A NOTE ON REORDERING ORDERED TOPOLOGICAL SPACES AND THE EXISTENCE OF CONTINUOUS, STRICTLY INCREASING FUNCTIONS

J. D. Mashburn To appear in Proceedings.

I. Introduction The origin of this paper is in a question that was asked of the author by Michael Wellman, a computer scientist who works in artificial intelligence at Wright Patterson Air Force in Dayton, Ohio. He wanted to know if, starting with Rn and its usual topology and product partial order, he could linearly reorder every finite subset and still obtain a continuous from Rn into R that was strictly increasing with respect to the new order imposed on Rn. It is the purpose of this paper to explore the structural characteristics of ordered topological spaces which have this kind of behavior. In order to state more clearly the questions that we will consider, we must define some terms. By an order we mean a relation that is transitive and asymmetric (p < q q ≮ p). Unless otherwise stated, we will denote the order of an ordered X as <. Two distinct elements,⇒ p and q, of X are comparable if p < q or q < p, and are incomparable otherwise. An order is called a linear order if every two distinct elements of the set are comparable. A chain is a subset of an ordered set that is linearly ordered. An antichain of an ordered set X is a subset of X in which every two distinct elements are incomparable. If p and q are elements of X then [p, q] = r X : p r q . A subset A of X is convex if [p, q] A for all p, q X. { ∈ ≤ ≤ } ⊆ An∈ ordered topological space is an ordered set with a topology. There need be no connection between the order and the topology. So R2 with its usual ordering (i.e. a, b c, d a c and b d) is an ordered topological space, as is R2 with the usual topology and the lexicographich i ≤ h i ordering, ⇔ ≤ and R2≤with the discrete topology and the usual ordering. We will use the word space when discussing an ordered set with a topology, and the word set when only an order is involved. Let X,

The author wishes to thank the referee and Professor Herden for their suggestions improving the quality of this paper.

Typeset by -TEX AMS 1 The remainder of this introduction is a review of terms and theorems that we will use in exploring the questions stated above. In Section III, we will extend our results for ordered spaces to preordered spaces. Notions of separability in ordered sets will play an important role in our discussions. The most basic that we will use is the concept of order separability that can be found in Birkhoff’s book [B]. An ordered set X is order separable if there is countable subset C of X such that for every p, q X with p < q there is r C such that p r q. This is all we need to know about a linearly ordered space∈ to be sure that it looks∈ like a subset of the≤ real≤ line. Theorem 1 (Birkhoff, 1948). A linearly ordered set is order isomorphic to a subset of R if and only if it is order separable. We want to know what sorts of ordered spaces not only have continuous strictly increasing functions into R, but have the much stronger property that every time we linearly reorder them in a way that is consistent with the existing order, we can still get such a function. For which ordered topological spaces, X, is there a continuous, strictly increasing function f : X R? We need some definitions and notation before giving the answers to this question. Let X be an ordered→ set. For every p X we will set d(p) = q X : q p and i(p) = q X : p q . We will use this older version of the more∈ modern notation {(p)∈ and (≤p)} to better match{ ∈ the notation≤ } D(A) and I(A) given ↓ ↑ below. Also, d0(p) = d(p) p = q X : q < p and i0(p) = i(p) p = q X : p < q . In order to differentiate between these\{ types} of{ sets∈ for different} orders, we will\{ sometimes} { write∈ the order} in current use as a subscript, such as d (p) or d<(p). So, if Y is a subset of an ordered set X and Y is an extension ≺ ≺ of

Lemma 6. Let X be an ordered set and Y X. If Y extends

Example. Let X = Q (P 2) where 2 = 0, 1 , and let <0 be the usual order on R. Define < on X by setting x < y if and only∪ if one× of the following{ properties} holds.

