In this note we give a new constructive proof of a theorem of Julian and Richman that pinpoints the role of the statement

THE FAN THEOREM AND POSITIVE-VALUED UNIFORMLY CONTINUOUS FUNCTIONS ON COMPACT INTERVALS

Josef Berger and Douglas Bridges

(Received June 2006)

Abstract. Julian and Richman showed that the statement “every uniformly continuous, positive-valued function on [0, 1] has a positive infimum” is con- structively equivalent to Brouwer’s fan theorem for detachable bars. In this note we give an alternative proof of their result, a proof based on an idea of Aberth in recursive function theory.

In this note we give a new constructive proof of a theorem of Julian and Richman that pinpoints the role of the statement

(*) Every uniformly continuous, positive-valued function on a compact interval has positive infimum

in constructive mathematics (that is, mathematics carried out with intuitionistic logic, an appropriate foundational system such as Aczel’s CZF [2], and, where needed, countable or dependent choice). The original proof of this theorem in [12]

(see also Chapter 6 of [7]) depends on a series of rather complicated lemmas. Our proof, the first part of which adapts Aberth’s construction, in [1] (pages 70–71), of a recursive counterexample to (*), is at least as natural as theirs and, we believe, is worth presenting as an alternative. It is part of an ongoing development of constructive reverse mathematics; see [3, 4, 5, 9, 10].

We begin with some background definitions and constructions. Let a, b be dis- tinct integers, and let {a, b} ^∗ denote the set of all finite sequences in {a, b} . Let α = (α ₁ , α ₂ , . . .) be a finite or infinite sequence of elements of {a, b} . Then for each applicable n,

αn = (α 1 , . . . , α n )

is called a restriction of α. By a path in {a, b} ^∗ we mean a finite or infinite sequence in {a, b}. We say that a path α is blocked by a subset B of {a, b} ^∗ if some restriction of α is in B. A subset B of {a, b} ^∗ is called a bar for {a, b} ^∗ if each infinite path of {a, b} ^∗ is blocked by B; a bar B for {a, b} ^∗ is uniform if there exists a positive integer n such that each finite path of length n is blocked by B.

A subset S of a set X is said to be detachable from X if

∀ _x∈X (x ∈ S ∨ x / ∈ S) .

One version of Brouwer’s fan theorem for detachable bars states that every

detachable bar for {a, b} ^∗ is uniform.

We aim to prove the Julian–Richman theorem:

Theorem 1. Let B be a detachable subset of {0, 1} ^∗ . Then there exists a uniformly continuous function f : [0, 1] → R such that

(a) f (x) > 0 for each x ∈ [0, 1] if and only if B is a bar for {0, 1} ^∗ , and (b) inf f > 0 if and only if B is a uniform bar for {0, 1} ^∗ .

Conversely, if f : [0, 1] → R is a uniformly continuous function, then there exists a detachable subset B of {0, 1} ^∗ that satisfies (a) and (b).

The following definitions and lemmas are necessary for our proof.

For each a ∈ R and all real numbers b, c with 0 6 b < c, define the corresponding trapezoidal function tr(a, b, c, ·) : R → R by

tr(a, b, c, x) = max



min  c − |x − a|

c − b , 1

 , 0

 .

This function is uniformly continuous on R, vanishes outside the open interval (a − c, a + c) , equals 1 throughout the closed interval [a − b, a + b] , and is linear in each of the intervals [a − c, a − b] , [a + b, a + c] .

Define rational numbers r(u) for u ∈ {−1, 1} ^∗ inductively, as follows:

r(h i) = 0,

r(u ∗ k) = r(u) + k4 ^−|u|−1 (k = ±1) ,

where, as usual, |u| is the length of the string u. Then define the corresponding function w(u, .) : [−1, 1] → R ⁰⁺ by

w(u, x) = 1 2 4 ^−|u| tr

 r(u), 1

2 4 ^−|u| , 4 ^−|u| , x

 .

This function vanishes outside the interval r(u) − 4 ^−|u| , r(u) + 4 ^−|u|
and equals

1

2 4 ^−|u| on the interval r(u) − ¹ ₂ 4 ^−|u| , r(u) + ¹ ₂ 4 ^−|u|  . Its graph has slope 1 between r(u) − 4 ^−|u| and r(u) − ¹ ₂ 4 ^−|u| , and slope −1 between r(u) + ¹ ₂ 4 ^−|u| and r(u) + 4 ^−|u| .

