A proofâtechnique in uniform space theory - Semantic Scholar
A proof–technique in uniform space theory Douglas Bridges and Luminit¸a Vˆıt¸a ˘ Abstract. In the constructive theory of uniform spaces there occurs a technique of proof in which the application of a weak form of the law of excluded middle is circumvented by purely analytic means. The essence of this proof–technique is extracted and then applied in several different situations.
The theory of apartness spaces, a counterpart of the classical theory of proximity spaces, appears promising as a foundation for constructive1 topology. In several of the papers dealing with metric and uniform apartness spaces [5, 8, 11], we have used ad hoc variants of what appears to be a general proof–technique, which we now outline. ∞ We have two sequences (xn )∞ n=1 and (yn )n=1 in a uniform space (X, U) , and two properties P (x, y) and Q(x, y) of elements (x, y) of X × X. We want to prove that P (xn , yn ) holds eventually—that is for all sufficiently large n. We first use additional information, such as the strong continuity (see below) of some mapping from X into a uniform space Y, to establish that either P (xn , yn ) holds eventually or else Q(xn , yn ) holds infinitely often. In order to rule out the second alternative, we then show that it implies a weak form of the law of excluded middle, and that if this weak form of excluded middle is added to intuitionistic logic, then it is contradictory that Q(xn , yn ) hold infinitely often. It follows from all this that P (xn , yn ) holds eventually. In this note we extract the essence of this argument and then show how the general technique applies to various situations in apartness–space theory. We require only minimal knowledge of the constructive theory of uniform spaces (as found in [9]). However, to assist the reader, we now give some of the basic definitions of that theory. Let X be a nonempty set, and let U, V subsets of the Cartesian product X × X. We define certain associated subsets as follows: U ◦ V = {(x, y) : ∃z ∈ X ((x, z) ∈ U ∧ (z, y) ∈ V )}, U 1 = U, U
−1
U n+1 = U ◦ U n
(n = 1, 2, . . . ),
= {(x, y) : (y, x) ∈ U }.
We say that U is symmetric if U = U −1 . The diagonal of X × X is the set ∆ = {(x, x) : x ∈ X}. A family U of subsets of X × X is called a uniform structure, or uniformity, on X if the following conditions hold. 1
By constructive mathematics we mean mathematics with intuitionistic logic; see [1, 2, 4, 10]
U1 I Every finite intersection of sets in U belongs to U. II Every subset of X × X that contains a member of U is in U. U2 Every member of U contains both the diagonal ∆ and a symmetric member of U. U3 For each U ∈ U there exists V ∈ U such that V 2 ⊂ U . U4 For each U ∈ U there exists V ∈ U such that ∀x ∈ X × X (x ∈ U ∨ x ∈ / V ). The elements of U are called the entourages of (the uniform structure on) X. Condition U1, and the fact that, by U2, each element of U is nonempty, show that U is a filter on X × X. Classically axiom U4 is superfluous (we simply take V = U ); but constructively it is essential for the development of the theory. A good reference for the classical theory of uniform spaces is [3]. The motivating example of a uniform space is a metric space (X, ρ) , in which the unique uniform structure has a basis of sets of the form {(x, y) ∈ X × X : ρ (x, y) ≤ ε}
(1)
with ε > 0. The (uniform) topology on a uniform space (X, U) is the one in which a base of neighbourhoods of a point x ∈ X consists of the sets U [x] = {y ∈ X : (x, y) ∈ U } (U ∈ U) . A topology τ on a set X is said to be given by the uniform structure U on X if it coincides with the uniform topology arising from U. We define a canonical inequality on a uniform space (X, U) by x 6= y ⇔ ∃U ∈ U ((x, y) ∈ / U) . Note that, by axioms U1II and U2, if U ∈ U,then U −1 ∈ U. It follows that if x 6= y, then y 6= x. Moreover, since U contains ∆, if x 6= y, then ¬ (x = y) . Thus 6= has the two properties that define an inequality relation in constructive mathematics. In turn, the inequality on X induces an associated inequality on X × X in the usual way. We say that two subsets A, B of a uniform space (X, U) are apart, and we write A o n B, if there exists an entourage U such that A × B ⊂∼U = {x ∈ X × X : ∀y ∈ U (x 6= y)} . A mapping f : X → Y between uniform spaces is said to be strongly continuous if for all subsets A, B of X, f (A) ./ f (B) ⇒ A ./ B. We now establish some technical lemmas whose metric–space counterparts are found in [8]. Our first lemma is a peculiarly constructive one that takes the sting out of a number of succeeding proofs and is made necessary by the constructive failure of what Bishop called the limited principle of omniscience (LPO):
For each binary sequence (λn )∞ n=1 either λn = 0 for all n, or else there exists n such that λn = 1. In its recursive interpretation (with classical logic), LPO entails the decidability of the halting problem ([4], Chapter 3). Lemma 1. Let X, Y be uniform spaces, f : X → Y a strongly continuous function, and V an entourage of Y . Let (λn )∞ n=1 be an increasing binary sequence, ∞ ∞ and (An )n=1 , (Bn )n=1 sequences of subsets of X such that B for each entourage U of X there exists N such that for each n > N, either An × Bn = ∅ or else An × Bn intersects U ; B if λn = 0, then An = ∅; and B if λn = 1 − λn−1 , then An 6= ∅, Bn 6= ∅, f (An ) × f (Bn ) ⊂∼V, and Ak = ∅ for all k > n. Then there exists N such that λn = λN for all n > N. Proof. Note that if λ1 = 1, then there is nothing to prove. Writing A=
[∞
B=
[
n=1
An ,
{Bn : λn = 1 − λn−1 } ,
we see that f (A) × f (B) ⊂∼ V : for if x ∈ A, then there exists n such that A = An , B = Bn , and f (An )×f (Bn ) ⊂∼V. Hence f (A) ./ f (B) and therefore, by the strong continuity of f, there exists an entourage U of X such that A×B ⊂∼U. Choose N such that for each n > N, either An ×Bn = ∅ or else An ×Bn intersects U. Either λN = 1 and therefore λn = 1 for all n > N, or else λN = 0. In the latter case, if λm = 1 − λm−1 for some m > N, then A = Am 6= ∅, B = Bm 6= ∅, and A × B intersects U. This contradicts our choice of U, and so ensures that λn = 0 for all n > N.