(1) x, y Q and x <0 y. ∈ (2) x Q, y = z, j P 2, and x <0 z. ∈ h i ∈ × (3) x = z, j P 2, y Q, and z <0 y. h i ∈ × ∈ (4) x = w, i P 2, y = z, j P 2, and w <0 z. h i ∈ × h i ∈ × Then X is a weakly order separable set, but is not pliable. Proof. The set Q is weakly order dense in X. Let be any linear extension of <. The sets z, 0 , z, 1 for z P form an uncountable collection of pairwise≺ disjoint intervals, so there can be no strictly{h i-increasingh i} function∈ from X, into R. ≺ h ≺i It is to avoid these large collections of pairwise disjoint intervals that we introduce the notion of a collection of nonoverlapping sets, similar to the notion of cellularity. A relation R will be called order-consistent if TC[R] is an order. A collection of subsets of an ordered set X is called a collection of nonoverlapping subsets of X if and only if it satisfiesA the following conditions. (1) is a collection of pairwise disjoint sets, each having at least two elements. (2)A The relation A, B 2 : A = B and p A q B(p < q) is order- consistent.  h i ∈ A 6 ∃ ∈ 4 ∃ ∈ If we wish, we may assume that the elements of the collection each have exactly two elements. The previous example shows how these collections of nonoverlappingA sets can be used to create collections of pairwise disjoint intervals. If every linear extension of the existing order is to be order isomorphic to a subset of R, we must keep these collections countable. We will do that by means of a cardinal function. For any ordered set X, ν(X) is the supremum of the of all nonoverlapping collections of subsets of X. When the order of X is to be emphasized, it can included as in ν(X, <). Note that in a linearly ordered set, saying that ν(X) ω is equivalent to saying that every collection of pairwise disjoint intervals is at most countable. ≤ Lemma 10. Let X, < be an ordered set and let be a linear extension of <. If ν ( X, < ) ω then ν ( X, ) ω. h i ≺ h i ≤ h ≺i ≤ Proof. If is a collection of nonoverlapping subsets of X under , then it is also such a collection under <. A ≺ Lemma 11. If an ordered set X is pliable, then ν(X) ω. ≤ Proof. Denote the order of X by <. Assume that ν(X) > ω. Let be an uncountable collection of A nonoverlapping subsets of X. We may assume that A = 2 for every A . The transitive closure <1 of 2 | | ∈ A the relation A, B : A = B p A q B(p < q) is an order. Now let <2 be a linear ordering of h i ∈ A 6 ∧ ∃ ∈ ∃ ∈ A that extends <1. Define <3 on Y = by p <3 q if and only if p < q or there are A, B such that p A, ∪A ∈ A ∈ q B, and A <2 B. It is easy to show that <3 is an order that extends

Proof. Assume that X is not order separable. Let β ω1 and assume that for every α β, Yα is a ∈ ∈ countable subset of X. Since X is not order separable, there are pβ, qβ X such that pβ < qβ and ∈ [pβ, qβ] [ α βYα] = . Set Yβ = pβ, qβ and set Y = α ω1 Yα. ∩ ∪ ∈ ∅ { } ∪ ∈ Now define a linear extension of 0 and assume that β: β γ satisfies the following properties {≺ ∈ } (1) β γ, β is a linear extension of

III. Pliable Preordered Spaces In the applications of the social sciences, a preorder rather than an order is used. A preorder is a relation that is reflexive and transitive. If . is a preorder on a set X and p, q X, then p < q means that p . q and q p. If p q and q p then we say that p q. The definitions given∈ in the introduction can be restated in 6. . . ∼ terms of a preordered, rather than an ordered, set or space by replacing by .. We do want to require that a strictly increasing function on a preordered set also be increasing. The≤ definition of a extremely continuous ordered space can similarly be extended to one for a extremely continuous preordered space. It is easy to see that if X is a preordered set, f : X R is increasing, and p, q X such that p q, then f(p) = f(q). This collapse of a set onto a single point→ allows us to use ordered sets∈ to obtain corresponding∼ theorems for preordered sets. This principle, used in this section to obtain a characterization of pliable preordered sets and spaces, can also be used to extend the results of sections IV and V to preordered sets. We will first redefine a collection of nonoverlapping sets for preordered sets, as it is slightly different than the corresponding definition in ordered sets. Let X be a preordered space and let be a collection of subsets of X. is a collection of nonoverlapping subsets of X if and only if the followingA conditions are satisfied. A (1) is a collection of pairwise disjoint subsets of X, each of which has at least two elements. (2)A The relation A, B 2 : A = B p A q B(p < q) is order consistent. (3) A, B (p hA, q i ∈B, A p = q 6 p ∧ ∃q).∈ ∃ ∈ ∀ ∈ A ∈ ∈ 6 ⇒ The last condition prohibits sets containing elements which, as far as the preorder and strictly increasing functions are concerned, are really the same. To construct an ordered set from a preordered set, we need only follow the example set by a strictly increasing function and collapse the equivalence classes under the relation to a single point. Let X, , τ ∼ h . i be a preordered topological space and, for every p X, letp ˆ = q X : p q . The quotient X of X ∈ { ∈ ∼ } generated by is the ordered space X, <, σ where b ∼ h i (1) X = pˆ : p X , b { ∈ } (2)

