A proof-technique in uniform space theory

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 constructive 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 = U, U = U ◦ U (n = 1, 2, . . . ), U = {(x, y) : (y, x) ∈ U}. We say that U is symmetric if U = U. 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 ∈ 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 on 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