Let S be a set of positive integers. We say that n ∈ S eventually if there exists N such that n ∈ S for all n ≥ N, and that n ∈ S infinitely often if we can construct a strictly increasing sequence (nk )∞ k=1 of positive integers such that nk ∈ P for each k. ∞ ∞ Two sequences (xn )n=1 and (x0n )n=1 in a uniform space (X, U) are said to be eventually close if for each U ∈ U we have (xn , x0n ) ∈ U eventually. Lemma 2. Let X, Y be uniform apartness spaces, f : X → Y a strongly contin0 ∞ uous mapping, and (xn )∞ n=1 , (xn )n=1 sequences in X that are eventually close. Let P, Q be sets of positive integers such that N+ = P ∪ Q, let V be an entourage of Y, and suppose that (f (xn ), f (x0n )) ∈ ∼ V for all n ∈ Q. Then either n ∈ P for all n or else there exists n ∈ Q.
Proof. Construct2 an increasing binary sequence (λn ) such that λn = 0 =⇒ ∀k 6 n (k ∈ P ) λn = 1 − λn−1 =⇒ n ∈ Q. We may assume that λ1 = 0. If λn = 0, set An = Bn = ∅. If λn = 1 − λn−1 , set An = {xn } and Bn = {x0n } , and note that, as n ∈ Q, we have f (An )×f (Bn ) ⊂∼ V ; in this case, also set Ak = ∅ = Bk for each k > n. Now consider any entourage U of X. There exists ν such that (xn , x0n ) ∈ U for all n ≥ ν. For each such n, if λn = 0 or λn−1 = 1, then An × Bn = ∅. On the other hand, if λn = 1 − λn−1 , then An × Bn = {(xn , x0n )} ⊂ U. Thus the hypotheses of Lemma 1 are satisfied. Applying that lemma, we produce N such that λn = λN for all n > N. If λN = 0, then n ∈ P for all n; whereas if λN = 1, there exists k 6 n such that k ∈ Q. Lemma 3. Under the hypotheses of Lemma 2, either n ∈ P eventually or else n ∈ Q infinitely often. Proof. In view of Lemma 2 we may assume that there exists n1 ∈ Q; we then set λ0 = λ1 = 0. Using dependent choice, we construct inductively an increasing binary sequence (λk ) and a strictly increasing sequence (nk ) of positive integers such that for each k > 1, – if λk = 0, then nk ∈ Q, and – if λk = 1 − λk−1 , then n ∈ P for all n > nk−1 . Suppose we have found λk−1 and nk−1 . We may assume that λk−1 = 0. Define P 0 = {j > 1 : nk−1 + j ∈ P }, Q0 = {j > 1 : nk−1 + j ∈ Q}. Then N+ = P 0 ∪ Q0 . Applying Lemma 2 with P, Q, xj and x0j replaced by P 0 , Q0 , xnk−1 +j and x0nk−1 +j respectively, we see that either j ∈ P 0 for all j ≥ 1 or else there exists m ∈ Q0 . In the first case we have n ∈ P for all n > nk−1 ; for each j ≥ k we then set λj = 1 and nj = nk−1 + j. In the second case we set λk = 0 and nk = nk−1 + m. This completes our inductive construction. Now let P 00 = k ∈ N+ : λk = 0 ∨ λk−1 = 1 , Q00 = k ∈ N+ : λk = 1 − λk−1 . Then N+ = P 00 ∪ Q00 . If k ∈ Q00 , then nk−1 ∈ Q, f (xnk−1 ), f (x0nk−1 ) ∈∼V, and n ∈ P for all n > nk−1 . Applying Lemma 2 with P, Q, xk , x0k replaced by P 00 , Q00 , xnk−1 , x0nk−1 respectively, we see that either k ∈ P 00 for all k or else there exists K ∈ Q00 . In the first case, if λj = 1 − λj−1 for some j, then j ∈ / P 00 , ∞ a contradiction; whence λk = 0 for all k, and therefore (nk )k=1 is a strictly increasing sequence of elements of Q. In the second case, n ∈ P for all n > nK−1 . 2
The construction of the sequence (λn ) requires the weak choice principle discussed in [7].
Our next lemma may seem bizarre, since it shows that under certain hypotheses, LPO holds. However, it enables us to use LPO to rule out one of two alternatives in the proof of Proposition 1. Lemma 4. Under the hypotheses of Lemma 2, if n ∈ Q infinitely often, then LPO holds. ∞
Proof. Choose a strictly increasing sequence (nk )k=1 in Q, and consider any 0 increasing binary sequence (λk )∞ k=1 . Applying Lemma 2 with P, Q, xk , and xk replaced by P 0 = k ∈ N + : λk = 0 , Q 0 = k ∈ N + : λk = 1 , xnk , and x0nk respectively, we find that either k ∈ P 0 for all k or else there exists k ∈ Q0 . Proposition 1. Let X, Y be uniform apartness spaces, f : X → Y a strongly ∞ ∞ continuous mapping, and (xn )n=1 , (x0n )n=1 sequences in X that are eventually close. Let P, Q be sets of positive integers such that N+ = P ∪ Q, let V be an entourage of Y, and suppose that (f (xn ), f (x0n )) ∈ ∼ V for all n ∈ Q. Suppose also that LPO ⇒ ¬∀n∃k > n (k ∈ Q) . Then n ∈ P eventually. Proof. By Lemma 3, either n ∈ P eventually or else n ∈ Q infinitely often. In the second case, Lemma 4 shows that LPO holds, which, in view of our final hypothesis, is absurd. For our first application of this very general proposition we introduce another type of continuity. A mapping f : X → Y between uniform spaces is said to be 0 ∞ uniformly sequentially continuous if for all sequences (xn )∞ n=1 , (xn )n=1 that ∞ ∞ 0 are eventually close in X, the sequences (f (xn ))n=1 , (f (xn ))n=1 are eventually close in Y. Proposition 2. A strongly continuous mapping of a uniform space into a uniform space is uniformly sequentially continuous. The proof requires a definition and two more lemmas. For each positive integer n we define an n–chain of entourages of X to be an n–tuple (U1 , . . . , Un ) of entourages such that for each applicable k, Uk is symmetric, Uk2 ⊂ Uk−1 , and ∀x ∈ X × X (x ∈ Uk−1 ∨ x ∈∼ Uk ) . Axiom U3 ensures that for each U ∈ U and each positive integer n there exists an n–chain (U1 , . . . , Un ) of entourages with U1 = U.
