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The intermediate value theorem in constructive mathematics without choice
Annals of Pure and Applied Logic 163 (2012) 1050–1056
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The intermediate value theorem in constructive mathematics without choice Matthew Hendtlass Department of Pure Mathematics, University of Leeds, Leeds LS2 9JT, England, United Kingdom
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Article history: Available online 10 January 2012 Keywords: Constructive mathematics Intermediate value Countable choice
abstract In Bishop’s constructive mathematics without choice axioms, it seems that in order to construct an object you require it to satisfy some (strong) uniqueness property. As a consequence, it seems unlikely that the standard constructive version of the intermediate value theorem—requiring only the additional hypothesis that the function is locally nonzero—carries over to the setting of Bishop’s constructive mathematics without choice. We give a weaker version of the intermediate value theorem which does hold without choice; the uniqueness of the root constructed follows from it being minimal. © 2011 Elsevier B.V. All rights reserved.
The intermediate value theorem (IVT) Let f : [a, b] → R be a continuous function such that f (a) 6 0 6 f (b). Then there exists an x ∈ [a, b] such that f (x) = 0.1 has a surprisingly rich history within Bishop’s constructive mathematics2 (the framework in which this paper is set). There are two standard classical proofs of this result. In the first we define x = sup{y : f (y) 6 0} and show that this x is a root of f . Constructively, however, this supremum may not exist; the supremum of a subset X of R exists constructively if and only if X is upper order located: for all α, β ∈ Q, if α < β , then either x < β for all x ∈ X or there exists x ∈ X with α < x. In the second proof we ‘construct’ a sequence of nested intervals [an , bn ], whose lengths tends to zero, such that f (an ) 6 0 6 f (bn ) for all n. This is done using interval halving: set a1 = a and b1 = b. Suppose that we have constructed −1 n −1 (ak )nk= 1 and (bk )k=1 , and let m be the midpoint of [an−1 , bn−1 ]. Either f (m) > 0 or f (m) 6 0. In the first case we set an = an−1 , bn = m, and in the second we set an = m, bn = bn−1 . The desired root is then given by the shared limit of (an )n>1 and (bn )n>1 , which are nondecreasing and nonincreasing, respectively. The problem with this argument constructively is in deciding, for each n, whether the value of f at the midpoint of [an−1 , bn−1 ] is greater than or equal to zero, or less than or equal to zero: the statement
∀x∈R x > 0 ∧ x 6 0 implies the essentially nonconstructive principle LLPO: For each binary sequence (an )n>1 with at most one term equal to 1, either an = 0 for all even n or else an = 0 for all odd n. So neither of these proofs is constructive, and in fact IVT implies LLPO, and in the presence of the axiom of countable choice,
E-mail address: [email protected]. 1 Throughout this paper, when we refer to a continuous function f we mean that f is pointwise continuous rather than continuous in the sense of Bishop. 2 That is, mathematics with intuitionistic logic and an appropriate set-theoretic foundation such as those in [1,4,7,8]. For more on Bishop’s constructive mathematics (BISH) and other varieties of constructive mathematics with choice, see [3,5,6]. For BISH without choice see [9–13]. 0168-0072/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.apal.2011.12.026
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CC: If Y is an inhabited set, S a subset of N+ × Y , and for each positive integer n there exists y ∈ Y such that (n, y) ∈ S, then there is a function f : N+ → Y such that (n, f (n)) ∈ S for each n ∈ N+ . IVT and LLPO are equivalent. The first constructive version of IVT, the approximate intermediate value theorem, was given by Bishop in [2] and showed, using choice, the existence of approximate solutions of arbitrary precision: (aIVT) Let f : [a, b] → R be a uniformly continuous function such that f (a) 6 0 6 f (b). Then for all ε > 0 there exists x ∈ [a, b] such that |f (x)| < ε . The (constructive) proof of aIVT essentially uses the classical interval halving argument above, stopping when the value of |f | at the midpoint of [an−1 , bn−1 ] is known to be less than ε . In [5, Chapter 3, Theorem 2.5], Bridges and Richman gave an exact constructive version of the intermediate value theorem3 requiring only the additional hypothesis that f be locally nonzero: for each x ∈ (a, b) and each ε ∈ (0, min{x − a, b − x}), there exists y ∈ (x − ε, x + ε) such that f (y) ̸= 0. Once again the proof proceeds by an interval halving argument, this time with the midpoint of [an−1 , bn−1 ] perturbed slightly, if necessary, so that the function evaluation is nonzero. These two versions of the intermediate value theorem suffice for most applications in analysis. However, since their proofs use interval halving they require the axiom of dependent choice. This axiom and the weaker axiom of countable choice (see below), though widely used in the literature on constructive mathematics, are rejected by some constructive mathematicians on either philosophical or pragmatic grounds (see for instance [10]). In the absence of any choice axioms, there are fewer Cauchy reals than Dedekind reals. Moreover, proving that the Cauchy reals are complete, one of the characterising properties of the reals, requires the axiom of countable choice. Wanting to conserve the completeness of the reals, when working in BISH without choice, we use the Dedekind reals (see [12]), which given CC are isomorphic to the Cauchy reals, but which are complete in the absence of choice. The first choice free proof of aIVT, for uniformly continuous functions, was given by Fred Richman. In [13], Schuster gave a choice free constructive proof of IVT for strictly increasing functions; [13] also contains several more general choice free constructive versions of aIVT. In this paper we extend on these results, proving IVT for functions with well behaved roots. First, for completeness, we give a choice free constructive proof of aIVT. Proposition 1. Let f : [a, b] → R be a uniformly continuous function such that f (a) 6 0 6 f (b). Then for all ε > 0 there exists x ∈ [a, b] such that |f (x)| < ε . Proof. Fix ε > 0 and let δ > 0 be such that for all x, y ∈ [a, b], if |x − y| < 2δ , then |f (x) − f (y)| < ε ; without loss of generality δ divides b − a. Set S = {a, a + δ, . . . , b − δ, b} , and suppose that min {|f (x)| : x ∈ S } > ε/2. Then S is the disjoint union of S − = {x ∈ S : f (x) < 0}
and S + = {x ∈ S : f (x) > 0} .
By a simple contradiction argument there exists x ∈ S such that x ∈ S − and x + δ ∈ S + . Then
|f (x) − f (x + δ)| = f (x + δ) − f (x) > ε/2 − (−ε/2) = ε, but |x − (x + δ)| < 2δ —a contradiction. Hence min {|f (x)| : x ∈ S } < ε , so there exists x ∈ S such that |f (x)| < ε .
