Abstract

We show that Brouwer’s fixed point theorem with isolated fixed points is equivalent to Brouwer’s fan theorem.

1. Introduction

It is well known that Brouwer's fixed point theorem cannot be constructively proved.

Kellogg et al. [1] provided a constructive proof of Brouwer's fixed point theorem. But it is not constructive from the view point of constructive mathematics á la Bishop. It is sufficient to say that one-dimensional case of Brouwer's fixed point theorem, that is, the intermediate value theorem is nonconstructive (see [2, 3]).

Sperner's lemma which is used to prove Brouwer's theorem, however, can be constructively proved. Some authors have presented an approximate version of Brouwer's theorem using Sperner's Lemma (see [3, 4]). Thus, Brouwer's fixed point theorem is constructively, in the sense of constructive mathematics á la Bishop, proved in its approximate version.

Recently Berger and Ishihara [5] showed that the following theorem is equivalent to Brouwer's fan theorem.

Each uniformly continuous function from a compact metric space into itself with at most one fixed point and approximate fixed points has a fixed point.

In this paper we require a more general condition that each uniformly continuous function from a compact metric space into itself may have only isolated fixed points and show that the proposition that such a function has a fixed point is equivalent to Brouwer's fan theorem.

In another paper we have shown that if a uniformly continuous function in a compact metric space satisfies stronger condition, sequential local non-constancy, then without the fan theorem we can constructively show that it has an exact fixed point (see [6]).

2. Brouwer's Fixed Point Theorem with Isolated Fixed Points and His Fan Theorem

Let 𝑋 be a compact (totally bounded and complete) metric space, 𝑥 be a point in 𝑋, and consider a uniformly continuous function 𝑓 from 𝑋 into itself.

According to [3, 4] 𝑓 has an approximate fixed point. It means the following, Foreach𝜀>0thereexists𝑥𝑋suchthat||𝑥𝑓(𝑥)||<𝜀.(1)

Since 𝜀>0 is arbitrary, inf𝑥𝑋||𝑥𝑓(𝑥)||=0.(2)

The notion that 𝑓 has at most one fixed point in [5] is defined as follows.

Definition 1 (at most one fixed point). For all 𝑥,𝑦𝑋, if 𝑥𝑦, then 𝑓(𝑥)𝑥 or 𝑓(𝑦)𝑦.

Now we consider a condition that 𝑓 may have only isolated fixed points. First we recapitulate the compactness of a set in constructive mathematics. We say that 𝑋 is totally bounded if for each 𝜀>0 there exists a finitely enumerable 𝜀-approximation to 𝑋. (A set 𝑆 is finitely enumerable if there exist a natural number 𝑁 and a mapping of the set {1,2,,𝑁} onto 𝑆.) An 𝜀-approximation to 𝑋 is a subset of 𝑋 such that for each 𝑥𝑋 there exists 𝑦 in that 𝜀-approximation with |𝑥𝑦|<𝜀. According to Corollary 2.2.12 of [7], about totally bounded set we have the following result.

Lemma 2. If 𝑋 is totally bounded, for each 𝜀>0 there exist totally bounded sets 𝐻1,,𝐻𝑛, each of diameter less than or equal to 𝜀, such that 𝑋=𝑛𝑖=1𝐻𝑖.

Since inf𝑥𝑋|𝑥𝑓(𝑥)|=0, we have inf𝑥𝐻𝑖|𝑥𝑓(𝑥)|=0 for some 𝐻𝑖𝑋 such that 𝑋=𝑛𝑖=1𝐻𝑖.

The definition that a function may have only isolated fixed points is as follows.

Definition 3 (isolated fixed points). There exists 𝜀>0 with the following property. For each 𝜀>0 less than or equal to 𝜀, there exist totally bounded sets 𝐻1,,𝐻𝑛, each of diameter less than or equal to 𝜀, such that 𝑋=𝑛𝑖=1𝐻𝑖, and in each 𝐻𝑖 if 𝑥𝑦, then 𝑓(𝑥)𝑥 or 𝑓(𝑦)𝑦.

In each 𝐻𝑖, 𝑓 has at most one fixed point. Now we show the following lemma, which is based on Lemma 2 of [8].