Lemma 5. Let Y be a uniform space, let (an ) , (bn ) be sequences in Y, and let (V1 , V2 , V3 ) be a 3–chain of entourages of Y such that (an , bn ) ∈∼V1 for each n. Then it is impossible that for each n, (an , bk ) ∈ V3 for all sufficiently large k. Proof. Suppose that for each n we have (an , bk ) ∈ V3 for all sufficiently large k. Choose N such that (a1 , bk ) ∈ V3 for all k > N. By our supposition, there exists M such that (aN , bk ) ∈ V3 for all k > M. Take K = max {M, N } . Then (aN , bK ) ∈ V3 , (a1 , bK ) ∈ V3 , and (a1 , bN ) ∈ V3 ; whence (aN , bN ) ∈ V33 ⊂ V22 ⊂ V1 , a contradiction. Lemma 6. Assuming LPO, let (an ) and (bn ) be sequences in a uniform space Y, and let (V1 , . . . , V4 ) be a 4–chain of entourages of Y, such that (an , bn ) ∈∼ V1 ∞ for each n. Then there exists a strictly increasing sequence (nk )k=1 of positive integers such that (anj , bnk ) ∈∼ V4 for all j and k. Proof. Set n0 = 0 and k0,j = j (j > 1) . Assume that we have constructed the ∞ natural number ni and an associated strictly increasing sequence (ki,j )j=1 of positive integers ≥ ni . Let P = m ∈ N+ : aki,m , bki,j ∈ V3 for all sufficiently large j , Q = m ∈ N+ : aki,m , bki,j ∈∼V4 for infinitely many j . Since we are assuming LPO, N+ =P ∪ Q. Moreover, LPO ensures that either there exists j 0 such that aki,j0 , bki,j ∈∼V4 for infinitely many j, or else for each j 0 , aki,j0 , bki,j ∈ V3 for all sufficiently large j. The second case is ruled out by Lemma 5. Hence we can find ni+1 > ni , and a strictly increasing subsequence ∞ such that k = n and a , b of (k ) (ki+1,j )∞ i+1,1 i+1 ni+1 ki+1,j ∈∼V4 for each i,j j=1 j=1 j. This describes the inductive construction of an infinite sequence of positive integers n1 < n2 < . . . such that anj , bnk ∈∼ V4 for all j < k. Note that as V4 ⊂ V1 , we actually have anj , bnk ∈∼ V4 for all j ≤ k. Passing to a subsequence, we may assume that (ai , aj ) ∈∼ V4 whenever i ≤ j. ∞ ∞ Now apply the foregoing argument to the sequences (bk )k=1 and (ak )k=1 , in that ∞ order. We obtain a strictly increasing sequence(nk )k=1 of positive integers such that bnk, anj ∈∼ V4 , and therefore anj , bnk ∈∼ V4 , for all k ≤ j. Since also anj , bnk ∈∼ V4 for all j ≤ k, we conclude that anj , bnk ∈∼ V4 for all j and k. We now prove Proposition 2. ∞
0 Proof. Let the sequences (xn )∞ n=1 and (xn )n=1 in X be eventually close. Let V be an entourage of Y, and construct a 5–chain (V1 , . . . , V5 ) of entourages of Y with V1 = V. Define
P = {n : (f (xn ), f (x0n )) ∈ V }, Q = {n : (f (xn ), f (x0n )) ∈∼V2 }.
Then N+ = P ∪ Q. By Lemma 3, either n ∈ P eventually or else n ∈ Q infinitely often. It remains to rule out the latter alternative. Supposing that n ∈ Q infinitely often, we see from Lemma 4 that LPO holds. Thus, by Lemma ∞ 6, there exists a strictly increasing sequence (nk )k=1 of positive integers such 0 that f (xnj ), f (xnk ) ∈∼ V5 for all j, k. Writing A = xnj : j > 1 , B = x0nk : k > 1 , we obtain f (A) × f (B) ⊂∼ V5 ; so, as f is strongly continuous, there exists an entourage U of X such that A×B ⊂∼ U. But this is absurd, since xnk , x0nk ∈ U eventually. It is shown in [8] that Proposition 2 is the best we can produce constructively without introducing a principle that, while valid in the standard models (intuitionistic, recursive, classical) of constructive mathematics, appears not to be derivable using only intuitionistic logic and dependent choice. Under certain additional hypotheses, Proposition 2 can be strengthened to produce the uniform continuity of the strongly continuous function f ; see [5]. The uniform continuity theorem for continuous mappings from compact metric spaces into metric spaces follows easily from Proposition 2 with classical logic. In the same way, a standard classical result about uniform convergence can be obtained from our next application of Proposition 1. This needs more definitions. ∞ Let S be a set, X a uniform space, (φn )n=1 a sequence of mappings of S into X, and φ a mapping of S into X. We say that (φn ) is proximally convergent to φ if for all A ⊂ S and B ⊂ X, φ(A) ./ B ⇒ ∃N ∀n ≥ N (φn (A) ./ B) ; and that (φn ) is uniformly sequentially convergent to φ if for each sequence ∞ ∞ ∞ (xn )n=1 in S, the sequences (φn (xn ))n=1 and (φ(xn ))n=1 are eventually close. Proposition 3. Let S be a set, X a uniform space, and (φn )∞ n=1 a sequence of mappings of S into X that converges proximally to φ : S → X. Then (φn ) is uniformly sequentially convergent to φ. ∞
Proof. Given a sequence (xn )n=1 in S and an entourage U of X, choose a symmetric entourage V of X such that V 2 ⊂ U and X = U ∪ ∼V. Write N+ as a union of subsets P, Q such that if n ∈ P, then (φn (xn ), φ(xn )) ∈ U, and if n ∈ Q, then (φn (xn ), φ(xn )) ∈∼V. Adapting arguments in [11] to the uniform space context, we can show that ∀n∃k > n (k ∈ Q) ⇒ LPO. It follows from Proposition 1, applied with X = Y, f the identity mapping of X into itself, and the sequences xn , x0n replaced by φn (xn ), φ(xn ) respectively, that (φn (xn ), φ(xn )) ∈ U eventually.
