Munich Personal RePEc Archive
On the extension of a preorder under
translation invariance
Mabrouk, Mohamed
Ecole Superieure de Statistique et d’Ananlyse de l’Information de
Tunis
19 April 2018
On the extension of a preorder
under translation invariance
Mohamed ben Ridha Mabrouk1
Version: April 19, 2018
Abstract
This paper proves the existence, for a translation-invariant preorder on a divisible commutative group, of a complete preorder ext ending the preorder in
question and satisfyingtranslation invariance. We also prove that the extension
may inherit a property of continuity. As an application, we prove the existence
of a complete translation-invariant strict preorder on R which t ransgressesscalar
invarianceand also the existence of a complete translation-invariant preorder
sat isfying the social choice axiomsstrong Paretoand…xed–step-anonymity on a
setN0whereis a divisiblecommutative group. Moreover, the two extension
results areused to makescalar invarianceappear as a consequenceoftranslation
invarianceunder a continuity requirement.
1- Introduction
(Szpilrajn 1930) extension theorem may be st ated as follows. For any re‡ex-ive and transitre‡ex-ive binary relation (i.e. a preorder) on a gre‡ex-iven set, there exists a complete preorder which is an ext ension of the given preorder. Szpilrajn theo-rem proved of great utility in mathematical social choice theory as in some other branches of mathematics. There exists today stronger versions of Szpilrajn the-orem, requiring weaker assumptions on the initial binary relation or imposing additional conditions on the relation extension. We refer to (Alcantud-Diaz 2014) for an overview on the applicat ions and extensions of Szpilrajn t heorem.
The present paper establishes the exist ence, for any preorder on a divisible
commutative group satisfying translation invariance, of a complete preorder
extending the given preorder and satisfyingtranslation invariance (sect ion 3,
t heorem 1). In (Demuynck-Lauwers 2009) the existence of an extension under
t he conditionstranslation invarianceandscalar invarianceis proven. However,
t he result proved here is stronger in the sense that it is freed from thescalar
invarianceassumption. The proof of theorem 1 follows the same diagram as
t he proof of Szpilrajn theorem. Starting from a preorder satisfyingtranslation
invariance, one adds comparisons on some pairs of alt ernatives in such a way
t hat translation invariance remains satis…ed. Then, an argument based on
is used to " clean" the extended preorder given by t heorem 1 from undesirable rankings that transgress t he continuity requirement.
As an application, we give two examples, t he …rst of which shows the exis-t ence of a compleexis-te exis-t ranslaexis-tion-invarianexis-t sexis-tricexis-t preorder on R which exis-transgresses
scalar invarianceand the second shows the existence of a complete
translation-invariant preorder satisfying t he social choice axiomsstrong Pareto and …xed–
step-anonymityon a set N0whereis a divisible commutative group.
Moreover, theorems 1 and 2 are used t o makescalar invarianceappear as a
consequence oftranslation invarianceunder a continuity requirement (Corollary
2, section 5) or under a Paret o axiom (Theorem 3, section 6).
2- Preliminaries
N0 is the set of positive int egers. is the set of rational numbers. (+ )
is a divisible commutative group. being a binary relation on and two
elements of, is denoted% ,[and non()] is denotedÂ
and [and ] is denoted». The symbols · ¸ are used for the
natural order on R, except in example 1, section 4. A re‡exive and transitive
binary relation on is a preorder on. If, on t op of that, for all either
% or- , it is a complete preorder. A binary relation1is said t o be
a subrelation t o a binary relation2 , or 2 an extension of 1if for all
in
%1 =) %2
and
Â1 =) Â2
A xiom t ranslat i on i nvar i ance ( T I ) A preorder satis…es translation
invarianceif:
8( ) 2 £ 82 [%) +%+ ]
A xiomdi vi si on i nvar i ance(D I ) A preordersatis…esdivision-invariance
if:
82 82 N
·
%)
1
%
1
¸
Lem m a 1 If a preorder onsat is…es T I , then there exists a preorder b
on of whichis a subrelation and such that bsatis…es T I and D I .
