1 + Vn is identically zero.Thus, the fact that (P+)(Pn_1+lXn+V n) is identically zero implies that Pn-1 + ]n-1 is identically zero.Hence, n is identically zero.Since n represents an arbitrary integer larger than 2, we conclude that the vector v defined in (2.15) is of the form v (0,1,0,0 where P and 1 are identically zero.We conclude that 2nk-1 (2.16) < n r.Presumably VO is not identically zero and,, consequently, Vl # Ker(B).Let supp( I) denote the support of 11" Then we observe that L2v E and that there is some v I in E such that L2v v (O,Vl,O ).
We observe that L 2 is a one-to-one mappirg.Hence Ll({v}) contains only a single element.Observe that if the support of vI is defined by (2.16), then L)(v)--(0,,,0,0, 0,...) (2.17 Now we observe that since PI Observe that the support of STn 1 is given by 22nn1-1 supp(sTn, q 22nn -1 r ,q (2.22) Observe that B(STnI I O Thus, we can say that the vector space generated by the elements we know to be in E has the property that Now the support of SmsTn@ 1 is given by 2m(2.2nni-12m(2.2nnr-1)supp (SmSTn I q q (2.24) Now there must be a vector w in E of the fom w (0,i,2, n"'" such that L l(w) (O,Vl,O ,0,...) Since W is a vector space and (-0,0,0 ,0,...) is in W we may assume that w is actually of the form (0,i, 2 Cn ), where I Ker(B).But 1 Ker(B) and (P+)1 0 implies that (0,-1,0 ,0 W Thus, we may assume that w (0,0, 2 Cn"'" We can show that if n > 2, that Cn is identically zero.Thus, we may assume that Since (P+-P) 2 w (0,0,2,0, 0,...) 0 we deduce that 2 1" Hence, we deduce that (0,1,0,...,0,...), (O,O,Vl,0 0,...) are all contained in E. Thus, the set of functions in E whose coordinate functions are in P(Ker(B)) or else are modulo a function in P(Ker(B)) a function with support sufficiently far out forms a proper subspace of E which is mapped onto itself by L 1 and L 2 3.

THE MINIMAL SPACE OF SURJECTIVITY PROBLEM FOR VECTOR SPACES AND LINEAR TRANSFORMATIONS
Let L be a linear transformation of a vector space V onto itself, and let W be a subspace of V such that L(W) is a proper subspace of W. The solvability of the MSS problem for the category of vector spaces and linear transformations is expressed in the fol owing theorem.Theorem 3.1.Let L, V, and W be as defined in the introduction to this section.
Then there is a subspace E of V containinq W such that L(E) E and such that if E 1 is a proper subspace of E containinq W, then L(E) # Ez.
Proof of Theorem 3.1.Let W 1 be a subspace of W such that W L(W) ) W 1. Let B(W I) be a bases for W 1. Let 'I(W) be a set consisting of precisely one men)er from each of the sets in the family {L-l(w) w B(W1)} The axiom of choice tells us that this set exists.We need the following result.Lemma 3.1.Let L V V be a linear transformation of V onto itself.Let T be a linearl>, independent subset of V. I__f s is a set consistin9 of precisely one member from each of the sets in the family {L-1(t) t T} (3.2) then S is a linearly independent set.
Proof of Lemma 3.1.Let v v m be an arbitrary finite subset ofs.Let c 1, c m be scalars.Then ClV + + CmVm 0 implies ciL(v1) + + CmL(Vm) 0. But {L(v I) L(Vm)} is an m-element subset of T and is, therefore, linearly independent.Hence, c c 2 c m 0.
Lemma 3.2.Let L, T, and S be as defined in Lemma 3.1.If [9] and Is] denote the vector spaces generated by and %, r.esp.e.ctively, and if B([S]) is a basis for IS], then L(B[S] is a bases for [7]. Proof of Lemma 3.2.Let {u I u m} be an m-element subset of B([%]).Write P uj 1= a(k'j)Vk (J I, m) Vp} is a p-element subset of % and the matrix a( a(p,1) a(p,m) is a ene-to-one linear transformation from m-dimensional space to p-dimensional space.
Then L is a linear implies P L(uj) k--a(k'j)L(Vk) (j 1, m) (.5) Now suppose that we had m cjL(uj) 0 (3.6) j=l Then interchanging summation signs we deduce from combining 3.6 and 3.5 that p m (1"= a(k'j)cj) L(vk) 0 By linearity of L and (3.10) we see that L(v) is a linear combination of a finite number of elements in L(B[S]).
