Order extension of order monomorphisms on a preordered topological space

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VOL. 16 NO. 4 (1993) 663-668

ORDER

EXTENSION OF ORDER

MONOMORPHISMS ON A PREORDERED

TOPOLOGICAL SPACE

GHANSHYAMMEHTA

Department

ofEconomics

The University ofQueensland Qld. 4072,Australia

(Received

April 29,

1992)

ABSTRACT. This paper proves Nachbin-type extensiontheorems for infinitelymany functions on atopologicalspace equippedwithapreorder.

KEYWORDS AND PHRASES. Orderextension, order monomorphism. 1991AMS SUBJECT

CLASSIFICATION

CODES. 54F05,90A10.

1.

INTRODUCTION.

The well-knownextensiontheorem of Nachbin

[i,

p.

36]

gves

conditionsunder whicha real-valued continuous order-homomorphism defined on a closed subset of a normally preordered

space E can be extendedto a real-valued continuous order homomorphismon the whole space.

In

the particular case in which the preorder on the space is the discrete order, the Nachbin

extensiontheorem reduces to the Urysohn-Tietzeextensiontheoremonnormalspaces.

The objective of this paper is to study the following variant of the extension problem considered by Nachbin.

Suppose

that instead of a single real-valued continuous order homomorphism, one has a collection F=

{fi)iEl

of real-valued continuous order monomorphisms, where

fi

is defined on a closed subset

o

ofa preordered topological space E.

Then theproblemis tofinda real-vluedcontinuousorder-monomorphism

y

onthe wholespace

/ that is an order-extension of each order monomorphism

y

inthe collection F.

In

this paper,

wegivesufficient conditionsfor theexistenceof sucha’universal’ order-extension.

A

special caseof the problemstudied in thispaper arises inmathematicaleconomics inthe context of the Euclidean distance approach used by

Arrow

and

"I-Iahn

[2]

to construct a continuous order monomorphism on a convex subset X of Rn _{equipped with} _a _total _preorder.

The Arrow-Hahn method

[2,

pp.

82-87]

consists of taking a point x₀ and then by defining the ’utility’ ofz to be the Euclidean distancefrom _z0 to the upper section ofx. This Arrow-Hahn

function can be shown to be continuous under certainconditions.

In

this way, one is given a collection of continuous order monomorphisms {yz:x D} defined on the upper sections of the points x in D. Now the problem is to

’construct’

a continuous order monomorphism

I

on the wholespace

x.

It should be observed that the problemisnot merelytoshow theexistenceofa continuousorder monomorphism since, inthefinite-dimensionalArrow-Hahn context,this canbe deduced concisely fromgeneral topologicaltheorems suchasEilenberg’stheorem

([3]

orCorollary 1 of

[4]).

The problem is to somehow

’construct’

the function _f as an order-extension of the

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2. PRELIMINARIES.

Let E be a set.

A

preorder < on E is areflexive and transitive binary relation on E.

A

preordered set is a set equipped with a preorder.

A

preorder _< on E gives rise to two other

binary relationson E asfollows. Ifz,y gthenz _Iifand onlyift_< and _I_<z. The relation

is anequivalencerelation. Ifz,y Ethen < if andonlyif _<yand noty<z. The relation

< is irreflexiveandtransitive.

A

preorder < is said tobeanorder ifit isantisymmetric.

A

preorder _< issaid to betotal if for every two elementsz, geither _<yory_<z.

A

subset X of a preordered set g _is said to be decreasingif b X, a_<b imply that a X.

Each subset Xof Edetermines, in aunique manner, adecreasing set d(X) whichis thesmallest

oneamong thedecreasingsetscontainingX. Dually,onedefinesthe conceptsofanincreaaingset and thesmallest increasingset i(x)containingagiven subset

x.

Let

Ebeatotally preorderedset. Considerthe followingcollectionofsubsetsofE:

(1)

(a,b)

(2) (a,

b0]

{z E:a<z<b) ifb₀isthelast element ofE;

(3)

[a0,b)

{z E:a

o

<z< b} if_a0isthe first element ofE.

Thenthis collectionofsubsets ofE isabasisforatopologyonEcalled <-ordertopology.

Let

E bea topological spaceequippedwitha preorder _<. Each subset

x

ofE determines uniquely a closed decreasing set D(X) which is the smallest among the closed decreasing sets containing

x.

