Fixed Point Theory and Applications
Volume 2010 (2010), Article ID 716971, 14 pages
doi:10.1155/2010/716971
Research Article

Convergence of the Sequence of Successive Approximations to a Fixed Point

Tomonari Suzuki

Department of Basic Sciences, Kyushu Institute of Technology, Tobata, Kitakyushu 804-8550, Japan

Received 29 September 2009; Accepted 21 December 2009

Academic Editor: Mohamed A. Khamsi

Copyright © 2010 Tomonari Suzuki. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

If ( 𝑋 , 𝑑 ) is a complete metric space and 𝑇 is a contraction on 𝑋 , then the conclusion of the Banach-Caccioppoli contraction principle is that the sequence of successive approximations { 𝑇 𝑛 𝑥 } of 𝑇 starting from any point 𝑥 ∈ 𝑋 converges to a unique fixed point. In this paper, using the concept of 𝜏 -distance, we obtain simple, sufficient, and necessary conditions of the above conclusion.

1. Introduction

The following famous theorem is referred to as the Banach-Caccioppoli contraction principle. This theorem is very forceful and simple, and it became a classical tool in nonlinear analysis.

Theorem 1.1 (see Banach [1] and Caccioppoli [2]). Let ( 𝑋 , 𝑑 ) be a complete metric space and let 𝑇 be a self contraction on 𝑋 , that is, there exists 𝑟 ∈ [ 0 , 1 ) such that 𝑑 ( 𝑇 𝑥 , 𝑇 𝑦 ) ≤ 𝑟 𝑑 ( 𝑥 , 𝑦 ) for all 𝑥 , 𝑦 ∈ 𝑋 . Then the following holds. (A) 𝑇 has a unique fixed point 𝑧 , and { 𝑇 𝑛 𝑥 } converges to 𝑧 for any 𝑥 ∈ 𝑋 .

We note that the conclusion of Kannan's fixed point theorem [3] is also (A). See Kirk's survey [4]. Recently, we obtained that (A) holds if and only if 𝑇 is a strong Leader mapping [5, 6].

Theorem 1.2 (see [6]). Let 𝑇 be a mapping on a complete metric space ( 𝑋 , 𝑑 ) . Then the following are equivalent. (i)(A) holds.(ii) 𝑇 is a strong Leader mapping, that is, the following hold.(a)For 𝑥 , 𝑦 ∈ 𝑋 and 𝜀 > 0 , there exist 𝛿 > 0 and 𝜈 ∈ ℕ such that 𝑑 𝑇 𝑖 𝑥 , 𝑇 𝑗 𝑦 𝑇 < 𝜀 + 𝛿 ⟹ 𝑑 𝑖 + 𝜈 𝑥 , 𝑇 𝑗 + 𝜈 𝑦 < 𝜀 , ( 1 . 1 ) for all 𝑖 , 𝑗 ∈ ℕ ∪ { 0 } , where 𝑇 0 is the identity mapping on 𝑋 .(b)For 𝑥 , 𝑦 ∈ 𝑋 , there exist 𝜈 ∈ ℕ and a sequence { 𝛼 𝑛 } in ( 0 , ∞ ) such that 𝑑 𝑇 𝑖 𝑥 , 𝑇 𝑗 𝑦 < 𝛼 𝑛 𝑇 ⟹ 𝑑 𝑖 + 𝜈 𝑥 , 𝑇 𝑗 + 𝜈 𝑦 < 1 𝑛 , ( 1 . 2 ) for all 𝑖 , 𝑗 ∈ ℕ ∪ { 0 } and 𝑛 ∈ ℕ .

The following theorem is proved in [7, 8].

Theorem 1.3 (see Rus [7] and Subrahmanyam [8]). Let ( 𝑋 , 𝑑 ) be a complete metric space and let 𝑇 be a continuous mapping on 𝑋 . Assume that there exists 𝑟 ∈ [ 0 , 1 ) satisfying 𝑑 ( 𝑇 𝑥 , 𝑇 2 𝑥 ) ≤ 𝑟 𝑑 ( 𝑥 , 𝑇 𝑥 ) for all 𝑥 ∈ 𝑋 . Then the following holds. (B) { 𝑇 𝑛 𝑥 } converges to a fixed point for every 𝑥 ∈ 𝑋 .

We obtained a condition equivalent to (B) in [9].

Theorem 1.4 (see [9]). Let 𝑇 be a mapping on a complete metric space ( 𝑋 , 𝑑 ) . Then the following are equivalent. (i)(B) holds.(ii)The following hold.(a)For 𝑥 ∈ 𝑋 and 𝜀 > 0 , there exist 𝛿 > 0 and 𝜈 ∈ ℕ such that 𝑑 𝑇 𝑖 𝑥 , 𝑇 𝑗 𝑥 𝑇 < 𝜀 + 𝛿 ⟹ 𝑑 𝑖 + 𝜈 𝑥 , 𝑇 𝑗 + 𝜈 𝑥 < 𝜀 , ( 1 . 3 ) for all 𝑖 , 𝑗 ∈ ℕ ∪ { 0 } .(b)For 𝑥 , 𝑦 ∈ 𝑋 , there exist 𝜈 ∈ ℕ and a sequence { 𝛼 𝑛 } in ( 0 , ∞ ) such that 𝑑 𝑇 𝑖 𝑥 , 𝑇 𝑗 𝑦 < 𝛼 𝑛 𝑇 ⟹ 𝑑 𝑖 + 𝜈 𝑥 , 𝑇 𝑗 + 𝜈 𝑦 < 1 𝑛 , ( 1 . 4 ) for all 𝑖 , 𝑗 ∈ ℕ ∪ { 0 } and 𝑛 ∈ ℕ .