Our final application of Proposition 1 needs one more lemma and some preliminary explanation. Lemma 7. LPO implies that a compact (that is, total bounded, complete) metric space is sequentially compact. Proof. The proof is relatively straightforward and is omitted. In the classical theory of uniform spaces it is well known that the topology on a compact Hausdorff space X is given by a unique uniform structure U, whose entourages are the neighbourhoods of the diagonal in X × X. A base of entourages in U consists of sets of the form [ν (Un × Un ) (2) n=1
where {U1 , . . . , Un } is a finite open cover of X. If (X, ρ) is a compact metric space, then the unique uniform structure compatible with the topology on X is the metric one, for which the sets given by (1), with ε > 0, form a basis. Then for each finite open cover {U1 , . . . , Un } of X there exists ε > 0 such that the set given by (1) is contained in the set given by (2). Our next result is a weak constructive version of this observation. Recall that a subset S of a metric space X is located if ρ(x, S) = inf {ρ(x, y) : y ∈ S} exists for each x in X. Proposition 4. Let (X, ρ) be a compact metric space, and (rn )∞ n=1 a strictly decreasing sequence of positive numbers converging to 0, such that for each n the set Vn = {(x, y) ∈ X × X : ρ (x, y) ≤ rn } is compact. Let {U1 , . . . , Uν } be a finite open cover of X, and set U=
ν [
(Un × Un ) .
n=1
Suppose that U is located, and for each n write sn = sup {ρ ((x, y) , U ) : (x, y) ∈ Vn } . Then sn → 0 as n → ∞. Proof. Note first that the existence of such a sequence (rn )∞ n=1 follows from Theorem (4.9) on page 98 of [2]. Given ε > 0, write N+ as a union of subsets P, Q such that if n ∈ P, then sk < ε for all k ≤ n, and if n ∈ Q, then sk > ε/2 for some k ≤ n. For each n, if n ∈ P, then we choose any (xn , yn ) in Vn ; otherwise,
n ∈ Q and we can find (xn , yn ) in Vn such that ρ((xn , yn ), U ) > ε/2. Suppose that n ∈ Q infinitely often; passing to a subsequence if necessary, we may assume without loss of generality that n ∈ Q for each n. Then LPO holds, by Lemma ∞ 4; so, by Lemma 7, X is sequentially compact and the sequence (xn )n=1 has ∞ ∞ a subsequence (xnk )k=1 that converges to a limit x∞ ∈ X. Then (ynk )k=1 also converges to x∞ , since ρ (xnk , ynk ) ≤ rnk ; whence lim ρ((xnk , ynk ), U ) = ρ ((x∞ , x∞ ) , U ) = 0.
k→∞
This contradicts the fact that nk ∈ Q for each k. We now apply Proposition 1 to show that n ∈ P eventually. Acknowledgement: The authors thank the Marsden Fund of the Royal Society of New Zealand for generously supporting their research since 1997. They also thank Jeremy Clarke for suggesting corrections and improvements to the paper.
References 1. M.J. Beeson, Foundations of Constructive Mathematics, Springer-Verlag, Heidelberg, 1985. 2. E. Bishop and D.S. Bridges, Constructive Analysis, Grundlagen der Math. Wiss. 279, Springer–Verlag, Heidelberg, 1985. 3. Nicolas Bourbaki, General Topology (Part 1), Addison–Wesley, Reading, MA, 1966. 4. D.S. Bridges and F. Richman, Varieties of Constructive Mathematics, London Math. Soc. Lecture Notes 95, Cambridge Univ. Press, London, 1987. 5. D.S. Bridges and L.S. Vˆıt¸a ˘, ‘Strong continuity and uniform continuity: the uniform space case’, preprint, University of Canterbury, Christchurch, New Zealand, 2001. 6. D.S. Bridges and L.S. Vˆıt¸a ˘, ‘Apartness spaces as a framework for constructive topology’, Ann. Pure Appl. Logic. 119(1–3), 61–83, 2002. 7. Douglas Bridges, Fred Richman and Peter Schuster, ‘A weak countable choice principle’, Proc. Amer. Math. Soc. 128(9), 2749–2752, 2000. 8. D.S. Bridges, H. Ishihara, P.M. Schuster and L.S. Vˆıt¸a ˘, ‘Strong Continuity Implies Uniform Sequential Continuity’, preprint, Ludwig–Maximilians Universit¨ at, M¨ unchen, 2001. 9. P.M. Schuster, D.S. Bridges and L.S. Vˆıt¸a ˘, ‘Apartness as a relation between subsets’, in: Combinatorics, Computability and Logic (Proceedings of DMTCS’01, Constant¸a, Romania, 2–6 July 2001; C.S. Calude, M.J. Dinneen, S. Sburlan (eds.)), 203–214, DMTCS Series 17, Springer–Verlag, London, 2001. 10. A.S. Troelstra and D. van Dalen, Constructivism in Mathematics (Vols. I, II), North–Holland Publ. Co., Amsterdam, 1988. 11. L.S. Vˆıt¸a ˘, ‘Proximal and uniform convergence’, to appear in Sci. Math. Japonicae.