Lemma 23. If there is a distinguishing countable collection n : n ω of countable linear separable {E ∈ } systems on an ordered space X then there is a splitting collection n : n ω of separable systems on X. {F ∈ } 2 Proof. For every n ω, let n = En,r : r Q . Let P be the set of ordered pairs r, s Q such that ∈ E { ∈ } h i ∈ r < s and, for every n ω and r, s P set n,r,s = En,t : r < t < s . ∈ h i ∈ F { } We will first show that n,r,s is a separable system. Let t, u (r, s) Q such that t < u. Then F ∈ ∩ En,t,En,u n,r,s and En,t En,u. Now let En,t,En,u n,r,s with En,t En,u. If t < u then ∈ F ⊆ ∈ F ⊆ En,t En,u. If u < t, then, choosing u v t, we have En,t En,u En,v and En,v En,t En,u. Thus ⊆ ≤ ≤ ⊆ ⊆ ⊆ ⊆ n,r,s satisfies both conditions for being a separable system. F Now let p, q X with p < q. There is n ω and there is r, s P such that p En,r and q / En,s. If ∈ ∈ h i ∈ ∈ ∈ En,t n,r,s then p En,r En,t and q / En,s En,t. Thus n,r,s : n ω and r, s P is the desired splitting∈ F collection of∈ separable⊆ systems ∈ ⊇ {F ∈ h i ∈ }

