On the Uniform Continuity in G-Metric Spaces

Baekjin Kim
designeralice@daum.net

A thesis submitted to the committee of Graduate School, Kongju National University in partial fulfillment of the requirements for the degree of Master of Arts Conferred in February 2009.

RISS: http://www.riss.kr/link?id=T11546441

Cite this thesis as:
Baekjin Kim (2009). On the Uniform Continuity in G-Metric Spaces (Master's thesis, Kongju National University, Korea). Retrieved from Research Information Sharing Service: http://www.riss.kr/link?id=T11546441.

ABSTRACT

In 2006, Z. Mustafa and B. Sims introduced G-metric which is a new generalized metric. In this paper, we define the uniform continuity in G-metric spaces and prove some properties of the uniform continuity in G-metric spaces.

Keywords: G-metric space, uniform continuity, Cauchy sequence

I. INTRODUCTION

In 1963, the notion of 2-metric space was introduced by S. Gähler [6, 7] as a generalization of usual notion of metric space \((X,\,d).\) The definition of 2-metric space is as follows.

But different authors proved that there is no relation between these two functions, for instance, K. S. Ha et al. [8] showed that the 2-metric is not always a continuous function, further there is no easy relationship between results obtained from the two metrics.

In 1992, B. Dhage in his Ph. D. thesis [1] introduced a new class of generalized metric space called D-metric spaces.

In a subsequent series of papers [2, 3, 4], B. Dhage attempted to develop topological structures in such spaces. He claimed that D-metrics provide a generalization of ordinary metric functions.

But in 2004, Z. Mustafa and B. Sims demonstrated that most of the claims concerning the fundamental topological structure of D-metric space are incorrect (See [10, 13]), and in 2006, they introduced more appropriate notion of generalized metric space as follows (See [11, 12]).

EXAMPLE 1.4. Immediate examples of such a function are \[G(x,\,y,\,z) := \max \left\{ d(x,\,y) ,\, d(y,\,z) ,\, d(z,\,x) \right\}\] and \[G(x,\,y,\,z) := d(x,\,y) + d(y,\,z) + d(z,\,x)\] where \(d\) is an ordinary metric function.

EXAMPLE 1.5. If \(X = \mathbb{R}^n\) then a function \(G\) defined by \[G(x,\,y,\,z) := ( \lVert x-y \rVert ^p + \lVert y-z \rVert^p + \lVert z-x \rVert^p )^{\frac{1}{p}}\] is a G-metric for \(p > 0.\)

EXAMPLE 1.6. If \(X = \mathbb{R}^{+}\) then a function \(G\) defined by \[G(x,\,y,\,z) := \begin{cases} 0 \quad & \mathrm{if} \,\, x=y=z, \\[8pt] \max\left\{x,\,y,\,z\right\} \quad & \text{otherwise} \end{cases}\] is a G-metric.

II. PRELIMINARIES

We introduce some concepts and properties of sequences and functions on G-metric spaces as preliminaries. To find the proofs of propositions omitted in this chapter, see [11, 12, 14].

To avoid overusing the prefix 'G-' we adopt the convention that the prefix is not denoted at convergence, continuity, completeness if it doesn't make any confusions.

PROPOSITION 2.2.  (Equivalent Conditions for Convergence of a Sequence)
Let \((X,\,G)\) be a G-metric space, then the following four statements are equivalent.

• \(\left\{ x_n \right\}\) is G-convergent to \(x.\)
• \(G(x_n ,\,x_n ,\,x ) \to 0\) as \(n \to +\infty .\)
• \(G(x_n ,\,x,\,x) \to 0\) as \(n \to +\infty .\)
• \(G(x_m ,\, x_n ,\, x) \to 0\) as \(m,\,n \to +\infty .\)

PROPOSITION 2.3.  (Uniqueness of the Limit of a Sequence)
Let \((X,\,G)\) be a G-metric space. If sequence \(\left\{ x_n \right\}\) in \(X\) converges to \(x\), then \(x\) is unique.