Authors’ address: Department of Mathematics & Statistics, University of Canterbury, Private Bag 4800, Christchurch, New Zealand email: [email protected] AMS Subject Classifications (2000): 03B60, 03F60 Revised 6 January 2003
The theory of apartness spaces, a counterpart of the classical theory of proximity spaces, appears promising as a foundation for constructive1 topology. In several of the papers dealing with metric and uniform apartness spaces [5, 8, 11], we have used ad hoc variants of what appears to be a general proof–technique, which we now outline. ∞ We have two sequences (xn )∞ n=1 and (yn )n=1 in a uniform space (X, U) , and two properties P (x, y) and Q(x, y) of elements (x, y) of X × X. We want to prove that P (xn , yn ) holds eventually—that is for all sufficiently large n. We first use additional information, such as the strong continuity (see below) of some mapping from X into a uniform space Y, to establish that either P (xn , yn ) holds eventually or else Q(xn , yn ) holds infinitely often. In order to rule out the second alternative, we then show that it implies a weak form of the law of excluded middle, and that if this weak form of excluded middle is added to intuitionistic logic, then it is contradictory that Q(xn , yn ) hold infinitely often. It follows from all this that P (xn , yn ) holds eventually. In this note we extract the essence of this argument and then show how the general technique applies to various situations in apartness–space theory. We require only minimal knowledge of the constructive theory of uniform spaces (as found in [9]). However, to assist the reader, we now give some of the basic definitions of that theory. Let X be a nonempty set, and let U, V subsets of the Cartesian product X × X. We define certain associated subsets as follows: U ◦ V = {(x, y) : ∃z ∈ X ((x, z) ∈ U ∧ (z, y) ∈ V )}, U 1 = U, U
−1
U n+1 = U ◦ U n
(n = 1, 2, . . . ),
= {(x, y) : (y, x) ∈ U }.
We say that U is symmetric if U = U −1 . The diagonal of X × X is the set ∆ = {(x, x) : x ∈ X}. A family U of subsets of X × X is called a uniform structure, or uniformity, on X if the following conditions hold. 1
By constructive mathematics we mean mathematics with intuitionistic logic; see [1, 2, 4, 10]
U1 I Every finite intersection of sets in U belongs to U. II Every subset of X × X that contains a member of U is in U. U2 Every member of U contains both the diagonal ∆ and a symmetric member of U. U3 For each U ∈ U there exists V ∈ U such that V 2 ⊂ U . U4 For each U ∈ U there exists V ∈ U such that ∀x ∈ X × X (x ∈ U ∨ x ∈ / V ). The elements of U are called the entourages of (the uniform structure on) X. Condition U1, and the fact that, by U2, each element of U is nonempty, show that U is a filter on X × X. Classically axiom U4 is superfluous (we simply take V = U ); but constructively it is essential for the development of the theory. A good reference for the classical theory of uniform spaces is [3]. The motivating example of a uniform space is a metric space (X, ρ) , in which the unique uniform structure has a basis of sets of the form {(x, y) ∈ X × X : ρ (x, y) ≤ ε}
(1)
with ε > 0. The (uniform) topology on a uniform space (X, U) is the one in which a base of neighbourhoods of a point x ∈ X consists of the sets U [x] = {y ∈ X : (x, y) ∈ U } (U ∈ U) . A topology τ on a set X is said to be given by the uniform structure U on X if it coincides with the uniform topology arising from U. We define a canonical inequality on a uniform space (X, U) by x 6= y ⇔ ∃U ∈ U ((x, y) ∈ / U) . Note that, by axioms U1II and U2, if U ∈ U,then U −1 ∈ U. It follows that if x 6= y, then y 6= x. Moreover, since U contains ∆, if x 6= y, then ¬ (x = y) . Thus 6= has the two properties that define an inequality relation in constructive mathematics. In turn, the inequality on X induces an associated inequality on X × X in the usual way. We say that two subsets A, B of a uniform space (X, U) are apart, and we write A o n B, if there exists an entourage U such that A × B ⊂∼U = {x ∈ X × X : ∀y ∈ U (x 6= y)} . A mapping f : X → Y between uniform spaces is said to be strongly continuous if for all subsets A, B of X, f (A) ./ f (B) ⇒ A ./ B. We now establish some technical lemmas whose metric–space counterparts are found in [8]. Our first lemma is a peculiarly constructive one that takes the sting out of a number of succeeding proofs and is made necessary by the constructive failure of what Bishop called the limited principle of omniscience (LPO):
For each binary sequence (λn )∞ n=1 either λn = 0 for all n, or else there exists n such that λn = 1. In its recursive interpretation (with classical logic), LPO entails the decidability of the halting problem ([4], Chapter 3). Lemma 1. Let X, Y be uniform spaces, f : X → Y a strongly continuous function, and V an entourage of Y . Let (λn )∞ n=1 be an increasing binary sequence, ∞ ∞ and (An )n=1 , (Bn )n=1 sequences of subsets of X such that B for each entourage U of X there exists N such that for each n > N, either An × Bn = ∅ or else An × Bn intersects U ; B if λn = 0, then An = ∅; and B if λn = 1 − λn−1 , then An 6= ∅, Bn 6= ∅, f (An ) × f (Bn ) ⊂∼V, and Ak = ∅ for all k > n. Then there exists N such that λn = λN for all n > N. Proof. Note that if λ1 = 1, then there is nothing to prove. Writing A=
[∞
B=
[
n=1
An ,
{Bn : λn = 1 − λn−1 } ,
we see that f (A) × f (B) ⊂∼ V : for if x ∈ A, then there exists n such that A = An , B = Bn , and f (An )×f (Bn ) ⊂∼V. Hence f (A) ./ f (B) and therefore, by the strong continuity of f, there exists an entourage U of X such that A×B ⊂∼U. Choose N such that for each n > N, either An ×Bn = ∅ or else An ×Bn intersects U. Either λN = 1 and therefore λn = 1 for all n > N, or else λN = 0. In the latter case, if λm = 1 − λm−1 for some m > N, then A = Am 6= ∅, B = Bm 6= ∅, and A × B intersects U. This contradicts our choice of U, and so ensures that λn = 0 for all n > N.