The function

G : {−1, 1} ^N

→



− 1 3 , 1

3

 ,

as now defined, and its range S will play an important part in the proof of our main result:

G(α) =

∞

X

n=1

α n 4 ⁻ⁿ 

α ∈ {−1, 1} ^N

 ,

S = ( _∞

X

n=1

α n 4 ⁻ⁿ : α ∈ {−1, 1} ^N

)

.

Lemma 2. The set S is compact (that is, totally bounded and complete).

Proof. First note that for each α ∈ {−1, 1} ^N

and each positive integer N, 

∞

X

n=1

α n 4 ⁻ⁿ − r(αN )     

=     

∞

X

n=N +1

α n 4 ⁻ⁿ     

6

∞

X

n=N +1

4 ⁻ⁿ = 1 3 4 ^−N . Hence

r(u) : u ∈ {−1, 1} ^∗ , |u| = N

is a ¹ ₃ 4 ^−N -approximation to S. Since N is arbitrary, it follows that S is totally bounded. To show that S is complete and therefore compact, we first note that if x = P ∞

n=1 α n 4 ⁻ⁿ and y = P ∞

n=1 β _n 4 ⁻ⁿ , where α n , β _n ∈ {−1, 1} for each n and α _m 6= β _m , then |x − y| > ¹ ₃ 4 ^−m+1 . This is proved by elementary estimation with infinite series.

Now let (x _n ) _n>1 be a Cauchy sequence in S. Write x k =

∞

X

n=1

α nk 4 ⁻ⁿ ,

where α nk ∈ {−1, 1} for all n and k. Compute a strictly increasing sequence (n i ) _i>1 of positive integers such that |x _j − x _k | < ¹ ₃ 4 ⁻ⁱ⁺¹ for each i and all j, k > n i . For such i, j, k we have α _ij = α _ik ; for if α _ij 6= α ik , then by the foregoing, |x _j − x k | > ¹ ₃ 4 ⁻ⁱ⁺¹ , a contradiction. Let α _i denote the common value of the coefficients α _ij for j > n i , and let

ξ =

∞

X

n=1

α _n 4 ⁻ⁿ ∈ S.

Then for all j > n ⁱ we have

|ξ − x j | 6

i

X

n=1

|α n − α nj | 4 ⁻ⁿ +

∞

X

n=i+1

|α n − α nj | 4 ⁻ⁿ

6 0 +

∞

X

n=i+1

2 × 4 ⁻ⁿ

(since j > n ^k for all k 6 i)

= 2 3 4 ⁻ⁱ .

Hence the sequence (x k ) _k>1 converges to ξ; so S is complete.  Lemma 3. For each x ∈ [−1, 1] there exists α ∈ {−1, 1} ^N

such that if x 6= G(α), then

0 < ρ(x, S) = inf {|x − y| : y ∈ S} .

Proof. In view of Lemma 2, this is a special case of Bishop’s lemma. See [6] (page

92, (3.8)) or [8] (Proposition 3.1.1). 

We can now give our proof of Theorem 1.

Proof. For the first part of the theorem, a routine translation enables us to replace {0, 1} ^∗ and [0, 1] by {−1, 1} ^∗ and [−1, 1] , respectively. Let B be a detachable subset of C = {−1, 1} ^∗ . Without loss of generality, we may assume that B is closed under extensions; see [11]. Clearly, we lose nothing by also assuming that B does not contain the empty string over {−1, 1} . For each n define a uniformly continuous function w n : [−1, 1] → R ⁰⁺ by

w n (x) = X

|u|=n

χ(u)w(u, x),

where χ : C → {0, 1} is defined by

χ(u) =







0 if u ∈ B 1 if u / ∈ B.

Then define w : [−1, 1] → R ⁰⁺ by w =

∞

X

n=0

w _n .

The series on the right converges uniformly on [−1, 1] , by comparison with P ∞

n=0 1

2 4 ⁻ⁿ ; so w is uniformly continuous. In fact (see [1], pages 70–71),

|w(x) − w(y)| 6 |x − y| (x, y ∈ [−1, 1]) . (1) We prove the following:

(i) If x ∈ [−1, 1] and ρ(x, S) > ¹ ₃ 4 ^−ν+1 , then w(x) 6 ² ₃ − ¹ ₂ 4 ^−ν . (ii) If B is a bar for C, then w(x) < ² ₃ for each x ∈ [−1, 1] . (iii) If B is a uniform bar for C, then sup w < ² ₃ .