Let U W and let E be the vector space generated by (U I) and U 1, where (U 1) is an image of the choice function on the family of sets {L-1(t)'t B(W1 )}" Then we where [S] denotes the vector space generated by S for all subsets S of the vector space V. Lemma 3.3.uppose L is a linear transformation of V into itself and U is a subspace of V such that L(U I) is a proper subspace of U I. Suppose E(U I) is the set obtained by taking one member from each set in.the family {L-l(t) t e B(WI)} (3.11) where U L(U I) ( Then E I is a proper subspace of E 1 containinq U implies that L(E1) does not contain UI" Proof of Lemma 3.3.Let 71 be a projection of E onto [(UI)].Then I(E I) must be a proper subspace of [E(U1)] since the definition of direct sum and the fact that E 1 is a proper subspace of E 1 containing U implies E 1 U 1 I(E I) (

3.13)
We denote by EI(u I) a basis for [(UI)] which contains B(I(E1)), a basis for I(EI).By Lema 3.2 we know that L('I(u1 )) is a basis for W 1. Since I(E 1) is a proper subspace of [(UI)], it follows that B(I(EI)) must be a proper subset of I(u1).and consequently that L(E I) is a proper subspace L(U1)Q W 1 U 1.
Hence, L(E1) does not contain U I. We use transfinite induction to construct for every ordinal a less than 6 + I, where 6 is the ordinality of a basis for V, a set E which is minimal with respect to the property that and Us W (u{Ey -y < } c L(E) We have constructed E for 1.Thus, we suppose that E has been constructed for all B < a. Then define U as in (3.16).It is clear that Us is a vector space, since W and each Ey is a vector space and y1 < Y2 implies E Let W be a subspace of V such that Y 2 Let B(W) be a basis for W:.Let (U) be defined to be the set obtained by taking one element frn each set in the family

{L-I(w) w B(W)} (3.20)
We let Es be the vector space generated by Us and(U).Then the Lemmas 3.1, 3.2 and the previous argument show that if E satisfies (3.18), then L(E 1) E 1. Since E But the L(V) V, it is clear that there is some ordinal s O such that L(O SO ordinals are well ordered.Thus, we may suppose min(y L(Ey) Ey} (3.21)Indeed, it is easy to see that B is the first infinite ordinal.
The following lemma gives important information about the space E B-Lemma 3.4.l__f B is defined bj (3.21), then U B E B.
Proof of Lemma 3.4.Suppose U B # E B. Then L(U B) would be a proper subspace of U B. It is easy to show by transfinite induction that L(U)C U for every ordinal s.
Then we can write  (3.28) and for every, vector v in V v w + (v) (1< < a + 1) (3.29) where w is a member of W.
Proof of Lemma 3.5.To prove (3.28) note that if v [E(U)] n U then L(v) L(U) n We {0}.Thus, v ClV1 + + CmVm for some elements v 1, Vm in (U) and some scalars c 1, c m.But {L(Vl),...,L(Vm)} is an m-element subset of B(W) a basis for W and is therefore a linearly independent set.Hence, L(v) 0 implies v O.In other words (3.28) is valid.Next we show that < B implies In order that L(E I) E 1 we must have in particular that UC L(E 1).But U: L(U) L([E(U)]) (3.38) Now (I-)(El)) n E 1 is a proper subspace of (I)(V).We want to show that L(E 1) could not possibly contain U. Suppose L(E I) did contain U. Then L(E I) n U L(U) T U. LetB(l) be a basis for.ThenB(T)c L(EI).Thus, E'c E B implies that But L is one-to-one on each of the subspaces [E (U)] and is consequently one-to-one Y on their direct sum.Thus, v 0 for all ordinals y such that s < y < B, where the Hence, E n E E which contradicts (3.34). 4.

NONUNIQUENESS OF THE MSS PROBLEM FOR THE CATEGORY OF VECTOR SPACES AND LINEAR TRANSFORMATIONS
In this section we prove that if V is a vector space with a subspace W and L is an epimorphism of V which maps W into a proper subspace of itself, then there does not necessarily exist a unique subspace E of V containing W such that L(E) E and L(E I) # E 1 for all proper subspaces E 1 of E containing W. We also define the semigroup s(V, L, W) of endomorphisms of V which commute with L and leave W fixed. Theorem 4.1.Le___t F be a field.Then there exists a pair of vector spaces V and W such that both V an___d W have countable dimension, with W being a subspace of V, such that there exists a linear transformation L o__f V onto itself which maps W into but not onto itself.Furthermore, V, W, and L described above can be constructed in such a way that there exists a sequence of spaces {V 'm'{ m N} such that V V '0'{ V '1'{ V2Jl W, L maps each V mjl linearly onto itself, and V (m) W (4.1) m=0 In addition there exists an uncountable family of subspaces {E A} of V containing W such that L maps each E linearly onto itself and such that if E Is a subspace of V satisfyin9 then L(E 1) is not equal to E Furthermore we may construct the minimal spaces of suriectivitj E so that if I =2 then 1 2 if we ask onl), that A be countable.