Let S be a set of positive integers. We say that n ∈ S eventually if there exists N such that n ∈ S for all n ≥ N, and that n ∈ S infinitely often if we can construct a strictly increasing sequence (nk )∞ k=1 of positive integers such that nk ∈ P for each k. ∞ ∞ Two sequences (xn )n=1 and (x0n )n=1 in a uniform space (X, U) are said to be eventually close if for each U ∈ U we have (xn , x0n ) ∈ U eventually. Lemma 2. Let X, Y be uniform apartness spaces, f : X → Y a strongly contin0 ∞ uous mapping, and (xn )∞ n=1 , (xn )n=1 sequences in X that are eventually close. Let P, Q be sets of positive integers such that N+ = P ∪ Q, let V be an entourage of Y, and suppose that (f (xn ), f (x0n )) ∈ ∼ V for all n ∈ Q. Then either n ∈ P for all n or else there exists n ∈ Q.
Proof. Construct2 an increasing binary sequence (λn ) such that λn = 0 =⇒ ∀k 6 n (k ∈ P ) λn = 1 − λn−1 =⇒ n ∈ Q. We may assume that λ1 = 0. If λn = 0, set An = Bn = ∅. If λn = 1 − λn−1 , set An = {xn } and Bn = {x0n } , and note that, as n ∈ Q, we have f (An )×f (Bn ) ⊂∼ V ; in this case, also set Ak = ∅ = Bk for each k > n. Now consider any entourage U of X. There exists ν such that (xn , x0n ) ∈ U for all n ≥ ν. For each such n, if λn = 0 or λn−1 = 1, then An × Bn = ∅. On the other hand, if λn = 1 − λn−1 , then An × Bn = {(xn , x0n )} ⊂ U. Thus the hypotheses of Lemma 1 are satisfied. Applying that lemma, we produce N such that λn = λN for all n > N. If λN = 0, then n ∈ P for all n; whereas if λN = 1, there exists k 6 n such that k ∈ Q. Lemma 3. Under the hypotheses of Lemma 2, either n ∈ P eventually or else n ∈ Q infinitely often. Proof. In view of Lemma 2 we may assume that there exists n1 ∈ Q; we then set λ0 = λ1 = 0. Using dependent choice, we construct inductively an increasing binary sequence (λk ) and a strictly increasing sequence (nk ) of positive integers such that for each k > 1, – if λk = 0, then nk ∈ Q, and – if λk = 1 − λk−1 , then n ∈ P for all n > nk−1 . Suppose we have found λk−1 and nk−1 . We may assume that λk−1 = 0. Define P 0 = {j > 1 : nk−1 + j ∈ P }, Q0 = {j > 1 : nk−1 + j ∈ Q}. Then N+ = P 0 ∪ Q0 . Applying Lemma 2 with P, Q, xj and x0j replaced by P 0 , Q0 , xnk−1 +j and x0nk−1 +j respectively, we see that either j ∈ P 0 for all j ≥ 1 or else there exists m ∈ Q0 . In the first case we have n ∈ P for all n > nk−1 ; for each j ≥ k we then set λj = 1 and nj = nk−1 + j. In the second case we set λk = 0 and nk = nk−1 + m. This completes our inductive construction. Now let P 00 = k ∈ N+ : λk = 0 ∨ λk−1 = 1 , Q00 = k ∈ N+ : λk = 1 − λk−1 . Then N+ = P 00 ∪ Q00 . If k ∈ Q00 , then nk−1 ∈ Q, f (xnk−1 ), f (x0nk−1 ) ∈∼V, and n ∈ P for all n > nk−1 . Applying Lemma 2 with P, Q, xk , x0k replaced by P 00 , Q00 , xnk−1 , x0nk−1 respectively, we see that either k ∈ P 00 for all k or else there exists K ∈ Q00 . In the first case, if λj = 1 − λj−1 for some j, then j ∈ / P 00 , ∞ a contradiction; whence λk = 0 for all k, and therefore (nk )k=1 is a strictly increasing sequence of elements of Q. In the second case, n ∈ P for all n > nK−1 . 2
The construction of the sequence (λn ) requires the weak choice principle discussed in [7].
Our next lemma may seem bizarre, since it shows that under certain hypotheses, LPO holds. However, it enables us to use LPO to rule out one of two alternatives in the proof of Proposition 1. Lemma 4. Under the hypotheses of Lemma 2, if n ∈ Q infinitely often, then LPO holds. ∞
Proof. Choose a strictly increasing sequence (nk )k=1 in Q, and consider any 0 increasing binary sequence (λk )∞ k=1 . Applying Lemma 2 with P, Q, xk , and xk replaced by P 0 = k ∈ N + : λk = 0 , Q 0 = k ∈ N + : λk = 1 , xnk , and x0nk respectively, we find that either k ∈ P 0 for all k or else there exists k ∈ Q0 . Proposition 1. Let X, Y be uniform apartness spaces, f : X → Y a strongly ∞ ∞ continuous mapping, and (xn )n=1 , (x0n )n=1 sequences in X that are eventually close. Let P, Q be sets of positive integers such that N+ = P ∪ Q, let V be an entourage of Y, and suppose that (f (xn ), f (x0n )) ∈ ∼ V for all n ∈ Q. Suppose also that LPO ⇒ ¬∀n∃k > n (k ∈ Q) . Then n ∈ P eventually. Proof. By Lemma 3, either n ∈ P eventually or else n ∈ Q infinitely often. In the second case, Lemma 4 shows that LPO holds, which, in view of our final hypothesis, is absurd. For our first application of this very general proposition we introduce another type of continuity. A mapping f : X → Y between uniform spaces is said to be 0 ∞ uniformly sequentially continuous if for all sequences (xn )∞ n=1 , (xn )n=1 that ∞ ∞ 0 are eventually close in X, the sequences (f (xn ))n=1 , (f (xn ))n=1 are eventually close in Y. Proposition 2. A strongly continuous mapping of a uniform space into a uniform space is uniformly sequentially continuous. The proof requires a definition and two more lemmas. For each positive integer n we define an n–chain of entourages of X to be an n–tuple (U1 , . . . , Un ) of entourages such that for each applicable k, Uk is symmetric, Uk2 ⊂ Uk−1 , and ∀x ∈ X × X (x ∈ Uk−1 ∨ x ∈∼ Uk ) . Axiom U3 ensures that for each U ∈ U and each positive integer n there exists an n–chain (U1 , . . . , Un ) of entourages with U1 = U.