Pr oof: First, notice that under , it is possible to sum inequalities. Indeed,
by T I , if are such that % and% then+ % + and
+ % + By transitivity, + % + For each positive integer ,
consider the binary relation de…ned by
% i¤%
If are such t hat % we can sum times this inequality. Thus,
is a subrelat ion to Moreover, is re‡exive and transitive. It is easily
checked that satis…es T I .
Consider the binary relation b
= [ 2 N0
de…ned onby % i¤ there issuch that %.
is a subrelat ion to b. Moreover, b is re‡exive and transitive. It is a
preorder. Since for each positive int eger , satis…es T I , we deduce that b
sat is…es T I . The lemma is proved if we show that b satis…es D I . Let be a
positiveinteger, and such that%Thereexistsa positiveintegersuch
t hat % . Thus% . We can write that as(
1
) % (
1
).
Thus1
% 1
, what implies
1
%
1
b satis…es DI.¥
Rem ar k 1 (1) It is easily seen that b is the minimal preorder satisfying
T I and D I , of which is a subrelat ion. (2) If is complete, since is a
subrelation to b, we have necessarily= b. This shows t hat if the preorder is complete, T I implies D I .
3- The translation-invariant extension theorem
T heor em 1 Let be a preorder on satisfying T I . Then there exist s a
complete preorder onsatisfying T I , of which is a subrelation.
Pr oof: If is a complete preorder, there is nothing to prove. Suppose
t hat is not complete. Consider the preorder b built in the proof of lemma
1, and the set < of all preorders on satisfying T I and D I , and of which
is a subrelat ion. < is not empty since b 2 < . Let () be a chain in < , i.e.
for any 0 is a subrelation to0 or0is a subrelation to. Notice
t hat (1) the relation [() de…ned on by: [[()] i¤ there issuch
t hat is a preorder, (2) it satis…es T I and D I , (3) is a subrelation to
[ () (4) for all , is a subrelat ion to [ () Hence, in t he set < ,
every chain admits an upper bound. According t o Zorn’s lemma, < admits at
least a maximal element. Denot e such a maximal element in <. Suppose we
can prove t he following claim:
Claim 1 For any non complete0in < and any pair of 0-incomparable
alternatives (0 0)thereexists a preorder01in < t o which 0is a subrelation
and such that 0 and0 are01¡ comparable.
Then, ifwere not complete, there would exist a preorder in < to which
is a strict subrelation. This would contradict thatis maximal in <Therefore,
if the claim holds, would be necessarily complete. would be the preorder
we are looking for.
What remains of the proof is devoted to establish claim 1. This is done t hrough the following 6 steps.
If there is no non complete preorder in < , the theorem is proved since <
is not empty. Let 0be a non complete preorder in < and 0 0 be two
0¡ incomparable elements of
Consider the binary relation on: % i¤ either %0 or t here is
We prove successively that the two clauses of the de…nition ofare exclusive
(step 1), t hat t heindi¤erencerelations areequal (step 2), that0is a subrelation
t o (step 3), that is weakly acyclic (this prepares for transit ivity) (step 4),
t hat0is a subrelation t o the transitive closure of(step 5), that the transitive
closure of satis…es T I and D I (step 6). The transitive closure of is then
t he required preorder.
St ep 1: t he t wo clauses ar e ex clusive. If there is a positive rational
such that ¡ = (0¡ 0)then are0¡ incomparable. Suppose not. For
instance suppose%0By T I ,¡ %00By D I , for all positive integer,
1
(¡ ) %0 0 Recall that it is possible to sum inequalities (see the proof of
lemma 1). We sumt imes the inequality 1
(¡ ) %00 being a positive
integer. We obtain
(¡ ) %00 Take
= It gives0¡ 0%00, what
cont radicts0 0 being incomparable. The case%0is similar.
St ep 2: equivalence of indi¤er ences. Clearly, »0 ) »
We show now that » entails»0 According t o the de…nition of ,
it is enough to prove that and are necessarily 0¡ comparable. Suppose
not. Then % implies that there is some posit ive rational such t hat
¡ = (0¡ 0)We have also % Thus, for some positive rational 0,
¡ = 0(0¡ 0) We see that this gives0(0¡ 0) = ¡ (0¡ 0) what
implies0¡ 0= 0 because 0are both positive. But that contradicts0 0
being0¡ incomparable.