Proof. Let \(x_n \to y\) and \(y \neq x.\) Since \(\left\{ x_n \right\}\) converges to \(x\) and \(y,\) for each \(\epsilon > 0\) there exist \(N_1 \in \mathbb{N}\) and \(N_2 \in \mathbb{N}\) such that for every \(n \ge N_1 ,\) \(G(x,\,x_n ,\,x_n ) < \frac{\epsilon}{2}\) holds and for every \(n \ge N_2 ,\) \(G(y,\,y,\,x_n ) < \frac{\epsilon}{2}\) holds. If \(N := \max \left\{ N_1 ,\,N_2 \right\}\) then for every \(n \ge N\) we have \[G(x,\,y,\,y) \le G(x,\,x_n ,\,x_n ) + G(x_n ,\,y,\,y) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon .\] Hence \(G(x,\,y,\,y) =0,\) which is a contradiction. So, \(x=y.\)

PROPOSITION 2.5.  (A Condition for a Sequence to be Cauchy)
If \((X,\,G)\) is a G-metric space, then the following statements are equivalent.

• The sequence \(\left\{ x_n \right\}\) is G-Cauchy.
• For every \(\epsilon > 0,\) there exists \(N \in \mathbb{N}\) such that \(G(x_n ,\,x_m ,\, x_m ) < \epsilon \) for all \(n,\,m \ge N.\)

PROPOSITION 2.6.  (Relation between Convergent and Cauchy)
Let \((X,\,G)\) be a G-metric space. If a sequence \(\left\{ x_n \right\}\) in \(X\) converges to \(x,\) then the sequence \(\left\{ x_n \right\}\) is a Cauchy sequence.

Proof. Since \(x_n \to x ,\) for each \(\epsilon > 0\) there exists \(N_1 ,\, N_2 \in \mathbb{N}\) such that \[n \ge N_1 \,\, \text{implies} \,\, G(x_n ,\, x,\,x) < \frac{\epsilon}{2}\] and \[n \ge N_2 \,\, \text{implies} \,\, G(x ,\, x_m ,\,x_m ) < \frac{\epsilon}{2} .\] If \(N := \max \left\{ N_1 ,\,N_2 \right\} ,\) then for every \(n,\,m \ge N\) we have \[G(x_n ,\, x_m ,\, x_m ) \leq G(x_n ,\, x,\,x ) + G(x,\, x_m ,\, x_m ) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon .\] Hence \(\left\{ x_n \right\}\) is a Cauchy sequence.

PROPOSITION 2.9.  (Relation between Continuity and Sequential Continuity)
Let \((X,\,G)\) and \((Y,\,H)\) be G-metric spaces. Then a function \(f : (X,\,G) \to (Y,\,H)\) is continuous at a point \(x\in X\) if and only if it is G-sequentially continuous at \(x,\) i.e., whenever \(\left\{ x_n \right\}\) is G-convergent to \(x,\) \(\left\{ f(x_n ) \right\}\) is G-convergent to \(f(x).\)

PROPOSITION 2.11.  (Continuity of a Symmetric G-metric)
Let \((X,\,G)\) be a G-metric space, then the function \(G(x,\,y,\,z)\) is jointly continuous in all three of its variables.

PROPOSITION 2.12.  (Relation between D-Metric and G-Metric)
Every G-metric space \((X,\,G)\) defines a metric space \((X,\,d_G )\) by \[d_G (x,\,y) := \frac{1}{2} (G(x,\,y,\,y) + G(y,\,x,\,x)) \tag{2.1}\] for all \(x,\,y\in X.\)

Note that if \((X,\,G)\) is a symmetric G-metric space, then \[d_G (x,\,y) = G(x,\,y,\,y) \tag{2.2}\] for all \(x,\,y\in X .\) However, if \((X,\,G)\) is not symmetric, then it holds by the G-metric properties that \[\frac{3}{4} G(x,\,y,\,y) \leq d_G (x,\,y) \leq \frac{3}{2} G(x,\,y,\,y) \tag{2.3}\] for all \(x,\,y\in X\) and in general these inequalities cannot be improved.

