Natural compactifications of lattices

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63 DEFINITION 6.23. Suppose that Z is an admissible subbase for X. Then for each id e XX, we let XiD,iE) = {x e X: e^(x) < £} and X(I,^) = {x e X: .^ ^ e^(x)}. REMARK 6.24. It is easily seen from Definition 6.23 that for each ;£ e X„X, V e^[X(D,:f)] < Jf < A e^[X(I,;^)]. Hence exactly one of the following statements is valid : (i) V ej.[X(D,:£)] = ;^ = A e^[X(I,£)] (ii) V e^[X(D,;f)] = / < A e^[X(I,i6)] (iii) V e^[X(D,;d)] < £ = a e^CX(I,;t)] (iv) V e^CX(D,;C)] < :d < A ej,[X(I,;d)]. Clearly for each £ e AX, X(D,;^) is a closed ideal of X which may or may not be in E and X(l,t) is a closed dual ideal of X which may or may not be in E. DEFINITION 6.25. If E is an admissible subbase for X, then E will be called thick if and only if for each pair /, W e AX such that i. satisfies condition (ii) in Remark 6.24 and ^ satisfies condition (iii) in Remark 6.24, it follows that G = E U{X(D,jf), X(I,?r)} is an admissible subbase for X and there exists an iseomorphism f : A X -> AX such that ^ ° ^E = ^0-

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54 THEOREM 5.26. If. ^ i£ a thick , admissible subbase for X, then X is_ lattice dense in AX. PROOF. To see that X is lattice dense in XX, let J^, /? e AX such that I e e^CX]^^ e^[X]^,;>fe ^^^'^^^ " ^z""^^^' ^""^ ^^ ^^ ^^"^^ X = V {e^(x): X e X and e^(x) < X.} = v e^CX(D,Jd)] (5.23). Since £. i e [X] , it must be the case that ^ < A {e^(x): X e X and j2 < e^(x)} = A e^[X(I,^)] (6.23). Hence •£ satisfies condition (ii) of Remark 5.2M-. By a dual argument we see that "fli satisfies condition (iii) of Remark 5.24. Thus since E is thick, AX is iseomorphic with AX where = E U{X(D,^), X(I,/<)} (5.25) Let f: AX-)AX denote the iseomorphism. Since f is a lattice isomorphism, f o e = e (5.25), and X(D,;^) e , it follows that fC^) = f(v e^[X(D,/)]) = V f[e^[X(D,i)]] (2.2) V eQ[X(D,sd)] = V X(D,X)^ (4.10 (iv)). Dually, since X(I,;^) e , f(>?j) = A X(I,%)^. Thus X(D,;^) e f(jf) and X(I,/^) e f(/Y) (4.10 (vi)). Since f preserves the order of AX (2.5) and /^ < vd, it follows that f{n) < f(if) and so x(i,;7z) e fim) n 0^ c f (^) n 0^ (4.7); therefore X(D,;f), Xil,Pt) e f(it) and so X(D,:d) H X(I,^) ^ 0. Let X e X(D,;d) D Xil, 7n). Then f(^) < e^ix) < f(jd). Since f o e = e , we have fiH) < f(.e (x)) < f(iC), and since f is a lattice isomorphism, yn< e (x) < id. Finally, since X'i^ e [X] and Jd I e [X], it follows that /? < e (x) < *f . Consequently, X is lattice dense in A X.

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65 PROPOSITION 6.27. If Y is_ a topological lattice completion of X with respect to i : X -* Y and X is_ lattice dense in Y , then I = {i [S]: SeZ ^i^^ thick , admissible subbase for X. PROOF. Recall that Z is an admissible subbase for X (6.17 (ii)). By Theorem 6.22, Y is iseomorphic with XX. To see that Z is a thick subbase for X, suppose that i£ , M e AX such that t satisfies condition (ii) ,of Remark 6.24and ^ satisfies condition (iii) of Remark 6.24. Since V {ej,(x): X e X and e^(x) <^} = X. < A {e^(x): X e X and ^ <.e^(x)} (6.23, 6.24), it follows that £ e e^CX]"*"^ e^[X]"^. Thus {Q e XY: C < £] e E^^^ (6.15), L L ' L ' D and so X(D,^) = i"^[{^e A^X: ^ < t)'] e l^. Dually, /Tie e^[X]^^ e^CX]^, {jl e Aj,X: yn.< p e l^^^\ and so X(I,»z) = i~^[{ ^ e A^X: )^ < ^}] £ l^. Consequently, = Z u{X(D,Jl), X(I,»£)} = E is an admissible subbase for X, and there is an iseomorphism f : A X ->• AX, namely, the identity map. such that f o e = e . Hence I is thick, PROPOSITION 6.28. Th£. usual subbase T and the_ subbase V are both thick. PROOF. To see that T is thick, we recall that AX is the completion by cuts of X so that AX = e [X] . Thus it follows that every member of AX satisfies condition (i) in Remark 6.24. Hence T vacuously satisfies the definition of a thick subbase.

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55 To see that r is thick, suppose that S^, />^e AX such that «£ satisfies condition (ii) of Remark 5.2'+ and Z'? satisfies condition (iii) of Remark 5.24. Since X(D,/) is a closed ideal of X and X(l,Pt) is a closed dual of X, then X(D,;t) e r and X(I,;>^ e r (4.1 (ii)). Thus = r u{X(D,;C), X(I,/»)} = ris an admissible subbase for X (4.4-) and the identity map f : XX -^ AX is an iseomorphism such that f o e = e . COROLLARY 6.29. X is lattice dense in AX and AX. PROOF. Since both T and r are thick (5.28), admissible subbases for X (4.4), the proof is immediate from Theorem 5.25. All of the admissible subbases which are considered in the examples (Chapter 8) are thick subbases for X. Whether or not every admissible subbase for X is thick remains an open question. THEOREM 6.30. Suppose that X is a topological lattice and is an admissible subbase for X. Then X is lattice dense in AX if and only if there exists a thick admissible subbase E for X such that Ax is iseomorphic with Ax via an iseomorphism f : A„X -> A^X with the property that f o e = e . PROOF. Let be an admissible subbase for X. Thus AX is a topological lattice completion of X with respect to e^: X ^ AgX (5.22).

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67 -1 ^^0^ IfX is lattice dense in X^X, then Z = {e^ [S] : S e L } is a thick admissible subbase for X (6.27) and XX is iseomorphic with XX via an iseomorphism f : X X ^ XX with the property that f o e^. = eg (6.20). Conversely, if there is some thick admissible subbase E for X and an iseomorphism f : XX -> XX such that f ° e = e , then since X is lattice dense in XX (6.26), it follows that X is lattice dense in XX.