We sometimes call a mapping satisfying (A) a Picard operator [10]. We also call a mapping satisfying (B) a weakly Picard operator [11–13].

We cannot tell that the conditions (ii) of Theorems 1.2 and 1.4 are simple. Motivated by this, we obtain simpler conditions which are equivalent to Conditions (A) and (B).

2. Preliminaries

Throughout this paper, we denote by ℕ , ℤ , and ℝ the sets of positive integers, integers and real numbers, respectively.

In 2001, Suzuki introduced the concept of 𝜏 -distance in order to improve results in Tataru [14], Zhong [15, 16], and others. See also [17].

Definition 2.1 (see [18]). Let ( 𝑋 , 𝑑 ) be a metric space. Then a function 𝑝 from 𝑋 × 𝑋 into [ 0 , ∞ ) is called a 𝜏 -distance on 𝑋 if there exists a function 𝜂 from 𝑋 × [ 0 , ∞ ) into [ 0 , ∞ ) and the following are satisfied:( 𝜏 1 ) 𝑝 ( 𝑥 , 𝑧 ) ≤ 𝑝 ( 𝑥 , 𝑦 ) + 𝑝 ( 𝑦 , 𝑧 ) for all 𝑥 , 𝑦 , 𝑧 ∈ 𝑋 ,( 𝜏 2 ) 𝜂 ( 𝑥 , 0 ) = 0 and 𝜂 ( 𝑥 , 𝑡 ) ≥ 𝑡 for all 𝑥 ∈ 𝑋 and 𝑡 ∈ [ 0 , ∞ ) , and 𝜂 is concave and continuous in its second variable,( 𝜏 3 ) l i m 𝑛 𝑥 𝑛 = 𝑥 and l i m 𝑛 s u p { 𝜂 ( 𝑧 𝑛 , 𝑝 ( 𝑧 𝑛 , 𝑥 𝑚 ) ) ∶ 𝑚 ≥ 𝑛 } = 0 imply 𝑝 ( 𝑤 , 𝑥 ) ≤ l i m i n f 𝑛 𝑝 ( 𝑤 , 𝑥 𝑛 ) for all 𝑤 ∈ 𝑋 ,( 𝜏 4 ) l i m 𝑛 s u p { 𝑝 ( 𝑥 𝑛 , 𝑦 𝑚 ) ∶ 𝑚 ≥ 𝑛 } = 0 and l i m 𝑛 𝜂 ( 𝑥 𝑛 , 𝑡 𝑛 ) = 0 imply that l i m 𝑛 𝜂 ( 𝑦 𝑛 , 𝑡 𝑛 ) = 0 ,( 𝜏 5 ) l i m 𝑛 𝜂 ( 𝑧 𝑛 , 𝑝 ( 𝑧 𝑛 , 𝑥 𝑛 ) ) = 0 and l i m 𝑛 𝜂 ( 𝑧 𝑛 , 𝑝 ( 𝑧 𝑛 , 𝑦 𝑛 ) ) = 0 imply that l i m 𝑛 𝑑 ( 𝑥 𝑛 , 𝑦 𝑛 ) = 0 .

The metric 𝑑 is a 𝜏 -distance on 𝑋 . Many useful examples and propositions are stated in [9, 18–23] and references therein. The following fixed point theorems are proved in [18].

Theorem 2.2 (see [18]). Let 𝑋 be a complete metric space and let 𝑇 be a mapping on 𝑋 . Assume that there exist a 𝜏 -distance 𝑝 and 𝑟 ∈ [ 0 , 1 ) such that 𝑝 ( 𝑇 𝑥 , 𝑇 2 𝑥 ) ≤ 𝑟 𝑝 ( 𝑥 , 𝑇 𝑥 ) for all 𝑥 ∈ 𝑋 . Assume the following. (i)If l i m 𝑛 s u p { 𝑝 ( 𝑥 𝑛 , 𝑥 𝑚 ) ∶ 𝑚 > 𝑛 } = 0 , l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑇 𝑥 𝑛 ) = 0 , and l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑦 ) = 0 , then 𝑇 𝑦 = 𝑦 .Then (B) holds. Moreover, if 𝑇 𝑧 = 𝑧 , then 𝑝 ( 𝑧 , 𝑧 ) = 0 .

Theorem 2.3 (see[18]). Let 𝑋 be a complete metric space and let 𝑇 be a mapping on 𝑋 . Assume that 𝑇 is a contraction with respect to some 𝜏 -distance 𝑝 , that is, there exist a 𝜏 -distance 𝑝 and 𝑟 ∈ [ 0 , 1 ) such that 𝑝 ( 𝑇 𝑥 , 𝑇 𝑦 ) ≤ 𝑟 𝑝 ( 𝑥 , 𝑦 ) , ( 2 . 1 ) for all 𝑥 , 𝑦 ∈ 𝑋 . Then (A) and 𝑝 ( 𝑧 , 𝑧 ) = 0 hold.

