Rudin s Lemma and Reverse Mathematics

Transcription

1 Annals of the Japan Association for Philosophy of Science Vol25 (2017) Rudin s Lemma and Reverse Mathematics Gaolin Li, Junren Ru and Guohua Wu Abstract Domain theory formalizes the intuitive ideas of approximation and convergence in a very general way, and provides a fundamental tool in the study of computing theory and computability theory In search of complete lattices on which the Laswon topology is Hausdorff, Gierz, Lawson and Stralka introduced in [3] quasicontinuous lattices, which inherit many good properties of domains Gierz, et al pointed in [3] that Rudin s Lemma for finding a cross-section of certain descending family of sets plays a central role in the development of the whole theory of quasicontinuous lattices In this paper, we study Rudin s Lemma from reverse mathematics point of view and prove that the Rudin s Lemma is equivalent to ACA 0 over RCA 0 1 Introduction Domain theory formalizes the intuitive ideas of approximation and convergence in a very general way It provides a fundamental tool in the study of computing theory and computability theory The idea was proposed independently by Dana Scott in paper [6] and Yuri Ershov in paper [1] around 1970s Domain theory studies various topologies on partially ordered sets, such as Alexandrov topology, Scott topology, and Lawson topology, etc A domain refers to a continuous dcpos (directly complete partial orders) In their work of generalizing the theory of continuous dcpos to general ordered structures, and also in search of complete lattices on which the Laswon topology is Hausdorff, Gierz, Lawson and Stralka introduced in [3] quasicontinuous lattices, School of Mathematics and Statistics, Yancheng Teachers University, Yancheng, Jiangsu Province , China Division of Mathematical Sciences, School of Physical & Mathematical Sciences, Nanyang Technological University, Singapore Acknowledgement of grants: G Li is partially suported by Jiangsu Overseas Reserach and Training Program for University Prominent Young and Middle-aged Teachers and Presidents and Natiural Science Foundation of Yancheng teachers university (Grant No13YSYJB0109) and Jiangsu Higer Education Institutions of China (Grant No14KJD110008) G Wu is partially supported by AcRF grants MOE2016-T from Ministry of Education of Singapore, RG29/14, M and RG32/16, M from Ministry of Education of Singapore 57

2 58 Gaolin Li, Junren Ru and Guohua Wu Vol 25 where many good properties of domains remain valid In this theory, one of the milestone results is Rudin s Lemma, a lemma enables us to find a cross-section of certain descending family of sets Rudin s Lemma[2]: Given a poset (P, ), and {F i } i I a -directed family of nonempty finite subsets of P, there is a -directed subset D i I F i that meets all F i Rudin s Lemma implies that the Scott topology on quasicontinuous domain is locally compact and sober, and Lawson topology on quasicontinuous domain is regular and Hausdorff In this paper, we study Rudin s Lemma from reverse mathematics point of view and prove that the Rudin s Lemma is equivalent to ACA 0 over RCA 0 We organize the paper as follows In Section 2, we introduce basic definitions about domain theory and reverse mathematics We will show in this section that RCA 0 can prove the equivalence between chain complete and directed complete In Section 3, we prove the equivalence between Rudin s Lemma and ACA 0 over RCA 0 In Section 4, we study a stronger version of Rudin s Lemma, and prove that this strengthened version is equivalent to ACA 0 over RCA 0 2 Preliminaries Let (X, ) be a partially ordered set, and P f (X) be the collection of all finite subsets of X For A X, denote A = {x X :( a A)[a x]}, the upper set of A Definition 1 For A, B P f (X), define on P f (X) is called the Smyth preorder A B B A It is obvious that A B ( b B) ( a A) [a b] Definition 2 For an ordered set (X, ), a subset D of X is directed if x, y D, there exists z D such that x, y z So, in Rudin s Lemma, {F i } i I is a -directed family means for all F i and F j, there exists F k such that F i,f j F k Definition 3 For an ordered set (X, ), 1 X is directed complete, if for all directed subsets D X, the least upper bound of D exists in X 58