Throughout this paper we use I to denote either a closed bounded interval or the entire real line. Let f be a continuous real valued function on I. In [14], Taylor calls a root x of f stable if
∀ε>0 ∃y,z ∈(x−ε,x+ε)∩I (f (y) < 0 < f (z )); a root x of f is stable if f crosses the horizontal axis arbitrarily close to x. Since the property of all the roots of a function f being stable only gives us positive information about f when we already have a root of f , it seems unlikely that this condition is strong enough to prove a constructive version of the intermediate value theorem, even when we allow choice. So we need a strengthening of the notion of a function having only stable roots. A function f : I → R is said to have uniformly stable roots if
∀ε>0 ∃δ>0 ∀x∈I (|f (x)| < δ ⇒ ∃y,z ∈(x−ε,x+ε)∩I f (y) < 0 < f (z )). It is easy to see that if f has uniformly stable roots, then f is locally nonzero and has only stable roots. Since, given a function f : [a, b] → R with f (a) 6 0 and f (b) > 0, we need only construct the smallest root of f , we only require f to have uniformly upper-stable roots:
∀ε>0 ∃δ>0 ∀x∈(a,b) (|f (x)| < δ ⇒ ∃z ∈(x−ε,x+ε) f (z ) > 0). Our choice free constructive version of the intermediate value theorem is the following.
3 In the presence of countable choice, we only require f to be sequentially continuous for both aIVT and this exact version of the intermediate value theorem (see exercise 11, page 21 of [6]).
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Proposition 2. Let f : [a, b] → R be a uniformly continuous function with uniformly upper-stable roots such that f (a) 6 0 and f (b) > 0. Then there exists x ∈ [a, b] such that f (x) = 0 and f (y) < 0 for all y ∈ [0, x). Before proving Proposition 2, we give some examples of functions with uniformly stable roots. If f : R → R is a uniformly continuous, continuously differentiable function which only has roots of multiplicity one in the following strong sense There exists r > 0 such that max{|f (x)|, |f ′ (x)|} > r for all x ∈ R. then f has uniformly stable roots. To see this, fix ε > 0 and let r > 0 be such that max{|f (x)|, |f ′ (x)|} > r for all x ∈ R; without loss of generality ε < 1/2 and r < 1. Since f is uniformly continuous, there exists δ1 ∈ (0, ε) such that if |x − y| < δ1 , then |f (x) − f (y)| < r /2. Set δ = δ1 r and fix x ∈ R such that |f (x)| < δ . Then for all y ∈ (x − δ1 , x + δ1 )
|f (y)| 6 |f (x)| + |f (y) − f (x)| < δ + r /2 < r, so, by our choice of r, |f ′ (y)| > r. Suppose that thereexist y,y′ ∈(x − δ1 , x + δ1 ) such that f ′ (y) < 0 < f ′ (y′ ). Applying aIVT to f ′ |(min{y,y′ },max{y,y′ }) produces z ∈ (min y, y′ , max y, y′ ) such that |f ′ (z )| < r, which is a contradiction. Hence either f ′ (y) > r for all y ∈ (x − δ1 , x + δ1 ) or else f ′ (y) 6 −r for all y ∈ (x − δ1 , x + δ1 ). In the first case f (x + δ1 ) > r δ1 + f (x) = δ + f (x) > 0, and f (x − δ1 ) 6 −r δ1 + f (x) = −δ + f (x) < 0. In the second case f (x + δ1 ) < 0 < f (x − δ1 ). Since x + δ1 , x − δ1 ∈ (x − ε, x + ε), f has uniformly stable roots. If f is a polynomial, then we can do better. First we require a few notions from [9]. We denote by πn (R) the set of monic polynomials of degree n with real coefficients given any of the standard metrics. An n-multiset is an image of {1, 2, . . . , n}. The distance between two n-multisets X = {x1 , x2 , . . . , xn } and Y = {y1 , y2 , . . . , yn } is given by
ρn (X , Y ) = inf sup{|xi − yσ (i) | : 1 6 i 6 n}, σ ∈Sn
where Sn denotes the group of permutations on n elements; the set Mn (C) of n-multisets of complex numbers taken with ρn is a pseudometric space. In [9] it is shown that the natural map F from Mn (C) to πn (R) is one-one and uniformly bicontinuous. Let p(x) =
n (x − ri ) (ri ∈ C) i=1
be a polynomial4 such that ri ̸= rj for all 1 6 i < j 6 n. Then p has uniformly stable roots. We prove that there exists r > 0 such that for all x ∈ R either |p(x)| > r or |p′ (x)| > r; this is done by induction on the degree of p. The case where p is linear is trivial. Suppose that the result holds for polynomials of degree n − 1 and write p(x) = (x − r1 )
n (x − ri ) ≡ (x − r1 )q(x). i =2
Then by our induction hypothesis there exists r ′ ∈ (0, 1) such that for all x ∈ R either |q(x)| > r ′ or |q′ (x)| > r ′ . Fix x ∈ R and let R ∈ R be such that p′ (y) < R for all y ∈ [r1 − 1, r1 + 1]; without loss of generality r ′ < R. If |q(x)| > r ′ , then either |x − r1 | > r ′ /3R and
|p(x)| = (x − r1 )q(x) >
r ′2 r′ ′ r = , 3R 3R
or |x − r1 | < 2r ′ /3R. In the latter case
|(x − r1 )q′ (x)| <
2r ′ 3R
R=
2r ′ 3
,
so
|p′ (x)| = |(x − r1 )q′ (x) + q(x)| >
r′ 3
.
It remains to consider the case |q′ (x)| > r ′ . Let ε > 0 be such that |r1 − ri | > ε for all i ∈ {2, . . . , n}. Using the uniform bicontinuity of the natural map F sending n-multisets of C to polynomials of degree n, there exists δ > 0 such that for all q′ ∈ πn (R) if ρ(q, q′ ) < δ , then ρn (F −1 (q), F −1 (q′ )) < ε . Suppose that |q(r1 )| < δ . Then ρ(q, q + q(r1 )) < δ , so ρn (F −1 (q), F −1 (q + q(r1 ))) < ε/2. Since q + q(r1 ) has a root at r1 , there exists i ∈ {1, . . . , n} such that |r1 − ri | < ε, which is
4 Of course, the application of Proposition 2 to such polynomials is not very interesting.
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absurd. It follows that |q(r1 )| > δ > 0. By the continuity of q at r1 , there exists δ ′ ∈ (0, min{r ′ /R, r ′ }) such that for all x ∈ R either
|x − r1 | > δ ′ or |q(x)| > δ ′ . In the first case either |q(x)| > r ′ δ ′ /3 and
|p(x)| = |(x − r1 )q(x)| >
r ′ δ ′2 r ′ δ′ ′ δ = , 3 3
or |q(x)| < 2r δ ′ /3. If |q(x)| < 2r ′ δ ′ /3, then |(x − r1 )q′ (x)| > r ′ δ ′ and
|p′ (x)| = |(x − r1 )q′ (x) + q(x)| >
r ′ δ′ 3
>
r ′ δ ′2 3
.