Analogously, we define the concept of the smallest closed increasing set l(X) containing a given subset X. If A and B are two subsets of g wewrite A<B toindicate that D(A)nI(B)-.

Let

E be a topological spaceequipped with apreorder _<. The preorder _< is said to be continuous ifd(y) (z E:z_<y}, the lower sectionofy, andi(y) {z E:y_<z}, theuppersection

of y, areclosed in Efor every y E. Thepreorderis saidtobe

strongly

continuousif d(X) and

i(X) areclosed inEfor every closed subset X ofE.

A

topological space E equipped with a preorder _< is said to be nominallypreordered if for every two disjoint closed subsets F₀ and F of E, F₀ being decreasingand F increasing, there

exist two disjoint open subsets A₀ and A such that A₀ contains F₀ and is decresing and A contains F andisincreasing.

Let

E and E₂ be two preordered sets.

A

function ] on E to

E2.

is said to be an order homamorphism

(or

isotone)

ifz,1 E andz_< imply thaty(z)_<l(v).

A

function

y

onE to E₂

is said to be an order monomorphism if it is anorder homomorphism and z_{< u implies} that f(=)<f().

Let

E,H be two preordered sets and F a subset of E.

Let

:F--,// be an order monomorphism. Thenanordermonomorphism f:E-ltissaid tobeanorder extension of0ifit

satisfiesthefollowingconditions:

(i)

0(a)<0(b) anda,b Fimplythat f(a)<_f(b)

(ii)

(a)<g(b)anda,b Fimply that f(a)<l(b).

3.

ORDER EXTENSION

OF ORDER

MONOMORPHISMS.

The following fundamental theoremon theextension ofan order homomorphism is due to Nachbin

[1,

p.

36].

(Nachbin’s

Extension

Theorem)

LetEbeanormallypreordered spaceF C_ Ea closedsubset and

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where is areal number. Thenin order that thefunction/"maybeextended to Einsuchaway

as tobecome abounded, continuousand isotone real functionon E it is necessary and sufficient that <r"implies A(r)<B(r’).

We now use Nachbin’s extension theorem to prove the following theorem on extensions of order monomorphisms.

THEOREM 1. Let E be a T topological spaceequippedwith a preorder < such that the followingconditionsaresatisfied:

(i)

Eisnormallypreordered;

(ii)

The preorder < isstronglycontinuous;

(iii)

ForeachzeE, theset{aeE:a<x} is open inE;

(iv)

F {Dn}

n=

is a countable family of closed increasing subsets of E such that for each n 1,2,. thereexistsareal-valuedcontinuousorder-monomorphism

In

definedon

Dn;

(v)

There exists acountable and topologicallydense subset Z

{Zn}

n=

of/ withtheproperty that _neDnfor alln.

Then there exists a real-valued continuous order monomorphism /’ on E that is an order

extensionof

fn

for each positive integern.

PROOF.

By

hypothesis, for each n 1,2,... there exists a real-valued continuous order

.monomorphism

In

on Dn. Since the extended realline isorderhomeomorphicto[0,1],wemay assumewithoutloss ofgeneralitythat for each positive integern,

fn:

Dn--.[0,1].

We prove next that for each n 1,2,. the conditions ofNachbin’s extension theorem

[9,

p.36]

aresatisfied for thefunction

In.

To this end,let

r,r"

be tworealnumbers such thatr<r’.

Define An(r) {.e

Dn:

In(Z)

<-

r} and

Bn(r"

{ze

Dn:

In(Z)

>=

r’}.

We claim that

D(An(r))NI(Bn(r’))

b foreverypositive integern.

We prove first that d(An(r))tqi(Bn(r))=tk for every positive integer n. Let m be a fixed positive integer. If either Am(r or

Bm(r"

is empty the result holds.

So

we may assume that Am(r and

Bin(r"

are nonempty. Now suppose that

d(Am(r))t’li(Brn(r’))eb.

Then ze

d(Am(r))f3i(Bm(r’))

for some zeE. Since zed(Am(r)) there exists a_{e Am(r)} such that z<a.

