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1 Electronic Notes in Theoretical Computer Science 6 (1997) URL: 9 pages Computational Models for Ultrametric Spaces Bob Flagg Department of Mathematics and Statistics University of Southern Maine Portland ME , USA flaggusm.maine.edu Ralph Kopperman Department of Mathematics City College of CUNY New York, NY 10031, USA rdkcccunyvm.cuny.edu Abstract For every ultrametric space, the set of closed balls of radius 0 or 2 n for some n, form an algebraic poset under reverse inclusion. If the ultrametric space is complete separable, then they form a Scott computational model for it. Conversely, every topological space with an algebraic computational model is a complete separable ultrametrizable space. 1 Introduction A computational model [8] for a topological space X is an!-continuous dcpo P together with an embedding i : X! Max(P )suchthat (i) the restrictions of the Scott topology,, and the Lawson topology,, to Max(P ), are equal and (ii) i : X! (Max(P ) jmax(p )) is a homeomorphism. In the seminal work [2,3] Edalat initiated the study of classical mathematical structures via computational models. He constructed computational models for metrizable locally compact Hausdor spaces and used them in a wide variety of applications involving measures, dynamical systems and iterated function systems and fractals. The success of Edalat's program focused attention on the question of which topological spaces have computational models. Lawson [8] settled this question by showing that a topological space has a computational model i it is a Polish space (that is, a complete, separable metrizable space). Using formal c1997 Published by Elsevier Science B. V. Open access under CC BY-NC-ND license.

2 balls, Edalat and Heckmann [4] provided a simple explicit construction of a computational model for a Polish space. They also gave domain theoretic proofs of several classical results on Polish spaces and constructed computational models for measure theory over these spaces. Since the domains used in denotational semantics are almost always algebraic, it is natural to ask which spaces have algebraic computational models that is, which spaces have computational models (P ) with P an!-algebraic dcpo. Kamimura and Tang [6] have shown that a space X has a bounded complete algebraic computational model i it has a clopen countable base C closed under consistent nite intersections such that every ltered subset of C has a nonempty intersection. Such spaces are easily seen to be complete separable ultrametic spaces. Morover, if X is a complete separable ultrametic space with countable dense subset Y,then C = fn 2 n(y)jn 2 N & y 2 Y g is a clopen countable base for X closed under consistent nite intersections every ltered subset of which has a nonempty intersection. Thus X has a bounded complete algebraic computational model i X is a complete separable ultrametic space. In this note we remove the condition of bounded completeness from this characterization and give a simple explicit construction { via closed balls under reverse inclusion { of a bounded complete algebraic computational model for complete separable ultrametic space. 2 Ultrametric Spaces and Scott Domains By an ultrametric space we mean a pair (X d), where X is a set and d : X X! [0 1] satises the following conditions for all x y z 2 X: (i) d(x x) =0 (ii) d(x y) =d(y x) (iii) d(x y) _ d(y z) d(x z) and (iv) d(x y) = 0 implies x = y. For x 2 X p > 0, N p (x) =fy j d(x y) pg and B p (x) =fy j d(x y) <pg. Let A = f 1 2 n jn 2 Ng [f0g and for X an ultrametric space let Partially order AX by reverse inclusion: For 2 AX, let AX = fn p (x)jx 2 X&p 2 Ag: v () =minfp 2 Aj9x: = N p (x)g: Ultrametric spaces are unusual in that each element of a ball is a center of it that is: if y 2 N p (x) then N p (y) =N p (x). This is immediate from the 2