3 Rudin s Lemma and Reverse Mathematics 59 2 X is chain complete, if for all chains {x i } i Λ X, the least upper bound of the chain exists in X In [5], Markowsky proved that assuming well-ordering principle, directed completeness and chain-completeness are equivalent We will have a look at this equivalence from reverse mathematics point of view and show that this it can be proved in RCA 0 Reverse mathematics is a program devoted to determine the strength of mathematical theorems by calibrating the precise set existence axioms necessary and sufficient to carry out their proofs in second-order arithmetic Definition 4 (Simpson [7]) The axioms of second order arithmetic consist of the universal closures of the following L 2 -formulas: (1) basic axioms: n +1 0, m +1=n +1 m = n, m +0=m, m +(n +1)=(m + n)+1, m 0=0, m (n +1)=(m n)+m, m <0, m<n+1 (m <n m = n) (2) induction axiom: (0 X n (n X n +1 X)) n (n X) (3) comprehension scheme: X n (n X φ(n)), where φ(n) is any formula of L 2 in which X does not occur freely (4) By second order arithmetic, denoted as Z 2, we mean the formal system in the language L 2 which are deducible from those axioms by means of the usual logical axioms and rules of inference By a subsystem of Z 2, we mean a formal system in the language L 2 each of whose axioms is a theorem of Z 2 In this paper, we will be mainly focused on two subsystems: RCA 0 and ACA 0 Definition 5 1 The system RCA 0 consists of basic axioms together with schemes of Σ 0 1 induction and Δ 0 1 comprehension 59

4 60 Gaolin Li, Junren Ru and Guohua Wu Vol 25 2 The system ACA 0 consists of basic axioms and the induction axiom together with the arithmetic comprehension RCA 0 is a kind of formalized recursive mathematics, as the development of ordinary mathematics within RCA 0 corresponds to the positive content of computable mathematics In reverse mathematics, RCA 0 also plays a role of a weak base theory on which many equivalences between mathematical statements and stronger subsystems are established Here we prove in RCA 0 two basic facts about partial orders Proposition 1 (RCA 0 ) Let (X, ) be a countable poset Then X is directed complete if and only if X is chain complete Proof One direction is trivial Since any chain must be directed subset, then directed complete implies chain complete For another direction, let X be chain complete, D X be a directed subset Let D = {d 0,d 1,,d n,} Define a chain E D as following: e 0 = d 0 ; e 1 Dsuchthate 0,d 1 e 1 ; e 2 Dsuchthate 1,d 2 e 2 ; e n Dsuchthate n 1,d n e n ; e n above exists because D is directed The chain E = {e 0,e 1,,e n,} exists since d n E m n[d n = e m ] It is easy to check that sup D =supe SinceX is chain-complete, sup E exists, and hence sup D exists The main part of this paper is to prove the equivalence between Rudin s Lemma and ACA 0 Below is an easy fact about finite directed sets provable within RCA 0 Proposition 2 (RCA 0 ) Let (X, ) be a countable poset A P f (X), thena is directed if and only if A contains a greatest element of A Proof Suppose A = {a 0,a 1,,a n }IfA is directed, let b 1 Asuchthata 0,a 1 b 1 60

5 Rudin s Lemma and Reverse Mathematics 61 b 2 Asuchthatb 1,a 2 b 2 b n Asuchthatb n 1,a n b n so b n A is the greatest element of A For the other direction, if A contains a greatest element b of A, then certainly any two elements of A are less than or equal to b, and hence A is directed In Proposition 2, the finiteness of A is necessary for the forward direction A simple counterexample is to let X be N under inclusion, A be the set of all finite subsets of N, thena is directed but has no greatest element The following are well-known facts about ACA 0 Proposition 3 (Simpson [7]) The following are equivalent over RCA 0 : 1 ACA 0 ; 2 For all one to one function f : N N, there exists a set X N such that n(n X m(f(m) =n)); 3 König lemma: Every infinite, finitely branching tree T N <N has at least one infinite path 3 Rudin s Lemma and ACA 0 In this section, we show the equivalence between Rudin s lemma and ACA 0 over RCA 0 Theorem 1 The following are equivalent over RCA 0 : 1 ACA 0 ; 2 Rudin s Lemma: Given a countable poset (X, ), and F = F i : i N is a -directed family of nonempty finite subsets of X, then there exists a -directed subset D i N F i that meets all F i Proof We first prove Rudin s Lemma within ACA 0 LetX N We define a function f : N N as following: f is well-defined since F is -directed f(0) = 0, f(n +1) = μk[f n+1 F k,f f(n) F k ] 61