REMARK 2.13. The original definition of \(d_G\) by Z. Mustafa and B. Sims was \[d_G (x,\,y) := G(x,\,x,\,y) + G(x,\,y,\,y) .\] However, since metric \(d_D\) induced by D-metric is defined by \[d_D (x,\,y) := D(x,\,x,\,y) = D(x,\,y,\,y) ,\] it makes sense to define \(d_G\) by (2.1). (Note that every D-metric is symmetric.)

PROPOSITION 2.14.  (Relation between Completeness of G-Metric and induced D-Metric Spaces)
A G-metric space \((X,\,G)\) is G-complete if and only if \((X,\,d_G )\) is a complete metric space.

EXAMPLE 2.16. Let \(X = \mathbb{R} .\) Denote \[G(x,\,y,\,z) = \lvert x-y \rvert + \lvert y-z \rvert + \lvert z-x \rvert\] for all \(x,\,y,\,z \in \mathbb{R}.\) Thus \[\begin{eqnarray} B_G (1,\,2) &=& \left\{ y\in \mathbb{R} \,\vert\, G(1,\,y,\,y) + G(y,\,1,\,1) < 4 \right\} \\[8pt] &=& \left\{ y\in \mathbb{R} \,\vert\, \lvert y-1 \rvert < 1 \right\} = (0,\,2). \end{eqnarray}\] This example shows that an open ball in a G-metric space is a generalization of an open interval in a usual metric space.

Let \(\tau\) be the set of all \(A\subseteq X\) with \(x\in A\) if and only if there exists \(r > 0\) such that \(B_G (x,\,r) \subseteq A.\) Then \(\tau\) is a topology on \(X\) (induced by the G-metric \(G\)).

PROPOSITION 2.17.  (The Openness of Open Balls)
Let \((X,\,G)\) be a G-metric space. If \(r > 0 ,\) then a ball \(B_G (x,\,r)\) with center \(x\in X\) and radius \(r\) is an open set.

Proof. Let \(z\in B_G (x,\,r) ,\) hence \(G(x,\,z,\,z) + G(z,\,x,\,x) < 2r .\) If we set \[\delta := G(x,\,z,\,z) + G(z,\,x,\,x) \quad \text{and} \quad r ' := r-\delta \] then we prove that \(B_G (z,\,r ' ) \subseteq B_G (x,\,r).\) Let \(y\in B_G (z,\,r ' ),\) by definition of open ball, we have \[G(y,\,z,\,z) + G(z,\,y,\,y) < 2r ' .\] By (G5), we have \[\begin{eqnarray} G(y,\,x,\,x) + G(x,\,y,\,y) & \leq & G(y,\,z,\,z) + G(z,\,x,\,x) + G(x,\,z,\,z) + G(z,\,y,\,y) \\[8pt] & < & \delta + 2r ' = \delta + 2r - 2 \delta \leq 2r . \end{eqnarray} \] Hence \(B_G (z,\,r ' ) \subseteq B_G (x,\,r).\) That is, the ball \(B_G (x,\,r) \) is an open set.

PROPOSITION 2.19.  (Properties of a Closure)
Let \(S\) be a subset of a metric space \(X.\) Then following relations hold.

• \(\overline{S} \setminus S \subseteq \partial S .\)
• \(\overline{S} \setminus S \subseteq S ' .\)
• \(\overline{S} = S \cup S ' .\)

Proof. See [9] $17.