In the second case |x − r1 | > δ ′ /3R and so |p(x)| > δ ′2 /3R, or |x − r1 | < δ ′ /2R and |p′ (x)| > δ ′ /2 > δ ′2 /3R. Given that δ ′ < r ′ < 1, it just remains to set
r = min
r ′ δ ′2 δ ′2 3
,
3R
;
then for all x ∈ R either |p(x)| > r or |p′ (x)| > r. The next lemma is a (very) slight generalisation of a result in [13] which considers strictly increasing functions. Lemma 3. Let f be a locally nonzero nondecreasing real valued function on [a, b] such that f (a) 6 0 and f (b) > 0. Then there exists a unique x ∈ [a, b] with f (x) = 0. Proof. Without loss of generality a = 0 and b = 1. Define L = {y ∈ [0, 1] : f (y) 6 0} ; it is easily shown that the required x is given by the supremum of L provided that this supremum exists. Since f (0) 6 0, 0 ∈ L so L is nonempty. Let α, β ∈ Q be such that α < β; without loss of generality α,β ∈ [0, 1]. Since f is locally nonzero, there exist α ′ , β ′ such that α < α ′ < β ′ < β and f α ′ and f β ′ are nonzero. If f α ′ > 0 and f β ′ < 0, then f β ′ < 0 < f α ′ , which contradicts that f is nondecreasing. Hence either f α ′ < 0 or f β ′ > 0. If f α ′ < 0, then
f (α) < 0 and α ∈ L. If f β ′ > 0, then f (y) < f β ′ for each y ∈ L, so, since f is nondecreasing, y 6 β for each y ∈ L and β is an upper bound for L. Hence L is upper order located and therefore x = sup(L) exists. It remains to show that this x is unique. To this end, let x, y ∈ [0, 1] be such that f (x) = f (y) = 0, and suppose that x ̸= y. Without loss of generality x < y. Since f is locally nonzero, there exists z ∈ (x, y) such that f (z ) ̸= 0; either f (z ) > 0 or f (z ) < 0. In the first case we have that f (z ) < f (x) and x < z, contradicting that f is nondecreasing. In the second case, f (y) < f (z ) and z < y, which again gives a contradiction. Hence x = y and f has a unique root.
Here is the proof of Proposition 2: Proof. We may assume that a = 0 and b = 1. Define a nondecreasing function g : [0, 1] → R by5 g (x) = sup {f (y) : y ∈ [0, x]} ; it is easily shown that g is uniformly continuous. Then any zero of g is a zero of f : let x ∈ [0, 1] be a zero of g and suppose that f (x) ̸= 0. If f (x) > 0, then g (x) > 0, which is absurd. Therefore f (x) < 0. Using the continuity of f at x, pick r > 0 such that f (y) < f (x)/2 for all y ∈ (x − r , x). Let δ ∈ (0, |f (x)|/2) be such that for all x′ ∈ [0, 1] if |f (x′ )| < δ , then there exists z ∈ (x′ − r , x′ + r ) with f (z ) > 0. Since g (x) = sup {f (y) : y ∈ [0, x]} = 0, there exists x′ ∈ [0, x] such that |f (x′ )| < δ ; moreover, since δ < |f (x)|/2, x′ 6 x − r. Then, by our choice of δ , there exists z < x′ + r 6 x such that f (z ) > 0, and so g (x) > 0. This final contradiction ensures that f (x) = 0. Thus in order to show that f has a zero we need only show that g has a zero; in turn, by Lemma 3, it suffices to show that g is locally nonzero. To this end, fix x ∈ [0, 1] and ε > 0; without loss of generality x ∈ / {0, 1} and ε < min{x, 1 − x}. Using that f has uniformly upper-stable roots, pick δ > 0 such that for all x ∈ [0, 1], if |f (x)| < δ , then there exists z ∈ (x −ε, x +ε) such that f (z ) > 0. Either |g (x)| > 0 or, as we may assume, |g (x)| < δ . Then there exists y ∈ [0, x] such that |f (y)| < δ ; whence, by our choice of δ , there exists z ∈ (y − ε, y + ε) such that f (z ) > 0. Thus g (x + ε) > f (z ) > 0. It now follows from the continuity of g at x + ε that g is locally nonzero. 5 The supremum of {f (y) : y ∈ [0, x]} is given by q ∈ Q : ∃y∈[0,x] (q < f (y)) , q ∈ Q : ∀y∈[0,x] (f (y) < q)
.
This is a Dedekind real because f is uniformly continuous and [0, x] is totally bounded.
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Let x be the unique root of g; consider y ∈ [0, x) and set ε ′ = y − x. Since f has uniformly upper-stable roots, there exists δ > 0 such that for all x′ ∈ [0, 1] ′
|f (x)| < δ ′ ⇒ ∃z ′ ∈(x′ −ε′ ,x′ +ε′ )∩[0,1] f (z ′ ) > 0. Either f (y) < 0 or f (y) > −δ ′ . In the latter case, there exists z < y + ε ′ = x such that f (z ) > 0. This contradicts our construction of x; whence f (y) < 0. In [13], Schuster put forward the ‘guarded thesis’ that in order to construct an object in BISH without choice we require that object to be unique (in some strong sense). In Proposition 2 this uniqueness is given by the produced x being the minimal root of f . Finding the minimal root of a locally nonconstant uniformly continuous function f : [a, b] → R with f (a) 6 0 and f (b) > 0 is not possible in general. For example, let a be a real number close to 0 and define f : [−1, 2] → R by f (x) = x2 (x − 1) + a. Then, for sufficiently small a, f has minimal root less than 1/2 if and only if a > 0 and has minimal root greater than 0 if and only if a < 0. Knowing that f has only stable roots does not help, since there exists a stable root of f less than 1/2 if and only if a > 0. This example and the principle of unique existence suggest that the standard constructive version of the intermediate value theorem (in which f is locally nonzero) is likely to be unprovable in BISH without choice. We finish by presenting some examples showing the relationship between the various properties on a function f . There exists a function f : R → R that has only stable roots, for which it is impossible not to have a root, and for which we can, in general, only find a root if Markov’s Principle MP: For any binary sequence (an )n>1 , if it is impossible for an = 0 for all n, then there exists n such that an = 1. holds, but which is only locally nonzero if MP holds. Let a ∈ R be such that ¬(a = 0) and define f : R → R by6
f (x) =
1
1 −x − a + a
a>
−x + + a a
a6 <0 x otherwise.
1
a
x 1
>0
To see that f has only stable roots, let x ∈ R be a root of f . Note that if a ̸= 0, then f is locally linear at x with gradient −1, so f has only stable roots. Pick N > 0 such that |x| < N. If |a| < 1/N, then |x| > N, which is absurd. Hence |a| > 0 and f has only stable roots. Now suppose that f is locally nonzero. Then there exists x ∈ (−1, 1) with f (x) ̸= 0 and so |a| > min{1, |f (x)|} > 0. This example does not have a direct impact on whether having only stable roots is sufficient to prove a constructive intermediate value theorem (without choice), since it is necessary in this example that the domain of f is unbounded. It is possible to give an example of a function f : [0, 2] → R with compact domain such that f has only stable roots, f is only locally non-zero, in general, if MP holds, and it is impossible for f not to have a root: let a be a non-negative real number such that ¬(a = 1) and define 2x − 1 f (x) = 2(a − 1)x + 3 − 2a a
x ∈ [0, 1] x ∈ [1, 3/2] x ∈ [3/2, 2].