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7. LARGEST AND SMALLEST LATTICE SUPEREXTENSIONS In this chapter we will consider the possible existence of certain maps between different lattice superextensions of the same topological lattice. The results will then be used to establish the existence of maximal and minimal lattice superextensions for all topological lattices and in certain cases, largest and smallest lattice superextenions (Theorem 7.15). Thus in the realm of lattice superextensions we have analogues to the Stone-Cech and Alexandroff compactifications. Kaufman [12] obtained similar analogues for the totally ordered case. Theorem 7.15 is a generalization of Kaufman's results. In this chapter , as in the previous chapters , X will denote a topological lattice. Also, to avoid confusion, we will sometimes use the symbol """ in place of "+" to denote the subbasic closed sets of a lattice superextension; e.g., if is an admissible subbase for X, then for each T e 0, T''« = {t e XX: lei.) and 0''« = {P-: T e 0}. DEFINITION 7.1. Let and E be admissible subbases for X. 68

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69 (i) We say that XX algebraically dominates XX if and only if there is a surjective, order-preserving function f: XX -> XX such that the diagram \^ ^0^ commutes (i.e., f » e = e ) and such that f preserves the suprema of ideals in X and the infima of dual ideals in X; i.e., if I is an ideal of X and D is a dual ideal of X, then f(v e [I]) = v (f o e^)[I] and f(A e [D]) = a (f o e )[D]. Such a function is called a dominating function of XX over XX. ' ' Li ' W (ii) We say that the superextension XX dominates the superextension XX if and only if XX algebraically dominates XX via a continuous dominating function. It might be noted that although XX and XX are both lattices, a dominating function f is not required to preserve all suprema and infima of XX; i.e., f is not required to be a lattice homomorphism. The reason for this is apparent from Example 8.9 in which a topological lattice X is exhibited for which the lattice XX algebraically dominates the lattice XX (and also for which the superextension XX dominates

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70 the superextension AX), but for which there are i^, W e A X such that f(:^) A f (/%) i f{£. AfK). (See 8.9 for details.) Example 8.7 shows that algebraic domination does not necessarily imply superextension domination, for in this example there exists a unique algebraic dominating function f: AX * AX which is not continuous. PROPOSITION 7.2. Suppose that and E are admissible subbases for X such that AX algebraically dominates AX via a_ dominating function f : AX -> AX. Then (i) for each R e ^^ ( dually , R e E ), f( v r"*") = v e [R] ( dually , f(AR^) = aSqCR]), (ii) for each R e E Pi ( dually , R e E H ) , f ( v r"*") = v R>'( dually , f ( A R ) = A R--V), and (iii) for each ;C e A^X, iX. O 0) c f (jC) . PROOF. (i_) Suppose that R e E . Then by the hypotheses on f and since R is an ideal of X, it follows that f(vR^) = f(ve^[R]) (U.IO (iv)) = V (f o e^)[R] = V eg[R]. If R e ^j5 then a dual argument shows that f ( a R ) = a e [R]. (ii) Suppose that R e E H . Then since f ( v r"*") = v e [R] ((i) above) and v e [R] = v R" (i+.lO (iv)), we see that f(v r"^) = v R" . Dually, if.R e E Pi , then f(A r"*") = a R. ( iii ) Suppose that it e AX. Let Re {£ n 0j.). Thus jjif < v r"^ (4.10 (vi)), and since f is order-preserving, f(;^) < f(v R ) = v R" (Cii) above) ;i.e.. Re fU) (4.10 (vi)). Hence (^ H ) < f(/), and by a dual argument, (,^ n 0^) c f(^). Therefore, (;f r\ 0) c_f(/).

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71 LEMMA 7.3. ^ T. ^ an_ admissible subbase for X , then for each a e X, V e^^CXj^Ca)] = e^(a) and a e^CX^Ca)] = e^(a). PROOF. Let a e X. Then for each x e ^n^^)' x < a so that e (x) < e (a) since e is a lattice inclusion (4.9); therefore, V ej.[Xj^(a)] < e^(a). Since a e Xj^(a), it follows that e^(a) = v e^[Xj^(a)] A dual argument shows that e (a) = a e [X (a)], PROPOSITION 7.4. If_ and I are admissible subbases for X such that c^ E, then XX algebraically dominates AX. PROOF. We define f : A^X ^ X^X by f(;£) = V {a e [S]: S € (jC H Z )} for each £ e X^X. f ° e„ = e : Let x e X. For each See(x) H Z, xeSso that A e [S] < e (x); therefore, f(ej,(x)) = v {a e^ES]: S e e (x) n E } < e (x). Now suppose that T e e (x) H , Thus X e T and since T e £ E , it follows that T e e (x). Hence by Proposition 4.10 (iv) and the definition of f, a Ta e [T] < f(e (x))i i.e., T e f(e^(x)) H 0^ (4.10 (vi)) and so e^ix) < f(e^(x)) (4.7). It follows that f o e (x) = ^^^(x) for each x e X. f preserves the order of X X: Suppose that if, X"? e XX such that / < /r. Thus (;t n E ) ^ {rnr\ z^) (4.7) and so f(Af) = V {a eg[S]: S e £. n Ej} < V {A e^CS]: S e ;^ H E^} = fO^j).

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72 For each I e X ^ X, (;( D Q) E. Ht) : Let if e X^X. If T £ Jf n , then by the definition of f, A e [T] < f (/) . Since A e [T] A T" (4.10 (iv)), it follows that T e fOf) (4.10 (vi)), and so ( Jf H ) c f(iS). If T e (;f H 0^), then for each S e (/ H 0^), T n S ?! 0. Thus there is some x e S H T and so a e^ES] < e^Cx) < v e^CT] It follows that fU) = V {a e^CS]: S e (/ n E^)} < v e^CT] = v T'"' (4.10 (iv)); therefore, T e f(^) (4.10 (vi)) and so (^ H 0^^) c f (^) . f is surjective : If /"Z e ^(^> "then I'^is a linked system of E and hence is contained in some maximal linked system J^of E. Since fli<^ ii C\ 0) c f(^), it follows from the maximality of ^ that f(af) = /?! f preserves suprema of ideals and infima of dual ideals of X : Suppose that I is an ideal of X and ^ = v e„[I]. Since f preserves the order of XX, we see that for each x e I, f(e (x)) = e (x) ^ f('^) and so V e [I] < f(;f). If T e (V e [I] H ), then for each x e I, T e e (x) (4.8) and so x e T ; consequently, I £T. Since T e E , it follows that T e V e^[I] = ^ (4.8). Thus T e (/ H 0) c f{'£) and so V e [I] (^ 0^ £ f(jS). By the definition of the order on AX (4.7), f(;e) < V eg[I]; therefore f(^) = f(v e^[I]) = v SgCl] = v (f o e^)[I]. A dual argument shows that f preserves the infima of dual ideals of X. In view of the above arguments, XX algebraically dominates XX.

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73 PROPOSITION 7.5. Suppose that and E are admissible subbases for X such that c_ E. (i) If_ for each iteX^X, {£0 0) is_ contained in a unique maximal linked system of 0, then there is a_ unique dominating function f : XX -> XX. Moreover , f may be defined by f(i^) = {T e 0: (r n 0) u {T} is_ linked } for each / e XX. (ii) If for each ^ g X X,(Jf H 0) is a maximal linked system of , then there is a_ unique and continuous dominating function f: X^X . XgX. PROOF, (i^) Since c_ E , there is a dominating function f: XX -> XX (7.4). Suppose that g: XX -> XX is also a dominating Li \J tt \J function. Thus for each / e XX, (;^ H 0) c g(sd) and (if D 0) £ f(sJ.) (7.2 (iii)). Since (/ O 0) is contained in a unique maximal linked system of 0, it follows that g(;^) = fiat) for each Z e XX; i.e., f = g and f is unique. To 'See that f can be defined as stated, let / c XX and ^= {T e 0: (:tn 0) [j{T} is linked}. To see that ff is linked, let S, T e ^. Then U H 0) u{S} is linked and (if H 0) u{T} is linked. Since (Jf (^ 0) is contained in a unique maximal linked system y of Q, it' follows that (/HO) u{S} and (Jf H 0) u{T} are both subsets of ^ Thus S n T i' and ^ is linked. To see that ^ is a maximal linked system of 0, suppose that T e such that ^ u{T} is linked. Thus (it n 0) u(T} is linked, and so by the definition of ^, T e if. Since both ^ and f(J^) are maximal linked systems of which contain (Jf O 0) and (Jf H 0) is contained in a unique maximal linked system of 0, it follows that f(s^) = tf.

