Math. Log. Quart. 46 (2000) 1, 139 – 143

Mathematical LogicQuarterly

c© WILEY-VCH Verlag Berlin GmbH 2000

Sequential Continuity of Functions in Constructive Analysis

Douglas Bridges and Ayan Mahalanobis

Department of Mathematics and Statistics, University of Canterbury,Private Bag 4800, Christchurch, New Zealand1)

Abstract. It is shown that in any model of constructive mathematics in which a certainomniscience principle is false, for strongly extensional functions on an interval the distinctionbetween sequentially continuous and regulated disappears. It follows, without the use ofMarkov’s Principle, that any recursive function of bounded variation on a bounded closedinterval is recursively sequentially continuous.

Mathematics Subject Classification: 03F60, 26E40.

Keywords: Constructive analysis, Sequentially continuous, Regulated, Bounded variation.

A famous theorem of Kreisel-Lacombe-Schonfield-Ceitin (see [9], [6], and[5, pp. 67 – 71]) says that in recursive constructive mathematics – that is, recur-sive mathematics carried out with intuitionistic logic – every real-valued functionon a bounded closed interval is pointwise continuous.2) The proof of that theoremis known to require Markov’s Principle, a form of unbounded search that is inde-pendent of Heyting arithmetic (Peano arithmetic with intuitionistic logic) and that isviewed with at best suspicion by many practitioners of constructive mathematics. Themain result of our present note enables us to prove, recursively and without Markov’sPrinciple, the sequential continuity of functions of bounded variation on a boundedclosed interval. However, the main result itself is proved within Bishop’s constructivemathematics – mathematics with intuitionistic logic – and does not require Church’sThesis.

Among the several weak forms of the law of excluded middle that are independentof Heyting arithmetic, the one of most significance in the work below is the limitedprinciple of omniscience (LPO),

If (an) is a binary sequence, then either an = 0 for all n or else thereexists n with an = 1 ([5, pp. 3 – 4]).

This principle is easily shown to be equivalent to the proposition(∀ξ ∈ R) (ξ > 0 ∨ ξ ≤ 0),

which is false in intuitionistic mathematics and in recursive constructive mathematics(that is, mathematics with intuitionistic logic and Church’s Thesis) – see [5, p. 53and p. 108].

1)e-mail: d.bridges@math.canterbury.ac.nz2)But not necessarily uniformly continuous – see [5, Chapter 6].

140 Douglas Bridges and Ayan Mahalanobis

Throughout the paper, I will be a closed interval in R, and f a real-valued functionon I. Normally we shall postulate that f is strongly extensional in the sense that∀x∀y (f(x) 6= f(y) ⇒ x 6= y), where, for two real numbers x and y, x 6= y means|x−y| > 0. Note that in recursive constructive mathematics every real-valued functionon a closed interval is strongly extensional: this follows from [8, Proposition 2] and[7, Theorem 2].

A point x ∈ I is called a left interior point (respectively, right interior point) of Iif I ∩ (−∞, x) (respectively, I ∩ (x,∞)) is nonempty.3) Let x be a left interior pointof I; if limn→∞ f(xn) exists, and is independent of (xn), for each sequence (xn) inI ∩ (−∞, x) that converges to x, then we call that limit the sequential limit of f(t) ast tends to x from below, and we denote it by seq limt→x− f(t). We make the obviousanalogous definition of the sequential limit of f(t) as t tends to x from above for aright interior point x of I, and we denote it by seq limt→x+ f(t).

We say that f is regulated on I if seq limt→x− f(t) exists for each left interiorpoint x ∈ I and seq limt→x+ f(t) exists for each right interior point x ∈ I. Wesay that f is sequentially continuous on I if limn→∞ f(xn) = f(x) for each x ∈ Iand each sequence (xn) of points of I converging to x. A sequentially continuousfunction is clearly regulated. Classically, a step-function is regulated, but it fails tobe sequentially continuous at each of its proper jumps. Our aim in this paper is toprove that for strongly extensional functions in any model of Bishop’s mathematicsin which LPO is false, the distinction between sequentially continuous and regulateddisappears.

T h e o r em 1.

¬LPO ` Every strongly extensional regulated function on a closed intervalis sequentially continuous.

The proof of Theorem 1 depends on two lemmas of a common type in constructiveanalysis (cf. [7, Lemmas 1 and 2]).

L e mm a 2. Let g : I −→ R be strongly extensional on the closed interval I,let x ∈ I, and let (xn) be a sequence of points of I converging to x, such thatg(xn) ≥ t > g(x) ≥ 0 for all n. Then either xn < x for all n or else there exists nwith xn > x.