PROPOSITION 2.20.  (Relation between Convergence of a Sequence and Splitted Sequences)
Let \(\left\{ c_n \right\}\) be a sequence in a G-metric space \((X,\,G),\) \(M \cup K = \mathbb{N},\) \(M\) and \(K\) be disjoint and infinite, \(M = \left\{ m_k \right\},\) \(K = \left\{ n_k \right\},\) both \(\left\{ m_k \right\}\) and \(\left\{ n_k \right\}\) be strictly increasing sequences, and the limits of both \(\left\{ c_{n_k} \right\}\) and \(\left\{ c_{m_k} \right\}\) converge to a same point \(\lambda\) as \(k \to +\infty .\) Then \(\left\{ c_n \right\}\) converges to \(\lambda\) as \(n \to + \infty .\)

Proof. Let \(\epsilon > 0\) be given. There exists \(N_1 \in \mathbb{N}\) and \(N_2 \in \mathbb{N}\) such that \[ G(c_{n_k} ,\, \lambda ,\, \lambda ) < \epsilon \,\,\text{and}\,\, G(c_{m_k} ,\,\lambda ,\,\lambda ) < \epsilon \] for all \(k \ge N_1 \) and \(k \ge N_2 .\) Set \(N := \max \left\{ n_{N_1} ,\, m_{N_2} \right\}\) and let \(n > N\) then \(n=n_k \) or \(n=m_k\) for some \(k .\) Besides, we have \(k \ge N_1\) and \(k \ge N_2\) by monotonicity of \(\left\{ m_k \right\}\) and \(\left\{ n_k \right\}.\) Thus one of \[G(c_n ,\,\lambda,\,\lambda ) = G(c_{n_k} ,\, \lambda ,\,\lambda ) < \epsilon \] or \[G(c_n ,\,\lambda,\,\lambda ) = G(c_{m_k} ,\, \lambda ,\,\lambda ) < \epsilon \] holds and it completes the proof.

III. THE MAIN RESULT

One of the well-known properties of continuous functions is sequential continuity. That is, if\(f:D \to E\) is a continuous function and \(\left\{ x_n \right\}\) is a convergent sequence in \(D\) where \(D\) and \(E\) are metric spaces then the sequence \(\left\{ y_n \right\}\) defined by \(y_n := f(x_n )\) is also convergent. If \(D\) is a first-countable space, then the converse also holds: any function preserving sequential limit is continuous. When \(D\) and \(E\) are further complete, by definition of completeness, \(f : D \to E\) is continuous if and only if \(\left\{ f(x_n )\right\}\) becomes a Cauchy sequence whenever \(\left\{ x_n \right\}\) is a Cauchy sequence in \(D.\) Revising this equivalent statement, it is possible to derive a sequential condition for uniform continuity of \(f.\)

For a function between Euclidean spaces, uniform continuity can be defined in terms of how the function behaves on sequences (See [5]). More specifically, let \(A\) be a subset of \(\mathbb{R} ^n .\) A function \(f : A \to \mathbb{R}^m\) is uniformly continuous if and only if for every pair of sequences \(\left\{ x_n \right\}\) and \(\left\{ y_n \right\}\) such that \[\lim_{n\to\infty} \lVert x_n - y_n \rVert =0\] we have \[\lim_{n\to\infty} \lVert f(x_n ) - f(y_n ) \rVert =0 .\] Being motivated from Fitzpatrick's alternative definition of uniform continuity, an equivalent condition for uniform continuity related Cauchy sequence can be suggested: If \(f\) is defined on a subset of a compact set and preserves every Cauchy sequence to a Cauchy sequence, then \(f\) is uniformly continuous. We observe this property of uniform continuity in G-metric spaces.

Proof. It is trivial that the first statement implies the second one. Now we prove that the second statement implies the first one. Let \(\epsilon > 0\) be given. Then there exists \(\delta > 0\) such that \[H(f(s) ,\, f(t) ,\, f(t)) < \frac{\epsilon}{2}\] whenever \(G(s,\,t,\,t) < \delta .\) Let \(G(s,\,t,\,u) < \delta\) then we have \[G(s,\,t,\,t) \leq G(s,\,t,\,u) < \delta\] and \[G(t,\,t,\,u) \leq G(s,\,t,\,u) < \delta .\] Thus \[\begin{eqnarray} H(f(s),\,f(t),\,f(u)) & \leq & H(f(s),\,f(t),\,f(t)) + H(f(t),\,f(t),\,f(u)) \\[8pt] & < & \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{eqnarray}\] It completes the proof.