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74 (ii) If for each £e AX, (^ ^ 0) is a maximal linked system of 0, then (Jf H 0) is contained in a unique maximal linked system of 0. By (i) above, there is a unique dominating function f: X^X -> AgX defined by f(; XX such that f o e = e . ^ 2j y h \j PROPOSITION 7.7. If and E are admissible subbases for X such that each of XX and XX algebraically dominates the other , then and E are equivalent . Specifically , we will show that if f : XX -> XX and g: XX -> XX are dominating functions , then (i) g o f is_ the identity on_ X X and f o g is_ the identity on XX (ii) f and g are inject ive (iii ) f and g are closed maps (iv) f and g are homeomorphisms (v) f and g are lattice isomorphisms (vi) X„X is iseomorphic with X^X. L '

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75 PROOF. The hypotheses describe the following diagram: ^E^ 'O I XgX where g » e_ = e and f " e = e 0' (i^) Let R e I . Thus R is an ideal of X. By the properties of dominating functions (7.1 (i)) and Proposition 4.10 (iv), g 6 f(v r"^) = g o f(v ej,[R]) = g(v f o e^[R]) = g( V eg[R]) = V g o ggCR] = V ej,[R] = v R . Dually, for each ReL, g o fCAp"*") = ar , Now let ^^ e X^X. If R € X. C) Z , then / < v r'^ (4.10 (vi)), and so f(/) < f(v r"^) since f is' order-preserving. Thus g o f(^)
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75 ( iii ) Let R € E . If .;^ e r"^, then R e t and so t < v r"^ (4.10 (vi)) Since f is order-preserving, f(;£) < f(v r"*") = v e^ER] (7,2 (i)). Thus fCR"*"] c {j e AgX: ^ < ^ eg[R]}. Conversely, if ^ e XX such that fl. < v e [R], then since f is surjective, there is some Z e XX with f (i?) = ^ . Now since g is order-preserving and g ° f is the identity on XX, it follows that ^ = g o f(^) = g(jl) ^ g(v e [R]). Thus by the properties of g (7.1 (i)), 'i. < g(v eg[R]) = V (g o eg)[R] = v e^[R] = v r"^ (4.10 (iv)). Hence R e £ (4.10 (vi)) and so ;£ e R : therefore {J e XgX: ^ < V eg[R]} = fCR"^] = g"^CR"^]. A dual argument shows that if R e E , then -lr„+ g' [R^] f[R^] = {/ e XgX: A e^CR] < f }. e'since Z is a subbase for the closed sets of XX, it follows from the above equalities that f = g is continuous. Hence f is a closed map, A similar argument shows that g is a closed map. (iv) Since each of f and g is closed, continuous, and bijective, each must be a homeomorphism. (v_) Suppose that C is a non-void subset of XX and '^ ^ C . Since f preserves the order of XX, it is clear that iK^ {f(J): jl e C} < f(jd). To see that f(;(.) < />J let T e />Z H Thus >7? < v T='' (4.10 (vi)) so that for each j! e C , f{i) < v T". Since g is orderpreserving and g o f is the identity on XX, it follows that for each ^ = g o fip < g(v T'":) = V e^[T] (7.2 (i))

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77 Thus J^=v{ft:^eC}^v e [T]. Finally, since f is a dominating function (7.1 (i)), it follows that f(/) < f(v ej,CT]) = V f o e^m = v eg[T] = v T'*' (4.10 (iv)). Hence, T e f()t) (4.10 (vi)). Therefore, W H c_ f (it) so that ^= f(i^); i.e., f(v C) = v f[C]. A dual argument shows that f(A c) = A f[C] and so since f is bijective, f is a lattice isomorphism. A similar argument shows that g is a lattice isomorphism. (vi) That f and g are iseomorphisms follows immediately from (iv) and (v) above. COROLLARY 7.8. If_ and Y, are admissible subbases for X such that c^ E and AX algebraically dominates AX, then is_ equivalent to E. PROOF. Since £ E , it follows that AX algebraically dominates AX (7.4). By the hypotheses, AX algebraically dominates AX. Thus is equivalent with Z (7.7). PROPOSITION 7.9. Suppose that is_ an admissible subbase for X and E = u T. Then (i) E is_ an_ admissible subbase for X, and for each £ e AX, (JC n 0) is_ a^ maximal linked system of 0, and (ii) for each ii e AX and each a e X, i^ (:f H 0) < e (a), then X (a) e ii, and dually , i£ ^ (a) <. U H 0) , then X^(a) e .^.

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78 PROOF. (i_) Since Tc_Z=0 U T£r, Eisan admissible subbase for X (4.3). Let ^ e X X. Clearly i:C C) Q) is linked. To see that (<^ n 0) is a maximal linked system of 0, let T e such that CC n 0) u{t} is linked. If T ^ ;£ then there is some U e E H / such that T H U = 0. Since E = u T and {T} ij { XX which may be defined by f(jC) = (/ H 0) for each ^ e XX. Suppose that .^ e X X and a e X such that (.!f n 0) = fit) < CgCa). For each S e (X 1^ E^), a S^ < if (4.10 (vi)) then so that f(A s ) < f(J^) since f is order-preserving. If S e , f(A s"^) = A S" (7.2 (ii)) and since f(jC) < eg(a) , it follows that A S" < e (a); therefore a e S so that S H X (a) ^ 0. If S ii^ , then there is some b e X such that S = X (b). In this case f(A S^) = A SgLS] (7.2 (i)) = A eg[X^(b)] = e^ih) (7.3). Thus e (b) S fiX,) < e (a); therefore b < a since e is a lattice inclusion (4.9) and S H X (a) = X (b) H X (a) 7^ 0. It follows that X. U^X (a)} is linked and so X (a) e ^. A dual argument shows the other statement.

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79 PROPOSITION 7 . 10 . If_ is_ an admissible subbase for X , then is_ equivalent to Q \j T . PROOF. Let E = 9 U T. Then I is an admissible subbase for X (7.9 (i)). For each 1^ e XX, (:(. O 0) is a maximal linked system of (7.9 (i)) so that XX algebraically dominates XX via a unique, continuous function f : XX -> XX which may be defined by f(/) = CJi ^ Q) for each J6e XX (7.5). To see that f is an iseomorphism , we need only show that f is a closed, injective lattice homomorphism (2.10), f is injective : Suppose that *, ^'Ze XX such that * ^ /^ Then we may assume, without loss of generality, that there exist T e (^ H E ) ^ and S e (/T H e ) such that T H S = 0. If both T and S are in 0, then clearly fC^) = (rf H 0) M:^n 0) = fC^). Suppose that T ^ Q ; i.e., there is some b e X such that T = X (b). Note that e (b) ^ f(.^, for if eg(b) < f («?) , then T = X (b) e X>2(7.9 (ii)) — a contradiction. Thus e (b) i f(^) and there is some T' e such that T' e e (b) and T' ^ fO^) = (X7 n 0) (4.7). It follows that X (b) £ T', T' e (/ n 0) = f{it), and hence f{£) i f{Pl). In the case that T e 0^ and S i Q , a dual argument shows that there is some S' e such that S c S', S' e f(/7), SW (;f ^ 0) = f(lt), and hence f(J() i f(^. f is a lattice homomorphism : Suppose C is a non-void subset of XX. Since f is order-preserving, it follows that for each £e C, fCt) <. f(v C)i therefore, v f[c] < f(v C). If T e v f[C] D 0^, then

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80 for each ^ e C, T e f{;t) = (^ !^ 0) (4.8); hence T e
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81 THEOREM 7 . 11 . If. J^ is_ any admissible subbase for X , then XX algebraically dominates XX and is algebraically dominated by X„X, We consider now some situations in which one lattice superextension of X dominates another lattice superextension of X not only algebraically, but also as a superextension. PROPOSITION 7.12. If and Z are admissible subbases for X such that Q £_ I and is_ weakly-normal , then (i) for each £ e X^X, (Jl H 0) is_ contained in a_ unique maximal linked system of 0, and (ii) the superextension XX dominates the superextension XX. PROOF. (i^) Suppose that £ e XX and (.£ H 0) is not contained in a unique maximal linked system of 0. Then there exist /?7, /? e XX such that ^ ^ n, CC n 0) c /'Tand (it H 0) c /?. Since «?< /?, there exist T ei^tand S e /? such that S O T = (2.16 (ii)). By the weaknormality of 0, there is a finite screen {T,, T^, •••, T } <= of S and 1 z n — T (2.11 (iii)). Since B^X = X^X, it follows that It is prime (2.18 (iii)) and hence there is some j e {1, 2, •••, n} such that T. e ^. Therefore, T n T. i' and S r\ T . ^ 0—a contradiction to the fact that {T , •••, T } is a screen for S and T. Thus {£ H 0) is contained in a unique maximal linked system of 0.

