A generic result in vector optimization

(1)

ALEXANDER J. ZASLAVSKI

Received 17 November 2005; Revised 19 March 2006; Accepted 24 March 2006

We study a class of vector minimization problems on a complete metric space such that all its bounded closed subsets are compact. We show that for most (in the sense of Baire category) problems in the class the sets of minimal values are infinite.

Copyright © 2006 Alexander J. Zaslavski. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

1. Introduction

The study of vector optimization problems has recently been a rapidly growing area of research. See, for example, [1–5] and the references mentioned therein. In this paper, we study a class of vector minimization problems on a complete metric space such that all its bounded closed subsets are compact. This class of problems is associated with a complete metric space of continuous vector functionsᏭdefined below. For eachF from Ꮽ, we denote byv(F) the set of all minimal elements of the imageF(X)= {F(x) :x∈X}. In this paper, we will study the setsv(F) withF∈Ꮽ. It is clear that for a minimization problem with only one criterion the set of minimal values is a singleton. In the present paper, we will show that for mostF∈Ꮽ(in the sense of Baire category) the setsv(F) are infinite. Such approach is often used in many situations when a certain property is studied for the whole space rather than for a single element of the space. See, for example, [7,8] and the references mentioned there. Our results show that in general the setsv(F),F∈Ꮽ, are rather complicated. Note that in our paper as in many other works on optimization theory [1–6] inequalities are of great use.

In this paper, we use the convention that∞/∞ =1 and denote by Card(E) the cardi-nality of the setE.

LetRbe the set of real numbers and letnbe a natural number. Consider the finite-dimensional spaceRn_{with the Chebyshev norm}

x =x1,. . .,xn=maxxi:i=1,. . .,n

, x=x1,. . .,xn

∈Rn_. _(1.1)

Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 54027, Pages1–14

(2)

Let{e1,. . .,en}be the standard basis inRn:

e1=(1, 0,. . ., 0),. . .,en=(0,. . ., 0, 1). (1.2)

Letx=(x1,. . .,xn),y=(y1,. . .,yn)∈Rn. We equip the spaceRnwith the natural order

and say that

x≥y ifxi≥yi∀i∈ {1,. . .,n},

x > y ifx≥y,x=y,

xy ifxi> yi∀i∈ {1,. . .,n}.

(1.3)

We say thatx y(resp.,x < y,x≤y) ifyx(resp.,y > x,y≥x).

Let (X,ρ) be a complete metric space such that each of its bounded closed subsets is compact. Fixθ∈X.

Denote byᏭthe set of all continuous mappingsF=(f1,. . .,fn) :X→Rnsuch that for

alli∈ {1,. . .,n},

lim

ρ(x,θ)→∞fi(x)= ∞. (1.4)

For eachF=(f1,. . .,fn),G=(g1,. . .,gn)∈Ꮽ, set

d(F,G)=supfi(x)−gi(x):x∈X,i=1,. . .,n,

d(F,G)=d(F,G)1 +d(F,G)−1.

(1.5)

Clearly, the metric space (Ꮽ,d) is complete.

Note thatd(F,G)=sup{F(x)−G(x):x∈X}for allF,G∈Ꮽ.

LetA⊂Rn_{be a nonempty set. An element}_x_∈_A_{is called a minimal element of}_A_if

there is noy∈Afor whichy < x.

Let F∈Ꮽ. A pointx∈X is called a point of minimum of F ifF(x) is a minimal element ofF(X). Ifx∈Xis a point of minimum ofF, thenF(x) is called a minimal value ofF. Denote byM(F) the set of all points of minimum ofFand putv(F)=F(M(F)).

The following proposition will be proved inSection 2.

Proposition1.1. LetF=(f1,. . .,fn)∈Ꮽ. ThenM(F)is a nonempty bounded subset of

(X,ρ)and for eachz∈F(X)there isy∈v(F)such thaty≤z.

In the sequel we assume thatn≥2 and that the space (X,ρ) has no isolated points. The following theorem is our main result. It will be proved inSection 4.