We need a lemma of Heine-Cantor's theorem for G-metric spaces.

Proof. Let \(\epsilon > 0\) be given. Since \(f\) is continuous, we can associate to each point \(s\in X\) a positive number \(\delta _s\) such that \[t\in X,\, G(t,\,s,\,s) < \delta_s \,\,\text{implies}\,\, H(f(t),\,f(s),\,f(s)) < \frac{\epsilon}{2} . \tag{3.1}\] Let \(J_s\) be the set of all \(t\in X\) for which \[G(t,\,s,\,s) < \frac{1}{4} \delta_s .\] Since \(s\in J_s ,\) the collection of all the sets \(J_s\) is an open cover of \(X.\) Since \(X\) is compact, there is a finite set of points \(s_1 ,\) \(s_2 ,\) \(\cdots ,\) \(s_n\) in \(X\) such that \[X \subseteq J_{s_1} \cup J_{s_2} \cup \cdots \cup J_{s_n} . \tag{3.2}\] We put \[\delta := \frac{1}{4} \min \left\{ J_{s_1} ,\, J_{s_2} ,\, \cdots ,\, J_{s_n} \right\} .\] Then \(\delta > 0 .\) Now let \(t\) and \(s\) be points of \(X\) such that \(G(t,\,s,\,s) < \delta .\) By (3.2), there is an integer \(m\) such that \(1 \le m \le n\) and \(s\in J_{s_m} .\) Hence \[G(s_k ,\,s,\,s ) \leq \frac{1}{4} \delta_{s_m} \] and we also have \[\begin{eqnarray} G(t,\,s_m ,\, s_m ) & \leq & G(t,\,s,\,s) + G(s,\,s_m ,\, s_m ) \\[8pt] & \leq & G(t,\,s,\,s) + 3G(s,\,s,\,s_m ) \\[8pt] & < & \delta + \frac{3}{4} \delta_{s_m} \leq \delta_{s_m} . \end{eqnarray}\] Finally (3.1) shows that \[H(f(t),\,f(s),\,f(s)) \leq H(f(s) ,\, f(s) ,\,f(s_m )) + H(f(t),\,f(s_m ),\, f(s_m )) < \epsilon .\] This completes the proof.

The following theorem shows our main result.

Proof of the necessary condition. Assume that \(f\) is uniformly continuous on \(S.\) Suppose further that \(\left\{ s_n \right\}\) is a Cauchy sequence in \(S.\)

Let \(\epsilon > 0\) be given. The uniform continuity of \(f\) implies the existence of \(\delta > 0\) such that for all \(s,\,t,\,u\in S,\) \[H(f(s),\,f(t),\,f(u)) < \epsilon\] holds whenever \(G(s,\,t,\,u) < \delta .\) Since \(\left\{ s_n \right\}\) is a Cauchy sequence, there exists a positive integer \(N\) such that \[G(s_m ,\,s_n ,\,s_p ) < \delta\] for all \(m,\,n,\,p > N.\) Thus we have \[H(f(s_m ),\, f(s_n ),\, f(s_p )) < \epsilon \] for all \(m,\,n,\,p > N \) and we conclude that \(\left\{ f(s_n )\right\}\) is a Cauchy sequence.

Proof of the sufficiency condition. Assume that \(\left\{ f(s_n ) \right\}\) becomes a Cauchy sequence whenever \(\left\{ s_n \right\}\) is a Cauchy sequence in \(S.\) Assumption of theorem implies continuity of \(f.\) Suppose contrarily that \(f\) is not uniformly continuous on \(S.\) We need to prove the following claim.