Similarly, there exists be

Bin(r"

such that b<z. Since ae Am(r and be

Brn(r’)

we have frn(a)

<-

r<

r"

<=

.fm(b). On the other

hand,

b<z and z<a imply that b<a by transitivity of the preorder. Therefore,

Ira(b)<=

Ira(a) because /’m is an order homomorphism. This contradiction provesthat foreverypositive

n,d(An(r))Cli(Bn(r’))=

.

Now thepreorder < isstronglycontinuous. Thisimplies that d(An(r))isacloseddecreasing set in E and

i(Bn(r’))

is a closed increasing set in E for every positive integer n. Therefore, D(An(r))Cd(An(r))and

I(Bn(r’))

C

i(Bn(r’)). Hence,

O(An(r))OI(Bn(r’))=

for alln.

Thus all the conditions of Nachbin’s extension theorem

[1,

p.36]

are satisfied and wemay

conclude that thereexists a real-valuedcontinuous order homomorphicextension t/nof

In

toE. Clearly,wemayassumethatgn:E[O,2-n].

Define /’:ER by f()= gn() for zeE. _{Clearly, f} is a real-valued continuous order homomorphismonE.

To

provethat

I

isanorder monomorphism let z, ybetwoelementsin Ewithz<y.

Suppose

firstthat Ehas noleast element. Thisimplies that theset {aeE:a< z} isnonempty. It isopen bycondition

(iii).

Since gis densein E thereexists apositive integernsuch that znbelongs to {aeE:a<z}. Furthermore since gn is an order monomorphism on the increasing set D_n and zneD_n, wehavegn()<gn(Y)"

Hence,

f() < I(t).

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n such that _O~z_n because in that case there is nothing to be proved. If e

O<x

the set

{a

.

E:a<x} isnonempty andopen. Then arguingasbeforewehave ’(x)< f(v). Ifx ₀ theset

{a c=E:a< y} is nonempty andopen. Therefore, it contains an element _nof

z.

Again, without loss ofgenerality, wemay assumethat ₀<zn<!t. Since

n

is anorder homomorphism on Ewe

have

en(eO)

-<

en(zn). Nowbecausego isanorder monomorphismonthe increasingset D_nwehave 9n(Zn)< gn(Y)"

Hence,

gn(X)

9n(eo)

<=

gn(Zn)< gn(Y)- Thisimplies that l(z) <l(y).

We have proved that for allz,yin Esuch that z<ywehave l()<l(y). Consequently,)" is

acontinuousorder monomorphismon E. Finally, it isclearthat

I

is anorder extension of

In

for

alln. Thiscompletestheproofof thetheorem, q.e.d.

REMARK

1. It is worth observing that in a compact ordered space the preorder is

necessarilystronglycontinuous

[1,

p.44].

REMARK

2. The strong continuity condition has been used in

[9]

to provegeneralizations of

Katetov’s

theorem on the interpolation of a continuous function between semi-continuous

functions. Forfurther applications of this condition the readeris referred to

[10,

11].

A

[12,

13,

14].

Another important special case of the preceding theorem occurs when E is a totally preordered set with the order topology. Under these conditions, the preorder is strongly

continuousas wenowprove.

PROPOSITION 1.

Let

Ebeaset, _< atotalpreorder onE and assumethat Ehas the < ordertopology. Then thepreorderisstronglycontinuous.

PROOF. Let K beaclosed subset ofE. Weclaimthat d(K) isclosed. Ifd(K) is notclosed there exists a net

{zk,

k ED} in d(K) that converges to zfd(K). This implies that

z

K.

Therefore,thereexists an open set

v

such that zEVandVnK because K isclosed. Since E

has the ordertopologywemayconclude that thereisabasiselementBsuchz BCV. Wehave

toconsiderthefollowingthreecases: 1. z

.

B (a,b)

2. zEB (a,

b0]

3. z B

[a0,b

Observe,

in the first two cases, thata_<

ZkO

for some k0 EDbecause the net

{zk,

k D} converges

to z. Since

Zk-o

"

d(K),

Zko

<_vfor some v K.

Now

because thepreorder istotal either z_<t,or

v_<z. Ifz<vthen z d(K) becausevEK. Since z d(K) by hypothesis, we may conclude that v<z.

Hence,

we have a_<_zk _<v<z. This implies that v

v

which is contradiction because

0

Vf3K=.

It

remains toconsiderthe thirdcase.