6 62 Gaolin Li, Junren Ru and Guohua Wu Vol 25 Now, consider a new family B = B i : i N where B 0 = F f(0), B 1 = F 1, B 2 = F f(1), B 2n = F f(n), B 2n+1 = F n+1, It s obvious that B is also -directed and contains all F i,andthat i N B i = i N F isoitissufficienttofind -directed subset D i N B i that meets all B i Define a tree T ω <ω such that σ T n<lh(σ)[σ(n) B n ] T is infinite, finitely branching tree because every B n is finite set Now, we construct a subtree S T such that for any infinite path g of S, the nodes on g forms a -directed set, ie, {g(n) :n N} is -directed S is constructed as follows: For σ T, S; if lh(σ) =1,thenσ S; if lh(σ) =2n(n 1), then σ S η σ[η S]; if lh(σ) =2n +1(n 1), then σ S ( η σ[η S]) [σ(2n 2),σ(2n 1) σ(2n)] 62

7 Rudin s Lemma and Reverse Mathematics 63 From the last case, if σ S and lh(σ) > 2k, thenwehave σ(0),σ(1) σ(2), σ(2),σ(3) σ(4), σ(2k 2),σ(2k 1) σ(2k), so σ(2k) is the greatest element among {σ(n) :n 2k}, it implies {σ(n) :n 2k} is -directed by proposition 2 Obviously, S is a finitely branching tree We have the following claims for S: 1 n x B 2n, there exists σ S such that lh(σ) =2n +1andσ(2n) =x 2 S is infinite tree 3 For every infinite path g of S, theset{g(n) :n N} is -directed By Claim 2, with König lemma (which is equivalent to ACA 0 by proposition 3), S contains at least one infinite path g By Claim 3, with Σ 0 1-comprehension, D = {x : n[x = g(n)]} exists and forms a -directed set As D meets all B i, D is the desired set We prove Claim 1 by arithmetical induction Let φ(n) x[x B 2n σ S[lh(σ) =2n +1 σ(2n) =x]] It is obvious that φ(0) is true Suppose φ(k) is true for all k n Toshowφ(n +1) is true, we consider for any x B 2n+2 = F f(n+1),sinceb 2n+1 = F n+1 F f(n+1) 63

8 64 Gaolin Li, Junren Ru and Guohua Wu Vol 25 and B 2n = F f(n) F f(n+1), there is some y B 2n+1,andz B 2n such that y x and z x By induction, φ(n) holds, there is an η S such that lh(η) =2n +1and η(2n) =z By construction, we have η y S, and hence η y x S Sowechoose σ = η y x,andφ(n +1)holds Claim 2 follows directly from Claim 1 To show Claim 3, suppose that g is an infinite path of S Let D = {x : n[x = g(n)]} We will prove D is -directed For any x, y D, then exists l N such that x, y {g(0),g(1),,g(l)} Wolg, assume l is even, then g(l) is the greatest element among {g(0),g(1),,g(l)}, sox, y g(l) andd is directed We now prove the other direction We assume Rudin s Lemma, and let f : N N be a one to one function Consider (N, ) with the standard order, and define { {2i}, if k j such that f(k) =i; A i,j = {1}, otherwise where, is a standard coding from N 2 to N A i,j exists by Δ 0 1-comprehension: x A i,j (x =2i k j[f(k) =i]) or (x =1 k j[f(k) i]) A i,j : i, j N is -directed family because any two A i1,j 1,A i2,j 2 is - comparable By Rudin s Lemma, there exists D i,j N A i,j such that D meets all A i,j Let X = {i N : 2i D}: X exists by Δ 0 1-comprehension Then n(n X m(f(m) =n)), so ACA 0 holds 4 Further Research One may ask the following question: in Rudin s Lemma, is it possible to find a directed subset D of i I F i which intersects each F i at exactly one point? Heckmann and Keimel pointed out in [4] that in general, we cannot have a positive solution For example, if the family contains the subsets {1}, {2} and {1, 2}, then the directed set D must intersect {1, 2} at two points However, we can give a positive solution to this question if {F i } i I is a disjoint family In the following, we prove that this variant is equivalent to ACA 0 over RCA 0 Theorem 2 The following are equivalent over RCA 0 : 1 ACA 0 ; 2 Given a countable poset (X, ), and F = F i : i N a -directed family 64