St ep 3: 0is a subr elat ion t o This is a direct consequence of %0
=) % (de…nition of ) and» , »0 (step 2).
St ep 4: is weakly-acyclic. We show that for all in :%
and% ) % or non(%).
One of the four following cases is implied by% and%(1)%0
and%0(2) thereare 0such that¡ = (0¡ 0) and¡ = 0(0¡ 0)
(3)%0 and t here is0such that ¡ = 0(0¡ 0)(4) there issuch t hat
¡ = (0¡ 0) and%0Consider successively the four cases:
(1) By transitivity of 0:%0Thus,%
(2)¡ = (0¡ 0) and¡ = 0(0¡ 0) entails¡ = (+0)(0¡ 0).
Thus%
(3) Suppose we had % We would have either %0 or ¡ =
00(0¡ 0) Both possibilities cont radict %0 and ¡ = 0(0¡ 0)
Indeed, with%0 %0gives%0what contradicts¡ = 0(0¡ 0)
(step 1); whereas¡ = 0(0¡ 0) with¡ = 00(0¡ 0) implies¡ =
(0+00)(0¡ 0)what contradicts%0As a result, we have non(% )
(4) This case is similar to (3)
Rem ar k 2 Let be such that % and % . Weak acyclicity
entails that if one of the comparisons% and% is a strict preference,
t hen either the comparison on ( ) is orand are¡ incomparable.
St ep 5: 0is a subr elat ion t o t he t r ansit ive closur e of Consider
the transitive closure of de…ned by: % if there is a sequence ()= 1
such that % 1 1 % 2 and % It is clear that %0 implies
% (step 3: 0is a subrelation to). It is enough to prove t hat %
For a positive integer consider the statement : " If there is a sequence
()= 1 such that % 1 % 2 % % then non( Â0 )." Let ’s
prove by induction t hat is t rue for all positive integers. Notice that when
t he sequence () hasterms, there is+ 1 successive comparisons.
= 1 : We have% 1%By step 4, we have% or non(% ).
Both possibilit ies contradict Â0. So, we have non(Â0).
Suppose that is true and let ’s show t hat + 1 is true. Consider the
sequence of+ 2 comparisons: % 1%2%+ 1 %.
Each one of these comparisons comes either from t he clause %0 or
t he clause¡ = (0¡ 0) of the de…nition of If t here is two successive
comparisons coming from the clause%0 say%0+ 1 %0+ 2 (with
= 0 + 2 and t he convention: 0 = and + 2 = ), by t ransitivity
of 0we have: % % + 2 % which constitut es a sequence of
+ 1 comparisons. By we have non( Â0 ). If there is two successive
comparisons coming from the clause¡ = (0¡ 0)say% + 1%+ 2,
t hen¡ + 1 = (0¡ 0) and + 1¡ + 2 = 0(0¡ 0)Thus, ¡ + 2 =
(+ 0) (0¡ 0) so that % + 2 We have again reduced the number of
comparisons to+ 1 Thus, we have also non(Â0). It remains to consider
t he cases where the comparisons are alternat e. Two cases must be considered:
+ 2 even and+ 2 odd.
+ 2 even: The sequence of comparisons either begin or ends with a
com-parison from0 Suppose it begins with a comparison from 0: %0 1 %
2 %0 + 1 % Apply to 1 % 2 %0 + 1 % . It gives
non( Â0 1). Since %0 1, we cannot have Â0 If the sequence of
comparisons ends with a comparison from 0 the proof is similar. So it is
omitt ed.