Lemma 5. Let Y be a uniform space, let (an ) , (bn ) be sequences in Y, and let (V1 , V2 , V3 ) be a 3–chain of entourages of Y such that (an , bn ) ∈∼V1 for each n. Then it is impossible that for each n, (an , bk ) ∈ V3 for all sufficiently large k. Proof. Suppose that for each n we have (an , bk ) ∈ V3 for all sufficiently large k. Choose N such that (a1 , bk ) ∈ V3 for all k > N. By our supposition, there exists M such that (aN , bk ) ∈ V3 for all k > M. Take K = max {M, N } . Then (aN , bK ) ∈ V3 , (a1 , bK ) ∈ V3 , and (a1 , bN ) ∈ V3 ; whence (aN , bN ) ∈ V33 ⊂ V22 ⊂ V1 , a contradiction. Lemma 6. Assuming LPO, let (an ) and (bn ) be sequences in a uniform space Y, and let (V1 , . . . , V4 ) be a 4–chain of entourages of Y, such that (an , bn ) ∈∼ V1 ∞ for each n. Then there exists a strictly increasing sequence (nk )k=1 of positive integers such that (anj , bnk ) ∈∼ V4 for all j and k. Proof. Set n0 = 0 and k0,j = j (j > 1) . Assume that we have constructed the ∞ natural number ni and an associated strictly increasing sequence (ki,j )j=1 of positive integers ≥ ni . Let P = m ∈ N+ : aki,m , bki,j ∈ V3 for all sufficiently large j , Q = m ∈ N+ : aki,m , bki,j ∈∼V4 for infinitely many j . Since we are assuming LPO, N+ =P ∪ Q. Moreover, LPO ensures that either there exists j 0 such that aki,j0 , bki,j ∈∼V4 for infinitely many j, or else for each j 0 , aki,j0 , bki,j ∈ V3 for all sufficiently large j. The second case is ruled out by Lemma 5. Hence we can find ni+1 > ni , and a strictly increasing subsequence ∞ such that k = n and a , b of (k ) (ki+1,j )∞ i+1,1 i+1 ni+1 ki+1,j ∈∼V4 for each i,j j=1 j=1 j. This describes the inductive construction of an infinite sequence of positive integers n1 < n2 < . . . such that anj , bnk ∈∼ V4 for all j < k. Note that as V4 ⊂ V1 , we actually have anj , bnk ∈∼ V4 for all j ≤ k. Passing to a subsequence, we may assume that (ai , aj ) ∈∼ V4 whenever i ≤ j. ∞ ∞ Now apply the foregoing argument to the sequences (bk )k=1 and (ak )k=1 , in that ∞ order. We obtain a strictly increasing sequence(nk )k=1 of positive integers such that bnk, anj ∈∼ V4 , and therefore anj , bnk ∈∼ V4 , for all k ≤ j. Since also anj , bnk ∈∼ V4 for all j ≤ k, we conclude that anj , bnk ∈∼ V4 for all j and k. We now prove Proposition 2. ∞
0 Proof. Let the sequences (xn )∞ n=1 and (xn )n=1 in X be eventually close. Let V be an entourage of Y, and construct a 5–chain (V1 , . . . , V5 ) of entourages of Y with V1 = V. Define
P = {n : (f (xn ), f (x0n )) ∈ V }, Q = {n : (f (xn ), f (x0n )) ∈∼V2 }.
Then N+ = P ∪ Q. By Lemma 3, either n ∈ P eventually or else n ∈ Q infinitely often. It remains to rule out the latter alternative. Supposing that n ∈ Q infinitely often, we see from Lemma 4 that LPO holds. Thus, by Lemma ∞ 6, there exists a strictly increasing sequence (nk )k=1 of positive integers such 0 that f (xnj ), f (xnk ) ∈∼ V5 for all j, k. Writing A = xnj : j > 1 , B = x0nk : k > 1 , we obtain f (A) × f (B) ⊂∼ V5 ; so, as f is strongly continuous, there exists an entourage U of X such that A×B ⊂∼ U. But this is absurd, since xnk , x0nk ∈ U eventually. It is shown in [8] that Proposition 2 is the best we can produce constructively without introducing a principle that, while valid in the standard models (intuitionistic, recursive, classical) of constructive mathematics, appears not to be derivable using only intuitionistic logic and dependent choice. Under certain additional hypotheses, Proposition 2 can be strengthened to produce the uniform continuity of the strongly continuous function f ; see [5]. The uniform continuity theorem for continuous mappings from compact metric spaces into metric spaces follows easily from Proposition 2 with classical logic. In the same way, a standard classical result about uniform convergence can be obtained from our next application of Proposition 1. This needs more definitions. ∞ Let S be a set, X a uniform space, (φn )n=1 a sequence of mappings of S into X, and φ a mapping of S into X. We say that (φn ) is proximally convergent to φ if for all A ⊂ S and B ⊂ X, φ(A) ./ B ⇒ ∃N ∀n ≥ N (φn (A) ./ B) ; and that (φn ) is uniformly sequentially convergent to φ if for each sequence ∞ ∞ ∞ (xn )n=1 in S, the sequences (φn (xn ))n=1 and (φ(xn ))n=1 are eventually close. Proposition 3. Let S be a set, X a uniform space, and (φn )∞ n=1 a sequence of mappings of S into X that converges proximally to φ : S → X. Then (φn ) is uniformly sequentially convergent to φ. ∞
Proof. Given a sequence (xn )n=1 in S and an entourage U of X, choose a symmetric entourage V of X such that V 2 ⊂ U and X = U ∪ ∼V. Write N+ as a union of subsets P, Q such that if n ∈ P, then (φn (xn ), φ(xn )) ∈ U, and if n ∈ Q, then (φn (xn ), φ(xn )) ∈∼V. Adapting arguments in [11] to the uniform space context, we can show that ∀n∃k > n (k ∈ Q) ⇒ LPO. It follows from Proposition 1, applied with X = Y, f the identity mapping of X into itself, and the sequences xn , x0n replaced by φn (xn ), φ(xn ) respectively, that (φn (xn ), φ(xn )) ∈ U eventually.