The following lemmas are useful in our proofs.

Lemma 2.4 (see [18]). Let ( 𝑋 , 𝑑 ) be a metric space and let 𝑝 be a 𝜏 -distance on 𝑋 . If sequences { 𝑥 𝑛 } and { 𝑦 𝑛 } in 𝑋 satisfy l i m 𝑛 𝑝 ( 𝑧 , 𝑥 𝑛 ) = 0 and l i m 𝑛 𝑝 ( 𝑧 , 𝑦 𝑛 ) = 0 for some 𝑧 ∈ 𝑋 , then l i m 𝑛 𝑑 ( 𝑥 𝑛 , 𝑦 𝑛 ) = 0 . In particular for 𝑥 , 𝑦 , 𝑧 ∈ 𝑋 , 𝑝 ( 𝑧 , 𝑥 ) = 0 and 𝑝 ( 𝑧 , 𝑦 ) = 0 imply that 𝑥 = 𝑦 .

Lemma 2.5 (see [18]). Let ( 𝑋 , 𝑑 ) be a metric space and let 𝑝 be a 𝜏 -distance on 𝑋 . If a sequence { 𝑥 𝑛 } in 𝑋 satisfies l i m 𝑛 s u p { 𝑝 ( 𝑥 𝑛 , 𝑥 𝑚 ) ∶ 𝑚 > 𝑛 } = 0 , then { 𝑥 𝑛 } is a Cauchy sequence. Moreover if a sequence { 𝑦 𝑛 } in 𝑋 satisfies l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑦 𝑛 ) = 0 , then l i m 𝑛 𝑑 ( 𝑥 𝑛 , 𝑦 𝑛 ) = 0 .

The following lemmas are easily deduced from Lemmas 2.4 and 2.5.

Lemma 2.6. Let ( 𝑋 , 𝑑 ) be a metric space and let 𝑝 be a 𝜏 -distance on 𝑋 . Then for every 𝑧 ∈ 𝑋 and 𝜀 > 0 , there exists 𝛿 > 0 such that 𝑝 ( 𝑧 , 𝑥 ) ≤ 𝛿 and 𝑝 ( 𝑧 , 𝑦 ) ≤ 𝛿 imply that 𝑑 ( 𝑥 , 𝑦 ) ≤ 𝜀 .

Lemma 2.7. Let 𝑋 be a metric space and let 𝑝 be a 𝜏 -distance on 𝑋 . Assume that a sequence { 𝑥 𝑛 } in 𝑋 satisfies l i m 𝑛 s u p { 𝑝 ( 𝑥 𝑛 , 𝑥 𝑚 ) ∶ 𝑚 > 𝑛 } = 0 , l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑦 ) = 0 , and l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑧 ) = 0 . Then 𝑦 = 𝑧 .

The following is proved at Page 442 of [18]. However we give a proof because we use reductio ad absurdum in [18].

Lemma 2.8 (see [18]). Let 𝑔 be a nondecreasing function from [ 0 , ∞ ) into itself satisfying i n f { 𝑔 ( 𝑡 ) ∶ 𝑡 > 0 } = 0 . Define a function 𝑓 from [ 0 , ∞ ) into itself by 𝑓 ( 𝑡 ) = 𝑡 + s u p 𝑛 𝑖 = 1 𝛼 𝑖 𝑔 𝑠 m i n 𝑖 , 1 ∶ 𝑡 = 𝑛 𝑖 = 1 𝛼 𝑖 𝑠 𝑖 , 𝑠 𝑖 ≥ 0 , 𝛼 𝑖 > 0 , 𝑛 𝑖 = 1 𝛼 𝑖 = 1 . ( 2 . 2 ) Then 𝑓 ( 0 ) = 0 , 𝑓 ( 𝑡 ) ≥ 𝑡 + 𝑔 ( 𝑡 ) for all 𝑡 ∈ [ 0 , ∞ ) ; and 𝑓 is concave and continuous.