9 Theorem 24. Let X be an ordered space. A finite subset Y of X is a pliable subset of X if and only if the following properties are satisfied. (1) There is a distinguishing countable linear separable system on X. (2) For every subsets, A and B, of Y such that d(A) B = there is a countable linear separable ∩ ∅ system A,B on X such that for every K A,B, A K and B K = . K ∈ K ⊆ ∩ ∅ Proof. Since the theorem is trivially true when Y has less than two elements, we may assume that Y has at least two. Let Y be a linear extension of z + n then F 0 = Fu for some u Q (z + n, z + n + 1). Now i n β(u), so ∈ ∩ r+t r+s ∈ ∩ ≤ ≤ yi Fu. Thus yi Cr,s,t and p Cr,s,t. If r + t < z + 1 then F 0 = Fu for some u Q (z, z + 1). But ∈ ∈ ∈ r+t ∈ ∩ then β(u) = 0 < j, so yj / Fu, a contradiction. Let k Z such that z + k + 1 < r + t < z + k + 2. Then ∈ ∈ z + k < r + s because t s < 1. In order that yj F 0 , it must be true that j < k + 1. But then i < k, so − ∈ r+t yi Fr0+s. Therefore yi Cr,s,t, and hence p Cr,s,t. Thus Cr,s,t is -decreasing. ∈We already know that∈ is a countable∈ linear separable system≺ on X, . Now we will show that F h ≺i , , and s,t are all countable linear separable systems on X, . Let r, s Q such that r < s. Then A B C h ≺i ∈ Ar Er Gr Es Gs = As and Br = Er Hr Es Hs = Bs. Thus and are countable linear ⊆ ∩ ⊆ ∩ ∪ ⊆ ∪ A B separable systems on X, . Fix t, u M. Then Cr,t,u (Er F 0 ) F r+t (Es F 0 ) F 0 = Cs,t,u. h ≺i h i ∈ ⊆ ∩ r+u ∪ ⊆ ∩ s+u ∪ s+t So t,u is a countable linear separable system on X, . C h ≺i It remains to show that , , r,s : r, s M is distinguishing. Let p, q X with p q. If {F A B} ∪ {C h i ∈ } ∈ ≺ p ≮ q then there are i, j n + 2 such that i < j, p yi, and yj < q. By choosing r, s Q such that ∈ 10 ≤ ∈ z + i < r < s < z + i + 1 we get that p Fr and q / Fs. So we may assume that p < q. We may also ∈ ∈ assume that there are not t, u Q such that t < u, p F 0, and q / F 0 . Because is a distinguishing linear ∈ ∈ t ∈ u E separable system on X, < , we can find r, s Q such that r < s, p Er, and q / Es. There are three cases to consider. h i ∈ ∈ ∈ First, assume that for every t Q, q F 0. Then p F 0 for every t Q. It follows that p Er Gr = Ar ∈ ∈ t ∈ t ∈ ∈ ∩ and q / Es Gs = As. So distinguishes p and q. ∈ ∩ A Next, assume that for every t Q, p / F 0. Then for every t Q we have q / F 0. It follows that ∈ ∈ t ∈ ∈ t p Er Hr = Br and q / Es Hs = Bs. So distinguishes p and q. ∈ ∪ ∈ ∪ B Finally, assume that there are t, u Q such that q / F 0 and p F 0 . By our assumption above, t u. ∈ ∈ t ∈ u ≤ We also know that if v < t then p / F 0 , and if u < v then q F 0 . Let S = sup v Q : q / F 0 ∈ v ∈ v R{ ∈ ∈ v} and I = infR v Q : p F 0 . Then t S I u. But if S < I then there are v, w Q such that { ∈ ∈ v} ≤ ≤ ≤ ∈ S < v < w < I. This means that q Fv0 and p / Fw0 , a contradiction. Therefore S = I. Choose v, w M with r + v < I < r + w. Now pick∈x Q such∈ that r < x < s and x + v < I. Since I < r + wh , wei∈ have ∈ p F 0 . But we already know that p Er, so p (Er F 0 ) F 0 = Cr,v,w. Since x + v < S, we ∈ r+w ∈ ∈ ∩ r+w ∪ r+v have q / F 0 . But x < s and q / Es, so q / Ex. Therefore q / (Ex F 0 ) F 0 = Cx,v,w. Thus v,w ∈ x+v ∈ ∈ ∈ ∩ x+w ∪ x+v C separates p and q.  By giving X the added structure of a normally ordered space, we can greatly simplify the requirements for Y to be a pliable subset of X. Theorem 25. Let X be a normally ordered space. A finite subset Y of X is a pliable subset of X if and only if the following conditions are satisfied. (1) There is a distinguishing countable linear separable system on X. (2) For every subsets A and B of Y , if d(A) B = , then D(A) I(B) = . ∩ ∅ ∩ ∅

Proof. The normality of X can be used to define the separable systems A,B used in Theorem 24. K  The methods used in this section will not generalize to infinite subsets. It is still an open question whether similar characterizations can be given for infinite subsets. It is possible that no theorem can be obtained which improves on just saying that Y is pliable if and only if for every extension Y of < there is a splitting separable system on X with respect to . ≺ ≺ V. Pliable Subsets of Normally ordered Spaces In this section, we improve on the work in the previous section in the sense that we will obtain a charac- terization of pliable subsets without any restrictions on their . This is done at the cost of placing some strong restrictions on the type of space being considered, and how the subset sits in the space. Some of our restrictions will be necessary, anyway, while others occur naturally in familiar spaces. Rather than using Herden’s approach to the existence of continuous, strictly increasing functions, we will return to Nachbin’s approach. Our spaces will be normally ordered, the orders will be weakly continuous, and we will need a type of separability. We need to build these structures in such a way that they will be preserved when the space is reordered. Let us begin by considering the problem of making sure that X, is normally ordered h ≺i for every extension Y of

Proof. Part 1. Assume that A does not have a finite cofinal subset and B does not have a finite coinitial subset. Then by Proposition 18 there is a cofinal subset pn : n ω of A such that if m < n then { ∈ } pn  pm and there is a coinitial subset qn : n ω of B such that if m < n then qm  qn. Let Y { ∈ } ≺ be an extension of