3 fact that if y 2 N p (x) z 2 N p (y), then d(x z) maxfd(x y) d(y z)g p, so z 2 N p (x) thus N p (y) N p (x), but since symmetry implies that x 2 N p (y) in this situation, we have the reverse inclusion as well. As a result, for 2 = maxfd(y z) j y z 2 g: for let = (x) then for any y 2, = (y). Because A is well-ordered by, we can choose z 2 for which d(y z) ismaximum then certainly d(y z) =@. A standard example of ultrametric space is the Cantor space, given in the form C = f0 1g!, with d(x y) = 2 V fnjxn6=yng. d(x y) = maxfx yg if x 6= y. Another is simply A, with Lemma 2.1 Assume D AX is directed (byv). Then either D has a largest element or Proof. Suppose inff@j 2 Dg =0: q =inff@j 2 Dg > 0: Choose 2 D < 2q. Suppose v 2 D. Then and if But then, since for a b 2 A, b>a) b 2a, 2q>@ 2@ and so q>@,which is absurd. Thus =. It follows that is maximal in D and since D is directed, is the largest element ofd. 2 Theorem 2.2 For every directed subset D of AX there is an ascending sequence 1 v 2 v ::: of elements of D which has the same upper bounds as D. Proof. If D has a largest element,, then we can take i = for i = :::. Otherwise inff@j 2 Dg =0 and we canchoose for each n an element n 2 D with 1 2 n. Let 1 = 1 and for n>1let n be an upper bound of n 1 and n in D. Suppose is an upper bound of all the n and let = N p (x) 2 D. Suppose z 2. Let y 2 \ n. Then p n d(x y) _ d(y z) d(x z): n! 0, so p d(x z) that is, z 2. Hence v. 2 Proposition 2.3 Assume 1 v 2 v ::: is an ascending sequence inax and lim n =0. For 2 AX, the following are equivalent: (i) = W n n (ii) is an upper bound of ( n ) n (iii) For some x 2 X = fxg = T n n. Proof. (1) =) (2): Trivial. (2) =) (3): lim n = 0 and for all n, n, so@ =0. Thus there is an x 2 X such that fxg =, and clearly = fxg = T n n. (3) =) (1): This follows at once from the fact that v is reverse inclusion.2 3

4 Lemma 2.4 Every Cauchy sequence (x n ) n in X has a subsequence (x nk ) k such that (N 1 2 k (x nk )) k is ascending in AX. Proof. Let n 1 = 1 and for all k 2 N choose n k n k 1 so that for all i j 1 n k 2 k d(x i x j ). Then N 1 2 k (x nk ) v N 2k+1 1 (x nk+1 ). 2 Theorem 2.5 For an ultrametric space X, the following are equivalent: (i) X is Cauchy complete. (ii) AX isacpo. (iii) In AX every ascending sequence has a supremum. Proof. By Theorem 2.2, (2) and (3) are equivalent. Assume X is Cauchy complete. Let ( n ) n be an ascending sequence in AX. If ( n ) n is eventually constant in AX, then it trivially has a supremum. Otherwise, by Proposition 2.1, lim n = 0. Let x n 2 n for n = 1 2 :::. Then (x n ) n is a Cauchy sequence in X and if lim n!1 x n = x, thenfxg = T n n. By Proposition 2.3, N 0 (x) = W n n. Hence AX is a dcpo. Assume (3) holds. Let (x n ) n be a Cauchy sequence in X. By Lemma 2.4, (x n ) n has a subsequence (x n k ) k such that(n 1 2 k (x nk )) k is ascending in AX. By (3) this sequence has a supremum. Since lim 1 2 k (x nk ) = 0, by Proposition 2.3, there is an x such that T n N 1 2k (x nk ) =fxg. Clearly, lim k!1 x n k = x. Since (x n) n is Cauchy, lim n!1 x n = x. Thus X is Cauchy complete. 2 An element a in a poset P is compact if for every directed subset D of P which has a supremum, if a W D,thenforsomed 2 D, a d. Lemma 2.6 Assume 2 AX. Then is compact i there is a p > 0 and x 2 X such that = N p (x). In particular, the non-compact elements of AX are those of the form N 0 (x), forx a non-isolated point in X. Proof. If for all p>0andx 2 X, 6= N p (x), then = fxg for some x 2 X. In this case = W n N 2 n(x) and for all n, 6v N 2 n(x). Consequently, is not compact. Assume there is a p>0andx 2 X such that = N p (x) and suppose W v n n for an ascending sequence ( n ) n. If ( n ) n,iseventually constant, then trivially there is an n such that v n. Otherwise, by Proposition 2.1, lim n =0. By Proposition 2.3, W n n = T n n = fyg, for some y 2 X. Choose n so that n. For z 2 n p d(z y) _ d(y x) d(x z). Thus n thatis, v n. 2 Recall that a subset S of a poset P is a basis for P if for each x 2 P, the set fs 2 Sjs xg is directed and has supremum x. As is well know, P has at most one basis consisting of compact elements, namely, the set of all compact elements. A poset is algebraic if its compact elements form a basis, and!-algebraic if it is algebraic and its set of compact elements is countable. Theorem 2.7 If X is any ultrametric space, then C = fn p (x) j x 2 X p 2 A&p > 0g is the unique basis of AX consisting of compact elements. Further, 4