Theorem1.2. There exists a setᏲ⊂Ꮽwhich is a countable intersection of open everywhere dense subsets ofᏭsuch that for eachF∈Ᏺthe setv(F)is infinite.

(3)

2. Proof ofProposition 1.1

Letz∈F(X). We show that there isy∈v(F) such thaty≤z. Set

Ω0=h∈F(X) :h≤z. (2.1)

We consider the setΩ0with the natural order and show thatΩ0has a minimal element by using Zorn’s lemma.

Assume thatDis a nonempty subset ofΩ0such that for eachh1,h2∈Deitherh1≥h2 orh1≤h2. Since all bounded closed subsets ofXare compact, it follows from (1.4) that the setF(X) is bounded from below. Together with (2.1) this implies that the setD is bounded. For each integeri∈ {1,. . .,n}, set

¯

hi=infλ∈R: there isx=x1,. . .,xn∈Dfor whichxi=λ (2.2)

and set

¯

h=_h¯_1,_{. . .}_{, ¯}_h_n_. _(2.3)

Clearly, the vector ¯his well defined.

Let pbe a natural number. By (2.2) and (2.3) for each natural number j∈ {1,. . .,n} there exists

z(p,j)₌ _z(p,j) 1 ,. . .,z

(p,j)

n

∈D (2.4)

such that

¯

hj≥z(jp,j)−

1

p. (2.5)

It is clear that there is

z(p)_∈_z(p,j)_:_j₌_1,_{. . .}_,_n _(2.6)

such that

z(p)≤z(p,j) ∀j=1,. . .,n. (2.7)

It follows from (2.5), (2.7), (2.2), (2.6), and (2.4) that for eachj=1,. . .,n,

¯ hj≤z

(p)

j ≤h¯j+ 1_p. (2.8)

By (2.6), (2.4), and (2.1) for each integerp≥1, there isxp∈Xsuch that

Fxp

=z(p). (2.9)

If the sequence {xp}∞p=1 is unbounded, then in view of (2.9) and (1.4) the sequence {z(p)_}∞

(4)

is bounded. Since any bounded closed set in (X,ρ) is compact, there is a subsequence {xpi}∞i=1of the sequence{xp}∞p=1which converges to some point ¯x∈X. In view of (2.8) and (2.9)

F( ¯x)=lim

i→∞F

xpi

=lim

i→∞z

(pi)=_h¯ _(2.10)

and ¯h∈F(X). Together with (2.1) and (2.2) this implies that ¯h∈Ω0. Definition (2.2) implies that ¯h≤hfor allh∈D. By Zorn’s lemma there is a minimal elementy∈F(X) such thaty≤z. This completes the proof ofProposition 1.1.

3. Auxiliary results

Proposition3.1. LetF=(f1,. . .,fn)∈Ꮽand letCard(v(F))=p, where pis a natural number. Then there is a neighborhoodWofFin(Ꮽ,d)such thatCard(v(G))≥pfor each G=(g1,. . .,gn)∈W.

Proof. Let

y1,. . .,yp∈v(F), (3.1)

yi=yj for each (i,j)∈Ω:= {1,. . .,p} × {1,. . .,p} \

(i,i) :i=1,. . .,p. (3.2)

For eachi∈ {1,. . .,p}, there isxi∈Xsuch that

Fxi=yi. (3.3)

By (3.2) and (3.3) for each (i,j)∈Ω, there isp(i,j)∈ {1,. . .,n}such that

fp(i,j)

xi

> fp(i,j)

xj

. (3.4)

Choose>0 such that

fp(i,j)

xi

> fp(i,j)

xj

+ 4 (3.5)

for all (i,j)∈Ω. Set

W=G∈Ꮽ:d(G,F)≤. (3.6)

Let

G=g1,. . .,gn∈W. (3.7)

For eachi∈ {1,. . .,p}, we haveG(xi)∈G(X) and it follows fromProposition 1.1 that

there is

¯

yi∈v(G) (3.8)

such that

¯ yi≤G

xi

(5)