+ 2 odd: If t he sequence of comparisons begins with a comparison from
0 t he proof is also similar. So it is omitted. If the sequence of comparisons
begins with a comparison from the clause¡ = (0¡ 0)we have
%1%02%0+ 1% (1)
Denote ( 1) by (1 1)(2 3) by (2 2)
¡
2(¡ 1) 2¡ 1
¢
by¡
¢
with= 1 + 12 and the convention0= and+ 2= Since comparisons
% 1 2 % 3¡ 1 % + 1 % come from the clause¡ =
(0¡ 0)we have¡ = (0¡ 0) for= 1,...,+ 32 Moreover, according
t o (1),%0+ 1 for= 1 + 12 Thus
1¡ 1(0¡ 0) % 02
2¡ 2(0¡ 0) % 03
(+ 1)2¡ (+ 1)2(0¡ 0) % 0(+ 3)2
We obtain
1+ (X+ 1)2
2
¡
(X+ 1)2
2
(0¡ 0) %0
(X+ 1)2
2
+ (+ 3)2
By T I we obt ain
1¡
(X+ 1)2
1
(0¡ 0) %0(+ 3)2
But 1= 1 and(+ 3)2= Denote= P (1+ 1)2 Thus
¡ (0¡ 0) %0
By T I , ¡ %0 (0 ¡ 0) If we had Â0 it would give 0 Â0
¡ %0 (0¡ 0) By transitivity of 0and by T I , 0 and 0 would be
0¡ comparable, which is not the case. As a result, we have non( Â0 ).
St ep 5 is proved.
Rem ar k 30is a subrelation to , but is not.
St ep 6: sat is…es T I . As 0 is translation-invariant, is clearly
t ranslation-invariant. It is easily deduced that is also t ranslation-invariant.
Likewise, it is easily seen thatsatis…es D I . Thus,is the required preorder.¥
Cor ollar y 1 Let be a re‡exive binary relat ion satisfying T I . Then there
exists a complete preorder satisfying T I , of which is a subrelation, i¤ is a
subrelation to it s transitive closure.
Pr oof: Necessity: the condition that is a subrelation to its transitive
closure is necessary and su¢ cient for t he existence of a complete preorder of
which is a subrelation (Suzumura 1976, Bossert 2008). Su¢ ciency: denote
the transitive closure of It easily seen that is a preorder satisfying
T I . Apply t heorem 1 to to deduce t hat there exists a complete preorder
sat isfying T I , of which is a subrelation. Since is a subrelation to, it is
also a subrelation to the complete preorder.¥
4- Examples of application
Exam ple 1: A translation-invariant and complete strict preorder on R with
01
Notice that only in this example, the symbols · ¸ are used for
some-t hing else some-than some-t he nasome-t ural order on R Consider the following binary relation
- on R :
- if t here is two nonnegative rationals 0such that ¡ = ¡ + 0
- is re‡exive, transitive and satis…es T I . Moreover, - is a strict preorder,
which means that - and - implies = Indeed ¡ = ¡ + 0
Thus (+ 1) = (0+ 01). We must have0+ 01 = 0 otherwisewould be
rat ional. Thus we have also+ 1 = 0. Since 1 0 01 are nonnegative, we
have= 1= 0= 10= 0 and=
Theorem 1 assert s the existence of a translat ion-invariant and complete
pre-order, say · of which - is a subrelation. · is strict like - Observe t hat
· respects the natural order of rationals. But it does not coincide wit h the
natural order of reals. Moreover it does not satisfy scalar invariance since if
you multiply 01 bythe inequality is reversed. Finally, · is not continuous.
Consider a positive sequence of rational () such that lim = 1 T I allows
t o mult iply an inequality by a positive rational. Mult iplying 0 by yields
0 = 0 for all But lim = 1 0 A question then arises: can
scalar invariancestill be transgressed under T I and continuity? An answer is
provided in section 5.
Exam ple 2: Existence of a translation-invariant, strong-Pareto,
…xed–step-anonymous and complete preorder onN0whereis a divisible commutative
group equipped with a complete preorder satisfying T I .
It is possible to demonstrat e the exist ence of such a preorder using the ultra-…lter technique, as in (Fleurbaey-Michel 2003, Lauwers 2009). We demonstrate here this existence without using ultra…lters, which are highly nonconstructive objects. Although our t heorem 1 also makes use of the axiom of choice, one may consider that our method is nevertheless more constructive in the sense that it indicates the concret e steps of adding comparisons.
Let = N0, let 0be a preorder on We …rst give the following
de…ni-t ions:
Fixed-st ep per mut at ion: (Fleurbaey-Michel 2003) is a …xed-step
per-mutation if there exist 2 N0 such that for all 2 N0 (f 1 g) =
f 1 g.