5 if Y X, then Y is dense in X if and only if C = fn p (y)jy 2 Y p 2 A&p > 0g. Proof. In the last lemma, we showed that each element ofc is compact. To show that W it is a basis, it remains to be seen that for each N p (x) 2 AX, N p (x) = (# N p (x) \C) and this family is directed. This is clear if N p (x) is compact, for then N p (x) 2# N p (x) \C. But otherwise, p = 0 and then N 0 (x) = W N 2 k(x) by Proposition 2.3, each N 2 k(x) 2Cand the sequence is increasing. The remaining assertion of the theorem results from the fact that if Y is dense, then for each p > 0, N p (x) is nonempty and open, so it meets Y, and for y 2 N p (x) \ Y, N p (x) = N p (y), by the paragraph following the denition of ultrametric space if Y is not dense then some N p (x) \ Y =, so N p (x) 2CnfN p (y)jy 2 Y p 2 A&p >0g, thus by the above, the latter cannot be a basis. 2 Corollary 2.8 For every ultrametric space X, AX is an algebraic poset. AX is!-algebraic i X is separable. Let i : X! AX be dened by i(x) =fxg. This is clearly a one-to-one correspondence between X and the subset Max(AX) = ffxgjx 2 Xg of AX. Proposition 2.9 The elements of Max(AX) are precisely the maximal elements of AX. For N p (x) 2 AX with p 6= 0,i 1 " N p (x) =N p (x). Since the sets N p (x) for x 2 X and p>0 form a clopen basis for the metric topology on X and p<d(y x) implies N p (x) \ N p (y) =, the next result is clear. Theorem 2.10 Assume X is an ultrametric space. On the subset Max(AX), the relative Scott and the relative Lawson topologies agree. Moreover, the function i : X! AX is a homeomorphism from X with the metric topology onto Max(AX) with the relative Scott or Lawson topology. Since for an ultrametric space, if z 2 N p (x) \ N q (y), then N p^q (z) = N p (x) \ N q (y), AX is bounded complete. Moreover, by Theorem 2.7, if X is separable, AX has a countable basis. A computational model (P ) isscott if P is a Scott domain that is P is!-algebraic and bounded complete. Theorem 2.11 Assume X is a complete separable ultrametric space. Then X has a Scott computational model. 3 Algebraic Computational Models In this section we show that a space has an algebraic computational model i it is a complete separable ultrametric space. Suciency follows from Theorem 2.11 and necessity is obtained by modifying in an obvious way Law- 5

6 son's argument [La] characterizing spaces with computational models as Polish spaces. Denition 3.1 Assume P isaposet and d is an ultrametric on P. Then d is radially convex if whenever x y z, then d(x y) _ d(y z) =d(x z): A ultra-quasi-metric space is a pair (X d), where X is a set and d : X X! [0 1] satises the following conditions for all x y z 2 X: (i) d(x x) =0 (ii) d(x y) _ d(y z) d(x z) and (iii) d(x y) =0=d(y x) implies x = y. For x 2 X p > 0, N p (x) =fy j d(x y) pg and B p (x) =fy j d(x y) <pg. Given an ultra-quasi-metric space, d is the topology generated bythesetofall B p (x) x2 X p > 0, d is the ultra-quasi-metric dened byd (x y) =d(y x), and d s is the ultrametric d + d. Proposition 3.2 Assume P is an!-algebraic poset. Then there is an ultraquasi-metric d on P such that d =, d =!, d s = and d s is radially convex. Proof. Let B = fb n g n2n be a countable base for P consisting of compact elements. Dene d : P P! [0 1] by Then for any x y 2 P, d(x y) = inff2 n j8k <n:(b k x =) b k y)g: (1) 2 n >d(x y) () 8k n:(b k x =) b k y): From (1) it follows easily \ that d is an ultra-quasi-metric and \ for x 2 P, n 2 N B n(x) = d 2 " b k and B n(x) = d 2 Xn "b k : kn b k x kn b k 6x Hence d =, d =! and so d s =. It remains to show thatd s is radially convex. Suppose x y z. If2 n 6 d(y x), then there is a k<nsuch that b k y and b k 6 x. But then b k z and b k 6 x, so2 n 6 d(z x). It follows that d(z x) d(y x). Similarly, d(z x) d(z y). Thus d s (x y) _ d s (y z) =d(y x) _ d(z y) =d(z x) =d s (x z): 2 We now need an ultrametric version of Alexandro's Theorem. The proof below is a simple adaptation to the ultrametric space setting of the argument given in [1]. Theorem 3.3 The relative topology on a G space is induced byacomplete ultrametric. subset of a complete ultrametric Proof. Let A be a G subset of the complete ultrametric space X. Belowwe 6