Proof. It is clear that 𝑓 ( 0 ) = 0 , 𝑓 ( 𝑡 ) ≥ 𝑡 + 𝑔 ( 𝑡 ) , and 𝑓 is concave. We shall prove that 𝑓 is continuous at 0 . Fix 𝜀 > 0 . Then there exists 𝛿 > 0 such that 𝑔 ( 𝛿 ) ≤ 𝜀 . Choose 𝜏 > 0 with 𝜏 + 𝜏 / 𝛿 ≤ 𝜀 . Fix 𝑡 ∈ ( 0 , 𝜏 ) . Let 𝛼 1 , 𝛼 2 , … , 𝛼 𝑛 > 0 and 𝑠 1 , 𝑠 2 , … , 𝑠 𝑛 ≥ 0 such that ∑ 𝑡 = 𝑛 𝑖 = 1 𝛼 𝑖 𝑠 𝑖 and ∑ 𝑛 𝑖 = 1 𝛼 𝑖 = 1 . Since 𝛿 ∑ { 𝛼 𝑖 ∶ 𝑠 𝑖 ≥ 𝛿 } ≤ 𝑡 , we have 𝑡 + 𝑛 𝑖 = 1 𝛼 𝑖 𝑔 𝑠 m i n 𝑖 , 1 ≤ 𝑡 + 𝑠 𝑖 < 𝛿 𝛼 𝑖 𝑔 𝑠 𝑖 + 𝑠 𝑖 ≥ 𝛿 𝛼 𝑖 ≤ 𝑡 + 𝑠 𝑖 < 𝛿 𝛼 𝑖 𝜀 + 𝑠 𝑖 ≥ 𝛿 𝛼 𝑖 𝑡 ≤ 𝑡 + 𝜀 + 𝛿 𝜏 ≤ 𝜏 + 𝜀 + 𝛿 ≤ 2 𝜀 . ( 2 . 3 ) Since 𝛼 1 , 𝛼 2 , … , 𝛼 𝑛 > 0 and 𝑠 1 , 𝑠 2 , … , 𝑠 𝑛 ≥ 0 are arbitrary, we obtain 𝑓 ( 𝑡 ) ≤ 2 𝜀 . Thus, l i m 𝑡 → + 0 𝑓 ( 𝑡 ) = 0 = 𝑓 ( 0 ) .

The following is obvious.

Lemma 2.9. Let 𝑇 be a mapping on a set 𝑋 . Let 𝐴 0 be a subset of 𝑋 such that 𝑇 ( 𝐴 0 ) ⊂ 𝐴 0 . Define a sequence { 𝐴 𝑛 } of subsets of 𝑋 by 𝐴 1 = 𝑇 − 1 𝐴 0 ⧵ 𝐴 0 , 𝐴 𝑛 + 1 = 𝑇 − 1 𝐴 𝑛 . ( 2 . 4 ) Then the following hold. (i)For every 𝑛 ∈ ℕ and 𝑥 ∈ 𝑋 , 𝑥 ∈ 𝐴 𝑛 if and only if 𝑇 𝑗 𝑥 ∉ 𝐴 0 for 𝑗 = 0 , 1 , … , 𝑛 − 1 and 𝑇 𝑛 𝑥 ∈ 𝐴 0 .(ii) 𝐴 𝑚 ∩ 𝐴 𝑛 = ∅ for 𝑚 , 𝑛 ∈ ℕ ∪ { 0 } with 𝑚 ≠ 𝑛 .(iii) 𝑇 ( 𝐴 𝑛 + 1 ) = 𝐴 𝑛 for every 𝑛 ∈ ℕ .

3. Condition (B)

In this section, we discuss Condition (B).

Theorem 3.1. Let 𝑋 be a complete metric space and let 𝑇 be a mapping on 𝑋 . Assume that there exist a 𝜏 -distance 𝑝 , 𝑟 ∈ [ 0 , 1 ) , and 𝑀 ∈ [ 0 , ∞ ) such that 𝑝 𝑇 𝑥 , 𝑇 2 𝑥 ≤ 𝑟 𝑝 ( 𝑥 , 𝑇 𝑥 ) , 𝑝 ( 𝑇 𝑥 , 𝑇 𝑦 ) ≤ 𝑀 𝑝 ( 𝑥 , 𝑦 ) , ( 3 . 1 ) for all 𝑥 , 𝑦 ∈ 𝑋 . Then (B) holds. Moreover, if 𝑇 𝑧 = 𝑧 , then 𝑝 ( 𝑧 , 𝑧 ) = 0 .

Proof. Assume that l i m 𝑛 s u p { 𝑝 ( 𝑥 𝑛 , 𝑥 𝑚 ) ∶ 𝑚 > 𝑛 } = 0 , l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑇 𝑥 𝑛 ) = 0 , and l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑦 ) = 0 . Then we have 𝑝 𝑥 𝑛 𝑥 , 𝑇 𝑦 ≤ 𝑝 𝑛 , 𝑇 𝑥 𝑛 + 𝑝 𝑇 𝑥 𝑛 𝑥 , 𝑇 𝑦 ≤ 𝑝 𝑛 , 𝑇 𝑥 𝑛 𝑥 + 𝑀 𝑝 𝑛 , 𝑦 , ( 3 . 2 ) and hence, l i m 𝑛 𝑝 ( 𝑥 𝑛 , 𝑇 𝑦 ) = 0 . By Lemma 2.7, we obtain 𝑇 𝑦 = 𝑦 . By Theorem 2.2, we obtain the desired result.

As a direct consequence of Theorem 3.1, we obtain the following.

Corollary 3.2. Let 𝑋 be a complete metric space and let 𝑇 be a mapping on 𝑋 . Assume that there exist a 𝜏 -distance 𝑝 and 𝑟 ∈ ( 0 , 1 ) such that 𝑝 𝑇 𝑥 , 𝑇 2 𝑥 ≤ 𝑟 𝑝 ( 𝑥 , 𝑇 𝑥 ) , 𝑝 ( 𝑇 𝑥 , 𝑇 𝑦 ) ≤ 𝑝 ( 𝑥 , 𝑦 ) , ( 3 . 3 ) for all 𝑥 , 𝑦 ∈ 𝑋 . Then (B) holds.

Corollary 3.2 characterizes Condition (B).

