ORDER EXTENSION OF ORDER MONOMORPHISMS ON A PREORDERED TOPOLOGICAL SPACE

This paper proves Nachbin-type extension theorems for infinitely many functions on a topological space equipped with a preorder.

In the particular case in which the preorder on the space is the discrete order, the Nachbin extension theorem reduces to the Urysohn-Tietze extension theorem on normal spaces.
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 of a preordered topological space E.
Then the problem is to find a real-vlued continuous order-monomorphism y on the whole space / that is an order-extension of each order monomorphism y in the collection F. In this paper, we give sufficient conditions for the existence of such a 'universal' order-extension.
A special case of the problem studied in this paper arises in mathematical economics in the 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 R n equipped with a total preorder.
The Arrow-Hahn method [2, pp. 82-87]consists of taking a point x 0 and then by defining the 'utility' of z to be the Euclidean distance from z0 to the upper section of x.This Arrow-Hahn function can be shown to be continuous under certain conditions.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 whole space x.It should be observed that the problem is not merely to show the existence of a continuous order monomorphism since, in the finite-dimensional Arrow-Hahn context, this can be deduced concisely from general topological theorems such as Eilenberg's theorem ([3] or Corollary 1 of [4]).The problem is to somehow 'construct' the function f as an order-extension of the functions Ix for x in D. See Mehta [5,6] for further discussion of the Arrow-Hahn theorem.
For some recent work on the separation and extension theorems of Nachbin in preordered topological spaces the reader is referred to Herden [7,8].
2. PRELIMINARIES.Let E be a set.A preorder < on E is a reflexive 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 relations on E as follows.If z, y g then z I if and only if t _< and I _< z.The relation is an equivalence relation.If z, y E then < if and only if _< y and not y < z.The relation < is irreflexive and transitive.
A preorder < is said to be an order if it is antisymmetric.A preorder _< is said to be total if for every two elements z, g either _< y or y _< z.
A subset X of a preordered set g is said to be decreasing if b X, a _< b imply that a X.
Each subset X of E determines, in a unique manner, a decreasing set d(X) which is the smallest one among the decreasing sets containing X. Dually, one defines the concepts of an increaaing set and the smallest increasing set i(x) containing a given subset x.
Let E be a totally preordered set.Consider the following collection of subsets of E: (1) (a,b)   (2) (a, b0] {z E:a < z < b) if b 0 is the last element of E; ( 3) [a0,b) {z E:a o < z < b} if a 0 is the first element of E.
Then this collection of subsets of E is a basis for a topology on E called <-order topology.
Let E be a topological space equipped with a preorder _<.Each subset x of E 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 we write A < B to indicate that D(A)nI(B)-.
Let E be a topological space equipped with a preorder _<.The preorder _< is said to be continuous if d(y) (z E:z _< y}, the lower section of y, and i(y) {z E:y _< z}, the upper section of y, are closed in E for every y E. The preorder is said to be strongly continuous if d(X) and i(X) are closed in E for every closed subset X of E.
A topological space E equipped with a preorder _< is said to be nominally preordered if for every two disjoint closed subsets F 0 and F of E, F 0 being decreasing and F increasing, there exist two disjoint open subsets A 0 and A such that A 0 contains F 0 and is decresing and A contains F and is increasing.
Let E and E 2 be two preordered sets.A function ] on E to E2. is said to be an order homamorphism (or isotone) if z,1 E and z _< imply that y(z)_< l(v).A function y on E to E 2 is said to be an order monomorphism if it is an order 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.Then an order monomorphism f:E-lt is said to be an order extension of 0 if it satisfies the following conditions: (i) 0(a) < 0(b) and a,b F imply that f(a) <_ f(b) (ii) (a) < g(b) and a,b F imply that f(a) < l(b).

ORDER EXTENSION OF ORDER MONOMORPHISMS.