9 Rudin s Lemma and Reverse Mathematics 65 of disjoint nonempty finite subsets of X, then there exists a -directed subset E i N F i that each E F i is a singleton Proof We first assume ACA 0 The proof of (2) is quite similar to the proof of Theorem 1, where f will be constructed The following construction will ensure that each B i D is -linearly ordered, and following this, we can select the greatest element as the unique witness First, we define functions f : N N and g : N N simultaneously as follows: f(0) = 0,g(0) = 1, f(1) = μk[f f(0),f g(0) F k ], g(1) = μk[k / {f(0),g(0),f(1)}], f(n +1)=μk[F f(n),f g(n) F k ], g(n +1)=μk[k / {f(0),g(0),,f(n),g(n),f(n +1)}], Now, consider a new family B = B i : i N where B 0 = F f(0), B 1 = F g(0), B 2 = F f(1), B 3 = F g(1),, B 2n = F f(n),b 2n+1 = F g(n), Obviously, B is also a -directed family and contains all F i,and i N B i = i N F i, So it is sufficient to find -directed subset E i N B i that meets each B i at exactly one point, and hence meets each F i also at exactly one point For B, we have the following properties: for any i, j, either B i = B j or B i B j =, but for each n, B 2n+1 B 0,B 1,,B 2n We now construct T N <N and S T as Theorem 1 The same arguments help us find a directed set D which meets all B i We then construct a set E D such that: x E (x D) ( i y[x, y B i D y x]) which means that we choose the greatest element from B i D as the witness We have the following three claims 1 For all i, B i D is -linearly ordered So the greatest element of B i D exists in B i D 2 E intersects each B i at exactly one point 3 E is directed 65

10 66 Gaolin Li, Junren Ru and Guohua Wu Vol 25 For Claim 1, let D = {h(n) :n N}, whereh is an infinite path of tree S Now fix i, weconsiderb i D Assume B i repeats, like B i = B j1 = B j2 = B j3 = with i j 1 j 2 j 3, then by the construction, we have all j 1,j 2, are even number Hence h(i) h(j 1 ) h(j 2 ) h(j 3 ) SoB i D is -linearly ordered Because B i is finite, the greatest element of B i D exists Claim 2 follows directly from Claim 1 since the greatest element is unique For Claim 3, note that for all x, y E, x, y D AsD is directed, there exists some z D such that x, y z Assume z B i and w is the greatest element in B i D Then z w and w E, and hence w is an upper bound of x and y in E Thus, E is directed We now prove the other direction Assume f : N N is a one to one function Consider (N, ) with the standard order, and define A i,j = { {5 i+1 } if f(j) =i, {2 i 3 j } otherwise A i,j exists by Δ 0 1-comprehension, and they are disjoint because f is one to one A i,j : i, j N is also a -directed family because any two A i1,j 1,A i2,j 2 is -comparable By (2), there exists E i,j N A i,j such that E meets all A i,j Let X = {i N : 5 i+1 E}, X exists by Δ 0 1-comprehension Then n(n X m(f(m) =n)), so ACA 0 holds References [1] Y L Ershov Computable functionals of finite types Algebra and Logic, 11(4): , 1972 [2] G Gierz, KH Hofmann, K Keimel, JD Lawson, M Mislove, and D Scott Continuous Lattices and Domains Cambridge University Press, 2003 [3] G Gierz, J D Lawson, and A Stralka Quasicontinuous posets Houston J Math, 9(2): , 1983 [4] R Heckmann and K Keimel Quasicontinuous domains and the smyth powerdomain Electronic Notes in Theoretical Computer Science, 298: , 2013 [5] G Markowsky Chain-complete posets and directed sets with applications Algebra Universalis, 6(1): 53 68, 1976 [6] D Scott Continuous Lattices Springer, 1972 [7] S G Simpson Subsystems of Second Order Arithmetic Cambridge University Press, 2009 (Received ; Accepted ) 66