A xiom …xed-st ep-anonymi t y: Denote() t he sequence obtained by
permuting the components of 2 according to the permutation . 0is
…xed-step-anonymous if for all 2 and …xed-step permutation we have
»0()
A xiom st rong P aret o: 0isstrong Paretoif, for all 2 such t hat
82 N0% and  for some we haveÂ0 ( denote the
component of resp. ).
Pareto axioms capture the idea that an increase of the components of a vector must increase the ranking of t he vector. Anonymity axioms express a requirement of symmetry in the treatment of individuals or dates.
T he …xed-st ep cat ching-up For all 2 RN0 %
i¤ there
exist 2 N0 such that, for all 2 N0 wit h we have
X
= 1
¸
X
= 1
Pr oposit ion 1: There exists a t ranslation-invariant, strong-Pareto,
…xed-step-anonymous and complete preorder on RN0
Pr oof: Apply t heorem 1 to There exists a t ranslation-invariant and
complete preorder0on of whichis a subrelation. being a subrelation
t o0entails that 0satis…esstrong Paretoand…xed–step-anonymity. 0is the
required preorder.¥
5-
Scalar invariance
as a consequence of T I and a
continuity requirement
For a given nontrivial preorderon a divisible commutativegroup,+()
is t he associat ed upper-order-topology, i.e. the t opology generated by the base of open intervals: +() = f f2 :Ág 2 g
T heor em 2: Let be a preorder onsatisfying T I . Then there exists a
complete preorder0onsatisfying T I , of whichis a subrelation, and such
t hat +(0) ½+().
Pr oof: The following proof is an adaptation of the proof of (Ja¤ray 1975) to a t ranslation-invariant preorder. We start from a translation-invariant complete
preorder which ext ends, whose existence is guaranteed by t heorem 1. We
t hen apply a clause2 to " clean up" rankings t hat do not respect the
upper-order-topology. It turns out t hat this clause is also translation-invariant, which makes it possible to build t he desired preorder.
Let 1be a complete preorder extendingand satisfying T I . Let 2 .
Consider the following clause :
( ): " There exists2 +() containingsuch that, for all 02 +()
containingwe can …nd02 0such that for all 2 we haveÁ1 0"
Because1satis…es T I , it is easily seen that if( ) is t rue,(+ +)
is true for allinMoreover, if( ) is t rue, it is clear that we cannot have
( ) true. Thus, we can de…ne a asymmet ric relation 2 checking T I as
follows: Á2 i¤ ( ) is true.
We prove now that 2is negatively transitive, i.e.
not(Á2 ) and not(Á2 ) implies not(Á2 )
We have:
Not(Á2 ) ( ) for all12 +() containingthere exists102 +()
cont aining such that [for all 01in01, there exists001 in1 such that 001 %1
01].
Not(Á2 ) ( ) for all22 +() containingthere exists202 +()
cont aining such that [for all 02in02, there exists002 in2 such that 002 %1
02].
Let 1 be in +() containing and 10 be the int erval which existence
is asserted by the clause " not( Á2 )" . Take10 as the interval 2 of the
clause " not( Á3 )" . Thus, t here exist s202 +() cont ainingsuch t hat
[for all 02in 20, there exists002 in 10 such that 002 %1 02]. Now apply the
clause " not(Á2 )" for 200inst ead of 01 and deduce that there exists001 in
1 such that 001 %1 002By transitivity of2 100%1 002 and002 %1 02 gives
001 %1 02
Summing up: for some1in+() cont ainingwe have found202 +()
cont ainingsuch that [for all02in20thereexist s001in1such that001 %1 01].
This is exact ly the clause not(Á2 ).
Since asymmetry and negative transitivity imply transitivity, 2 is
transi-t ive.
Now let 0be the following binary relation:
. 0 i¤ [(Á2 ) or not (Â2 )]
The t ransitivity and negative t ransitivity of 2 implies the transitivity of
0Moreover,0is complete and satis…es T I .