The following fundamental theorem on the extension of an order homomorphism is due to Nachbin [1, p. 36].(Nachbin's Extension Theorem) Let E be a normally preordered space F C_ E a closed subset and ! a bounded real-valued function which is continuous and isotone on F. We shall indicate by A(r) the set of points where is a real number.Then in order that the function/" may be extended to E in such a way as to become a bounded, continuous and isotone real function on 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 space equipped with a preorder < such that the following conditions are satisfied: (i) E is normally preordered; (ii) The preorder < is strongly continuous; (iii) For each z e E, the set {a e E:a < x} is open in E; (iv) F {Dn} n= is a countable family of closed increasing subsets of E such that for each n 1,2,.
there exists a real-valued continuous order-monomorphism In defined on Dn; (v) There exists a countable and topologically dense subset Z {Zn} n= of/ with the property that n e D n for all n.
Then there exists a real-valued continuous order monomorphism /' on E that is an order extension of fn for each positive integer n.
PROOF.By hypothesis, for each n 1,2,... there exists a real-valued continuous order .monomorphism In on D n.Since the extended real line is order homeomorphic to [0,1], we may assume without loss of generality that for each positive integer n, fn: Dn--.[0,1].We prove next that for each n 1,2,.the conditions of Nachbin's extension theorem [9, p.36] are satisfied for the function In.To this end, let r,r" be two real numbers such that r < r'.
We claim that D(An(r))NI(Bn(r')) b for every positive integer n.
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 z e d(Am(r))f3i(Bm(r')) for some z e E. Since z e d(Am(r)) there exists a e Am(r) such that z < a.
Similarly, there exists be Bin(r" such that b<z.Since a e 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 proves that for every positive n,d(An(r))Cli(Bn(r'))= .
Now the preorder < is strongly continuous.This implies that d(An(r)) is a closed decreasing set in E and i(Bn(r')) is a closed increasing set in E for every positive integer n.Therefore, D(An(r)) C d(An(r)) and I(Bn(r')) C i(Bn(r')).Hence, O(An(r))OI(Bn(r'))= for all n.
Thus all the conditions of Nachbin's extension theorem [1, p.36] are satisfied and we may conclude that there exists a real-valued continuous order homomorphic extension t/n of In to E.
Define /':ER by f()= gn() for z e E. Clearly, f is a real-valued continuous order homomorphism on E.
To prove that I is an order monomorphism let z, y be two elements in E with z < y.Suppose first that E has no least element.This implies that the set {a e E:a < z} is nonempty.It is open by condition (iii).Since g is dense in E there exists a positive integer n such that z n belongs to {ae E:a < z}.Furthermore since gn is an order monomorphism on the increasing set D n and z n e D n, we have gn() < gn(Y)" Hence, f() < I(t).
Suppose now that E has a least element 0. We may assume that there is no positive integer n such that O~zn because in that case there is nothing to be proved.If e O<x the set {a .E:a<x } is nonempty and open.Then arguing as before we have '(x) < f(v).If x 0 the set {a c= E:a < y} is nonempty and open.Therefore, it contains an element n of z.Again, without loss of generality, we may assume that 0 < z n < !t.Since n is an order homomorphism on E we have en(eO) -< en(zn).Now because go is an order monomorphism on the increasing set D n we have 9n(Zn) < gn(Y)" Hence, gn(X) 9n(eo) <= gn(Zn) < gn(Y)-This implies that l(z) < l(y).
We have proved that for all z,y in E such that z < y we have l()< l(y).Consequently, )" is a continuous order monomorphism on E. Finally, it is clear that I is an order extension of In for all n.This completes the proof of the theorem, q.e.d.
REMARK 1.It is worth observing that in a compact ordered space the preorder is necessarily strongly continuous [1, p.44].REMARK 2. The strong continuity condition has been used in [9] to prove generalizations of Katetov's theorem on the interpolation of a continuous function between semi-continuous functions.For further applications of this condition the reader is referred to [10,11].A related condition has been used in the theory of order compactifications [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 continuous as we now prove.PROPOSITION 1.Let E be a set, _< a total preorder on E and assume that E has the < order topology.Then the preorder is strongly continuous.