6 The standard construction of functions on R by case does not require choice.
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However, in this case it is easy to find the root of f . The condition that f : I → R have at most one root
∀x,y∈I (x ̸= y ⇒ f (x) > 0 ∨ f (y) > 0) is not strong enough to prove that f has uniformly upper-stable roots, even when f is continuous, has only stable roots, and there exist a, b ∈ I such that f (a)f (b) < 0. To see this, recall that in Russian recursive mathematics there exists a continuous function g : [0, 1/2] → R with supremum 0 such that g (x) < 0 for each x ∈ [0, 1/2] (see [5, Chapter 6, Corollary 2.9]). The construction of g uses the existence of an enumeration of the set of partial functions from N into N with countable domain. Assuming the Church–Markov–Turing thesis, this enumeration is constructed using countable choice [15]. Define
f ( x) =
g (x) g (1/2) 4(1 − g (1/2))x − 3 + 4g (1/2)
x ∈ [0, 1/2] x ∈ [1/2, 3/4] x ∈ [3/4, 1].
Then f is a continuous real valued function on the unit interval, f (0) = g (0) < 0, and f (1) = 1 > 0. Since f is linear with gradient 4(1 − f (1/2)) on [3/4, 1] and since the only root of f is within this interval, all roots of f are stable. To see that f has at most one root, note that f (x) ̸= 0 for all x ∈ [0, 1] −
3 − 4f (1/2) 4(1 − f (1/2))
.
But f does not have uniformly upper-stable roots: let ε = 1/4. For all δ > 0 there exists x ∈ [0, 1/2] such that |f (x)| < δ , but f (y) ̸= 0 for all y ∈ (x − ε, x + ε) ∩ [0, 1]. If a continuous function f : [a, b] → R with f (a) < 0 and f (b) > 07 has uniformly at most one zero—that is,
∀ε>0 ∃δ>0 ∀x,y∈I (|f (x)| < δ ∧ |f (y)| < δ ⇒ |x − y| < ε) —then f has uniformly stable roots. To see this, let f : [a, b] → R be a continuous function with uniformly at most one root and fix ε > 0. Let δ ∈ (0, r ) be such that for all x, y ∈ [a, b]
|f (x)| < δ ∧ |f (y)| < δ ⇒ |x − y| < ε, and let x ∈ [a, b] be such that |f (x)| < δ . Using the continuity of f at a and b, we may assume that |f (x)| > δ for all x ∈ [a, a + ε) ∪ (b − ε, b]. Then x ∈ [a + ε, b − ε] and so, by continuity, it suffices to show that f (x − ε) < 0 < f (x + ε). Suppose that f (x + ε) < δ . If f (x + ε) > −δ , then |f (x + ε)| < δ , so |x − (x − ε)| < ε —a contradiction from which it follows that f (x + ε) < 0. Then by the approximate intermediate value theorem, there exists z ∈ (x + ε, 1) such that |f (z )| < δ , but
|z − x| > |x + ε − x| = ε, again contradicting our choice of δ . Thus f (x + ε) > δ > 0. A similar proof shows that f (x − ε) < 0. With countable choice any continuous function f : [a, b] → R with f (a) < 0 and f (b) > 0 that has at most one root has only stable roots. Let f : [a, b] → R be a continuous function with f (a) < 0 and f (b) > 0 that has at most one root, let x ∈ [a, b] be such that f (x) = 0, and let ε > 0. Then x ̸∈ {a, b} and without loss of generality (x − ε, x + ε) ⊂ [a, b]. Since f has at most one root, f (x + ε/2) ̸= 0 and f (x − ε/2) ̸= 0. Suppose that f (x + ε/2) < 0. Having at most one zero implies being locally nonconstant, so we can apply the standard constructive intermediate value theorem (which requires countable choice) to construct y ∈ [x + ε/2, 1] with f (y) = 0—a contradiction. Hence f (x + ε/2) > 0. A similar proof shows that f (x − ε/2) < 0, so x is a stable root and f has only stable roots. As you would expect, if f : R → R has uniformly stable roots and does not have two roots, then f has uniformly at most one root. Let ε > 0. Since f has uniformly stable roots, there exists δ < 0 such that for all x ∈ R if |f (x)| < δ , then there exist y, z ∈ (x − ε/2, x + ε/2) with f (y) < 0 < f (z ). Let x, x′ ∈ R be such that |f (x)| < δ and |f (x′ )| < δ , and suppose
7 We need this condition to rule out a function contained entirely above or below the x-axis (for example, f (x) = |x − 1/2| on [0, 1]).
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′ that |x − x′ | > ε . Then there exist y, z ∈ (x − ε/2, x + ε/2) and y′ , z ′ ∈ (x′ − ε/ 2, x + ε/2) such that ′f (y) < 0 < f (z ) ′ ′ ′ ′ ′ and f (y ) < 0 < f (z ). By Proposition 2, there exist w ∈ (y, z ) and w ∈ y , z such that f (w) = f (w ) = 0. But, since |x − x′ | > ε, w ̸= w′ —this contradicts that f does not have two roots. Hence |x − x′ | 6 ε , and f has uniformly at most one zero. The following diagram summarises the relationships between the various properties of a function f : R → R; all implications are strict.
Acknowledgements The author wishes to thank Peter Schuster for stimulating conversations regarding this work, and the University of Leeds for their support via a FFIR scholarship. The author also wishes to thank the referees for helping him to bring this paper into its final form, and, in particular, for pointing out a mistake in one of the examples. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
P.H.G. Aczel, M. Rathjen, Notes on constructive set theory, Report No. 40, Institut Mittag-Leffler, Royal Swedish Academy of Sciences, 2001. E. Bishop, Foundations of Constructive Analysis, McGraw-Hill, New York, 1967. E.A. Bishop, D.S. Bridges, Constructive Analysis, in: Grundlehren der Math. Wiss., vol. 279, Springer-Verlag, Heidelberg, 1985. D.S. Bridges, A constructive Morse theory of sets, in: D. Skordev (Ed.), Mathematical Logic and its Applications, Plenum Publishing Corp., New York, 1987, pp. 61–79. D.S. Bridges, F. Richman, Varieties of Constructive Mathematics, in: London Math. Soc. Lecture Notes, vol. 97, Cambridge Univ. Press, 1987. D.S. Bridges, L.S. Vîţă, Techniques of Constructive Analysis, Universitext, Springer, New-York, 2006. H.M. Friedman, Set theoretic foundations for constructive analysis, Ann. Math. 105 (1) (1977) 1–28. J. Myhill, Constructive set theory, J. Symbolic Logic 40 (3) (1975) 347–382. F. Richman, The fundamental theorem of algebra: a constructive development without choice, Pacific Journal of Mathematics 196 (2000) 213–230. F. Richman, Constructive mathematics without choice, in: schuster, et al. (Eds.), Reuniting the Antipodes—Constructive and Nonstandard Views of the Continuum, in: Synthese Library, vol. 306, Kluwer, 2001, pp. 199–205. F. Richman, Spreads and choice in constructive mathematics, Indagations Mathematicae 13 (2002) 259–267. F. Richman, Real numbers and other completions, Math. Logic Quarterly 54 (2008) 98–108. P. Schuster, Unique existence, approximate solutions, and countable choice, Theoretical Computer Science 305 (2003) 433–455. P. Taylor, A Lambda calculus for real analysis, Computability and Complexity in Analysis, Kyoto, 2005. A.S. Troelstra, D. van Dalen, Constructivism in mathematics, Vol. I. An introduction, in: Studies in Logic and the Foundations of Mathematics, vol. 121, North-Holland Publishing Co., Amsterdam, 1988.