For (i), let u ∈ C have length ν. If x ∈ (r(u) − 4 ^−ν , r(u) + 4 ^−ν ), then ρ(x, S) 6 |x − r(u)| +

∞

X

n=ν+1

4 ⁻ⁿ 6 4 ^−ν + 1

3 4 ^−ν = 1 3 4 ^−ν+1 ,

a contradiction. Hence |x − r(u)| > 4 ^−ν and so w(u, x) = 0. It follows that w ν (x) = 0 and therefore that

w(x) 6

∞

X

n=0,n6=ν

1

2 4 ⁻ⁿ = 2 3 − 1

2 4 ^−ν .

This proves (i). To prove (ii), suppose that B is a bar for C, and consider any x ∈ [−1, 1] . By Lemma 3, there exists α ∈ {−1, 1} ^N

such that if x 6= G(α), then ρ(x, S) > 0. Choose N such that αN ∈ B. Note that for each n we have

|G(α) − r(αn)| 6 4 ⁻ⁿ , so w(u, G(α)) = 0 whenever |u| = n and u 6= αn. Thus w n (G(α)) = χ(αn)w(αn, G(α)) = 0

for all n > N, and therefore w(G(α)) =

N −1

X

n=0

w _n (G(α)) 6

N −1

X

n=0

1

2 4 ⁻ⁿ = 2

3 1 − 4 ^−N
.

Either |x − G(α))| < ² ₃ 4 ^{−N −1} or x 6= G(α). In the first case, by the Lipschitz property (1),

w(x) 6 w(G(α)) + |x − G(α)|

< 2

3 (1 − 4 ^−N ) + 2

3 4 ^{−N −1} = 2

3 1 − 4 ^{−N −1}
.

In the second case, there exists ν such that ρ(x, S) > ¹ ₃ 4 ^−ν+1 ; whence, by (i),

w(x) 6 ² 3 − ¹ ₂ 4 ^−ν . This completes the proof of (ii). Furthermore, if B is a uniform

bar for C, then in the foregoing proof of (ii) we can choose N independent of α,

so w(G(α)) 6 ² 3 1 − 4 ^−N

for all α ∈ {−1, 1} ^N

. For each x ∈ [−1, 1] , either ρ(x, S) < ² ₃ 4 ^{−N −1} or ρ(x, S) > ¹ ₃ 4 ^{−N −1} . In the former case we have

w(x) 6 2

3 1 − 4 ^−N
+ ρ(x, S) < 2

3 1 − 4 ^{−N −1}
, whereas in the second, property (i) shows that w(x) 6 ² 3 − ¹ ₂ 4 ^{−N −2} .

Writing f = ² ₃ − w, we now obtain a uniformly continuous, nonnegative function f on [−1, 1] such that if B is a bar for C, then f (x) > 0 for all x ∈ [−1, 1] , and if B is a uniform bar for C, then inf f > 0.

Now suppose, conversely, that f is everywhere positive. Given α ∈ {−1, 1} ^N

, let x = G(α). There exists a positive integer N such that f (x) > ² ₃ 4 ^{−N −1} and therefore w(x) < ² ₃ 1 − 4 ^{−N −1}
. Suppose that αN / ∈ B. Then w n (x) = w(αn, x) = ¹ ₂ 4 ⁻ⁿ for each n 6 N, so

w(x) >

N

X

n=0

1

2 4 ⁻ⁿ = 2

3 (1 − 4 ^{−N −1} ),

a contradiction. Hence αN ∈ B. Moreover, if inf f > 0, then we can choose N independent of α, so B is a uniform bar for C. This completes the proof of the first part of Theorem 1.

For the second part, define the mapping F : {0, 1} ^N

∪ {0, 1} ^∗ → R by F (α) =

|α|

X

n=1

α _n 2 ⁻ⁿ ,

where we write ∞ for |α| when α ∈ {0, 1} ^N

. Note that if X is a subset of [0, 1]

with the following property

∀ α∈{0,1}

∃ _r>0 ∀ _x∈[0,1] (|x − F (α)| < r =⇒ x ∈ X),

then X = [0, 1]; see [3]. The remainder of the proof is similar to one used in [5].