PROOF.Let K be a closed subset of E. We claim that d(K) is closed.If d(K) is not closed there exists a net {zk, k E D} in d(K) that converges to z fd(K).This implies that z K.
Therefore, there exists an open set v such that z E V and V n K because K is closed.Since E has the order topology we may conclude that there is a basis element B such z B C V. We have to consider the following three cases: Observe, in the first two cases, that a < ZkO for some k 0 E D because the net {zk, k D} converges to z.Since Zk-o " d(K), Zko <_ v for some v K. Now because the preorder is total either z _< t, or v < z.If z < v then z d(K) because v E K. Since z d(K) by hypothesis, we may conclude that v < z.Hence, we have a < < v < z.This implies that v v which is contradiction because zk 0 Vf3K=.
It remains to consider the third case.So suppose that z B [a0, ).If a 0 < z < then we are in the case just considered above.So let z a 0. Since B is an open set there exists k such that Zkl G [a0,b ).This implies that a 0 < z kl because z k .d(K)andz.d(K).Hence, a 0 z < Zkl _< , for some v E K since Zkl G d(K).Therefore, z d(K) wich is a contradiction.
We have proved that in all cases, K closed in E implies that d(K) is also closed.Similarly, we prove that i(K) is closed.Therefore, the preorder < is strongly continuous, q.e.d.
Suppose now that (E,t) is a topological space and < is a continuous total preorder on E.
Since is, in general, finer than the order topology, the above method of proof does not work.
However, a simple direct argument based on the Gluing Lemma of topology may be used instead as we now prove.
THEOREM 2. Let E be a topological space equipped with a total preorder < such that the following conditions are satisfied: (i) The preorder < is continuous; (ii) F {Dn}n=l is a countable family of closed increasing subsets of E such that for each n 1, 2,.
there exists a real-valued continuous order-monomorphism/'n defined on Dn; (iii) There exists a countable and topologically dense subset Z {Zn}= of E with the property that n D n for all n.Then there exists a real-valued continuous order monomorphism j' on E that is an order extension of In for each positive integer n.
PROOF.Without loss of generality, we may assume that for all n, .fn:Dn.-.[O, ].Let.m be an arbitrary but fixed positive integer.We want to show that there exists a real-valued continuous homomorphism am on E such that am(r)=/'re(z) for all z 6 Dm.
To this end, we claim that either there exists some point Um E such that i(ym)= D m or E\D m is closed.So suppose that i(a)# D m for all a 6 E. We need to prove that E\D m is closed.
If E\D m is not closed there exists a net {Zd:d D} in E\D m that converges to z 6 Dm.Since i(z)//: D m there exists v D m such that v<z because the preorder _< is total.The set {a E:v < a} is increasing, open (because the preorder is continuous) and contains z.Therefore, there exists d o such that d O _< d implies that z d {a g:v < a} c Dm.This contradiction proves that E\D m is closed, and completes the proof of the claim.
To show that there is a continuous order-homomorphism am on E extending f, we have to consider the following cases.Suppose first that E\D m is closed.Define a real-valued function gm on E by am(Z) fro(z Since both D m and E\D m are closed, the Gluing Lemma [15, pp.14-15] implies that am is continuous.Clearly, am is an order-homomorphism on E. Suppose now that for some Ym E, Dm =i(ym)" Clearly, Ym q Din" Define a real-valued function am on E by am(Z) fro(z) Observe that for all z,y D 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 Lemma again implies that gm is continuous.Clearly, gm is an order homomorphism on g.
We have proved that for each n= 1,2,... there exists a rel-valued continuous order homomorphism gn on E such that gn(z)= fn(Z) for all z D n.We may assume without loss of generality that gn: E-,[0, 2 hi.
Finally, observe that sets of the form {a E:a < z} are open in E for each z g because the preorder is continuous and total.Hence, condition (iii) of Theorem 1 is satisfied.Arguing as in the last part of Theorem 1, we may conclude the proof of the theorem, q.e.d.REMARK 3. It should 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 the preorder is total.

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