P r o o f . Choose a strictly increasing sequence (nk)∞k=1 of positive integers suchthat |x− xn| < k−1 whenever n ≥ nk. Construct an increasing binary sequence (λk)such that

λk = 0⇒ (∀n ≤ nk) (xn < x),λk = 1− λk−1 ⇒ xn > x for some n ∈ (nk−1, nk].

We may assume that λ1 = 0. If λk = 0, set ξk = x; if λk = 1− λk−1, choose n suchthat nk−1 < n ≤ nk and xn > x, and set ξi = xn for all i ≥ k. Then (ξk) is a Cauchysequence: in fact, |ξi − ξj| ≤ 2k−1 for all i, j ≥ k. So (ξk) converges to a limit ξ ∈ I.Either g(ξ) < t or g(ξ) > g(x). In the former case we must have λk = 0, and thereforexk < x, for all k. In the case g(ξ) > g(x), as g is strongly extensional, we have ξ 6= x.Choosing N such that |x − xn| < |x − ξ| for all n ≥ N , suppose that λN = 0. If

3)In constructive mathematics a set S is nonempty if ∃x (x ∈ S); this is a stronger property than¬∀x (x /∈ S).

Sequential Continuity of Functions in Constructive Analysis 141

λk = 1 − λk−1 for some k > N , then ξ = xn for some n > nk ≥ N , which is absurd;hence λk = 0 for all k > N and therefore for all k. But this implies that ξ = x, acontradiction. We conclude that λN = 1, and hence that xn > x for some n ≤ nN . 2

L e mm a 3. Let g : I −→ R be strongly extensional on the closed interval I, letx be a point of I, and let (xn) be a sequence of points of I converging to x such thatg(xn) > t > g(x) ≥ 0 for all n. Then either xn > x for infinitely many n or elsexn < x for all sufficiently large n.

P r o o f . Choose a strictly increasing sequence (nk)∞k=1 of positive integers suchthat |x−xn| < k−1 whenever n ≥ nk. Using Lemma 2, construct an increasing binarysequence (λk)∞k=1 such that

λk = 0 ⇒ (∃n ≥ nk) (xn > x), λk = 1 ⇒ (∀n ≥ nk) (xn < x).We may assume that λ1 = 0. If λk = 0, set ξk = x; if λk = 1− λk−1, choose i > nk−1

such that xi > x, and set ξn = xi for all n ≥ k. Then (ξk) is a Cauchy sequence: infact, |ξi−ξj| ≤ (k−1)−1 whenever i, j ≥ k ≥ 2. Hence (ξk) converges to a limit ξ ∈ I.Either g(ξ) < t or g(ξ) > g(x). In the first case we have λk = 0, and therefore xk > x,for all k. In the second we have ξ 6= x, by the strong extensionality of g. ChoosingN so that |x− xn| < |x− ξ| for all n ≥ N , suppose that λN = 0. If λk = 1 − λk−1

for some k > N , then ξ = xi for some i > nk−1 ≥ N, which is absurd; whence λk = 0for all k > N and therefore for all k. But this implies that ξ = x, a contradiction.Hence, in fact, λN = 1, and so xn < x for all sufficiently large n. 2

P r o po s i t i o n 4. Let I be a closed interval, and let f : I −→ R be stronglyextensional. Then f is sequentially continuous on I if and only if the following twoconditions hold:

(i) seq limt→x− f(t) exists and equals f(x) for each left interior point x of I;(ii) seq limt→x+ f(t) exists and equals f(x) for each right interior point x of I.

P r o o f . It is clear that conditions (i) and (ii) are necessary for f to be sequentiallycontinuous. To prove that they are sufficient, assume both (i) and (ii). Let x ∈ I,let (xn) be any sequence of points of I converging to x, and let ε > 0. By [7, Lem-ma 2], either |f(xn) − f(x)| > ε/2 for infinitely many n or else |f(xn) − f(x)| < εfor all sufficiently large n. Suppose the former condition holds. Applying Lemma 3with g(t) = |f(t) − f(x)|, we see that either xn > x for infinitely many n or elsexn < x for all sufficiently large n. In the first case we can extract a strictly decreasingsubsequence (xnk)∞k=1 of (xn) such that |f(xnk) − f(x)| > ε/2 for all k; whence|seq limt→x+ f(t)−f(x)| ≥ ε/2, contradicting hypothesis (ii). A similar contradictioncan be obtained in the second case. We therefore conclude that |f(x) − f(xn)| < εfor all sufficiently large n. Hence f is sequentially continuous on I. 2

We now arrive at theP r o o f o f T h e o r e m 1. Assuming ¬LPO, let f be strongly extensional and

regulated on the closed interval I. Consider a left interior point x of I, and let (xn)be a strictly increasing sequence of points of I such that 0 < x−xn < n−1 for each n.Suppose that l = seq limt→x− f(t) 6= f(x). Let ε = |f(x) − l|/3, and choose N suchthat |l − f(xn)| < ε for all n ≥ N . Given ξ ∈ R, construct an increasing binarysequence (λn) such that

λn = 0 ⇒ ξ < x + n−1, λn = 1 ⇒ ξ > x.