Our final application of Proposition 1 needs one more lemma and some preliminary explanation. Lemma 7. LPO implies that a compact (that is, total bounded, complete) metric space is sequentially compact. Proof. The proof is relatively straightforward and is omitted. In the classical theory of uniform spaces it is well known that the topology on a compact Hausdorff space X is given by a unique uniform structure U, whose entourages are the neighbourhoods of the diagonal in X × X. A base of entourages in U consists of sets of the form [ν (Un × Un ) (2) n=1
where {U1 , . . . , Un } is a finite open cover of X. If (X, ρ) is a compact metric space, then the unique uniform structure compatible with the topology on X is the metric one, for which the sets given by (1), with ε > 0, form a basis. Then for each finite open cover {U1 , . . . , Un } of X there exists ε > 0 such that the set given by (1) is contained in the set given by (2). Our next result is a weak constructive version of this observation. Recall that a subset S of a metric space X is located if ρ(x, S) = inf {ρ(x, y) : y ∈ S} exists for each x in X. Proposition 4. Let (X, ρ) be a compact metric space, and (rn )∞ n=1 a strictly decreasing sequence of positive numbers converging to 0, such that for each n the set Vn = {(x, y) ∈ X × X : ρ (x, y) ≤ rn } is compact. Let {U1 , . . . , Uν } be a finite open cover of X, and set U=
ν [
(Un × Un ) .
n=1
Suppose that U is located, and for each n write sn = sup {ρ ((x, y) , U ) : (x, y) ∈ Vn } . Then sn → 0 as n → ∞. Proof. Note first that the existence of such a sequence (rn )∞ n=1 follows from Theorem (4.9) on page 98 of [2]. Given ε > 0, write N+ as a union of subsets P, Q such that if n ∈ P, then sk < ε for all k ≤ n, and if n ∈ Q, then sk > ε/2 for some k ≤ n. For each n, if n ∈ P, then we choose any (xn , yn ) in Vn ; otherwise,
n ∈ Q and we can find (xn , yn ) in Vn such that ρ((xn , yn ), U ) > ε/2. Suppose that n ∈ Q infinitely often; passing to a subsequence if necessary, we may assume without loss of generality that n ∈ Q for each n. Then LPO holds, by Lemma ∞ 4; so, by Lemma 7, X is sequentially compact and the sequence (xn )n=1 has ∞ ∞ a subsequence (xnk )k=1 that converges to a limit x∞ ∈ X. Then (ynk )k=1 also converges to x∞ , since ρ (xnk , ynk ) ≤ rnk ; whence lim ρ((xnk , ynk ), U ) = ρ ((x∞ , x∞ ) , U ) = 0.
k→∞
This contradicts the fact that nk ∈ Q for each k. We now apply Proposition 1 to show that n ∈ P eventually. Acknowledgement: The authors thank the Marsden Fund of the Royal Society of New Zealand for generously supporting their research since 1997. They also thank Jeremy Clarke for suggesting corrections and improvements to the paper.
References 1. M.J. Beeson, Foundations of Constructive Mathematics, Springer-Verlag, Heidelberg, 1985. 2. E. Bishop and D.S. Bridges, Constructive Analysis, Grundlagen der Math. Wiss. 279, Springer–Verlag, Heidelberg, 1985. 3. Nicolas Bourbaki, General Topology (Part 1), Addison–Wesley, Reading, MA, 1966. 4. D.S. Bridges and F. Richman, Varieties of Constructive Mathematics, London Math. Soc. Lecture Notes 95, Cambridge Univ. Press, London, 1987. 5. D.S. Bridges and L.S. Vˆıt¸a ˘, ‘Strong continuity and uniform continuity: the uniform space case’, preprint, University of Canterbury, Christchurch, New Zealand, 2001. 6. D.S. Bridges and L.S. Vˆıt¸a ˘, ‘Apartness spaces as a framework for constructive topology’, Ann. Pure Appl. Logic. 119(1–3), 61–83, 2002. 7. Douglas Bridges, Fred Richman and Peter Schuster, ‘A weak countable choice principle’, Proc. Amer. Math. Soc. 128(9), 2749–2752, 2000. 8. D.S. Bridges, H. Ishihara, P.M. Schuster and L.S. Vˆıt¸a ˘, ‘Strong Continuity Implies Uniform Sequential Continuity’, preprint, Ludwig–Maximilians Universit¨ at, M¨ unchen, 2001. 9. P.M. Schuster, D.S. Bridges and L.S. Vˆıt¸a ˘, ‘Apartness as a relation between subsets’, in: Combinatorics, Computability and Logic (Proceedings of DMTCS’01, Constant¸a, Romania, 2–6 July 2001; C.S. Calude, M.J. Dinneen, S. Sburlan (eds.)), 203–214, DMTCS Series 17, Springer–Verlag, London, 2001. 10. A.S. Troelstra and D. van Dalen, Constructivism in Mathematics (Vols. I, II), North–Holland Publ. Co., Amsterdam, 1988. 11. L.S. Vˆıt¸a ˘, ‘Proximal and uniform convergence’, to appear in Sci. Math. Japonicae.
Authors’ address: Department of Mathematics & Statistics, University of Canterbury, Private Bag 4800, Christchurch, New Zealand email: [email protected] AMS Subject Classifications (2000): 03B60, 03F60 Revised 6 January 2003