Theorem 3.3. Let 𝑇 be a mapping on a metric space ( 𝑋 , 𝑑 ) such that (B) holds. Then there exist a 𝜏 -distance 𝑝 and 𝑟 ∈ ( 0 , 1 ) satisfying (3.3).

Proof. Let 𝑟 ∈ ( 0 , 1 ) be fixed. We note that every periodic point is a fixed point. That is, if 𝑥 ∈ 𝑋 satisfies 𝑇 𝑛 𝑥 = 𝑥 for some 𝑛 ∈ ℕ , then 𝑇 𝑥 = 𝑥 . Define a mapping 𝑇 ∞ from 𝑋 onto 𝐹 ( 𝑇 ) by 𝑇 ∞ 𝑥 = l i m 𝑛 𝑇 𝑛 𝑥 for 𝑥 ∈ 𝑋 , where 𝐹 ( 𝑇 ) is the set of all fixed points of 𝑇 . Define a mapping 𝐶 from 𝑋 into the set of subsets of 𝑋 by 𝐶 𝑥 = 𝑇 𝑥 , 𝑇 2 𝑥 , 𝑇 3 𝑥 , … , 𝑇 ∞ 𝑥 . ( 3 . 4 ) Since 𝑇 ∞ 𝑥 is a fixed point of 𝑇 , we have 𝑦 ∈ 𝐶 𝑥 ⟹ 𝐶 𝑦 ⊂ 𝐶 𝑥 . ( 3 . 5 )
Next, we define a function 𝑓 from 𝑋 into ℤ ∪ { ∞ } satisfying 𝑓 ( 𝑇 𝑥 ) ≥ 𝑓 ( 𝑥 ) + 1 , 𝑓 ( 𝑥 ) = ∞ ⟺ 𝑇 𝑥 = 𝑥 , ( 3 . 6 ) for all 𝑥 ∈ 𝑋 . We put 𝑓 ( 𝑥 ) = ∞ for 𝑥 ∈ 𝐹 ( 𝑇 ) . It is obvious that 𝑓 ( 𝑇 𝑥 ) = 𝑓 ( 𝑥 ) = ∞ = 𝑓 ( 𝑥 ) + 1 for 𝑥 ∈ 𝐹 ( 𝑇 ) . Define a sequence { 𝐴 𝑛 } of subsets of 𝑋 by 𝐴 1 = 𝑇 − 1 ( 𝐹 ( 𝑇 ) ) ⧵ 𝐹 ( 𝑇 ) , 𝐴 𝑛 + 1 = 𝑇 − 1 𝐴 𝑛 . ( 3 . 7 ) Then by Lemma 2.9, 𝐹 ( 𝑇 ) ∩ 𝐴 𝑛 = ∅ , 𝐴 𝑚 ∩ 𝐴 𝑛 = ∅ , ( 3 . 8 ) for 𝑚 , 𝑛 ∈ ℕ with 𝑚 ≠ 𝑛 . We put 𝑓 ( 𝑥 ) = − 𝑛 for 𝑥 ∈ 𝐴 𝑛 . We note that 𝑓 ( 𝑇 𝑥 ) = ∞ i f 𝑥 ∈ 𝐴 1 , ⨆ 𝑓 ( 𝑥 ) + 1 i f 𝑥 ∈ ∞ 𝑛 = 2 𝐴 𝑛 . ( 3 . 9 ) Put 𝑌 = 𝑋 ⧵ 𝐹 ( 𝑇 ) ⊔ 𝑛 ∈ ℕ 𝐴 𝑛 . ( 3 . 1 0 ) It is obvious that 𝑇 ( 𝑌 ) ⊂ 𝑌 , 𝑇 − 1 ( 𝑌 ) = 𝑌 , and 𝑌 ∩ 𝐹 ( 𝑇 ) = ∅ . So, 𝑇 𝑚 𝑥 = 𝑇 𝑛 𝑥 ⟺ 𝑚 = 𝑛 , ( 3 . 1 1 ) for 𝑥 ∈ 𝑌 and 𝑚 , 𝑛 ∈ ℕ ∪ { 0 } . Define an equivalence relation ∼ on 𝑌 as follows: 𝑥 ∼ 𝑦 if and only if there exist 𝑚 , 𝑛 ∈ ℕ ∪ { 0 } such that 𝑇 𝑚 𝑥 = 𝑇 𝑛 𝑦 . By Axiom of Choice, there exists a mapping 𝐵 on 𝑌 such that 𝐵 𝑥 ∼ 𝑥 , 𝑥 ∼ 𝑦 ⟺ 𝐵 𝑥 = 𝐵 𝑦 . ( 3 . 