Proof.Let V denote the set of all functions from N {0,1,2 into F which vanish for almost all members of N for 1, where O is the zero element of V for be a bijection.Define LI"V V by the rule, ,0,v(4),0 ),0,0 Define L2:V2 V by the rule, L2(v where y is defined by (4.6).Define L3"V3 V by the rule which shows that L maps V (m) linearly onto itself.Let k'V V k be the natural projection.Lemma  Note that E 2 and hence every E k must be infinite dimensional, since by hypothesis E V 1.The proof we have just repeated shows that if we just require that supp(E2), the support of the functions in E 2, satisfy supp(E 2) N {y(n,O), y(n,l), y(n,2) for every n, and take E k E 2 for all k 3,4,5 then L will map E linearly onto itself.Also E will be minimal provided that suppE 2 F {y(n,O), y(n,1), y(n,2) (4.17) has just one element in it for every n v.There are clearly an uncountable number of ways of choosing E 2 so that the set (4.17) has just one element in it for every n J. Also, the number of ways of choosing E 2 so that condition (4.2) satisfied is at most countable since for each integer n it must be true that supp(2El F {y(n,O), y(n,l) is not equal to supp(2E2 F {y(n,O), y(n,1) Definition 4.1.Le__t L be an epimorphism of a vector space V which has a subspace W with the propert}, that L(W) .isa..prop.ersubspace of itself.Le.__t 7(V,L,W) be the s.emi-group of all endomorphisms of V which commute with L and leave elements of W fixed.
Theorem 4.2.If L, V, and W satisfy the conditions of Definition 4.1 then S(V,L,W) contains only one element if and only if E is a proper subspace of V containing W .implies L(E) E.  Proof.Suppose E were a subspace of V containing W. Let us write V Ker(L)( F and write E Ker(L) n E ( .Let B() be a basis for , and let B(F) be a basis for F containing B().Let B(Ker(L)) be a basis for Ker(L) containing B(Ker(L) n E), a basis for Ker(L)n E. Then each v in V may be written as  Then define (v) al1 + + app + c11 + + Crr We deduce from the definition (4.23) of and the fact that L(E) E, that L((v)) cIL(GI) + + CrL(r) (L(v)) (4.23) (4.24) Thus, the projector we have constructed belongs to S(V,L,W).Conversely, if belongs to S(V,L,W) then E (V) satisfies L(E) L((V)) (L(V)) E. It is clear that the identity transformation belongs to (V,L,W).From our construction it is clear that if E is a proper subspace of V such that L(E) E, then the projector we have constructed is distinct from the identity.This completes the proof of the Theorem.

5.
SURJECTIVITY OF DIFFERENTIAL OPERATORS ON LOCALLY CONVEX SPACES CONTAINING THE MEROMORPHIC FUNCTIONS Let B(Cn) denote the meromorphic functions of n complex variables.We construct a special locally-convex space E n) containing B(Cn) such that every nontrivial linear partial differential operator with n independent variables and constant coefficients maps E {nl continuously onto itself.
Proposition I.The dual space of E (n) is isomorphic to the space Y1 E (n) C[T X 1,...,Tnnxn of polynomials.Furthermore, Y is a reflexive Frechet space for all y i__n %, where _T is liven by (5.2).
Proof.It is well-known (e.g.Treves [2], page 266) that [ n) and C[X X n] are duals of one another.But if E'c E, where E is a topological vector space, and J E F is an isomorphism then F' J(E').Furthermore, if E is a Frechet space, F is a locally convex space and J E F is a topological isomporphism, then F is also a Frechet space.