So

suppose thatz B

[a0,

). Ifa

0<z< thenweare inthecasejustconsideredabove. Soletz a

0. Since B isanopen set thereexists k such that

Zkl

G

[a0,b

). Thisimplies that a

0<z

kl

becausezk

.

d(K) andz

.

d(K).

Hence,

a0 z<

Zkl

_<

,

for somevEKsince

Zkl

Gd(K).

Therefore,

z d(K)

wich

isacontradiction.

We

haveproved thatin all cases, Kclosed in E implies that d(K)isalso closed. Similarly,

weprove thati(K) isclosed. Therefore,thepreorder < isstronglycontinuous, q.e.d.

Suppose

now that (E,t) is a topological space and < is a continuous totalpreorder on E.

Since is, in general, finer than the order topology, the above method ofproofdoes not work.

However,

asimpledirectargument basedonthe Gluing Lemmaoftopologymaybeused instead

aswe nowprove.

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(i) Thepreorder < iscontinuous;

(ii) F {Dn}n=l is a countable family of closed increasing subsets of E such that for each

n 1,2,. thereexists areal-valuedcontinuousorder-monomorphism/’ndefinedon

Dn;

(iii)

Thereexists acountable and topologically densesubset Z

{Zn}=

ofE with theproperty

that _n D_nfor alln.

Then there exists a real-valued continuous order monomorphism j’ on E that is an order

extensionof

In

foreachpositiveintegern.

PROOF.

Without lossofgenerality, wemay assume that for alln, .f

n:

Dn.-.[O,

].

Let.

mbe an arbitrary but fixed positive integer. We want to show that there exists a real-valued

continuoushomomorphism

am

onEsuch thatam(r)=/’re(z)forallz6

Dm.

To this end, we claim that either there exists some point _Um E such that i(ym)=Dm or E\Dmis closed. Sosuppose that i(a)# Dmforall a6E. Weneedto prove that E\Dmis closed. If E\Dm is not closed there exists a net

{Zd:d

D} in E\Dm that converges to z 6

Dm.

Since

i(z)//: Dm there exists

v

Dm such that v<z because the preorder _< is total. The set {a E:v<a} is increasing, open

(because

the preorderis

continuous)

andcontains z. Therefore, there exists d

o

such that _dO_<d implies that _zd {a g:v_<_{a} c}

Dm.

_{This contradiction} _proves

that E\Dmisclosed, andcompletestheproofof the claim.

To show that there is a continuous order-homomorphism

am

on Eextending f, we have to

considerthefollowingcases.

Suppose

first thatE\Dmis closed. Defineareal-valuedfunctiongm onEby

am(Z) fro(z) ifz/D_m 0 ifzq_E\D_m

Since both D_m and E\Dm are closed, the Gluing Lemma

[15,

pp.14-15]

implies that

am

is

continuous. Clearly,

am

isanorder-homomorphismonE.

Suppose

now that for some _Ym E,

Dm

=i(ym)" Clearly, _Ymq

Din"

Define a real-valued

function

am

onEby

am(Z) fro(z) ifzfii(ym) fm(Ym) ifz d(Om)

Observe that for all z,yD_m such that z~o we have fro(z)= fro(Y) because

fm

is an order homomorphism. Therefore, for all z d(ym)fli(ym), we have fro(z)= fro(Y)" Consequently, the Gluing Lemmaagain implies thatgm is continuous. Clearly,gm isanorder homomorphismon g.

We have proved that for each n=1,2,... there exists a

rel-valued

continuous order homomorphism gnon E such that gn(z)=fn(Z) for all z Dn. We may assume without loss of generalitythatgn:E-,[0,2

hi.

Finally, observe that sets of the form {a E:a<z} areopen in E foreachz gbecausethe preorderis continuous and total.

Hence,

condition

(iii)

ofTheorem1 is satisfied. Arguingasin

the last part of Theorem 1, wemayconclude theproofof thetheorem, q.e.d.

REMARK

3. Itshould be noted that the conditions of Theorem 2 imply that g_is_normally

preordered

[4,

Proposition

1].

On the other hand, the assumption of strong continuity is not needed because thepreorderistotal.

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TopologyandOrder, Van Nostrand, Princeton,

N.J.,

1965.

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K.

&

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G.,

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