(1) If p d(A) then p Y r. ∈ ≺ (2) If q i(B) then s Y q. ∈ ≺ (3) If p Y [d(A) d(r)] then r Y p. ∈ \ ∪ ≺ (4) If q Y [i(B) i(s)] then q Y s. ∈ \ ∪ ≺ 1 1 Let f : X R be continuous and strictly Y - increasing. Set U = f − [( , f(r))] and V = f − [(f(s), )]. Then U, V→ is an open corner in X, <≺ that expands A, B . Let p−∞U Y . Then f(p) < f(r).∞ If h i h i h i ∈ ∩ p Y [d(A) d(r)] then r Y p, so f(r) < f(p). Thus p [d(A) d(r)] Y . Similarly, if q V Y then q ∈ [i(B\ ) i(s∪)] Y . ≺ ∈ ∪ ∩ ∈ ∩ ∈Part 3.∪ Assume∩ that A has two maximal elements r and s. By Part 2, A r ,B can be expanded h \{ } i to an open corner U1,V1 in X, and A s ,B can be expanded to an open corner U2,V2 in X. Then h i h \{ } i h i U1 U2,V1 V2 is an open corner in X that expands A, B . The dual case when B has two minimal elementsh ∪ is done∩ thei same way. h i r To establish the second claim of Part 3, define an extension Y of

n Example. Let X be the set consisting of the points 0, 0 and 1, 0 , together with the sequences 0, 2− : n ω m n h i h i {h i ∈ } and 2− , 2− : m, n ω . We will set a, b < c, d in X if and only if a = c = 0 and b < d. {h i ∈ } h i h i 2 Let be the topology on X 0, 0 , 1, 0 inherited from R . For every k ω let Ek = 0, 0 Gm n \ {h i h i} n ∈ {h i} ∪ 2− , 2− : m, n ω and m, n > k and Fk = 1, 0 0, 2− : n ω and n > k . Give X the topol- {h i ∈ } {h i} ∪ {h i ∈ n } ogy generated by Ek : k ω Fk : k ω . Set Y = 0, 0 0, 2− : n ω , A = 0, 0 , and n { ∈ } ∪ { ∈ } ∪ G {h i} ∪ {h i ∈ } {h i} B = 0, 2− : n ω . Then X is a normally ordered space, Y is a pliable subset of X, d(A) B = , and A, B{hcannoti be∈ expanded} to an open corner of X. ∩ ∅ h i Proof. Note that A has a maximum element, B has no minimal elements, and A B = Y . That d(A) B = is obvious from the definitions. We also have the stronger property that D(A)∪ B = and A I(∩B) = ∅, because D(A) = 0, 0 , 1, 0 and I(B) = B 1, 0 . That Y is a pliable subset∩ of∅ X follows∩ from the∅ facts that Y is already{h i ah chaini} in X and the function∪ {h i}f : X R given by f ( x, y ) = y is continuous and strictly increasing. → h i If U is an open subset of X with A U then there is k ω such that Ek U. Let V be an open subset ⊆ ∈ n k 1⊆ n k 1 of X such that B V . Then there is j ω such that if j < n then 2− , 2− − V . But 2− , 2− − Ek, so U V = . Thus⊆ A, B cannot be∈ expanded to an open cornerh in X. i ∈ h i ∈ To∩ show6 that∅ X ish normallyi ordered, let H,K be a closed corner in X. Then 1, 0 cannot be in both H and K. h i h i If 1, 0 / H then there is i ω such that Fi H = . But H is decreasing, so Y H = . It follows h i ∈ ∈ ∩ m n∅ ∩ ∅ that there is j ω such that if m > j then 2− , 2− / H for all n ω. Let W = 1, 0 Y m n ∈ h i ∈ ∈ {h i} ∪ ∪ 2− , 2− : m, n ω and m > j . Then W is