Proof of Claim. Suppose contrarily that the limit of \(f\) converges at arbitrary point \(a\in \partial S.\) Define \(\overline{f} : \overline{S} \to Y\) by \[\overline{f} (x) = \begin{cases} f(x) \quad & \text{if} \,\, x\in S \\[8pt] \lim_{t\to x} f(t) \quad & \text{if} \,\, x\in\overline{S} \setminus S . \end{cases} \tag{3.3} \] It is guaranteed that the limit of \(f\) at \(x\in \overline{S} \setminus S\) in (3.3) converges, for \(\overline{S} \setminus S \subseteq \partial S .\)

Let \(b\in \overline{S}\) be given. We will prove that \(\overline{f}\) is continuous at \(b.\) Note that the variable \(t\) of the limit in (3.3) is taken over \(S,\) not \(\overline{S} .\) Thus continuity of \(\overline{f}\) is not yet guaranteed.

CASE 1. If \(b\) is an isolated point of \(\overline{S},\) then \(\overline{f}\) is continuous at \(b\) by definition of continuity.

CASE 2. If \(b\) is not an isolated point of \(\overline{S} ,\) then \(b\) is a cluster point of \(\overline{S}.\)

Let \(\left\{ a_n \right\}\) be a sequence in \(S\) that converges to \(b.\) Continuity of \(f\) and definition of \(\overline{f}\) together imply that \(\left\{ \overline{f} (a_n ) \right\}\) converges to \(f(b).\)

Let \(\left\{ b_n \right\}\) be a sequence in \(\overline{S} \setminus S\) that converges to \(b.\) By definition of \(\overline{f} ,\) for arbitrary positive integer \(n,\) there exists a positive real number \(\delta_n\) such that \(\delta_n < \frac{1}{n}\) and \[H ( \overline{f} (t_n ) ,\, \overline{f} ( b_n ) ,\, \overline{f} ( b_n )) < \frac{1}{n} \tag{3.4}\] whenever \(G(t_n ,\,b_n ,\, b_n ) < \delta_n\) and \(t_n \in S.\) By the axiom of choice, \(t_n\) can be chosen for every \(n,\) and a sequence \(\left\{t_n \right\}\) is well-defined. (3.4) yields \[\begin{eqnarray} H(\overline{f} (b) ,\, \overline{f} (b_n ),\, \overline{f} (b_n )) & \leq & H(\overline{f} (b) ,\, \overline{f} (t_n ),\, \overline{f} (t_n )) + H( \overline{f} (t_n ),\, \overline{f} (b_n ),\, \overline{f} (b_n )) \\[8pt] & < & \frac{1}{n} + H(\overline{f} (b) ,\, \overline{f} (t_n ),\, \overline{f} (t_n )) \tag{3.5} \end{eqnarray} \] for every \(n.\) Here, since \(\left\{ t_n \right\}\) converges to \(b\) and \(t_n \in S\) for every \(n,\) we have \[\frac{1}{n} + H(\overline{f} (b) ,\, \overline{f} (t_n ),\, \overline{f} (t_n )) \to 0 \tag{3.6}\] as \(n \to +\infty .\) Thus (3.5) and (3.6) together imply \[\lim_{n\to\infty} H(\overline{f} (b) ,\, \overline{f} (b_n ),\, \overline{f} (b_n )) =0 ,\] which implies \(\overline{f} (b_n ) \to \overline{f} (b)\) as \(n \to + \infty .\)

Let \(\left\{c_n \right\}\) be a sequence in \(\overline{S}\) that converges to \(b.\) Without loss of generality, we can assume that both \(\left\{ c_n \right\} \cap S\) and \(\left\{ c_n \right\} \cap (\overline{S} \setminus S)\) are infinite, for if one of them were finite it would be enough to consider the infinite one only. Let subsequence \(\left\{ c_{n_k} \right\}\) be constructed of all the terms of \(\left\{c_n \right\}\) which is in \(S,\) and subsequence \(\left\{ c_{m_k} \right\}\) be constructed of all the terms of \(\left\{c_n \right\}\) which is in \(\overline{S} \setminus S .\) By the result of previous discussion for \(\left\{ a_n \right\}\) and \(\left\{ b_n \right\} ,\) both \(\left\{ \overline{f} (c_{n_k} )\right\}\) and \(\left\{ \overline{f} (c_{m_k} ) \right\}\) converge to \(\overline{f} (b) .\) Thus \(\left\{ \overline{f} (c_n )\right\}\) converges to \(\overline{f} (b)\) and \(\overline{f}\) is continuous at \(b.\)