1 2 ) Let 𝑢 ∈ 𝑌 with 𝐵 𝑢 = 𝑢 . Then we put 𝑓 ( 𝑇 𝑛 𝑢 ) = 𝑛 for 𝑛 ∈ ℕ ∪ { 0 } . Define a sequence { 𝐷 𝑛 } of subsets of 𝑌 by 𝐷 0 = 𝑢 , 𝑇 𝑢 , 𝑇 2 𝑢 , 𝑇 3 𝑢 , … , 𝐷 1 = 𝑇 − 1 𝐷 0 ⧵ 𝐷 0 , 𝐷 𝑛 + 1 = 𝑇 − 1 𝐷 𝑛 . ( 3 . 1 3 ) Then we have 𝐷 𝑚 ∩ 𝐷 𝑛 = ∅ for 𝑚 , 𝑛 ∈ ℕ ∪ { 0 } with 𝑚 ≠ 𝑛 ; and { 𝑥 ∈ 𝑌 ∶ 𝑥 ∼ 𝑢 } = 𝑛 ∈ ℕ ∪ { 0 } 𝐷 𝑛 . ( 3 . 1 4 ) We put 𝑓 ( 𝑥 ) = − 𝑛 for 𝑥 ∈ 𝑌 with 𝑛 ∈ ℕ and 𝑥 ∈ 𝐷 𝑛 . We have defined 𝑓 . We note that 𝑓 ( 𝑥 ) ∈ ℕ implies that 𝑥 ∈ 𝑌 .
Next, we define a 𝜏 -distance 𝑝 by ⎧ ⎪ ⎨ ⎪ ⎩ 𝑟 𝑝 ( 𝑥 , 𝑦 ) = 𝑓 ( 𝑥 ) + 𝑟 𝑓 ( 𝑦 ) 𝑟 i f 𝑦 ∈ 𝐶 𝑥 , 𝑓 ( 𝑥 ) + 𝑟 𝑓 ( 𝑦 ) + 1 i f 𝑦 ∉ 𝐶 𝑥 , ( 3 . 1 5 ) where 𝑟 ∞ = 0 . We note that 𝑝 ( 𝑥 , 𝑦 ) < 1 implies either of the following. (i) 𝑇 𝑥 = 𝑥 = 𝑦 .(ii)There exist 𝑢 ∈ 𝑌 , 𝑘 ∈ ℕ , and ℓ ∈ ℕ ∪ { ∞ } such that 𝐵 𝑢 = 𝑢 , 𝑘 < 𝑙 , 𝑥 = 𝑇 𝑘 𝑢 , and 𝑦 = 𝑇 ℓ 𝑢 . (In this case, 𝑥 ∈ 𝑌 , 𝑢 = 𝐵 𝑥 , 𝑓 ( 𝑥 ) = 𝑘 , and 𝑓 ( 𝑦 ) = ℓ hold.)We shall show that 𝑝 is a 𝜏 -distance. Let 𝑥 , 𝑦 , 𝑧 ∈ 𝑋 . If 𝑦 ∈ 𝐶 𝑥 and 𝑧 ∈ 𝐶 𝑦 , then 𝑧 ∈ 𝐶 𝑥 . So we have 𝑝 ( 𝑥 , 𝑧 ) = 𝑟 𝑓 ( 𝑥 ) + 𝑟 𝑓 ( 𝑧 ) ≤ 𝑟 𝑓 ( 𝑥 ) + 𝑟 𝑓 ( 𝑦 ) + 𝑟 𝑓 ( 𝑦 ) + 𝑟 𝑓 ( 𝑧 ) = 𝑝 ( 𝑥 , 𝑦 ) + 𝑝 ( 𝑦 , 𝑧 ) . ( 3 . 1 6 ) If 𝑦 ∉ 𝐶 𝑥 or 𝑧 ∉ 𝐶 𝑦 , then 𝑝 ( 𝑥 , 𝑧 ) ≤ 𝑟 𝑓 ( 𝑥 ) + 𝑟 𝑓 ( 𝑧 ) + 1 ≤ 𝑟 𝑓 ( 𝑥 ) + 𝑟 𝑓 ( 𝑦 ) + 𝑟 𝑓 ( 𝑦 ) + 𝑟 𝑓 ( 𝑧 ) + 1 ≤ 𝑝 ( 𝑥 , 𝑦 ) + 𝑝 ( 𝑦 , 𝑧 ) . ( 3 . 1 7 ) These imply ( 𝜏 1). We shall define a function 𝜂 from 𝑋 × [ 0 , ∞ ) into [ 0 , ∞ ) . For 𝑥 ∈ 𝑋 ⧵ 𝑌 , we put 𝜂 ( 𝑥 , 𝑡 ) = 𝑡 . For 𝑥 ∈ 𝑌 , we put 𝑢 = 𝐵 𝑥 . Since { 𝑇 𝑛 𝑢 } converges to 𝑇 ∞ 𝑢 , there exists a strictly increasing sequence { ℎ 𝑢 ( 𝑛 ) } in ℕ such that 𝑗 ≥ ℎ 𝑢 ( 𝑛 ) implies that 𝑑 ( 𝑇 𝑗 𝑢 , 𝑇 ∞ 𝑢 ) ≤ 1 / 𝑛 for 𝑗 ∈ ℕ ∪ { ∞ } . Since l i m 𝑛 ℎ 𝑢 ( 𝑛 ) = ∞ , we can define a nondecreasing function 𝑔 𝑢 from [ 0 , ∞ ) into [ 0 , 1 ] such that 𝑔 𝑢 ( 𝑟 ℎ 𝑢 ( 𝑛 ) ) = 1 / 𝑛 . It is obvious that 𝑔 𝑢 ( 0 ) = l i m 𝑡 → + 0 𝑔 𝑢 ( 𝑡 ) = 0 . Put 𝜂 ( 𝑥 , 𝑡 ) = 𝑡 + s u p 𝑛 𝑖 = 1 𝛼 𝑖 𝑔 𝑢 𝑠 𝑖 ∶ 𝑡 = 𝑛 𝑖 = 1 𝛼 𝑖 𝑠 𝑖 , 𝑠 𝑖 ≥ 0 , 𝛼 𝑖 > 0 , 𝑛 𝑖 = 1 𝛼 𝑖 = 1 . ( 3 . 