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The intermediate value theorem in constructive mathematics without choice Matthew Hendtlass Department of Pure Mathematics, University of Leeds, Leeds LS2 9JT, England, United Kingdom
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Article history: Available online 10 January 2012 Keywords: Constructive mathematics Intermediate value Countable choice
abstract In Bishop’s constructive mathematics without choice axioms, it seems that in order to construct an object you require it to satisfy some (strong) uniqueness property. As a consequence, it seems unlikely that the standard constructive version of the intermediate value theorem—requiring only the additional hypothesis that the function is locally nonzero—carries over to the setting of Bishop’s constructive mathematics without choice. We give a weaker version of the intermediate value theorem which does hold without choice; the uniqueness of the root constructed follows from it being minimal. © 2011 Elsevier B.V. All rights reserved.
The intermediate value theorem (IVT) Let f : [a, b] → R be a continuous function such that f (a) 6 0 6 f (b). Then there exists an x ∈ [a, b] such that f (x) = 0.1 has a surprisingly rich history within Bishop’s constructive mathematics2 (the framework in which this paper is set). There are two standard classical proofs of this result. In the first we define x = sup{y : f (y) 6 0} and show that this x is a root of f . Constructively, however, this supremum may not exist; the supremum of a subset X of R exists constructively if and only if X is upper order located: for all α, β ∈ Q, if α < β , then either x < β for all x ∈ X or there exists x ∈ X with α < x. In the second proof we ‘construct’ a sequence of nested intervals [an , bn ], whose lengths tends to zero, such that f (an ) 6 0 6 f (bn ) for all n. This is done using interval halving: set a1 = a and b1 = b. Suppose that we have constructed −1 n −1 (ak )nk= 1 and (bk )k=1 , and let m be the midpoint of [an−1 , bn−1 ]. Either f (m) > 0 or f (m) 6 0. In the first case we set an = an−1 , bn = m, and in the second we set an = m, bn = bn−1 . The desired root is then given by the shared limit of (an )n>1 and (bn )n>1 , which are nondecreasing and nonincreasing, respectively. The problem with this argument constructively is in deciding, for each n, whether the value of f at the midpoint of [an−1 , bn−1 ] is greater than or equal to zero, or less than or equal to zero: the statement
∀x∈R x > 0 ∧ x 6 0 implies the essentially nonconstructive principle LLPO: For each binary sequence (an )n>1 with at most one term equal to 1, either an = 0 for all even n or else an = 0 for all odd n. So neither of these proofs is constructive, and in fact IVT implies LLPO, and in the presence of the axiom of countable choice,
E-mail address: [email protected]. 1 Throughout this paper, when we refer to a continuous function f we mean that f is pointwise continuous rather than continuous in the sense of Bishop. 2 That is, mathematics with intuitionistic logic and an appropriate set-theoretic foundation such as those in [1,4,7,8]. For more on Bishop’s constructive mathematics (BISH) and other varieties of constructive mathematics with choice, see [3,5,6]. For BISH without choice see [9–13]. 0168-0072/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.apal.2011.12.026
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CC: If Y is an inhabited set, S a subset of N+ × Y , and for each positive integer n there exists y ∈ Y such that (n, y) ∈ S, then there is a function f : N+ → Y such that (n, f (n)) ∈ S for each n ∈ N+ . IVT and LLPO are equivalent. The first constructive version of IVT, the approximate intermediate value theorem, was given by Bishop in [2] and showed, using choice, the existence of approximate solutions of arbitrary precision: (aIVT) Let f : [a, b] → R be a uniformly continuous function such that f (a) 6 0 6 f (b). Then for all ε > 0 there exists x ∈ [a, b] such that |f (x)| < ε . The (constructive) proof of aIVT essentially uses the classical interval halving argument above, stopping when the value of |f | at the midpoint of [an−1 , bn−1 ] is known to be less than ε . In [5, Chapter 3, Theorem 2.5], Bridges and Richman gave an exact constructive version of the intermediate value theorem3 requiring only the additional hypothesis that f be locally nonzero: for each x ∈ (a, b) and each ε ∈ (0, min{x − a, b − x}), there exists y ∈ (x − ε, x + ε) such that f (y) ̸= 0. Once again the proof proceeds by an interval halving argument, this time with the midpoint of [an−1 , bn−1 ] perturbed slightly, if necessary, so that the function evaluation is nonzero. These two versions of the intermediate value theorem suffice for most applications in analysis. However, since their proofs use interval halving they require the axiom of dependent choice. This axiom and the weaker axiom of countable choice (see below), though widely used in the literature on constructive mathematics, are rejected by some constructive mathematicians on either philosophical or pragmatic grounds (see for instance [10]). In the absence of any choice axioms, there are fewer Cauchy reals than Dedekind reals. Moreover, proving that the Cauchy reals are complete, one of the characterising properties of the reals, requires the axiom of countable choice. Wanting to conserve the completeness of the reals, when working in BISH without choice, we use the Dedekind reals (see [12]), which given CC are isomorphic to the Cauchy reals, but which are complete in the absence of choice. The first choice free proof of aIVT, for uniformly continuous functions, was given by Fred Richman. In [13], Schuster gave a choice free constructive proof of IVT for strictly increasing functions; [13] also contains several more general choice free constructive versions of aIVT. In this paper we extend on these results, proving IVT for functions with well behaved roots. First, for completeness, we give a choice free constructive proof of aIVT. Proposition 1. Let f : [a, b] → R be a uniformly continuous function such that f (a) 6 0 6 f (b). Then for all ε > 0 there exists x ∈ [a, b] such that |f (x)| < ε . Proof. Fix ε > 0 and let δ > 0 be such that for all x, y ∈ [a, b], if |x − y| < 2δ , then |f (x) − f (y)| < ε ; without loss of generality δ divides b − a. Set S = {a, a + δ, . . . , b − δ, b} , and suppose that min {|f (x)| : x ∈ S } > ε/2. Then S is the disjoint union of S − = {x ∈ S : f (x) < 0}
and S + = {x ∈ S : f (x) > 0} .
By a simple contradiction argument there exists x ∈ S such that x ∈ S − and x + δ ∈ S + . Then
|f (x) − f (x + δ)| = f (x + δ) − f (x) > ε/2 − (−ε/2) = ε, but |x − (x + δ)| < 2δ —a contradiction. Hence min {|f (x)| : x ∈ S } < ε , so there exists x ∈ S such that |f (x)| < ε .
Throughout this paper we use I to denote either a closed bounded interval or the entire real line. Let f be a continuous real valued function on I. In [14], Taylor calls a root x of f stable if
∀ε>0 ∃y,z ∈(x−ε,x+ε)∩I (f (y) < 0 < f (z )); a root x of f is stable if f crosses the horizontal axis arbitrarily close to x. Since the property of all the roots of a function f being stable only gives us positive information about f when we already have a root of f , it seems unlikely that this condition is strong enough to prove a constructive version of the intermediate value theorem, even when we allow choice. So we need a strengthening of the notion of a function having only stable roots. A function f : I → R is said to have uniformly stable roots if
∀ε>0 ∃δ>0 ∀x∈I (|f (x)| < δ ⇒ ∃y,z ∈(x−ε,x+ε)∩I f (y) < 0 < f (z )). It is easy to see that if f has uniformly stable roots, then f is locally nonzero and has only stable roots. Since, given a function f : [a, b] → R with f (a) 6 0 and f (b) > 0, we need only construct the smallest root of f , we only require f to have uniformly upper-stable roots:
∀ε>0 ∃δ>0 ∀x∈(a,b) (|f (x)| < δ ⇒ ∃z ∈(x−ε,x+ε) f (z ) > 0). Our choice free constructive version of the intermediate value theorem is the following.