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82 (ii) .By (i) above and Proposition 7.5 (i), there is a unique dominating function f: AX-*XX which may be defined by f(j£) = {T e 0: (^ n 0) y{T} is linked} for each il e XX. To see that f is continuous, let T e and ^ e XX such that i. i f"-^[T-"]. Thus T i fW so that {T} u (i^ '^ 0) is not linked. Hence there is some S e (^ H 0) such that T H s = 0. By the weaknormality of 0, there is a finite screen H* c^ of S and T. Let C = u{W : W e y and W H s = 0}. Clearly C is a closed subset of XX. Since S e ,1^ and S H w = implies that s"^ H w"^ = (2.16 (i)), it follows that ^ i^ w"*" for each W e "i" with \^<~^S = 0; i.e.,X^C. If C. is any member of f [T"] , then since 6 e \ X ulWiWe'i'}, there is some W e H* such that QeW;i.e.,Wed. Since W e , we see that i5 e (0 n 0) c f(^) (7.2 (iii)), and since T e f(p-), it follows that T n W i^ 0; therefore, S H W = 0. Thus ^ £ i)"^ £ u iw"*" : W e H* and wns = 0} = C. It follows that f~"^CT''0 c c and ^ »^ C . Thus since » -1 -1 X was an arbitrary point not in f [T"], it follows that f [T"] is a closed subset of XX and so f is continuous. If f is a non-void collection of subsets of X, then 4* denotes the collection of arbitrary intersections of members of H* . PROPOSITION 7.13. If_ is_ an admissible subbase for X and r = tJ u tJ, then

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83 (i) for each ;C e XX, (^C C\ T) is_ a maximal linked system of T, and (ii) the superextension AX dominates the superextension X X. PROOF. (i) Let ;^ e X^X. Clearly (:f O T) is a linked system of T. If U e Tp such that (/ O T) \J {U} is linked, then either U = X € (f. n T) or there is some a e X such that U = ^r,^^)To see that Xj^(a) e ;^, let T e ;t n 0^. By the hypotheses , T = n{V e T : T c V}. For each V e Tj such that T c V, it follows that V e i£ C) T) , V n Xj^(a) ^ 0, and a e V; therefore a e T so that T D X (a) ^ 0. Since T was an arbitrary member of (Jf O Q ) , it follows that (<^ n 0j) u{>^Q(a)} is linked and hence :C uiX^Ca)} is linked. Consequently, X^Ca) e (^ n T). Dually, if U e T, and (^ O T) U{U} is linked, then U e CC H T). Thus (;f H t) is a maximal linked system of T. (ii) That the superextension AX dominates the superextension AX follows immediately from (i) above and Proposition 7.5 (ii). We say that the lattice superextension AX is "larger" than the lattice superextension A_X (or, equivalently , XX is "smaller" than XX) if and only if the superextension XX dominates the superextension XX. In the following theorem we sum up the preceding propositions obtaining two situations in which AX is actually the smallest lattice superextension of X. The proof is immediate from Propositions 7.12 and 7.13.

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84 THEOREM 7 . 14 . If. T is_ weakly-normal or if^ r = T u T , then XX is_ the smallest lattice superextension of X ; specifically , if Q is any admissible subbase , then the superextension XX dominates the superextension XX, In view of the relationship between T, r and any admissible subbase Z, namely, T £ E £ F, it is natural to ask if XX is always the smallest or if XX is always the largest lattice superextension of X. In general the answer is negative to both questions. As noted previously there is a topological lattice X for which the unique dominating function f: XX ^ XX is not continuous; i.e., XX is not smaller than XX and XX is not larger than XX (8.7). Even under the restriction that r = T u T , X X is not necessarily the largest. In Example (8.6, Part II) there is a topological lattice which satisfies this restriction, but for which there is an admissible subbase E such that there is no continuous dominating function f: XX ^ XX; i.e., XX is not the largest superextension of X. The theorem below follows immediately from Proposition 7.12 (ii) and Theorem 7.14. It is stated here for completeness. THEOREM 7.15. If_ all admissible subbases for X are weakly-normal , then XX is_ the largest lattice superextension of X and XX is_ the smallest lattice superextension of X.

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85 DEFINITION 7.16. (Kaufman [12]) Y is called an orderable compact ificat ion of a totally ordered topological lattice X if Y is a topological compactification of X and the order on X can be extended to Y so as to yield the given topology on Y as the order (interval) topology. COROLLARY 7.17. (Kaufman) If_ X is totally , then X has a largest and a_ smallest orderable compactification . PROOF. Suppose that X is totally ordered. Thus each admissible subbase for X is weakly-normal (4.12). By Theorem 7.15, XX is the largest and AX is the smallest lattice superextension of X. It is sho^^m in Appendix I that Kaufman's orderable compactif ications of X are precisely the lattice superextensions of X; therefore, AX is the largest orderable compactification of X and AX is the smallest orderable compactification of X. In view of Kaufman's method for obtaining orderable compactif ications, we may characterize the lattice superextensions of a totally ordered space X as the topological closures of h [X] in I where I is ill the closed interval [0,1], E is any subset of {f: f is a continuous increasing map of X into I such that a f[X] = and v f[X] =1} which separates the points of X, and h : X -> I is defined by it (h (x)) = f(x) Li X Li for each f e E and each x e X.

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8 . EXAMPLES EXAMPLE 8.1. Let X = (0,^) U (y,l) with the usual ordering. Then the completion of X by cuts, [0,1] with its usual ordering, is iseomorphic with XX (5.13), and the largest superextension XX is 1 3 iseomorphic with [0,—] U Ct;?!] with its usual ordering. (The maximal linked system of T defined by {(0,—)} u { (0 ,a] : — < a < 1} U {[a,l): < a < — } corresponds to the point — in XX, and the one defined by {(-,1)} u {(0,a]: y < a < 1} \j {[a,l): < a < -} corresponds to 3 . — m X^X. ) EXAMPLE 8.2. Let X = (0,1) with its usual ordering. Then T = r and consequently every lattice superextension of X is iseomorphic with [0,1] with its usual ordering. EXAMPLE 8.3.. Let n be a positive integer. For each j e {1, •••, n}, let I. = (0,1) with its usual ordering. Thus each I . is a topological lattice. Let Y be the topological product of {I.: j = 1, ••*, n}, and for each j = 1, •••, n, let Z. = T. (Note that for each I., T = r.) 85

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87 Thus E = {TrT"^[T]: T e E . , j e {1, •••, n}} is a T^-subbase for Y (3.1), and the superextension (de Groot compactification) of the product Y with respect to Z, AY (6-.Y), is homeomorphic with the product nf the superextensions {A I.: j = 1, •••, n} (3.3) (de Groot compact if icat ions J ^ {&y I.: j = 1, •••, n} (3.4)). Since each I. = (0,1), it follows from Example 8.2 that AY J " is homeomorphic with a closed n-cell in Euclidean n-space ; therefore , a de Groot compactification of an open n-cell in Euclidean n-space is the n-cell with boundary attached. Note also that since the results in Chapter 3 (Products of Superextensions) were given for arbitrary products of T -spaces , we can generalize this example to the case that Y is the product of an arbitrary collection of copies of (0,1). In particular, in view of Anderson's result [3], a de Groot compactification of Hilbert space is the Hilbert cube. EXAMPLE 8.4. Let I = [0,1] with its usual ordering, and let X = {x e I: X is rational}. Then under the usual ordering X is a topological lattice, and since I is the completion of X by cuts, AX is iseomorphic with I (5.13). For each irrational y e I, let y and y be distinct points not in X which correspond to y. Let Y = X u{y|,:yel and y is irrational} (j {y : y e I and y is irrational}.