1 8 ) Then 𝜂 ( 𝑥 , 𝑡 ) satisfies ( 𝜏 2) and 𝜂 ( 𝑥 , 𝑡 ) ≥ 𝑡 + 𝑔 𝑢 ( 𝑡 ) by Lemma 2.8. In order to show ( 𝜏 3), we assume that l i m 𝑛 𝑥 𝑛 = 𝑥 and l i m 𝑛 s u p { 𝜂 ( 𝑧 𝑛 , 𝑝 ( 𝑧 𝑛 , 𝑥 𝑚 ) ) ∶ 𝑚 ≥ 𝑛 } = 0 . Then without loss of generality, we may assume that s u p { 𝜂 ( 𝑧 𝑛 , 𝑝 ( 𝑧 𝑛 , 𝑥 𝑚 ) ) ∶ 𝑚 ≥ 𝑛 } < 1 . Thus s u p { 𝑝 ( 𝑧 𝑛 , 𝑥 𝑚 ) ∶ 𝑚 ≥ 𝑛 } < 1 for 𝑛 ∈ ℕ . It is obvious that 𝑥 𝑚 ∈ 𝐶 𝑧 𝑛 for 𝑚 , 𝑛 ∈ ℕ with 𝑚 ≥ 𝑛 . We consider the following two cases.(i)There exists 𝜈 ∈ ℕ such that 𝑥 𝑛 ∈ 𝐹 ( 𝑇 ) for 𝑛 ≥ 𝜈 .(ii)There exists a subsequence { 𝑥 𝑛 𝑗 } of { 𝑥 𝑛 } such that 𝑥 𝑛 𝑗 ∉ 𝐹 ( 𝑇 ) .In the first case, since 𝐹 ( 𝑇 ) ∩ 𝐶 𝑧 𝜈 exactly consists of one element and 𝑥 𝑛 ∈ 𝐹 ( 𝑇 ) ∩ 𝐶 𝑧 𝜈 for 𝑛 ≥ 𝜈 , 𝑥 𝑛 = 𝑥 𝜈 holds for all 𝑛 ≥ 𝜈 . So 𝑥 = 𝑥 𝜈 . Thus, 𝑝 ( 𝑤 , 𝑥 ) = l i m 𝑛 𝑝 ( 𝑤 , 𝑥 𝑛 ) holds for every 𝑤 ∈ 𝑋 . In the second case, we note that 𝑧 𝑛 ∉ 𝐹 ( 𝑇 ) for all 𝑛 ∈ ℕ . Hence 𝑧 𝑛 ∈ 𝑌 . Put 𝑢 = 𝑃 𝑧 1 . Since 𝑥 𝑛 ∈ 𝐶 𝑧 1 , there exists a sequence { ℓ 𝑛 } in ℕ ∪ { ∞ } such that 𝑥 𝑛 = 𝑇 ℓ 𝑛 𝑢 . Since ℓ 𝑛 𝑗 ∈ ℕ for all 𝑗 ∈ ℕ , there exists a sequence { 𝑘 𝑛 } in ℕ such that 𝑧 𝑛 = 𝑇 𝑘 𝑛 𝑢 . Since l i m 𝑛 𝑝 ( 𝑧 𝑛 , 𝑥 𝑛 ) = 0 , we have l i m 𝑛 𝑘 𝑛 = ∞ and l i m 𝑛 ℓ 𝑛 = ∞ . So we obtain 𝑥 = 𝑇 ∞ 𝑢 . We note that 𝑥 ∈ 𝐶 𝑥 𝑛 for all 𝑛 ∈ ℕ . Let 𝑤 ∈ 𝑋 . In the case where 𝑥 ∈ 𝐶 𝑤 , we have 𝑝 ( 𝑤 , 𝑥 ) = 𝑟 𝑓 ( 𝑤 ) = l i m 𝑛 → ∞ 𝑟 𝑓 ( 𝑤 ) + 𝑟 𝑓 ( 𝑥 𝑛 ) ≤ l i m i n f 𝑛 → ∞ 𝑝 𝑤 , 𝑥 𝑛 . ( 3 . 1 9 ) In the other case, where 𝑥 ∉ 𝐶 𝑤 , we have 𝑥 𝑛 ∉ 𝐶 𝑤 , and hence, 𝑝 ( 𝑤 , 𝑥 ) = 𝑟 𝑓 ( 𝑤 ) + 1 = l i m 𝑛 → ∞ 𝑟 𝑓 ( 𝑤 ) + 𝑟 𝑓 ( 𝑥 𝑛 ) + 1 = l i m 𝑛 → ∞ 𝑝 𝑤 , 𝑥 𝑛 . ( 3 . 