Leti∈ {1,. . .,p}. By (3.8) there is ¯xi∈Xsuch that

Gx¯i=y¯i. (3.10)

In view of (3.7) and (3.6)

_G ¯

xi−Fx¯i≤. (3.11)

It follows from (3.1), (3.2), the equality Card(v(F))=p, andProposition 1.1that there is k(i)∈ {1,. . .,p}such that

Fx¯i

≥yk(i). (3.12)

By (3.3), (3.12), (3.11), (3.10), and (3.9)

Fxk(i)

=yk(i)≤F

¯ xi

≤Gx¯i

+(1, 1,. . ., 1)

≤Gxi

+(1, 1,. . ., 1)≤Fxi

+ 2(1, 1,. . ., 1). (3.13)

Together with (3.5) this implies thatk(i)=i. Combined with (3.13), (3.10), and (3.3) this equality implies that

yi≤y¯i+(1, 1,. . ., 1)≤yi+ 2(1, 1,. . ., 1). (3.14)

It follows from this inequality, (3.5), and (3.3) that

¯

yi=y¯j ifi,j∈ {1,. . .,p}satisfyi=j. (3.15)

This completes the proof ofProposition 3.1.

Proposition3.2. Assume thatF=(f1,. . .,fn)∈Ꮽ,pis a natural number,Card(v(F))= pand that

v(F)=y1,. . .,yp

, xi∈X, F

xi

=yi, i=1,. . .,p,

yi=yj ∀i,j∈ {1,. . .,p}satisfyingi=j.

(3.16)

Then for eachi=1,. . .,pthe inequalityF(xi)≤F(x)holds for allxbelonging to a neighbor-hood ofxi.

Proof. It is suﬃcient to consider the case withi=1. Clearly, for eachj∈ {2,. . .,n}, there iss(j)∈ {1,. . .,n}such thatfs(j)(x1)< fs(j)(xj). Choose>0 such that

fs(j)

x1

< fs(j)

xj

−2 ∀j∈ {2,. . .,n}. (3.17)

There isδ >0 such that for eachx∈Xsatisfyingρ(x,x1)≤δwe have

_F₍_x₎₋_F

x1≤

(6)

Letx∈Xsatisfyρ(x,x1)≤δ. Then (3.18) is true. ByProposition 1.1there existsy∈v(F) such that

y≤F(x). (3.19)

In order to complete the proof it is suﬃcient to show thaty=F(x1).

Let us assume the converse. Then there is j∈ {2,. . .,n}such that y=yj=F(xj). By

this relation, (3.18), and (3.19)

Fxj

=yj=y≤F(x)≤F

x1

+

2

(1, 1,. . ., 1),

fs(j)xj≤fs(j)x1+ 2.

(3.20)

This contradicts (3.17). The contradiction we have reached provesProposition 3.2.

Proposition3.3. Assume thatF=(f1,. . .,fn)∈Ꮽ,>0,pis a natural number and that

Cardv(F)=p, x1,. . .,xp∈X, yi=F

xi

, i=1,. . .,p,

v(F)=yi:i=1,. . .,p

. (3.21)

Then there existsG=(g1,. . .,gn)∈Ꮽsuch that fi(x)≤gi(x), x∈X,i=1,. . .,n, gi

xj

=fi

xj

, i=1,. . .,n, j=1,. . .,p, (3.22)

d(F,G)≤, (3.23)

v(G)=Gxj:j=1,. . .,p (3.24)

and that for eachx∈X\ {x1,. . .,xp}there is j∈ {1,. . .,p}for which

G(x)≥Gxj

+min1,ρx,xi

:i=1,. . .,p(1, 1,. . ., 1). (3.25)

Proof. For eachx∈Xandi=1,. . .,n, set gi(x)=fi(x) +min

1,ρx,xj

:j=1,. . .,p (3.26)

and setG=(g1,. . .,gn). Clearly,G∈Ꮽ,

gi(x)≥fi(x), x∈X,i=1,. . .,n,

gi

xj

= fi

xj

for eachi∈ {1,. . .,n}and eachj∈ {1,. . .,p} (3.27)

andd(F,G)≤. Therefore (3.22) and (3.23) hold.