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88 We extend the ordering on X to one on Y in the following way : (a) If y c I and y is irrational, then for every x e X and x' e X such that x <^ y
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89 lattice inclusion. The equalities below follow readily from the definitions and some properties of the real numbers: (a) If y e I is irrational, then i~"'"[Y^(y,)] = i"^[Y^(y )] = r\ {X(x) : x € X and y <^ x} and D i Du D 'I i"^[Yj(yj^)] = i"^CYj(y^)] = n{X^(x): x e X and x <^ y}. (b) If X e X, then i~^[Yj^(i(x) )] = X^(x) and i'-'^CY^dCx) )] = X^Cx), It follows from (a) and (b) that i is continuous; i.e., i is a continuous lattice inclusion. (c) i[X] = i[X], i[X] ^ = {y : y e I and y is irrational} uiCJ<], and i[X] = {y : y e I and y is irrational} Ui[X]. It follows from (c) that Y = iCX]-"-^ UiCX]-"-^ £ i[X]^ c Y, i.e., Y = i[X]^. Thus Y is a topological lattice completion of X with respect to i (5.10). To see that X is lattice dense in Y, suppose that v, w e Y with V e i[X] i[X] , w e i[X] i[X] , and w < v. It follows from (c) that there exist y, z e I such that y and z are irrational with w = y and v = z.. Since y <„ z„, it must be the case that y <^ z u . £ u Y )l I and so there is some rational number x e X such that y < x <-r 2; i.e., w = y <,, i(x) <„ z. = v. Thus X is lattice dense in Y (6.18). -'u Y Y i It now follows that Y is iseomorphic with XX where E = {i [S]: S e Z } (6.20). To see that Y is iseomorphic with X„X, we need only

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90 show that Z = r. Recall that E^^-* = {Y} u (Yj^CKx)): X e X} u {Y^CKx)): x e X} y {Y (y,,): y e I snd y is irrational} \j {Y^(y ) : y el and y is irrational} (6.15). If T is a closed ideal of X, then either T = X (x) = i~ [Y (i(x))] for some xeXorT= {xeX:x
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91 Z A W = W A Z w A p (z) , if w e A and z e B wAz,ifw,zeA,w, zeB, orw,zeC w, if w e A u B and z e C. (i) The lattice described above can be visualized as follows; (0,0) A (1,0) (2,0) (3,0) C (4,0) (ii) AX can be visualized as follows ; point in XX which is not in e^[X]. e^EC]

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92 (iii) If = t" u t" then = T U (A U B , C} and XX can be visualized as follows : points in AX which are not in e [X] (iv) r = T U{A, AUB, C} and AX can be visualized as follows: points in AX which are not in e [X] e^[C] From (ii), (iii), and (iv) we see that three quite natural subbases yield different lattice superextensions. EXAMPLE 8.5. Suppose that L and M are lattices. We define the lattice direct product, X, of L and M to be the set L x M with the

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93 ordering defined by (a,b) < (c,d) if and only if a <, c and b < d. A L M It is easily seen that the lattice direct product is indeed a lattice, Part I Let X be the lattice direct product (0,1) x (0,1) where (0,1) is the open interval with the usual ordering. Let Y be the lattice direct product [0,1] x [0,1] , where[0,1] is the closed interval with the usual ordering. X may be visualized as follows: (0,1) (1,1) (0,0) (1,0) (i) X is_ a sublattice of Y and_ Y = X^ = X^^ X^^ , but_ the inclusion map i: X ->Y is_ not continuous . Thus Y is_ not a_ topological lattice completion of X with respect to i . PROOF. It is easily seen from the definition of lattice direct product that X is a sublattice of Y , and it follows readily that Yj_ c x^^u X^ c x^ c Y^ so that Y^ = X^ (= X^^Hx^^).

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94 To see that i is not continuous, let W = i C^ ( (1,—) )]. We will show that W is not a closed subset of X. Suppose that r e X such that r ^ W, and let V be any basic open (in the topology on X generated by {X U: U is in the usual subbase for X}) set containing r. It follows from the definitions that V = AOB where A = X U{Xj^( (a^,b^) ): i = 1, 2, •••, n} and B = X U{Xt.( (c.,d.) ): j = 1, 2, •••, m} for some finite set {(a^,b.), •••, (a ,b ), (c.,d-), ••, (c ,d )} c X. Let 11 nn 11 mm — a = v{a.:i = l, 2, •••,n} and d = A {d, : j = 1, 2 , • • • , m}. By the finiteness of n and m and the properties of (0,1), it is possible to find (p,q) e X where a < p < 1 and < q < d a — . It follows that (p,q) e WHaHB = WOV. Since V was an arbitrary basic open set about r, it follows that r e W W; consequently, W is not closed in X. Let Y^ = X U{(0,0), (1,1)}. Extend the order of X to Y by letting (0,0) = a X and (1,1) = v v. Y may be visualized as follows; (1,1) (0,0) *• h

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95 (ii) The topological lattice Y is_ iseomorphic with AX. PROOF. It is easily seen that X is a sublattice of Y and Y = X nx = X . Thus Y is the completion of X by cuts and so Y is iseomorphic with XX (5.13 and 6.13). ( iii ) In this example F = T; consequently , Y is_ the unique lattice superextension of X. PROOF. To see that F = T, let C be a closed ideal of X; i.e.jCeF. IfC = X, then C € T . If C ?< X, then since C is a closed subset of X, it follows as in the proof of (i) above that C cannot be of the form {(x,y) e X: x < a for some a e (0,1)} or of the form {(x,y) e X: y < b for some b e (0,1)}. Thus there must be some (a,b) € X such that C £ X ( (a,b) ). Now by the completeness of the real numbers, it is possible to find real numbers c and d such that c = V {x: (x,y) e C} and d = v {y: (x,y) e C}. Since C ?f and C £ X ( (a,b) ), it follows that < c < a < 1 and < d < b < 1. Hence (c,d) e X. 'It is clear from the choice of c and d that C £ X ( (c,d) ) Since C is closed in the interval topology on X, and consequently in the order topology on X [ 9 ; Theorem 12], it follows that (c,d) e C. Hence C = X ( (c,d) ) 6 T . A dual argument shows that F £ T . Hence F = T, and since if E is any admissible subbase for X, we may assume without loss of generality that T c_ i (7.10), it follows that Z = T and so A^X = A^X is iseomorphic with Y .

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96 Part II In this part, let X be the lattice direct product [(0,l)u(2,3)] X [(0,1)U (2,3)] where (0,l)u(2,3) has the usual order. X may be visualized as follows: (0,3) .....v:v.-.-...-.-.v.-. ,,,,.v.......v.v..:. x:::i (0,2) i:i (0,1) (0,0) -^m^: Siiiiiii (3,0) In this example it is easily seen that T = T ij T . (i_) The lattice superextension A X naturally fills in the deleted cross so that AX is_ iseomorphic with the topological lattice Y^ ( Part I). Let Y be the lattice direct product [ (0,1] u [2,3) ] X [ (0,1] u [2,3) ], and let Y^ = Y^ u {(0,0), (3,3)}. Extend the ordering of Y to Y by letting (0,0) = a Y and (3,3) = v Y .