open and increasing in X and H X W . Since X W is discrete,{h H,Ki W∈ is an open corner} in X that expands H,K . ⊆ \ \ h ∪ i h i If 1, 0 / K then there is i ω such that Fi K = . Since K is increasing, and H is decreasing, h i ∈ ∈ ∩ ∅ m n there are j, k ω such that j k and if m < j and k < n then 0, 2− / H and 0, 2− / K. Now ∈ ≤ m nh i ∈ h i ∈ we can find l ω such that if m > l and n < j then 2− , 2− / H and if m > l and k < n then m n ∈ n m n h i ∈ 2− , 2− / K. Set U = 0, 2− : n > k 2− , 2− : m > l and n > k and set V equal to the union h i ∈ n {h i m } ∪n {h i } of the sets 0, 2− : n < j and 2− , 2− : m > l and n < j . Then Fi U H,V K is an open corner in X{hthat expandsi H,K} . {h i } h ∪ ∪ ∪ i h i  Simply being able to expand a closed corner to an open corner is not sufficient. In going to the larger open corner, we may add elements of Y . These new elements could cause the open sets , which were decreasing and increasing under the original order of X, to be no longer decreasing and increasing under the reordering. One of the important points of Lemma 27 is that the open corners in cases 1-3 add no new elements of Y . But note that the pairs in Lemma 27 were not closed corners in X. When we expand one of these to an open corner, say U, V , we might even have p U Y and q V Y such that q Y p for some extension Y h i ∈ ∩ ∈ ∩ ≺ ≺ of 1 , S = 1 n : n N , and T = 1 n : n P . Set m < 1 m − } { ∈ 1 } {− − ∈ } { − ∈ } − − for all m N and set 1 n < n for all n P . Set m < n for all m N and all n P such that m = n. ∈ − ∈ 13 ∈ ∈ 6 − This is trivially an order since if p, q X with p < q then there is no r X with q < r. Let Y = N P . Then X is a normally ordered space,∈Y is a pliable subset of X which is∈ not narrow, and for every linear∪ extension Y of K then 1 − ∈ ∈ ∈ ∈ ∈ 1 + n / A. Since A is decreasing, A n P : K < n = . We can choose K large enough so that if there ∈ ∩ { ∈ } ∅ 1 is m P such that m m then m < K. Set B0 = B 1 + : n P and K < n i ( n P : K < n ). ∈ ≺ − ∪ { n ∈ } ∪ ≺ { ∈ } Then B0 is a closed -increasing subset of X and A B0 = . Also, Z B0 is a closed -decreasing subset ≺ ∩ ∅ \ ≺ of X. Thus A (Z B0),B0 is a closed corner in X, that expands A, B . Assume thath ∪ 1 / \B and i1 B. Since 1 Sh, there≺i is K N suchh thati if n N and n < K then 1 ∈ − ∈ 1 − ∈ ∈ ∈ 1 / A. Set B0 = B 1 : n N and n < K . Then B0 is a closed, -increasing subset of X and − − n ∈ ∪ {− − n ∈ } ≺ A B0 = . Also, Z B0 is a closed -decreasing subset of X. Thus A (Z B0),B0 is a closed corner in X,∩ that∅ expands\A, B . ≺ h ∪ \ i h Finally,≺i if 1, 1 h B theni 1, 1 A = , which is just the dual problem of our first case. {− } ⊆ {− } ∩ ∅ We can now show that Y is a pliable subset of X by showing that for every linear extension Y of