Lemma 4. Let 𝑓 be a uniformly continuous function from 𝑋 into itself. Assume inf𝑥𝐻𝑖𝑓(𝑥)=0 for some 𝐻𝑖𝑋 defined above. If the following property holds: for each 𝜀>0 there exists 𝛿>0 such that if 𝑥,𝑦𝐻𝑖, |𝑓(𝑥)𝑥|𝛿 and |𝑓(𝑦)𝑦|𝛿, then |𝑥𝑦|𝜀. Then, there exists a point 𝑧𝐻𝑖 such that 𝑓(𝑧)=𝑧, that is, 𝑓 has a fixed point.

Proof. Choose a sequence (𝑥𝑛)𝑛1 in 𝐻𝑖 such that |𝑓(𝑥𝑛)𝑥𝑛|0. Compute 𝑁 such that |𝑓(𝑥𝑛)𝑥𝑛|<𝛿 for all 𝑛𝑁. Then, for 𝑚,𝑛𝑁 we have |𝑥𝑚𝑥𝑛|𝜀. Since 𝜀>0 is arbitrary, (𝑥𝑛)𝑛1 is a Cauchy sequence in 𝐻𝑖 and converges to a limit 𝑧𝐻𝑖. The continuity of 𝑓 yields |𝑓(𝑧)𝑧|=0, that is, 𝑓(𝑧)=𝑧.

Let 𝑌={0,1}, the set of all binary sequences, {0,1}𝑛 with a finite natural number 𝑛 be the set of finite binary sequences with length 𝑛+1. We write 𝑥,𝑦,, for the elements (𝑥𝑛)𝑛0,(𝑦𝑛)𝑛0, of 𝑌. Also for each 𝑥𝑌 and each natural number 𝑛 we write 𝑥(𝑛)=𝑥0,𝑥1,,𝑥𝑛1.(3)𝑌 is compact under the metric defined by (see [2, 8]) ||x𝑦||=inf2𝑛𝑥(𝑛)=𝑦(𝑛).(4)

Let 𝐵 be a set of finite binary sequences. 𝐵 is(i)detachable if 𝑥𝑌𝑛𝑥(𝑛)𝐵𝑥(𝑛)𝐵;(5)(ii)a bar if 𝑥𝑌𝑛𝑥(𝑛)𝐵;(6)(iii)a uniform bar if 𝑁𝑥𝑌𝑛𝑁𝑥(𝑛)𝐵.(7)

In [8] the following lemma has been proved (their Lemma 4).

Lemma 5. Let 𝑌={0,1}, and 𝐵 a detachable bar for 𝑌. Then, for each 𝑥𝑌, 𝜎(𝑥)=inf𝑛𝑥(𝑛)𝐵(8) exists, and the mapping 𝑥4𝜎(𝑥) is uniformly continuous in 𝑌.

Brouwer's fan theorem is as follows.

Theorem 6. Every detachable bar for {0,1} is a uniform bar.

It has been shown in [2, 5] that this theorem is equivalent to the following theorem.

Theorem 7. Every positive-valued uniformly continuous function on a compact metric space has positive infimum.

Now, according to the Proof of Theorem 5 in [8] and the Proof of Proposition in [5], we show the following result.

Theorem 8. Brouwer's fixed point theorem with isolated fixed points in a compact metric space is equivalent to Brouwer's fan theorem.