We show now that is a subrelation to0 Indeed, let be such t hat
ÁIn the clause( ), take = f2 :ÁgWe have2 and
for all0containingwe haveÁ1 for all 2 Hence the clause( )
is true andÁ2 Consequently,Á0 If are such that » the
clause( ) cannot be satis…ed. To see it, it su¢ ces t o not ice that an interval
cont aining necessarily cont ains and vice versa. If we take0= in the
clause( ), there is no0insuch that for all2 wehaveÁ1 0Thus
we have not(Á2 ). In the same way, we have not( Á2 )Consequently,
»0
It remains to show that +(0) ½ +() Let 2 We show that any
subset in +(0), the base of open intervals generating+(0), is open with
respect to+()Let 2 = f2 :Á0g. By the de…nition of0there
is in+()containingsuch that for all2 +() containingwe can
…nd02 such that for all 2 we have Á1 0We can see that t his
implies that for all 2 we haveÁ0 Hence ½ Recap: for all
inwe foundin+() containingsuch t hat ½As a result , is a
union of open sets of+()It is thus an open set of+()¥
Rem ar k 5: Theorem 2 holds if we replace+() and +(0) respectively
by¡ () and¡ (0) the lower-order-topologies.
Rem ar k 6: The inclusion+(0) ½+() entails the upper semicont inuity
of the extension wit h respect to any topology onstronger than+(). Upper
semicont inuity is used here in t he sense that lower sections f2 :Ág
are open. But it is not necessary for the topology on t o be stronger t han
+() to have the upper semicontinuity of the extension. For more information
on this issue, see (Ja¤ray 1975), section 5.
A xiom scalar i nvar i ance: For all nonnegat ive real and vectors in
a real vect or space equipped with a preorder %=) %
Cor ollar y 2: Let be a real normed vector space. Denot ethe topology
induced by thenorm ofLetbea preorder on satisfying T I and+() ½
2, i.e. a complete preorder of which is a subrelation, sat isfying T I and such
t hat +(0) ½+(). Then0satis…esscalar invariance.
Pr oof: We have +(0) ½ . Let be a nonnegative real and two
vectors in such that %Using T I and D I we get (¡ ) %00 for any
nonnegative rational number. Let () be a nonnegat ive sequence of rat ionals
converging toThe sequence(¡ ) converges to(¡ ) On the other
hand,(¡ ) 2 + = f2 :%00g and+ is closed since+(0) ½.
Thus, the limit of the sequence ((¡ ))which is(¡ ), belongs to+.
As a result (¡ ) %00What yields, by T I ,%0¥
An immediate consequence of corollary 2 is t he following:
Cor ollar y 3: Let be a complete preorder on , a real normed vector
space, satisfying T I and+() ½ where is the topology induced by the
norm of. Thensatis…esscalar invariance.
Rem ar k 7: +() ½ is a continuity requirement. Under that cont inuity
requirement and T I ,scalar invarianceis, in a sense, satis…ed since every
com-plet e preorder ext ending the original preorder and satisfying t he same axiom of cont inuity and T I must satisfyscalar invariance.
Rem ar k 8: (Demuynck-Lauwers 2009) showed that a given preorder
sat-isfying T I and scalar invariance can be ext ended into a complete preorder
sat isfying T I and scalar invariance. Corollary 2 shows that if, in addit ion,
t he initial preorder satis…es upper semicontinuity, then it admits an extension
which also satis…es upper semicontinuity in addit ion to the axioms T I andscalar
invariance.
Rem ar k 9: WhileCorollary 3 presentsscalar invarianceas a consequenceof
T I and a condition of continuity, (Weibull 1985) theorem A has shown that
un-der conditions T I ,scalar invarianceand a cont inuity requirement calledscalar
continuity, a complet e preorder veri…es a stronger condition of continuity t hat
results in represent ability, i.e. the existence of a real-valued order-preserving
cont inuous function. For more information onscalar continuity and its
proper-t ies in proper-the conproper-texproper-t of a monoproper-tone order, see (Miproper-tra-Ozbek 2013).
6-
Scalar invariance
as a consequence of T I and a
weak Pareto axiom
Wearenow in thespace
1 =
©
(1 2 ) :2 R and sup jj¡ + 1
ª ,
whereis a nonnegat ive real. This space is suitable for st udying economic
de-cisions in discrete time, in…nite horizon and exponentially growing economy. If
= 0, the economy remains bounded.
A xiom super weak P ar et o: if inf(¡ )¡ 0 thenÂ
The following lemma is a slight strengthening of theorem 4 of (Mabrouk 2011). It will be used to prove theorem 3.