Let j∈ {1,. . .,p}. We will show thatG(xj)∈v(G). Assume thatx∈Xand

G(x)≤Gxj

(7)

By (3.22), (3.26), and (3.28)

F(x)≤F(x) +min1,ρx,xi:i=1,. . .,p(1, 1,. . ., 1)=G(x)≤Gxj=Fxj.

(3.29)

Together with (3.21) this relation implies that

F(x)=Fxj

, x∈xi:i=1,. . .,p

,x=xj. (3.30)

Thus

Gxj:j=1,. . .,p⊂v(G). (3.31)

Assume that

x∈X\x1,. . .,xp. (3.32)

ByProposition 1.1and (3.21) there is j∈ {1,. . .,p}such that

Fxj

≤F(x). (3.33)

Relations (3.22), (3.33), (3.26), and (3.32) imply that

Gxj

=Fxj

≤F(x)< F(x) +min1,ρx,xi

:i=1,. . .,p(1,. . ., 1)≤G(x). (3.34)

This relation implies that

G(x)> Gxj

+min1,ρx,xi

:i=1,. . .,p(1, 1,. . ., 1) (3.35)

andG(x)∈v(G). Together with (3.31) this relation implies (3.24). This completes the

proof ofProposition 3.3.

Proposition3.4. Assume thatF=(f1,. . .,fn)∈Ꮽ,pis a natural number,

Cardv(F)=p (3.36)

and that>0. Then there existsG∈Ꮽsuch thatd(F,G)≤andCard(v(G))=p+ 1. Proof. Let

v(F)=y1,. . .,yp

, (3.37)

wherey1,. . .,yp∈Rn. Clearly,

yi=yj for eachi,j∈ {1,. . .,p}such thati=j. (3.38)

For eachi∈ {1,. . .,p}, there isxi∈Xsuch that

Fxi

(8)

ByProposition 3.3there existsF(1)₌₍_f(1) 1 ,. . .,f

(n)

1 )∈Ꮽsuch that

f_i(1)(x)≥fi(x) ∀x∈X,i=1,. . .,n, (3.40)

fi(1)

xj

= fi

xj

, i=1,. . .,n, j=1,. . .,p, (3.41)

dF,F(1)_≤

4, (3.42)

vF(1)=F(1)xj

:j=1,. . .,p (3.43)

and that for eachx∈X\ {x1,. . .,xp}there isj∈ {1,. . .,p}such that

F(1)₍_x₎_≥_F(1)_x

j+min1,ρx,xi:i=1,. . .,p(1, 1,. . ., 1). (3.44)

It is clear that there exists a positive number

0<min{1,}

8 (3.45)

and that for eachi,j∈ {1,. . .,p}satisfyingi=jthere existss(i,j)∈ {1,. . .,n}such that

f_s(1)(i,j)

xi

< f_s(1)(i,j)

xj

−80. (3.46)

Chooseδ0∈(0, 1/8) such that

ρxi,xj

≥8δ0 for eachi,j∈ {1,. . .,p}satisfyingi=j. (3.47)

There isδ1∈(0,δ0/2) such that for eachx∈Xsatisfyingρ(x1,x)≤2δ1

_F(1)_x

1−F(1)(x)≤0

8. (3.48)

Put

1=0

4. (3.49)

There isx0∈Xsuch that

0< ρx0,x1

< δ1. (3.50)

By (3.50) and (3.47)

x0∈

xi:i=1,. . .,p

. (3.51)

Choose a positive number

2<min

0ρ

x0,x1

4 ,1

(9)

Choose a positive numberδ2such that

4δ2< ρ

x0,x1

, (3.53)

fi(1)

x0

−fi(1)(x)≤₄2 for eachi∈{1,. . .,n}and eachx∈Xsatisfyingρ

x,x0

≤4δ2.