Theorem 29. If Y is a narrow, nicely embedded subset of a normally ordered space X and Y is a linear ≺ extension of

Proposition 30. Let Y be a pliable subset of an ordered space X and let Y be an extension of

Example. Let X = ω1 + 1. Isolate all elements of the set ω1 and let the point ω1 have as its neighborhood base the collection of all sets of the form ω1 A where ω1 A is countable. Define an order < on X by { } ∪ \ setting α < ω1 for every α ω1. Then X is a continuous, order separable, normally ordered space, but there is no continuous, strictly increasing,∈ real-valued function on X.

Proof. If α ω1 then d(α) = α and i(α) = α ω1 , both of which are closed. Also, d(ω1) = X and ∈ { } { } ∪ { } i(ω1) = ω1 are closed, so X is continuous. The set ω1 is order dense in X, so X is order separable. Let A, B be{ a} closed corner in X. If B = then X, {is an} open corner of X that expands A, B . If B = h i ∅ h ∅i h i 6 ∅ then ω1 / A. Thus A must be countable. So A, X A is an open corner in X that expands A, B , and X is normally∈ ordered. h \ i h i Let f : X R be a strictly increasing function. We will show that it cannot be continuous. Now → 1 n f[ω1] ( , f(ω1)). We may assume that f(ω1) = 0. There must be n ω such that f − [( , 2− )] is ⊆ −∞ 1 n 1 ∈n −∞ − uncountable. But then X f − [( 2− , )] is uncountable, so f − [( 2− , )] is not open in X. Hence f \ − ∞ − ∞ is not continuous.  The consequence of all this is that none of our current separability conditions will do. The following new property will work for us. It differs from the previous properties in that they were totally order theoretic in nature, while our new condition combines the order and the topology. Because of this, it neither implies nor is implied by any of the previous conditions. An ordered space X is said to be quasiseparable if there is a countable subset C of X such that for every nonminimal (nonmaximal) p X and every open increasing (decreasing) neighborhood U of p there is r C U such that r p (p r).∈ The set C is said to be it quasidense in X. Then Qn is quasidense in Rn ∈ ∩ ≤ ≤ 16 with its usual order and topology. The condition that there be some q X with q < p (or p < q) is there only to allow the consideration of some spaces having large sets of maximal∈ or minimal elements. For example, x, y R2 : y x is quasiseparable with this condition, but not without it, because we cannot find a h i ∈ ≤ − countable number of elements of the space to go above the elements of the antichain x, y R2 : y = x of maximal elements. h i ∈ − Lemma 31. If the set A of minimal elements and the set B of maximal elements of a subset Y of a quasiseparable ordered space X are countable and if Y is an extension of

Proof. Since extends <, we have D (d Y (D(A))) D (A). In order to reverse the subset, we will show ≺ ≺ ⊆ ≺ that D (d Y (D(A))) is - decreasing. Let p X and q D (d Y (D(A))) such that p q. If p < q then ≺ ≺ ≺ ∈ 17 ∈ ≺ ≮ p D (d Y (D(A))), so assume that p q. That means that there are r, s Y such that p r s q and ≮∈ ≺ ∈ ≤ ≺ ≤ r s. Now s D (d Y (D(A))), so either s D [d Y (D(A)) Y ] or s D(A). ∈ ≺ ∈ ≺ ∩ ∈ If r / d Y (D(A)) Y then r is maximal in [d Y (D(A)) Y ] r , so, by Lemma 33, D ([d Y (D(A)) Y ] r ) ∈ ≺ ∩ ≺ ∩ ∪{ } ≺ ∩ ∪ { } ∩ Y = [d Y (D(A)) Y ] d Y (r). Thus, if s is an element of the set D (d Y (D(A)) Y ) then s d Y (D(A)). ≺ ∩ ∪ ≺ ≺ ∩ ∈ ≺ But if s D(A) then r d Y (D(A)). So in either case, r d Y (D(A)). Therefore p D (d Y (D(A))). ∈ ∈ ≺ ∈ ≺ ∈ ≺ The proof that I (A) = I (i Y (I(A))) is similar. ≺ ≺ Lemma 35. Let X be an ordered set and let A, B, E, F X. If A B, E F , and d(A B) (E F ) = , then there is an extension of < such that A is cofinal in⊆ A B, E↑ is coinitial↓ in E F∪, and,∩ if p∪ A B∅ and q E F , then p q.≺ ∪ ∪ ∈ ∪ ∈ ∪ ≺ Proof. Recall that A B means that there is an extension A B of

D (D(p) A) I (I(q) B) = D (d Y (D(p))) I (i Y (I(q))) ∪ ∩ ∪ ≺ ∩ ≺ = D Y (p) I Y (q) ≺ ∩ ≺ = . ∅  Let us say that X is inherently weakly continuous with respect to Y if D (D(p) A) I (I(q) B) = for ∪ ∩ ∪ ∅ every p, q X with p < q and every A, B Y such that d

(1) Y,

Proof. We have already seen that Properties 1-4 are necessary. It remains to show that they are sufficient. Assume that the properties are satisfied. Let Y be a linear extension of

D (p) I (q) = D (d Y (D(p))) I (i Y (I(q))) ≺ ∩ ≺ ≺ ∩ ≺ = D (D(p) A) I (I(q) B) ∪ ∩ ∪ = . ∅

If p ≮ q then there are r, s Y such that p r Y s q. By Lemma 33, D(r) Y = d< (r) and ∈ ≤ ≺ ≤ ∩ Y I(s) Y = i

References

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Department of Mathematics University of Dayton Dayton, Ohio 45469-2316 E-mail address: [email protected]

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