2 0 ) Therefore we have shown ( 𝜏 3). Let us prove ( 𝜏 4). We assume that l i m 𝑛 s u p { 𝑝 ( 𝑥 𝑛 , 𝑦 𝑚 ) ∶ 𝑚 ≥ 𝑛 } = 0 and l i m 𝑛 𝜂 ( 𝑥 𝑛 , 𝑡 𝑛 ) = 0 . Without loss of generality, we may assume that s u p { 𝑝 ( 𝑥 𝑛 , 𝑦 𝑚 ) ∶ 𝑚 ≥ 𝑛 } < 1 . We consider the following two cases.(i)There exists 𝜈 ∈ ℕ such that 𝑦 𝑛 ∈ 𝐹 ( 𝑇 ) for 𝑛 ≥ 𝜈 .(ii)There exists a subsequence { 𝑦 𝑛 𝑗 } of { 𝑦 𝑛 } such that 𝑦 𝑛 𝑗 ∉ 𝐹 ( 𝑇 ) .In the first case, we have l i m 𝑛 → ∞ 𝜂 𝑦 𝑛 , 𝑡 𝑛 = l i m 𝑛 → ∞ 𝑡 𝑛 ≤ l i m 𝑛 → ∞ 𝜂 𝑥 𝑛 , 𝑡 𝑛 = 0 . ( 3 . 2 1 ) In the second case, as in the proof of ( 𝜏 3), there exist 𝑢 ∈ 𝑌 , a sequence { 𝑘 𝑛 } in ℕ , and a sequence { ℓ 𝑛 } in ℕ ∪ { ∞ } such that 𝐵 𝑢 = 𝑢 , 𝑥 𝑛 = 𝑇 𝑘 𝑛 𝑢 , and 𝑦 𝑛 = 𝑇 ℓ 𝑛 𝑢 . We note that 𝜂 ( 𝑥 𝑛 , 𝑡 ) = 𝜂 ( 𝑢 , 𝑡 ) . If 𝑦 𝑛 ∈ 𝐹 ( 𝑇 ) , then 𝜂 ( 𝑦 𝑛 , 𝑡 ) = 𝑡 ≤ 𝜂 ( 𝑢 , 𝑡 ) . If 𝑦 𝑛 ∉ 𝐹 ( 𝑇 ) , then 𝜂 ( 𝑦 𝑛 , 𝑡 ) = 𝜂 ( 𝑢 , 𝑡 ) . Therefore l i m 𝑛 → ∞ 𝜂 𝑦 𝑛 , 𝑡 𝑛 ≤ l i m 𝑛 → ∞ 𝜂 𝑢 , 𝑡 𝑛 = l i m 𝑛 → ∞ 𝜂 𝑥 𝑛 , 𝑡 𝑛 = 0 . ( 3 . 2 2 ) Let us prove ( 𝜏 5). We assume that 𝜂 ( 𝑧 , 𝑝 ( 𝑧 , 𝑥 ) ) < 1 / 𝑛 . We note that 𝑝 ( 𝑧 , 𝑥 ) < 1 . In the case where 𝑇 𝑧 = 𝑧 = 𝑥 , we have 𝑑 ( 𝑧 , 𝑥 ) = 0 < 1 / 𝑛 . In the other case, where there exist 𝑢 ∈ 𝑌 , 𝑘 ∈ ℕ , and ℓ ∈ ℕ ∪ { ∞ } such that 𝐵 𝑢 = 𝑢 , 𝑘 < 𝑙 , 𝑧 = 𝑇 𝑘 𝑢 , and 𝑥 = 𝑇 ℓ 𝑢 , we have 1 𝜂 ( 𝑧 , 𝑝 ( 𝑧 , 𝑥 ) ) < 𝑛 = 𝑔 𝑢 𝑟 ℎ 𝑢 ( 𝑛 ) ≤ 𝜂 𝑧 , 𝑟 ℎ 𝑢 ( 𝑛 ) . ( 3 . 2 3 ) Hence 𝑟 𝑘 + 𝑟 ℓ = 𝑝 ( 𝑧 , 𝑥 ) < 𝑟 ℎ 𝑢 ( 𝑛 ) . ( 3 . 2 4 ) Thus, we obtain 𝑘 > ℎ 𝑢 ( 𝑛 ) and ℓ > ℎ 𝑢 ( 𝑛 ) . So we have 𝑇 𝑑 ( 𝑧 , 𝑥 ) = 𝑑 𝑘 𝑢 , 𝑇 ℓ 𝑢 𝑇 ≤ 𝑑 𝑘 𝑢 , 𝑇 ∞ 𝑢 𝑇 + 𝑑 ℓ 𝑢 , 𝑇 ∞ 𝑢 ≤ 1 𝑛 + 1 𝑛 = 2 𝑛 . ( 3 . 2 5 ) Therefore 1 𝜂 ( 𝑧 , 𝑝 ( 𝑧 , 𝑥 ) ) < 𝑛 1 , 𝜂 ( 𝑧 , 𝑝 ( 𝑧 , 𝑦 ) ) < 𝑛 4 ⟹ 𝑑 ( 𝑥 , 𝑦 ) ≤ 𝑛 , ( 3 . 2 6 ) which imply ( 𝜏 5). Therefore we have shown that 𝑝 is a 𝜏 -distance on 𝑋 .
We shall show (3.3). Let