(3.54)

Choose a positive numberλsuch that

λδ2>21+ 22. (3.55)

Set

g1(x)=f1(1)(x) for eachx∈Xsatisfyingρ

x,x0

>2δ2, (3.56)

g1(x)=minf1(1)(x),f (1) 1

x0−1+λρx,x0 for eachx∈Xsatisfyingρx,x0≤2δ2. (3.57)

Fori∈ {2,. . .,n}, set

gi(x)=fi(1)(x) for eachx∈Xsatisfyingρ

x,x0

>2δ2, (3.58)

gi(x)=min

f_i(1)(x),f_i(1)x0

−2+λρ

x,x0

for eachx∈Xsatisfyingρx,x0

≤2δ2. (3.59)

SetG=(g1,. . .,gn). By (3.54) and (3.55) for eachi∈ {1,. . .,n}and eachx∈Xsatisfying δ2≤ρ(x,x0)≤2δ2,

f_i(1)x0

−1+λρ

x,x0

≥f1

i

x0

−1+λδ2

≥fi1

x0

+1+ 22≥fi(1)(x)−2/4 +1+ 22.

(3.60)

In view of (3.60), (3.57), and (3.59) for eachi∈ {1,. . .,n}and eachx∈Xsatisfyingδ2≤ ρ(x,x0)≤2δ2,

gi(x)=fi(1)(x). (3.61)

Together with (3.56)–(3.59) this implies thatGis continuous. By (3.56) and (3.58)G∈Ꮽ. Relations (3.61), (3.56), and (3.58) imply that for eachx∈Xsatisfyingρ(x0,x)≥δ2we have

F(1)₍_x₎₌_G₍_x₎_. _(3.62)

By (3.57) and (3.59) for eachx∈Xsatisfyingρ(x0,x)≤2δ2, we have

G(x)≤F1₍_x₎_. _(3.63)

Letx∈Xsatisfy

ρx,x0

(10)

By (3.56)–(3.59), (3.52), (3.64), and (3.54) for eachi∈ {1,. . .,n},

f_i(1)(x)≥gi(x)≥min

f_i(1)(x),f_i(1)x0

−1

≥minf_i(1)(x),f_i(1)(x)−2 4 −1

≥f_i(1)(x)− ₅

4

1,

F(1)₍_x₎_≥_G₍_x₎_≥_F(1)₍_x₎₋

2

(1, 1,. . ., 1).

(3.65)

Together with (3.62) this inequality implies thatd(F(1)_,_G₎_≤_/_{2. Combined with (3.42)} this implies that

d(F,G)≤dF,F(1)₊_d_F(1)_,_G_< 4+

2. (3.66)

Letx∈X. We show that there existsj∈ {0,. . .,p}such thatG(x)≥G(xj). There are two

cases:

ρx,x0

≥δ2, (3.67)

ρx,x0

< δ2. (3.68)

Assume that (3.67) holds. Then by (3.62)G(x)=F(1)₍_x_{). In view of}_{Proposition 1.1}_and (3.43) there is j∈ {1,. . .,p}such that

F(1)xj

≤F(1)(x)=G(x). (3.69)

Ifj=1, then (3.53) implies that

ρxj,x0

=ρx0,x1

>4δ2. (3.70)

Ifj=1, then by (3.47) and (3.50)

ρxj,x0≥ρxj,x1−ρx1,x0≥8δ0−δ1≥7δ0>4δ2. (3.71)

Thus in both casesρ(xj,x0)>4δ2. In view of this inequality and (3.62),

F(1)_x

j

=Gxj

. (3.72)

Together with (3.69) this equality implies thatG(xj)≤G(x). Assume that (3.68) holds.