Observing two cases, \(\overline{f}\) is continuous at arbitrary point \(b\in \overline{S}.\) Since \(\overline{S}\) is compact, \(\overline{f}\) is uniformly continuous on \(\overline{S} .\) Since \(f\) is a restriction of \(\overline{f},\) \(f\) is also uniformly continuous on \(S.\) But it is contradict to assumption of \(f\) in the main proof. This proves Claim.

Now we go on to the main proof. By the claim, the limit of \(f\) diverges at some point \(a\in \partial S.\) Since \(a\in \partial S,\) \(a\) is a cluster point of \(S\) and there exists a sequence \(\left\{ s_n \right\}\) in \(S\) such that the limit of \(\left\{s_n \right\}\) converges to \(a\) and the limit of \(\left\{ f(s_n )\right\}\) diverges as \(n \to +\infty .\) Since \(X\) is complete, \(\left\{ x_n \right\}\) is a Cauchy sequence and \(\left\{ f(x_n )\right\}\) is also a Cauchy sequence. Completeness of \(Y\) implies the convergence of \(\left\{ f(s_n )\right\}.\) This contradiction yields uniform continuity of \(f\) on \(S.\)

REMARK 3.5. In Theorem 3.4, compactness of \(\overline{S}\) cannot be ommitted. For example, let \[G(x,\,y,\,z) = \lvert x-y \rvert + \lvert y-z \rvert + \lvert z-x \rvert\] and \(f : (\mathbb{R} ,\,G ) \to (\mathbb{R} ,\,G )\) be defined by \[f(x) = x^2 .\] Then for every Cauchy sequence \(\left\{ s_n \right\},\) \(\left\{ f(s_n ) \right\}\) is also a Cauchy sequence, while \(f\) is not uniformly continuous on \(\mathbb{R} .\)

Applying theorem 3.4 to functions in Euclidean spaces, we have the following result.

COROLLARY 3.6.  (Cauchy-Sequential Condition for Uniform Continuity)
Let \(S\) be a bounded subset of \(\mathbb{R}^m\) and \(f : S \to \mathbb{R}^m\) be a function. Then \(f\) is uniformly continuous on \(S\) if and only if \(\left\{ f(s_n ) \right\}\) becomes a Cauchy sequence whenever \(\left\{ s_n \right\}\) is a Cauchy sequence.

Theorem 3.4 can be used to show a given function is not uniformly continuous.

EXAMPLE 3.7. Let \(f : (0,\,1) \to \mathbb{R}\) and \(\left\{ s_n \right\}\) be defined by \[f(x) = \cos \frac{1}{x} \,\,\text{and}\,\, s_n = \frac{1}{n \pi} .\] Then the limit of \(\left\{s_n \right\}\) converges to \(0\) and \(\left\{s_n \right\}\) is a Cauchy sequence. But \(\left\{ f(s_n ) \right\}\) is not a Cauchy sequence, for \[f(s_n ) = \cos n \pi = \begin{cases} 1 \quad & \text{if} \,\, n \,\,\text{is even},\\[8pt] -1 \quad & \text{if} \,\, n \,\,\text{is odd} \end{cases} \] diverges as \(n \to +\infty.\) Thus \(f\) is not uniformly continuous.

REMARK 3.8. As we see in Example 3.7, the condition of 'Cauchy sequence' in theorem 3.4 cannot be weaken as 'sequence.'

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