7 dene an equivalent complete ultrametric on A. By denition, 1\ A = U i i=1 where each U i is open in X and, without loss of generality, wemay assume that U1 U2 :::. Dene f i : U i! (0 1) by f i (x) = 1 d(x X n U i ) : (Of course, since X n U i is closed, d(x X n U i ) 6= 0 if x 2 U i.) Now let i : U i U i! [0 1) be dened by (2) (3) i (x y)=0 if f i (x) =f i (y) i (x y)= p 1 _ p otherwise, where p = f i(x) _ f i (y): The function i will in general not be an ultrametric for U i, because it is possible to have i (x y) =0whenx6= y. However, i is a pseudo-ultrametric (that is, it satises the other ultrametric axioms). Clearly, i (x x) =0and i (x y) = i (y x). So we need only show that (4) i (x y) _ i (y z) i (x z) for all x y z 2 U i. If f i (x) =f i (y), f i (y) =f i (z) orf i (x) =f i (z), (4) follows trivially from the denition of i.sosupposef i (x) f i (y) and f i (z) are distinct and let Then p = f i (x) _ f i (y) q = f i (y) _ f i (z) and r = f i (x) _ f i (z): (5) p _ q r>0: From (5) it follows easily that p 1 _ p _ q 1 _ q r 1 _ r which is (4). The desired ultrametric for A is given by (x y) =d(x y) _ sup 2 n n (x y): n As a supremum of ultrametrics, is indeed an ultrametric on A. To see that A is complete in this metric, note rst that if fx n g n is a Cauchy sequence relative to, thenitiscauchy relative to the smaller d. Hence it has a limit, x, in (X d). If x 2 A, we now show thatx n! x relative to : for > 0 let 2 n < it is possible to choose m so that if p m then d(x x p ) 1(x x p ) ::: n (x x p ) 2 n, and since each k is bounded by 1, this implies that (x x p ) 2 n _ 2 n 1, as required. Thus, provided that we can show each such x 2 A, every Cauchy sequence in (A ) converges and = d ja. But if x 62 A, there is an integer N such that for all n>n, x 2 X nu n. Let x k be a term in the sequence (x n ) n>n and consider i (x k x k+j) for i>n. Since x k+j! x d(x k+j X n U i )! 0 and so f i (x k+j)!1. Thus 7

8 i (x k x k+j)! 1asj!1. But then inf k sup j (x k x k+j) 2 N and (x n ) n is not Cauchy, which is absurd. Thus x 2 A. 2 In [8], it is shown that if P is an!-continuous dcpo and the relative Scott topology and relative Lawson topologies agree on Max(P ), then Max(P ) with the relative Scott topology is a Polish space. The following proposition is the ultrametric version of this result. Proposition 3.4 Assume P is an!-algebraic poset such that the relative Scott topology and the relative Lawson topology agree onmax(p ). Then Max(P ) with the relative Scott topology is a complete separable ultrametric space. Proof. The poset P with the Lawson topology embeds as the spectrum of its lattice of Scott open sets [7]. Since P is!-algebraic, is as well. By Proposition 3.2, there is an ultrametric d on which induces the Lawson topology. Since the Lawson topology on is compact, ( d) is complete. Since P = spec() is a G subset of ([G&] Exercise III, 4.21), P with the relative topology, which is the Lawson topology, is also a complete ultrametric space. Finally, by Corollary 2.5 of [La], Max(P )is G in P andisthus a complete ultrametric space in the relative Lawson (or Scott) topology. 2 Theorem 3.5 A topological space X has an!-algebraic computational model i it is a complete separable ultrametric space. Acknowledgement The authors would like to thank David Briggs and Philipp Sunderhauf for nding and correcting an error in an earlier version of this paper. References [1] Hocking, John G. and Gail S. Young Topology. Dover, 1988, New York, reprinted and corrected from Addison-Wesley, Reading, MA, [2] A. Edalat. Domain theory and integration. Theoretical Computer Science, 151, 1995, pp. 163{193. [3] A. Edalat. Dynamical systems, measures and fractals via domain theory. Information and Computation, 120:32{48, July [4] A. Edalat and R. Heckmann. A computational model for Metric Spaces. Preprint, [5] G. Gierz, K. H. Hofmann, K. Keimel, J. D. Lawson, M. Mislove, D. S. Scott. A Compendium of Continuous Lattices. Springer-Verlag, [6] T. Kamimura and A. Tang. Total Objects of Domains. Theoretical Computer Science, 34, 1984, pp. 275{288. 8

9 [7] J. D. Lawson. The Duality of Continuous Posets. Houston J. Math. 5, 1979, pp. 357{394. [8] J. D. Lawson. Spaces of maximal points. Preprint,

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