142 Douglas Bridges and Ayan Mahalanobis

Since we aim to show that either ξ > x or ξ ≤ x, we may assume that λ1 = 0. Ifλn = 0, set ζn = x; if λn = 1−λn−1, set ζk = xn for all k ≥ n. Then (ζn) is a Cauchysequence in [x1, x] and so converges to a limit ζ ∈ [x1, x]. Either |f(ζ) − f(x)| > εor |f(ζ) − l| > ε. Consider the first case, in which, by strong extensionality, we haveζ 6= x and therefore ζ < x. Choosing ν such that ζ < xν , suppose that λν = 0.If λn = 1 − λn−1 for some n > ν, then ζ = xn > xν > ζ, a contradiction. Henceλn = 0 for all n > ν and therefore for all n; but this leads to the contradiction ζ = x.It follows that λν = 1 and therefore that ξ > x. On the other hand, in the case|f(ζ) − l| > ε, consider λN . If λn = 1 − λn−1 for some n ≥ N , then ζ = xn and so|f(ζ) − l| < ε, a contradiction. So if λN = 0, then λn = 0 for all n and thereforeξ ≤ x; on the other hand, if λN = 1, then ξ > x.

Thus if l 6= f(x), then (∀ξ ∈ R) (ξ > x ∨ ξ ≤ x), and so LPO holds – a con-tradiction. We conclude that l = f(x). Similarly, if x is a right interior point of I,then seq limt→x+ f(t) = f(x). It follows from Proposition 4 that f is sequentiallycontinuous on I. 2

By a function of bounded variation on I we mean a mapping f : I −→ R with thefollowing property: there exists M > 0 such that

∑n−1k=0 |f(xk+1) − f(xk)| < M for

all strictly increasing finite sequences x0 < x1 < · · · < xn of points of I. Note thatalthough, classically, such a function can always be expressed as a difference of twoincreasing functions, we do not know if this decomposition is possible constructivelyunless the variation of f on I,

sup{∑n−1k=0 |f(xk+1)− f(xk)| : x0, x1, . . . , xn ∈ I and x0 < x1 < · · · < xn},

exists (see [3]).C o r o l l a r y 5. In recursive constructive mathematics a real-valued function of

bounded variation on a bounded closed interval I ⊂ R is sequentially continuous on I.P r o o f . It is proved in [4] that a strongly extensional function of bounded varia-

tion is regulated. The result now follows from Theorem 1, since in recursive construc-tive mathematics every real-valued function on an interval is strongly extensional andLPO is false. 2

References

[1] Bishop, E., Foundations of Constructive Analysis. McGraw-Hill, New York 1967.

[2] Bishop, E., and D. Bridges, Constructive Analysis. Springer-Verlag, Berlin-Heidel-berg-New York 1985.

[3] Bridges, D., A constructive look at functions of bounded variation. To appear in Bull.London Math. Soc.

[4] Bridges, D., A. Mahalanobis, and W. Veldman, The constructive regularity of func-tions of bounded variation. To appear.

[5] Bridges, D., and F. Richman, Varieties of Constructive Mathematics. London Math.Soc. Lecture Notes 97, Cambridge Univ. Press, Cambridge 1987.

[6] Ceitin, G. S., Algorithmic operators in constructive complete separable metric spaces.Doklady Akad. Nauk 128 (1959), 49 – 53 (in Russian).

[7] Ishihara, H., Continuity and nondiscontinuity in constructive mathematics. J. Sym-bolic Logic 56 (1991), 1349 – 1354.

[8] Ishihara, H., Markov’s principle, Church’s thesis and Lindelof’s theorem. Proc. Ko-ninklijke Akad. Wetenschappen (Indag. Math., N. S.) 4 (1993), 321 – 325.

Sequential Continuity of Functions in Constructive Analysis 143

[9] Kreisel, G., D. Lacombe, and J. Schonfield, Partial recursive functions and effec-tive operations. In: Constructivity in Mathematics – Proceedings of the Colloquium atAmsterdam 1957 (A. Heyting, ed.), North-Holland Publ. Comp., Amsterdam 1959.

[10] Mandelkern, M., Continuity of monotone functions. Pacific J. Math. 99 (1982),413 – 418.

[11] Mandelkern, M., Constructive Continuity. Mem. Amer. Math. Soc. 227 (1983).

(Received: November 22, 1998)