We will show thatG(x0)≤G(x). Relations (3.57) and (3.59) imply that

Gx0

= f1(1)

x0

−1,f2(1)

x0

−2,. . .,fn(1)

x0

−2

=F(1)x0

−1,2,. . .,2

. (3.73)

By (3.68)

F(1)(x)≥F(1)x0

− 2 4

(11)

By (3.68), (3.57), (3.59), (3.74), and (3.52)

g1(x)≥minf1(1)

x0

−2 4,f

(1) 1

x0

−1

= f1(1)

x0

−1 (3.75)

and fori∈ {1,. . .,p} \ {1},

gi(x)≥min

f_i(1)x0

−2

4,f (1)

i

x0

−2

= f_i(1)x0

−2. (3.76)

Together with (3.73) these inequalities imply thatG(x)≥G(x0). Thus we have shown that for eachx∈Xthere isj∈ {0,. . .,p}such thatG(x)≥G(xj).

Now assume that j1,j2∈ {0,. . .,p}satisfy

Gxj1

≤Gxj2

. (3.77)

We will show that j1=j2. In view of (3.47) and (3.50) for eachi∈ {2,. . .,p},

ρxi,x0

≥ρxi,x1

−ρx0,x1

≥8δ0−δ1>7δ0>4δ2. (3.78)

By (3.53)ρ(x0,x1)>4δ2. Therefore, for eachi∈ {1,. . .,p},

ρxi,x0

>4δ2. (3.79)

Together with (3.56) and (3.58)

Gxi

=F(1)xi

, i=1,. . .,p. (3.80)

If j1,j2∈ {1,. . .,p}, then in view of (3.77), (3.80), and (3.46)F(1)(xj1)=F(1)(xj2) and j1=j2. Therefore we may consider only the case with 0∈ {j1,j2}. Leti∈ {1,. . .,p} \ {1}. By (3.80)

Gxi

=F(1)_x

i

. (3.81)

By (3.46)

f_s(1)(i,1)(xi)< fs(1)(i,1)

x1

−80, f_s(1)(i,1)

x1

< f_s(1)(1,i)

xi

−80. (3.82)

It follows from (3.81), (3.82), (3.48), (3.50), (3.49), (3.57), and (3.59) that

gs(i,1)xi= fs(1)(i,1)

xi< fs(1)(i,1)

x1−80

≤ f_s(1)₍_i_,1)(x0) +0

4 −80< f (1)

s(i,1)

x0

−21≤gs(i,1)

x0

−1,

gs(i,1)

xi

= f_s(1)(i,1)

xi

> f_s(1)(1,i)

x1

+ 80≥f_s(1)(i,1)

x0

−0

4 + 80> gs(1,i)

x0

+ 70. (3.83)

These inequalities imply that the inequalityG(xi)≤G(x0) does not hold and that the

inequalityG(x0)≤G(xi) does not hold too. Together with (3.77) and the inclusion 0∈ {j1,j2}this implies that

j1,j2

(12)

By (3.80)

Gx1

=F(1)x1

, gs

x1

=fs1

x1

, s=1,. . .,n. (3.85)

Relations (3.79) and (3.44) imply that there isq∈ {1,. . .,p}such that

F(1)_x 0

≥F(1)_x

q

+min1,ρx0,xi

:i=1,. . .,p(1, 1,. . ., 1). (3.86)

It follows from (3.50), (3.48), and (3.86) that

F(1)xq≤F(1)x0≤F(1)x1+0/8(1,. . ., 1). (3.87)

Together with (3.46) this implies thatq=1. Combined with (3.86) this equality implies that

F(1)_x 0

≥F(1)_x 1

+min1,ρx0,xi:i=1,. . .,p(1, 1,. . ., 1). (3.88)

By (3.47), (3.50), (3.88), and (3.85) fori∈ {1,. . .,p} \ {1},

ρx0,xi

≥ρx1,xi

−ρx1,x0

≥8δ0−δ1≥7δ0,

min1,ρx0,xi:i=1,. . .,p=ρx0,x1,

(3.89)

F(1)_x 0

≥F(1)_x 1

+ρx0,x1

(1, 1,. . ., 1)=G(1)_x 1

+ρx0,x1

(1, 1,. . ., 1). (3.90)

Relations (3.90), (3.59), and (3.52) imply that fori∈ {1,. . .,p} \ {1}

g_i(1)x1

+ρx0,x1

≤f_i(1)x0

=gi

x0

+2,

g_i(1)x1

≤gix0

+2−ρ

x0,x1

≤gix0

−2.