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97 Thus the map i: X -> Y defined by (x,y) n(x,y) is a continuous lattice inclusion and so X is a sublattice of Y . Y can be visualized as follows : (0,0) (1,2) •:::::::::::v:.:.:.:.:, (3,3) ''•'''•'•'-' 'I'l'' (2,1) 1 . . I i'^ I I I I I f V (ii) Y i_s a_ topological lattice completion of X with respect to i. PROOF. Since i is a continuous lattice inclusion, it follows from the equalities below that Y is a topological lattice completion of X with respect to i (6.10): i[X]^ = X u{(0,0), (3,3)}, iCX]^^ i[X]^ = {(x,y): x = 1 or y = 1} {(1,2), (2,1)}, iCX]^^ i[X]^ = {(x,y): x = 2 or y = 2} {(1,2), (2,1)}, and i[X]^ (i[X]^^ Ui[X]^^) = {(1,2), (2,1)}. Thus Y^ = i[X]'

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98 (iii) AX is iseomorphic with the topological lattice Y PROOF. Since each proper non-principal closed ideal of X is of the form i CY ( (a,b) )] where a = 1 or b = 1 and each proper non-principal closed dual ideal of X is of the form i tY ( (a,b) )] where a = 2 or b = 2 , it follows that r = {i~ [S] : S e E } . It is easily seen (using the equalities in (ii)) that X is lattice dense in Y (6.18). Thus Y is iseomorphic with AX (6.20). Let E = T U{R,Q} where R = i """[Y ( (1,1) )] and .-Ir Q = i CY^^( (2,2) )]. Since i is a continuous lattice inclusion, it follows that Z is an admissible subbase for X (4.3). The lattice superextension AX can be visualized as follows: V R h"" (iv) There is no continuous algebraic dominating function f: A^X -> A^X,

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99 PROOF. In order to simplify the notation in this proof, we recall that XX is iseomorphic with Y and show that there is no continuous dominating function from Y onto AX. To do this, let f be any algebraic dominating function from Y onto AX. Thus we have the following commutative diagram: Let p be the point (1,2) in Y . We now establish a few needed lemmas. (1) Either f(p) = v R or f(p) = a Q : For each (x,y) e R, i((x,y)) < (1,1) < (1,2) = p and since f is order preserving, it follows that f(i((x,y))) = e^((x,y)) < f(p). Thus v r"*" = v e^[R] < f(p). A dual argument shows that f(p) < a Q . Now, from the definition of Z it follows that there is no_i^ e AX such that v R < <^ < a Q ; consequently V r"*" = f(p) or A q"^ = f(p). Y, Y (2) p £ {(l,y): 2 < y < 3} n{(x,2): < x < 1} : It follows from the definition of the order on Y that v {(x,2): < x < 1} = (1,2) = p = A {(l,y): 2 < y < 3}. Hence by [9; Theorem 12], 4 Y Y p e {(x,2): < X < 1} and p e {(l,y): 2 < y < 3} (3) {(l,y): 2 < y < 3} c f~^[Q^] and {(x,2): < x < 1} £ f "^[r"^] : Let 2 < y < 3. It follows from the definition of E and the properties on f that f(l,y) = A {e_((a,b)): (l,y) < i((a,b))}. Also, if (a,b) e X A^X I Y^

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100 such that (l,y) <„ i((a,b)), then 1 < a and 2 < y < b. Thus (2,2) < i((a,b)) and so (a,b) e Q. (Recall that there is no element (a,b) e X with 1 < a < 2.) Hence for each (a,b) e X with (l,y) < i((a,b)), it follows that a Q < e ((a,b)); therefore aq"^ S a {e_((a,b)): (l,y) < i((a,b))} = f(l,y). Thus Q e f(l,y) (4.10 (vi)) and so (l,y) e f'^^CQ"*"]. It follows that {(l,y):2
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101 I Let X , X € X. n m (x V x') , in case n = m n I I X vx =x vx = WxVx'),in case n ^^ m and n = n m m n / m 1, in all, other cases I I X Ax = X Ax n m m n (x A x') , in case n = m or x' = 1 n (x A x'), in case n ^ m. X can be visualized as in the following diagram where the arrows denote lattice isomorphisms from I„ onto each I : On (i_) For each n e N {0}, I is a closed dual ideal of X. PROOF. Suppose v 6 X and x < y for some x e I . Then m n m n n y = X V y and since n ^ it follows from the definition of suprema ^m n "^m "^ in X that either n = m or y =1. In either case y e I and I is m m n n therefore increasing. It also follows from the definition of infima in X that I contains x Ay when x , y el. To see that I is a n n m n m n n

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102 closed subset of X, suppose that y ^ I . Thus n j^ m and y ?^ 1 . If m = 0, then we may pick x e (0,1) such that < x < y. It is easily seen that I c X^(x ) u Xy^^ ) ^'^'^ Vm ^ ^n^^r.^ ^ ^T^^n^' If "> ^ 0' n — un in m u ii x n then V is not comparable with any member of I {1} so let x e (0,1). Again it is easily seen that I^ L Xq(><^) ^ ^j^^n^ ^"^ ^m '^ ^D^^n^ ^ ^I^^n^' (ii) If D is a_ closed ideal of X, then D is_ a principal ideal . PROOF. If D = X, then D = X (1). If D ?! X, then it follows from the definition of suprema in X and the fact that D contains d^ v d^ whenever d, , d„ e D that D c I^ u I for some n e N. Since D O I ?* , 12 —Onn D n I is a non-void closed ideal of I , and since 1 <^ D H I , it n n n follows from the completeness of the reals that there is some x e (0,1) such that X = v^ (D fl I ). Since D Pi I is a closed subset of X, n I n n n it follows that x £ D H I [9; Theorem 12]. Thus X (x ) c D. If n n D n — d £ D, then either d £ I , in which case d < x , or d £ I. , in which n n u case there is some x' e (0,1] such that d = x . Thus t t d = x„ < x„ V X = (x' V x) £ D n I , and so d < (x' v x) < x ; n n n n n therefore, D c X^(x ) and so D = X^(x ). — D n D n ( iii ) I£ D is_ a proper closed dual ideal of X , then D is_ either a_ principal dual ideal or D = I for some n£N-{0}. ,^ PROOF. Suppose that D is a closed dual ideal of X which is not principal and not equal to I for any n £ N {0}. Thus D ^ I^^ ^°^ any n £ N {0}, and so D meets at least two of the I 's for n £ N {0}; ' n therefore, D H I ?^ 0. Since D is not principal, it follows that I £ D. Since D is increasing, X c_ D; i.e., D = X so that D is not proper.

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103 In view of (i), (ii), and (iii), r = Tu{l :neN-{0}}. n Let i£ = ^. ,,I for each n e N. Since for each pair n XX n *^ m,n€N-{0},I ^ 1 and I (z! I , it follows that no two members n — m m — n of {»f : n e N {0}} are comparable. Also, since "^p. = ^ e [X] = a XX, it follows that /„ < «f for each n. X„X may be visualized as follows: On r -^ (iv) For each ^e X^X, either if e e^CX] or Jf = a l"*" for s — I 1 — n ome n e N, Turning now to XX we note that for each n e N, the onlymember of T^ which contains I is X. Thus it follows from Theorem 4. I n

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lOU that for each n, a e^[I^] = {X} \j T^ and hence a e^[I^] = a X^X, The lattice AX may be visualized as follows: (v) If ^ e XX, then (i^ Pi T) is contained in a unique maximal linked system of T. PROOF. Let ^ e AX. If j^ e e^[X], then (^ Pi T) is a maximal linked system of T. If S^»^ e [X], then by (iv), ^ " ^ ^^ ~ "^n for some n e N. Thus, as was noted above, the only member of T containing I is X, and it follows that ii^ O T) = {X} U {X^(x ): ^ n n D n X e I } c {X} U T^ = a ax. If ^e AX such that a AX < W, then n n — D T 1 1 there is a principal dual ideal T e ()^ H T ) (a AX) (U.V). Clearly I 52! T, so there is some r el T. Hence X^(r ) e (^ H T); theren — n n D n n fore (^ n T) cannot be contained in ^. It follows that for each n ^€ AX, iX n T) is contained in a unique maximal linked system of T.