Proof. (1) Assume that each uniformly continuous function from a compact metric space into itself with isolated fixed points has a fixed point. It implies that each uniformly continuous function from a compact metric space into itself with at most one fixed point has a fixed point. Consider 𝑌={0,1} and a function 𝜑𝑌𝑌. Let 𝑥𝑌, and 𝑇 be an infinite tree with at most one infinite path (A tree is a detachable set in {0,1} which is closed under restriction.) and define 𝜑(𝑥)𝑛=𝑥𝑛if𝑥(𝑛)𝑇,1𝑥𝑛if𝑥(𝑛)𝑇.(9) Since𝑥(𝑛)=𝑦(𝑛) implies 𝜑(𝑥)(𝑛)=𝜑(𝑦)(𝑛), 𝜑 is uniformly continuous. Thus, 𝜑 has a fixed point. From the definition of 𝜑 its fixed print is an infinite branch. Thus, 𝑇 has an infinite branch. Let 𝐵 be a detachable bar and set𝐵=𝑥𝑛𝑥(𝑛)𝐵.(10) Then, 𝐵 is also a detachable bar. For 𝑥=(𝑥0,,𝑥𝑛1) and 𝑦=(𝑦0,,𝑦𝑚1) set 𝑥𝑦=𝑥0,,𝑥𝑛1,𝑦0,,𝑦𝑚1.(11) If 𝑥𝐵, then 𝑥𝑦𝐵. Consider a tree {0,1}𝐵. Define for each 𝑛 a 𝑢𝑛{0,1}𝑛 by the following procedure. If {0,1}𝑛𝐵, let 𝑢𝑛 be any element of {0,1}𝑛𝐵. If{0,1}𝑛𝐵, let 𝑘 be the largest number such that {0,1}𝑘𝐵 and define 𝑢𝑛=𝑢k0,,0𝑛𝑘times.(12) Set  𝑇={0,1}𝐵𝑢𝑛{0,1}𝑛𝐵.(13) Then, 𝑇 is an infinite tree since it contains each 𝑢𝑛. For all 𝑥 with length 𝑛 we have 𝑥𝐵𝑇𝑥=𝑢𝑛.(14) Let 𝑥,𝑦{0,1} and suppose 𝑥𝑦. Then, there is 𝑛 such that 𝑥(𝑛)𝐵, 𝑦(𝑛)𝐵, and 𝑥(𝑛)𝑦(𝑛). Thus, 𝑥(𝑛)𝑢𝑛 or 𝑦(𝑛)𝑢𝑛, and so 𝑥(𝑛)𝑇 or 𝑦(𝑛)𝑇. Therefore, 𝑇 has at most one infinite branch. From the argument above it has an infinite branch 𝑥. Since 𝐵 is a bar, there is 𝑚 such that 𝑥(𝑚)𝐵. Thus, 𝑥(𝑚)𝐵𝑇, and so 𝑥(𝑚)=𝑢𝑚. Therefore, {0,1}𝑚𝐵, and 𝐵 is a uniform bar. It means that 𝐵 is also a uniform bar.
(2) Assume Brouwer's Fan theorem. Consider a compact metric space 𝑋 and a uniformly continuous function 𝑓 from 𝑋 into itself with isolated fixed points. Then, |𝑥𝑓(𝑥)| is uniformly continuous. Let 𝑥𝑋, and 𝐻𝑖, 𝑖=1,,𝑛 be totally bounded subsets of 𝑋, each of diameter less than or equal to 𝜀 in Definition 3, such that 𝑋=𝑛𝑖=1𝐻𝑖. Given 𝜀>0 assume that the set 𝐾=(𝑥,𝑦)𝐻𝑖×𝐻𝑖||𝑥𝑦||𝜀,(15) is nonempty and compact (see Theorem 2.2.13 of [7]). For 𝑥,𝑦𝐻𝑖 let 𝐹(𝑥,𝑦)=||𝑥𝑓(𝑥)||+||𝑦𝑓(𝑦)||.(16) Then, 𝐹 is uniformly continuous and positive-valued on 𝐾. So, by Theorem 70<𝛿=13inf{𝐹(𝑥,𝑦)(𝑥,𝑦)𝐾}.(17) For each (𝑥,𝑦)𝐾 we have ||𝑥𝑓(𝑥)||+||𝑦𝑓(𝑦)||=𝐹(𝑥,𝑦)>2𝛿.(18) Thus, either |𝑥𝑓(𝑥)|>𝛿 or |𝑦𝑓(𝑦)|>𝛿. It follows that if 𝑥,𝑦𝐻𝑖, |𝑥𝑓(𝑥)|𝛿 and |𝑦𝑓(𝑦)|𝛿, then (𝑥,𝑦)𝐾 and so |𝑥𝑦|𝜀. Then, from Lemma 4 there exists a fixed point of 𝑓 in 𝐻𝑖𝑋 such that inf𝑥𝐻𝑖|𝑥𝑓(𝑥)|=0. Thus, Brouwer's fan theorem implies his fixed point theorem for uniformly continuous functions with isolated fixed points.

Acknowledgment

This research was partially supported by the Ministry of Education, Science, Sports and Culture of Japan, Grant-in-Aid for Scientific Research (C), 20530165.