(3.91)

Relations (3.90) and (3.57) imply that

g1

x1

+ρx0,x1

≤f1(1)

x0

=g1

x0

+1. (3.92)

By this relation, (3.50), (3.48), (3.49), and (3.85)

g1

x0

= f1(1)

x0

−1≤f1(1)

x1

+0

8 −1=f (1) 1

x1

−0

8 =g1

x1

−0

8. (3.93) Then each of the inequalitiesG(x0)≤G(x1),G(x1)≤G(x0) does not hold. Together with (3.84), the inclusion 0∈ {j1,j2}and (3.77) this implies that j1=j2=0. Thus we have shown that ifj1,j2∈{0,. . .,p}and ifG(xj1)≤G(xj2), thenj1=j2. Therefore Card(v(G))=

p+ 1.Proposition 3.4is proved.

4. Proof ofTheorem 1.2

Lemma 4.1. Let F∈Ꮽ, p≥1 be an integer and let >0. Then there exists an open nonempty set

ᐁ⊂H∈Ꮽ:d(F,H)≤ (4.1)

(13)

Proof. If for eachG∈Ꮽsatisfyingd(F,G)<we have Card(v(G))≥p+ 1, then put

ᐁ=H∈Ꮽ:d(F,H)<. (4.2)

Assume that there isG0∈Ꮽsuch that

dF,G0

<, CardvG0

≤p. (4.3)

ByProposition 3.4there existsG1∈Ꮽsuch that

dF,G1

<, CardvG1

=p+ 1. (4.4)

ByProposition 3.1there exists an open neighborhoodᐁofG1inᏭsuch that

ᐁ⊂H∈Ꮽ:d(H,G)< (4.5)

and that for eachG∈ᐁthe inequality Card(v(G))≥p+ 1 holds.Lemma 4.1is proved.

Proof ofTheorem 1.2. Letpbe a natural number. ByLemma 4.1for eachF∈Ꮽand each integeri≥1 there exists an open nonempty set

ᐁ(F,i,p)⊂H∈Ꮽ:d(F,H)≤(2i)−1 (4.6)

such that for eachH∈ᐁ(F,i,p) the inequality Card(v(H))≥pholds. Define

Ᏺ= ∩∞

p=1

ᐁ(F,i,p) :F∈Ꮽ,i≥1 is an integer. (4.7)

It is clear thatᏲis a countable intersection of open everywhere dense subsets ofᏭand that for eachG∈Ᏺthe setv(G) is infinite.Theorem 1.2is proved.

References

[1] G.-Y. Chen, X. Huang, and X. Yang,Vector Optimization, Lecture Notes in Economics and Math-ematical Systems, vol. 541, Springer, Berlin, 2005.

[2] J. P. Dauer and R. J. Gallagher,Positive proper eﬃcient points and related cone results in vector optimization theory, SIAM Journal on Control and Optimization28(1990), no. 1, 158–172. [3] M. Ehrgott and X. Gandibleux (eds.),Multiple Criteria Optimization: State of the Art

Anno-tated Bibliographic Surveys, International Series in Operations Research & Management Science, vol. 52, Kluwer Academic, Massachusetts, 2002.

[4] J. Jahn,Vector Optimization. Theory, Applications, and Extensions, Springer, Berlin, 2004. [5] T. Tanino,Stability and sensitivity analysis in convex vector optimization, SIAM Journal on

Con-trol and Optimization26(1988), no. 3, 521–536.

(14)

[7] A. J. Zaslavski,Generic existence of solutions of nonconvex optimal control problems, Abstract and Applied Analysis2005(2005), no. 4, 375–421.

[8] ,Turnpike Properties in the Calculus of Variations and Optimal Control, Nonconvex Op-timization and Its Applications, vol. 80, Springer, New York, 2006.

Alexander J. Zaslavski: Department of Mathematics, The Technion-Israel Institute of Technology, 32000 Haifa, Israel

10.1155/JIA/2006/54027