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105 In view of (v), XX algebraically dominates XX via a unique dominating function f : X X ^ XX which may be defined by f(^) = {U € T: U n T) \j{U) is linked} for eachJt e X X (7.5 (i)), (vi) The dominating function f : X X -> XX is_ not continuous . PROOF. Let t £ (0,1). Then t^ e ^q -^' ^"^ ^D^^O^ ^ '''d * We will show that f [(X (t ))"] is not a closed subset of X X : To see this, let r e (t,l). It is clear that e (r ) i f~ [(X (t ))-] and we will show that there is no finite subset of r whose union covers f [(X (t ))"] and misses e (r ) . This follows from the following statements : (1) For each n e N , ^^.^ £ f""^[ (X (tQ ))'''] : If n £ N, then as was shown in the proof of (v) above, f(^ ) = a XX < v (X (t ))". Thus X_(t^) £ f(^ ) (4.10 (vi)). DO n (2) If n, m £ N {0}and n ^ m, then iC y £ = e„(l): Clearly . -^ fj, p t y £ < e(l) = V AX. If S £ i? v ^ n r , then by (ii) above nmr r nmD S = X (x, ) for some x^ £ X , and it must be the case that SHI ^ and S n I ^ 0. Hence x, £ I H I ; i.e., x, = 1 and S = X. Thus m k n m k S 6 e„(l) and so e (1) < / v ^ . Hence e^(l) = / v / . r r m n r n m (3) If X„(x, ) £ ;£ and n ?! 0, then k = n or x = 1 : If X„(x, ) e H , D— k n D T< n then X (x^) Pi I ^ and x^ £ I . Thus k = n or x = 1.

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106 (4) If S £ r and f"'''[(X (t ))"] c s"^, then S = X : Suppose that 5 e r^ and f~"^[ (X^^Ct^) )"] c_ s"^. By (1) , j^^^ e s"^ for each n e N, Thus by (2) and since S is an ideal of AX (4-. 10 (i)), e (1) e S ; therefore, S = \^^'^ = ^• ( 5 ) If S e r {X}, then S contains at most two members of {;f : n e N}: If S e T^ {X}, then by (ii) above S = X (x ) for some X, e X. Since S ?^ X , x ?^ 1 , and so it follows from (3) that K K S+ r){£^: n e N) = {^q.J^}. (6.) If S e r {X}, then S contains at most one member of {/ : n e N}: Suppose that S e r^ {X}. By (iii) either S = X^(x, ) — n I Ik for some x, e X or S = I, for some keN-{0}. In the first case k k S"*" n{<;;f : n e N} = and in the second case S H Lsf : n e N} = {^ } . n n k We have shown in (5) and (5) that each member of (l {X}) contains at most two members of {^^ : n e N}. Since {/ : n e N} is an infinite subset of f [(X (t ))"], it follows that no finite subcollection of r"^ can cover f~ [(X (t ))-] and miss e (r ). Thus f~ [(X (t ))-] is not closed, and so f is not continuous. EXAMPLE 8.8. Here we exhibit a topological lattice which has a topological lattice completion in which the given lattice is not lattice dense (6.18). Let each of A, B, C, D, £, and F be the

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107 interval (0,1) with its usual ordering. Let X be made up of these six pieces in the manner shown in the following diagram where the p.'s, i = 1,2,3,4, are lattice isomorphisms between the various parts: All elements of B are less than all elements of D \j E U F. All elements of A are less than all elements of F U E. All elements of C are less than all elements of E. An element b e B is less than p (b) e A, and an element a e A is less than p (a) e C. Dually, an J. w element d ^ D is less than p (d) e F, and an element f e F is less than p (f) € E. Finally, note that a c = and v n = .

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108 Let Y contain X as in the following diagram: We extend the order on X to Y by letting 0=ax=aY,I=vx=vY, and G Y is a continuous lattice inclusion. Thus Y is a topological lattice completion of X with respect to i (6.10). It is clear from the diagram that H e .[X]"""^ .[X]^, J e .[X]"^^ .[X]''", 1 11 1 and H < J. However, {x e X: H < i(x) < J} = 0; i.e., X is not lattice dense in Y (6.18). EXAMPLE 8.9. In this example we exhibit a topological lattice X for which A. X algebraically dominates AX (and even dominates it as a superextension), but for which the dominating function of AX over A is not a lattice homomorphism (2.2 (i)).

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109 Let A, B, and C be copies of [0,1) with the usual ordering, and let X be made up of A , B , and C as in the following diagram where p and p^ are lattice isomorphisms between the indicated parts: An element a e A is less than p (a) e B, and an element c e C is less than p (c) e B. (i_) AX can be visualized as follows : A X (ii) r = T U{A,C} and AX can be visualized as follows : A X = G It is easily seen that T is a weakly-normal admissible subbase for X. Thus the superextension AX dominates the superextension A X via a unique dominating function f: A X -> AX (7.5 and 7.12).

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110 ( iii ) The dominating function f : XX -> AX is_ not a_ lattice homomorphism. PROOF. The points D and F in XX are not in e [X], and by Theorem 4.8, Daf=G=aX. Since clearly f(D) = f(F) = H in XX, it follows that f(D) a f(F) = H; therefore, f(D ^^ X ^-^ ^ ^^°^ ""x X ^^^^• Hence f is not a lattice homomorphism (cf . , 2.2 (i)).

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APPENDIX I Kaufman's Orderable Compact ificat ions The purpose of this Appendix is to show that for totally ordered topological lattices, Kaufman's [12] orderable compactifications (7.16) are precisely the lattice superextensions. Throughout this Appendix X will denote a totally ordered topological lattice. Frink [ 9 ; Theorem 3] has shown that the order and interval topologies coincide for totally ordered spaces. Kaufman [12; Theorem 2] has shown that the orderable compactif ications of X are also totally ordered spaces; hence the order and the interval topologies in an orderable compactif ication of X coincide. PROPOSITION 1(1). Every lattice superextension of X is an orderable compact ificat ion of X. PROOF. Let E be any admissible subbase for X. Then ej,[X] is iseomorphic with X (2.16 (v) and 4-. 9) so that we may identify X with e [X]. Since XX = 3 X, X is topologically dense in XX, and since Lt Li L, Li e is a lattice inclusion (4.9), it follows that the ordering on XX is an extension of that on X. Finally, we recall that the topology on XX generated by I is the interval topology (4.11). Hence XX is a Kaufman orderable compactification (cf . , 7.16). Ill