3 In the presence of countable choice, we only require f to be sequentially continuous for both aIVT and this exact version of the intermediate value theorem (see exercise 11, page 21 of [6]).
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Proposition 2. Let f : [a, b] → R be a uniformly continuous function with uniformly upper-stable roots such that f (a) 6 0 and f (b) > 0. Then there exists x ∈ [a, b] such that f (x) = 0 and f (y) < 0 for all y ∈ [0, x). Before proving Proposition 2, we give some examples of functions with uniformly stable roots. If f : R → R is a uniformly continuous, continuously differentiable function which only has roots of multiplicity one in the following strong sense There exists r > 0 such that max{|f (x)|, |f ′ (x)|} > r for all x ∈ R. then f has uniformly stable roots. To see this, fix ε > 0 and let r > 0 be such that max{|f (x)|, |f ′ (x)|} > r for all x ∈ R; without loss of generality ε < 1/2 and r < 1. Since f is uniformly continuous, there exists δ1 ∈ (0, ε) such that if |x − y| < δ1 , then |f (x) − f (y)| < r /2. Set δ = δ1 r and fix x ∈ R such that |f (x)| < δ . Then for all y ∈ (x − δ1 , x + δ1 )
|f (y)| 6 |f (x)| + |f (y) − f (x)| < δ + r /2 < r, so, by our choice of r, |f ′ (y)| > r. Suppose that thereexist y,y′ ∈(x − δ1 , x + δ1 ) such that f ′ (y) < 0 < f ′ (y′ ). Applying aIVT to f ′ |(min{y,y′ },max{y,y′ }) produces z ∈ (min y, y′ , max y, y′ ) such that |f ′ (z )| < r, which is a contradiction. Hence either f ′ (y) > r for all y ∈ (x − δ1 , x + δ1 ) or else f ′ (y) 6 −r for all y ∈ (x − δ1 , x + δ1 ). In the first case f (x + δ1 ) > r δ1 + f (x) = δ + f (x) > 0, and f (x − δ1 ) 6 −r δ1 + f (x) = −δ + f (x) < 0. In the second case f (x + δ1 ) < 0 < f (x − δ1 ). Since x + δ1 , x − δ1 ∈ (x − ε, x + ε), f has uniformly stable roots. If f is a polynomial, then we can do better. First we require a few notions from [9]. We denote by πn (R) the set of monic polynomials of degree n with real coefficients given any of the standard metrics. An n-multiset is an image of {1, 2, . . . , n}. The distance between two n-multisets X = {x1 , x2 , . . . , xn } and Y = {y1 , y2 , . . . , yn } is given by
ρn (X , Y ) = inf sup{|xi − yσ (i) | : 1 6 i 6 n}, σ ∈Sn
where Sn denotes the group of permutations on n elements; the set Mn (C) of n-multisets of complex numbers taken with ρn is a pseudometric space. In [9] it is shown that the natural map F from Mn (C) to πn (R) is one-one and uniformly bicontinuous. Let p(x) =
n (x − ri ) (ri ∈ C) i=1
be a polynomial4 such that ri ̸= rj for all 1 6 i < j 6 n. Then p has uniformly stable roots. We prove that there exists r > 0 such that for all x ∈ R either |p(x)| > r or |p′ (x)| > r; this is done by induction on the degree of p. The case where p is linear is trivial. Suppose that the result holds for polynomials of degree n − 1 and write p(x) = (x − r1 )
n (x − ri ) ≡ (x − r1 )q(x). i =2
Then by our induction hypothesis there exists r ′ ∈ (0, 1) such that for all x ∈ R either |q(x)| > r ′ or |q′ (x)| > r ′ . Fix x ∈ R and let R ∈ R be such that p′ (y) < R for all y ∈ [r1 − 1, r1 + 1]; without loss of generality r ′ < R. If |q(x)| > r ′ , then either |x − r1 | > r ′ /3R and
|p(x)| = (x − r1 )q(x) >
r ′2 r′ ′ r = , 3R 3R
or |x − r1 | < 2r ′ /3R. In the latter case
|(x − r1 )q′ (x)| <
2r ′ 3R
R=
2r ′ 3
,
so
|p′ (x)| = |(x − r1 )q′ (x) + q(x)| >
r′ 3
.
It remains to consider the case |q′ (x)| > r ′ . Let ε > 0 be such that |r1 − ri | > ε for all i ∈ {2, . . . , n}. Using the uniform bicontinuity of the natural map F sending n-multisets of C to polynomials of degree n, there exists δ > 0 such that for all q′ ∈ πn (R) if ρ(q, q′ ) < δ , then ρn (F −1 (q), F −1 (q′ )) < ε . Suppose that |q(r1 )| < δ . Then ρ(q, q + q(r1 )) < δ , so ρn (F −1 (q), F −1 (q + q(r1 ))) < ε/2. Since q + q(r1 ) has a root at r1 , there exists i ∈ {1, . . . , n} such that |r1 − ri | < ε, which is
4 Of course, the application of Proposition 2 to such polynomials is not very interesting.
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absurd. It follows that |q(r1 )| > δ > 0. By the continuity of q at r1 , there exists δ ′ ∈ (0, min{r ′ /R, r ′ }) such that for all x ∈ R either
|x − r1 | > δ ′ or |q(x)| > δ ′ . In the first case either |q(x)| > r ′ δ ′ /3 and
|p(x)| = |(x − r1 )q(x)| >
r ′ δ ′2 r ′ δ′ ′ δ = , 3 3
or |q(x)| < 2r δ ′ /3. If |q(x)| < 2r ′ δ ′ /3, then |(x − r1 )q′ (x)| > r ′ δ ′ and
|p′ (x)| = |(x − r1 )q′ (x) + q(x)| >
r ′ δ′ 3
>
r ′ δ ′2 3
.
