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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