Using countable choice, construct a strictly increasing sequence (n k ) _k>1 of positive integers such that for all α ∈ {0, 1} ^N

and all k ∈ N ⁺ ,

|f (F (α)) − f (F (αn k ))| < 2 ^−k .

Again using countable choice, construct a family (λ _u ) _u∈{0,1}

in {0, 1} such that for each u ∈ {0, 1} ^∗ ,

λ _u = 0 ⇒ ∀k (|u| 6= n k ) ∨ ∃k (|u| = n _k ∧ f (F (u)) < 2 ^−k+2 ), λ _u = 1 ⇒ ∃k |u| = n k ∧ f ((F (u)) > 2 ^−k+1
.

Let

B = u ∈ {0, 1} ^∗ : λ u = 1 ,

which is a detachable subset of {0, 1} ^∗ . Consider α ∈ {0, 1} ^N

and k ∈ N ⁺ . If

f (F (αn k )) < 2 ^−k+3 , then f (F (α)) < 2 ^−k+4 . Thus if f (F (α)) > 2 ^−k+4 , then

f (F (αn k )) > 2 ^−k+2 and therefore λ αn

6= 0; whence αn k ∈ B. It follows that

if f (x) > 0 for each x ∈ [0, 1] , then B is a bar for {0, 1} ^∗ and that if inf f > 0, then

B is a uniform bar for {0, 1} ^∗ .

On the other hand, if αn k ∈ B, then f (F (αn k )) > 2 ^−k+1 and so f (F (α)) > 2 ^−k . It follows from this and the remark immediately after our definition of the function F that if B is a bar for {0, 1} ^∗ , then f (x) > 0 for each x ∈ [0, 1] . Moreover, if B is a uniform bar for {0, 1} ^∗ , then there exists K such that for each α ∈ {0, 1} ^N

there exists k 6 K such that f (F (αn ^k )) > 2 ^−k+1 ; whence

f (F (α)) > f (F (αn ^k )) − 2 ^−k > 2 ^−k > 2 ^−K .

Since the range of F is dense in [0, 1] and f is continuous, it follows that inf f >

2 ^−K . 

For the sake of completeness, we end with two consequences of Theorem 1. The proof of the first of these is clear.

Corollary 4. If the fan theorem for detachable bars holds, then every uniformly continuous, positive-valued map of [0, 1] into R has a positive infimum.

Corollary 5. If the Church–Markov–Turing thesis holds, then there exists a uni- formly continuous, positive-valued map of [0, 1] into R whose infimum is 0.

Proof. According to Proposition (3.1) in Chapter 5 of [7], under the Church–

Markov–Turing thesis there exists a detachable bar for {0, 1} ^N

that is not uniform.

The function associated with this bar according to Theorem 1 has the desired

properties. 

Acknowledgement. The authors thank The Royal Society of New Zealand for supporting Berger as a Marsden Postdoctoral Research Fellow at the University of Canterbury while this paper was written.

References

[1] O. Aberth, Computable Analysis, McGraw–Hill, New York, 1980.

[2] P. Aczel and M.J. Rathjen, Notes on Constructive Set Theory, Report No. 40, Institut Mittag–Leffler, Royal Swedish Academy of Sciences, 2001.

[3] J. Berger and D.S. Bridges, A fan-theoretic equivalent of the antithesis of Specker’s theorem, preprint, University of Canterbury, 2006.

[4] J. Berger, D.S. Bridges, and P.M. Schuster, The fan theorem and unique exis- tence of maxima, J. Symbolic Logic 71(2), 713–720, 2006.

[5] J. Berger and H. Ishihara, Brouwer’s fan theorem and unique existence in constructive analysis, Math. Logic Quarterly 51(4), 360–364, 2005.

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Wiss. 279, Springer-Verlag, Heidelberg, 1985.

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[8] D.S. Bridges and L.S. Vˆıt¸˘ a, Techniques of Constructive Analysis, Universitext,

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[10] H. Ishihara, Reverse mathematics in Bishop’s constructive mathematics, Philosophia Scientiae, forthcoming

[11] H. Ishihara, Weak K¨ onig’s lemma implies Brouwer’s fan theorem: a direct proof, Notre Dame J. Formal Logic, forthcoming.

[12] W.H. Julian and F. Richman, A uniformly continuous function on [0, 1] that is everywhere different from its infimum, Pacific J. Math. 111 (1984), 333–340.

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