In the second case |x − r1 | > δ ′ /3R and so |p(x)| > δ ′2 /3R, or |x − r1 | < δ ′ /2R and |p′ (x)| > δ ′ /2 > δ ′2 /3R. Given that δ ′ < r ′ < 1, it just remains to set
r = min
r ′ δ ′2 δ ′2 3
,
3R
;
then for all x ∈ R either |p(x)| > r or |p′ (x)| > r. The next lemma is a (very) slight generalisation of a result in [13] which considers strictly increasing functions. Lemma 3. Let f be a locally nonzero nondecreasing real valued function on [a, b] such that f (a) 6 0 and f (b) > 0. Then there exists a unique x ∈ [a, b] with f (x) = 0. Proof. Without loss of generality a = 0 and b = 1. Define L = {y ∈ [0, 1] : f (y) 6 0} ; it is easily shown that the required x is given by the supremum of L provided that this supremum exists. Since f (0) 6 0, 0 ∈ L so L is nonempty. Let α, β ∈ Q be such that α < β; without loss of generality α,β ∈ [0, 1]. Since f is locally nonzero, there exist α ′ , β ′ such that α < α ′ < β ′ < β and f α ′ and f β ′ are nonzero. If f α ′ > 0 and f β ′ < 0, then f β ′ < 0 < f α ′ , which contradicts that f is nondecreasing. Hence either f α ′ < 0 or f β ′ > 0. If f α ′ < 0, then
f (α) < 0 and α ∈ L. If f β ′ > 0, then f (y) < f β ′ for each y ∈ L, so, since f is nondecreasing, y 6 β for each y ∈ L and β is an upper bound for L. Hence L is upper order located and therefore x = sup(L) exists. It remains to show that this x is unique. To this end, let x, y ∈ [0, 1] be such that f (x) = f (y) = 0, and suppose that x ̸= y. Without loss of generality x < y. Since f is locally nonzero, there exists z ∈ (x, y) such that f (z ) ̸= 0; either f (z ) > 0 or f (z ) < 0. In the first case we have that f (z ) < f (x) and x < z, contradicting that f is nondecreasing. In the second case, f (y) < f (z ) and z < y, which again gives a contradiction. Hence x = y and f has a unique root.
Here is the proof of Proposition 2: Proof. We may assume that a = 0 and b = 1. Define a nondecreasing function g : [0, 1] → R by5 g (x) = sup {f (y) : y ∈ [0, x]} ; it is easily shown that g is uniformly continuous. Then any zero of g is a zero of f : let x ∈ [0, 1] be a zero of g and suppose that f (x) ̸= 0. If f (x) > 0, then g (x) > 0, which is absurd. Therefore f (x) < 0. Using the continuity of f at x, pick r > 0 such that f (y) < f (x)/2 for all y ∈ (x − r , x). Let δ ∈ (0, |f (x)|/2) be such that for all x′ ∈ [0, 1] if |f (x′ )| < δ , then there exists z ∈ (x′ − r , x′ + r ) with f (z ) > 0. Since g (x) = sup {f (y) : y ∈ [0, x]} = 0, there exists x′ ∈ [0, x] such that |f (x′ )| < δ ; moreover, since δ < |f (x)|/2, x′ 6 x − r. Then, by our choice of δ , there exists z < x′ + r 6 x such that f (z ) > 0, and so g (x) > 0. This final contradiction ensures that f (x) = 0. Thus in order to show that f has a zero we need only show that g has a zero; in turn, by Lemma 3, it suffices to show that g is locally nonzero. To this end, fix x ∈ [0, 1] and ε > 0; without loss of generality x ∈ / {0, 1} and ε < min{x, 1 − x}. Using that f has uniformly upper-stable roots, pick δ > 0 such that for all x ∈ [0, 1], if |f (x)| < δ , then there exists z ∈ (x −ε, x +ε) such that f (z ) > 0. Either |g (x)| > 0 or, as we may assume, |g (x)| < δ . Then there exists y ∈ [0, x] such that |f (y)| < δ ; whence, by our choice of δ , there exists z ∈ (y − ε, y + ε) such that f (z ) > 0. Thus g (x + ε) > f (z ) > 0. It now follows from the continuity of g at x + ε that g is locally nonzero. 5 The supremum of {f (y) : y ∈ [0, x]} is given by q ∈ Q : ∃y∈[0,x] (q < f (y)) , q ∈ Q : ∀y∈[0,x] (f (y) < q)
.
This is a Dedekind real because f is uniformly continuous and [0, x] is totally bounded.
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Let x be the unique root of g; consider y ∈ [0, x) and set ε ′ = y − x. Since f has uniformly upper-stable roots, there exists δ > 0 such that for all x′ ∈ [0, 1] ′
|f (x)| < δ ′ ⇒ ∃z ′ ∈(x′ −ε′ ,x′ +ε′ )∩[0,1] f (z ′ ) > 0. Either f (y) < 0 or f (y) > −δ ′ . In the latter case, there exists z < y + ε ′ = x such that f (z ) > 0. This contradicts our construction of x; whence f (y) < 0. In [13], Schuster put forward the ‘guarded thesis’ that in order to construct an object in BISH without choice we require that object to be unique (in some strong sense). In Proposition 2 this uniqueness is given by the produced x being the minimal root of f . Finding the minimal root of a locally nonconstant uniformly continuous function f : [a, b] → R with f (a) 6 0 and f (b) > 0 is not possible in general. For example, let a be a real number close to 0 and define f : [−1, 2] → R by f (x) = x2 (x − 1) + a. Then, for sufficiently small a, f has minimal root less than 1/2 if and only if a > 0 and has minimal root greater than 0 if and only if a < 0. Knowing that f has only stable roots does not help, since there exists a stable root of f less than 1/2 if and only if a > 0. This example and the principle of unique existence suggest that the standard constructive version of the intermediate value theorem (in which f is locally nonzero) is likely to be unprovable in BISH without choice. We finish by presenting some examples showing the relationship between the various properties on a function f . There exists a function f : R → R that has only stable roots, for which it is impossible not to have a root, and for which we can, in general, only find a root if Markov’s Principle MP: For any binary sequence (an )n>1 , if it is impossible for an = 0 for all n, then there exists n such that an = 1. holds, but which is only locally nonzero if MP holds. Let a ∈ R be such that ¬(a = 0) and define f : R → R by6
f (x) =
1
1 −x − a + a
a>
−x + + a a
a6 <0 x otherwise.
1
a
x 1
>0
To see that f has only stable roots, let x ∈ R be a root of f . Note that if a ̸= 0, then f is locally linear at x with gradient −1, so f has only stable roots. Pick N > 0 such that |x| < N. If |a| < 1/N, then |x| > N, which is absurd. Hence |a| > 0 and f has only stable roots. Now suppose that f is locally nonzero. Then there exists x ∈ (−1, 1) with f (x) ̸= 0 and so |a| > min{1, |f (x)|} > 0. This example does not have a direct impact on whether having only stable roots is sufficient to prove a constructive intermediate value theorem (without choice), since it is necessary in this example that the domain of f is unbounded. It is possible to give an example of a function f : [0, 2] → R with compact domain such that f has only stable roots, f is only locally non-zero, in general, if MP holds, and it is impossible for f not to have a root: let a be a non-negative real number such that ¬(a = 1) and define 2x − 1 f (x) = 2(a − 1)x + 3 − 2a a
x ∈ [0, 1] x ∈ [1, 3/2] x ∈ [3/2, 2].