Definition 1.The duality bracket between a po..lynomial P i_n C[X X n] and a formal power series u i__n C The duality bracketr between a polynomial Qy in the dual of E 'n'Y and a formal power series v in E nj is given by <Qy,Vy> ] n (T) [(@/@x) Qy]x:y [(@/@x) Vy]x= Y (5.5) It is easy to see that if P(B/BX) -]'n (II!)[(@I@X)P(X)]x:y (@/@X) (5.6) and u E n), then <P,u> [P()/@X)u]xi=Y {I n} (5.7) It is similarly easy to check that if Qy(@/BX) ]n (1/!)[(@IBX)TYQy(X)]x=y (B/BX) (5.8) and v E(n) then {1 ,n} (5.9) <Qy,vy> [Qy (@/ X) Vy]x :Y By E. Borel's Theorem (e.g.Treves [3], Theorem 18.1) if u belongs to E n), there is a in C(R n) such that the coefficients of the Taylor series expansion for about X (0) (0 0) are identical to the coefficients of u.Thus, we may rewrite (7) as <P,u> <P(-BIBX)6,> (5.10) where 6 is the Dirac delta function.Again by E. Borel's Theorem (e.g.Treves [3], (n) a Oy in (pn) such that the coefficients Theorem 18.1) there is for every vy in Ey C O of the Taylor series expansion of o(J) about y are equal to the corresponding coefficients of vj, where X (Y) y.Thus, we may rewrite (5.9) as <Qy,Vy> <Qy(-/X)(X-X(Y)),y> (5.11) Let L L(B/BX) denote a linear partial differential operator with constant can use m (5.10) and (5.11) to determine the action to tL on E n)" coefficients.We U and E j)' for all y in I, where I is given by (5.2).It is well known that Y <P,L(alBX)u> <P(-alBX),L(alaX)> (5.12) implies that the transpose of L is one-to-one, since it follows that tLP(X) L(X)P(X) <Qy,L(/BX)u> <L(-/@X)P(-@/BX),y> for all y in I.By Theorem 28.1 of Treves [3] it follows that L(X)E nj' is a closed Y subspace of E nj' for every y in I. This, in view of a classical theorem due Y essentially to S. Banach, which states that a continuous linear map L of one Frechet space E into another Frechet space F is an epimorphism if and only if its transpose is one-to-one and weakly closed, implies that L E (n) E (n) is an isomorphism for all y in/.
Let V (n) be the subspace of E (n) consisting of all members of E (n) which may be identified with a member of ] E, n) (exterior direct sum).Then L is still an y'y (n)/v(n) since L(V (n)) c V (n) Note that V (n)  (5.16)Thus, R(Cn) can, when regarded as a vector space, be given a locally convex topology in a natural way, namely the one induced by E n).
Let f(z I Zn)/g(z z n) be a member of R(cn).Let j'C n C n-1 be defined by the rule, (3.1) the fact that the matrix A defined by 3.4 is one-to-fact that (3.6) implies (3.9) for all m-element subsets {u u m} of B([S]) tells us that L(B[S]) is a linearly independent set.To see that L(B[S]) generates [T] we let L(v) denote an arbitrary element of T, where v S. Then there exist u For if B(,I(E1)) were equal to Zl(U1), then we would have I(E 1) [B(I(E1)] [1(UI) [(UI) (3.14) which contradicts the supposition that I(E I) is a proper subspace of [(U1)].Thus, (3.14) and the definition of 71 imply that L(E I) L(U I) Q [(I(EI)), (3.15) EI) does not contain U B) W B(3.22)   Then we would letB(WB) be a basis for W B and letE(U B) be the set consisting of one member from each set in the family, {L-l(w) w B(W B)} (3.23)Then E B is the vector space generated U B and E(UB) and clearly L(E B) c UB.This contradicts the supposition that L(W) be a basis for W let E(U) be a set consistinq, of preciseljl one element from each set in the famil), {L'I(w) w B(

( 3 .
27)be a projector of V onto the vector space [(U)] generated by (U).Then[(U)] n U {0} 30) and by (3.28) U B n [(U B)] {0} (3.31) Combining (3.30) and (3.31) we deduce that [E(U)] n [E(UB)] {0} if and B are ordinalm and < B. Then (3.29) is an immediate consequence of the fact that v w (R) since [EI(u )] U {0} is consequently a subspace of [E(U )].Sut this would imply that U [EI(u )] E n E is a subspace of E with the property that U c L(U [EI(u )]).This implies by Lema 3.2 that U $ [El(us)] U [E(U )] ) can be identified with a subspace ofE (n) ER)IvR) ER)Iv n), E n) be nonempty.Let EI(u be a set consisting of one member from each set in the family (3.39).Then[EI(u ]El E V Let y be an ordinal ntunber larger than Then each set in the family {L-I(w) n E I w B($1)} (3.39) must 4.1.We may write E 1E 2 E 3 E The proof of Lemma 4.1 is immediate.Let E k ,k E for k 1,2