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112 PROPOSITION 1(2). Every orderable compact if icat ion of X is_ a_ lattice superextension of X. PROOF. Let Y be an orderable compact if icat ion of X. We will show that Y must be a topological lattice completion of X in which X is lattice dense. The desired conclusion will then follow from Theorem 6.20. Y is a topological lattice completion of X : Since Y is compact in the interval topology, it is complete (2.9). Since X is contained in Y as both a subspace and a sublattice, it follows that the identity 2 map i: X -* Y is a continuous lattice inclusion. To see that Y = X , let y £ Y. It is easily seen that z=v{x6X: x
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APPENDIX II Lattice Closures The purpose of this Appendix is to show how Definition 5.1 can be extended to each ordinal in order to define the "lattice closure" of sets in a complete lattice and to show that this "lattice closure" operator is a Moore closure operator, or completion-operator as in Definition 5.1. Throughout this Appendix, Y will denote a complete lattice and Z will denote a subset of Y. Note that if a is an ordinal, then a is an element of exactly one of the classes N, L, and M where N = {1,2,''«}, L is the class of limit ordinals, and M= {a+n: ae L and n e N}. Keeping this in mind we state the following inductive definition: NOTATION IKI). Suppose that Z 7! . (i) For each ordinal a, =0 =0. (ii) Let Z,Z^,Z ,Z,Z^,Z , and Z be as defined in 6.1. (iii) If n € N is even and we have defined Z and Z , then we let let If n e N is odd and we have defined Z and Z , then we ^(ntDa __ (^na^lb ^^^ ^^n+Db __ (^^b^la^ 113

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114 (iv) If a e L and we have defined Z and Z for each ordinal g < a, then let Z"^ = U{Z^^: 6 < a) and z"^ = U{Z^'': g < ex). (v) If a € L and we have defined Z and Z , then let ^(a+Da ^ (2«a^la ^^^ ^^a+Db ^ (^a^^lb^ (vi) If a + n € M, n is odd, and we have defined Z and „(a+n)b ., ^ . Z , then let (a+n+l)a _ . (a+n)a.lb (a+n+l)lb _ . (a+n)b.la If a + n e M, n is even, and we have defined Z and „(a+n)b ., T . Z , then let (vii) For each ordinal a, let Z = Z Pi Z PROPOSITION 11(2). For each ordinal g. z" c z"^ u z"^ c z"'*""''. By an argument similar to the one in the proof of Proposition 6.2 (iv) where it is shown that Z ij Z £ Z , the analogous conclusion is valid for each ordinal ct. COROLLARY 11(3). For each ordinal a, z" £ z" for each ordinal a such that a < a. The proof is immediate from Proposition 11(2) and Definition 11(1) (iv) and (vii). PROPOSITION 11(4). If Z £ W c Y, then z"^ c w"^, z"^ c w"^ , and Z c W for each ordinal a.

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115 PROOF. If a is a non-limit ordinal, then the conclusion follows from an argument analogous to that in the proof of Proposition 6.2. If a is any limit ordinal and we have shown that Z c_ W and Z c_ W for each ordinal 3 < a, then Z"^ = uiZ^^: 3 < a} c uiW^^: 3 < a} = w"^ and Z"^ = U{Z^^ 3 < a} c u{W^^ 3 < a} = w'^^ therefore, z" = z'^^ H z"^ c w"^ H w"^ = W^ PROPOSITION 11(5). There exists a least ordinal a such that Z" = Z^"^^\ PROOF. Since Y is a set, it has some cardinality. Hence the process of adding new elements of Y to Z must eventually stop; i.e., there Ct Ct'i'l is some ordinal a for which Z = Z . Since the class of ordinal numbers is well-ordered, there is some least ordinal a for which Z'^ = Z^^\ In view of Proposition 11(5), we may define the lattice closure of Z in Y unambiguously as follows : DEFINITION 11(5). If Z is a subset of Y, then the lattice ex ot ct+1 closure of Z in_ Y is Z where a is the least ordinal such that Z = Z If Z is complete, then it follows that the lattice closure of Z • „ • r,0 m Y IS Z = Z.

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116 Y Y PROPOSITION 11(7). The map C: 2-^2 which takes each subset of Y onto its lattice closure in Y is_ a_ completion operator on Y . PROOF. CI. Let A c Y. Then since A = A , it follows from Corollary 11(3) that A c C(A). C2 . Let A c^ Y. Then there is some least ordinal a such that ^a ^ ^a+1 ^ Q(^)^ Itfollows from 11(1) that C(A)^ = (a")"'= (A^^^)^ c A*^^^ = a"* = C(A) = C(A)° and hence C(C(A)) = C(A)° = C(A). C3 . Suppose that A £ B c_ Y. It follows from Proposition 11(4) that C(A) c c(B). Thus the map C is a completion operator on Y (cf. Definition 5.1) <

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BIBLIOGRAPHY [1] Alo, R. A., and Frink, 0. "Topologies of Lattice Products," Canadian Journal of Mathematics , 18(1966), 1004-1014. [2] Alo, R. A., and Shapiro, H. L. "Normal Base Compactifications ," Mathematische Annalen , 175(1968), 337-340. [3] Anderson, R. D. "Topological Properties of the Hilbert Cube and the Infinite Product of Open Intervals," Transactions of the American Mathematical Society , 126(1967), 200-216. [4] Banaschewski , B. "On Homeomorphisms between Extension Spaces," Canadian Journal of Mathematics , 12(1960), 252-262. [5] Birkhoff, G. Lattice Theory . Colloquim Publications Vol. XXV. Providence, R. I.: American Mathematical Society, 1967. [6] de Groot , J., and Aarts, J. M. "Complete Regularity as a Separation Axiom," Canadian Journal of Mathematics , 21(1969), 96-105. [7] de Groot, J., Jensen, G. A., and Verbeek , A. Superextensions . ZW 1968-017. Amsterdam: Mathematisch Centrum, 1968. [8] Frink, 0. "Compactifications and Semi -normal Spaces," American Journal of Mathematics , 86(1964), 602-607. [9] . "Topology in Lattices," Transactions of the American Mathematical Society , 51(1942), 569-582. [10] Gillman , L., and Jerison, M. Rings of Continuous Functions . Princeton, N. J.: Van Nostrand, 1960. [11] Glicksburg, I. "Stone-Cech Compactif ication of Products," Transactions of the American Mathematical Society , 90(1959), 369-382. [12] Kaufman, R. "Ordered Sets and Compact Spaces," Colloquium Mathematicum, 17(1967), 35-39. 117

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118 [13] Kelley, J. L. General Topology . Princeton, N. J.: Van Nostrand, 1955. [14] Njastad, 0. "On Wallman-type Compactif ications ," Hathematische Zeitschrift , 91(1956), 267-276. [15] Ward, M. "The Closure Operators of a Lattice," Annals of Mathematics, 43(1942), 191-196.

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BIOGRAPHICAL SKETCH Linda Joe Wahl Smith was born May 8 , 1944 , in Jacksonville , Florida. In June, 1962, she graduated from Quincy High School, Quincy, Florida. In December, 1965, she received the degree of Bachelor of Science from Florida State University. In May, 1966, she entered the Graduate School of the University of Florida. She worked as a graduate assistant at the University Computing Center during the Spring Trimester 1965-66, and from September, 1956 through August, 1968 she was a National Science Foundation Trainee. In December, 1967, she received the degree of Master of Science from the University of Florida. From that time until the present she has worked toward the degree of Doctor of Philosophy. In September, 1968, she moved to Tallahassee, Florida, where she has worked as a part-time instructor of mathematics at Florida State University and Tallahassee Junior College . Linda Smith is the wife of Nevins Carson Smith, Jr. and the daughter of Mr. and Mrs. J. Kermit Wahl. She is a member of Phi Beta Kappa, Phi Kappa Phi, Pi Mu Epsilon, Kappa Delta Pi, Alpha Lambda Delta, and the Florida Council of Teachers of Mathematics. 119

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This dissertation was prepared .under the direction of the chairman of the candidate's supervisory committee and has been approved by all members of that committee. It was submitted to the Dean of the College of Arts and Sciences and to the Graduate Council, and was approved as partial fulfillment of the requirements for the degree of Doctor of Philosophy. December, 1969 Dean, Colleg nd Scxfences Dean , Graduate School Supervi^spry Committee (aui^ ; \.:.iP^

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