6 The standard construction of functions on R by case does not require choice.
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However, in this case it is easy to find the root of f . The condition that f : I → R have at most one root
∀x,y∈I (x ̸= y ⇒ f (x) > 0 ∨ f (y) > 0) is not strong enough to prove that f has uniformly upper-stable roots, even when f is continuous, has only stable roots, and there exist a, b ∈ I such that f (a)f (b) < 0. To see this, recall that in Russian recursive mathematics there exists a continuous function g : [0, 1/2] → R with supremum 0 such that g (x) < 0 for each x ∈ [0, 1/2] (see [5, Chapter 6, Corollary 2.9]). The construction of g uses the existence of an enumeration of the set of partial functions from N into N with countable domain. Assuming the Church–Markov–Turing thesis, this enumeration is constructed using countable choice [15]. Define
f ( x) =
g (x) g (1/2) 4(1 − g (1/2))x − 3 + 4g (1/2)
x ∈ [0, 1/2] x ∈ [1/2, 3/4] x ∈ [3/4, 1].
Then f is a continuous real valued function on the unit interval, f (0) = g (0) < 0, and f (1) = 1 > 0. Since f is linear with gradient 4(1 − f (1/2)) on [3/4, 1] and since the only root of f is within this interval, all roots of f are stable. To see that f has at most one root, note that f (x) ̸= 0 for all x ∈ [0, 1] −
3 − 4f (1/2) 4(1 − f (1/2))
.
But f does not have uniformly upper-stable roots: let ε = 1/4. For all δ > 0 there exists x ∈ [0, 1/2] such that |f (x)| < δ , but f (y) ̸= 0 for all y ∈ (x − ε, x + ε) ∩ [0, 1]. If a continuous function f : [a, b] → R with f (a) < 0 and f (b) > 07 has uniformly at most one zero—that is,
∀ε>0 ∃δ>0 ∀x,y∈I (|f (x)| < δ ∧ |f (y)| < δ ⇒ |x − y| < ε) —then f has uniformly stable roots. To see this, let f : [a, b] → R be a continuous function with uniformly at most one root and fix ε > 0. Let δ ∈ (0, r ) be such that for all x, y ∈ [a, b]
|f (x)| < δ ∧ |f (y)| < δ ⇒ |x − y| < ε, and let x ∈ [a, b] be such that |f (x)| < δ . Using the continuity of f at a and b, we may assume that |f (x)| > δ for all x ∈ [a, a + ε) ∪ (b − ε, b]. Then x ∈ [a + ε, b − ε] and so, by continuity, it suffices to show that f (x − ε) < 0 < f (x + ε). Suppose that f (x + ε) < δ . If f (x + ε) > −δ , then |f (x + ε)| < δ , so |x − (x − ε)| < ε —a contradiction from which it follows that f (x + ε) < 0. Then by the approximate intermediate value theorem, there exists z ∈ (x + ε, 1) such that |f (z )| < δ , but
|z − x| > |x + ε − x| = ε, again contradicting our choice of δ . Thus f (x + ε) > δ > 0. A similar proof shows that f (x − ε) < 0. With countable choice any continuous function f : [a, b] → R with f (a) < 0 and f (b) > 0 that has at most one root has only stable roots. Let f : [a, b] → R be a continuous function with f (a) < 0 and f (b) > 0 that has at most one root, let x ∈ [a, b] be such that f (x) = 0, and let ε > 0. Then x ̸∈ {a, b} and without loss of generality (x − ε, x + ε) ⊂ [a, b]. Since f has at most one root, f (x + ε/2) ̸= 0 and f (x − ε/2) ̸= 0. Suppose that f (x + ε/2) < 0. Having at most one zero implies being locally nonconstant, so we can apply the standard constructive intermediate value theorem (which requires countable choice) to construct y ∈ [x + ε/2, 1] with f (y) = 0—a contradiction. Hence f (x + ε/2) > 0. A similar proof shows that f (x − ε/2) < 0, so x is a stable root and f has only stable roots. As you would expect, if f : R → R has uniformly stable roots and does not have two roots, then f has uniformly at most one root. Let ε > 0. Since f has uniformly stable roots, there exists δ < 0 such that for all x ∈ R if |f (x)| < δ , then there exist y, z ∈ (x − ε/2, x + ε/2) with f (y) < 0 < f (z ). Let x, x′ ∈ R be such that |f (x)| < δ and |f (x′ )| < δ , and suppose
7 We need this condition to rule out a function contained entirely above or below the x-axis (for example, f (x) = |x − 1/2| on [0, 1]).
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′ that |x − x′ | > ε . Then there exist y, z ∈ (x − ε/2, x + ε/2) and y′ , z ′ ∈ (x′ − ε/ 2, x + ε/2) such that ′f (y) < 0 < f (z ) ′ ′ ′ ′ ′ and f (y ) < 0 < f (z ). By Proposition 2, there exist w ∈ (y, z ) and w ∈ y , z such that f (w) = f (w ) = 0. But, since |x − x′ | > ε, w ̸= w′ —this contradicts that f does not have two roots. Hence |x − x′ | 6 ε , and f has uniformly at most one zero. The following diagram summarises the relationships between the various properties of a function f : R → R; all implications are strict.
Acknowledgements The author wishes to thank Peter Schuster for stimulating conversations regarding this work, and the University of Leeds for their support via a FFIR scholarship. The author also wishes to thank the referees for helping him to bring this paper into its final form, and, in particular, for pointing out a mistake in one of the examples. References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
P.H.G. Aczel, M. Rathjen, Notes on constructive set theory, Report No. 40, Institut Mittag-Leffler, Royal Swedish Academy of Sciences, 2001. E. Bishop, Foundations of Constructive Analysis, McGraw-Hill, New York, 1967. E.A. Bishop, D.S. Bridges, Constructive Analysis, in: Grundlehren der Math. Wiss., vol. 279, Springer-Verlag, Heidelberg, 1985. D.S. Bridges, A constructive Morse theory of sets, in: D. Skordev (Ed.), Mathematical Logic and its Applications, Plenum Publishing Corp., New York, 1987, pp. 61–79. D.S. Bridges, F. Richman, Varieties of Constructive Mathematics, in: London Math. Soc. Lecture Notes, vol. 97, Cambridge Univ. Press, 1987. D.S. Bridges, L.S. Vîţă, Techniques of Constructive Analysis, Universitext, Springer, New-York, 2006. H.M. Friedman, Set theoretic foundations for constructive analysis, Ann. Math. 105 (1) (1977) 1–28. J. Myhill, Constructive set theory, J. Symbolic Logic 40 (3) (1975) 347–382. F. Richman, The fundamental theorem of algebra: a constructive development without choice, Pacific Journal of Mathematics 196 (2000) 213–230. F. Richman, Constructive mathematics without choice, in: schuster, et al. (Eds.), Reuniting the Antipodes—Constructive and Nonstandard Views of the Continuum, in: Synthese Library, vol. 306, Kluwer, 2001, pp. 199–205. F. Richman, Spreads and choice in constructive mathematics, Indagations Mathematicae 13 (2002) 259–267. F. Richman, Real numbers and other completions, Math. Logic Quarterly 54 (2008) 98–108. P. Schuster, Unique existence, approximate solutions, and countable choice, Theoretical Computer Science 305 (2003) 433–455. P. Taylor, A Lambda calculus for real analysis, Computability and Complexity in Analysis, Kyoto, 2005. A.S. Troelstra, D. van Dalen, Constructivism in mathematics, Vol. I. An introduction, in: Studies in Logic and the Foundations of Mathematics, vol. 121, North-Holland Publishing Co., Amsterdam, 1988.