Lem m a 3: (wrong) If a complete preorder 0on
1 satis…essuper weak
Paretoand T I , then for every2
1 such that Â00 there exists a non-zero,
cont inuous, positive (in the sense that if ¸ 0 for all then() ¸ 0) linear
functional on1 such that () () ) Â0and()0
called " weak inv (+)" was used in (Mabrouk 2011)3. For the convenience of
t hereader, werecall somede…nit ions and results: ±
1 + + = f2 1 : inf¡
0g = f 2
1 %0 0g and =
©
2
1 := + 2 2
±
1 + +
ª . In
t he proof of t heorem 4 of (Mabrouk 2011), is proved to be open and convex
and t o have the following properties: (i) 02(ii) 2 whenever2 and
is a posit ive real.
Now let be in
1 such that Â0 0. The idea is to consider the
con-vex hull 0of the set and vector inst ead of t he set . We have0 =
f02
1 : 9( ) 2 [01] £ 0= + (1 ¡ )g problem: 0is not open!!...
We show that 02 0 Suppose not. There would exist in [01] wit h +
(1 ¡ )= 0 Since 0 2 and Â0 0 we have6= 0 and 6= 1 Thus we
would have
1¡ + = 0But
1¡ 2 . Thus
1¡ Â0 0. SinceÂ0 0
by T I we would have
1¡ + Â0 0 A contradiction. Since 02 0, thanks
t o Hahn–Banach t heorem, there exist a non-zero continuous linear functional
supporting0. This is written: for all 0in0, (0) 0 In particular, ()0One shows, literally as in t he proof of theorem 4 of (Mabrouk 2011),
t hat for all in
1 ,() () ) Â0and that is positive.¥
T heor em 3: Let be a preorder on
1 satisfying T I and super weak
Pareto. Let 0be one of the complete preorders which existence is assert ed by
t heorem 1, i.e. a complete preorder of which is a subrelat ion and satisfying
T I . Then0satis…esscalar invariance.
Pr oof: Sinceisa subrelation to0,0also sat is…essuper weak Pareto. Let
2
1 such thatÂ0. Denote= ¡ We haveÂ00. Apply lemma
3. There exists a non-zero, continuous, positive linear functionalon1 such
t hat 80 0 2 1 (0) (0) ) 0Â0 0and () 0Let be a
positive real. Multiplying this inequality byone gets() = ()0
Replaceby ¡ . Then() = ((¡ )) = (¡ ) = () ¡
()0Hence,() () andÂ0We have shown that for
all positive real and 2
1 Â0 ) Â0 Moreover, »0
implies»0 (since if we had for exampleÂ0 we could multiply
t his last inequality by 1
and get Â0a contradiction). This provesscalar
invariance.¥
Theorem 3 indicat es t hat scalar invariance is satis…ed under T I andsuper
weak Paretoin thesamesenseas in remark 7. T I together withscalar invariance
is called strong invariance in the terminology of (Mitra-Ozbek 2013)4. If we
accept this justi…cation of scalar invarianceby T I , we are led to admit that,
undersuper weak Pareto, the axiom ,strong invarianceis in a way a consequence
of the axiom T I .
An immediate consequence of theorem 3 is the following:
Cor ollar y 4: Every complete preorder 0sat isfying T I and super weak
Pareto, satis…esscalar invariance.
3T he de…nit ion of " weak inv (+ )" is: 8 2 [Â
) +%+ ]Of
course, lemma 2 holds wit h weak inv (+ ) inst ead of T I .
4I n t he t erminology of (D’A spremont -Gevers 2002), it is calledi nvar i ance wi th respect to
Rem ar k 10: Since theorem 1 and lemma 3 hold in …nite dimension, it is also the case for theorem 3 and corollary 4. Consequently, when the preorder is
complete and super-weak Paret o,strong invarianceis equivalent to T I . Hence,
t heorem 18 of (D’Aspremont -Gevers 2002) or example 2 of (Mitra-Ozbek 2013) asserting the linear representability of a complete preorder respecting T I ,scalar invariance, weak Paretoand another axiom, hold without imposingscalar in-variance.
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