Answer to Question #217022 in Real Analysis for Prathibha Rose

Question

Prove that a uniformly continuous function from metric space to another metric space takes cauchy to cauchy sequences

Answer 1

Let $f: X\to Y$ be a uniformly continuous function from a metric space $(X,d_X)$ to another metric space $(Y,d_Y).$ Let $(x_n)_{n=1}^{\infty}$ be a Cauchy sequence. Let $\varepsilon$ be arbitrary positive real number. Since the function $f$ is uniformly continuous, there exists $\delta>0$ such that for any $x,x'\in X$ the inequality $d_X(x,x')<\delta$ implies $d_Y(f(x),f(x'))<\varepsilon.$ Since $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence, for this $\delta$ there exists $n\in\N$ such that $d_X(x_k,x_m)<\delta$ for any $k\ge n, m\ge n.$ It follows that $d_Y(f(x_k),f(x_m))<\varepsilon$ for any $k\ge n, m\ge n.$ We conclude that for any $\varepsilon>0$ there exists $n\in\N$ such that $d_Y(f(x_k),f(x_m))<\varepsilon$ for any $k\ge n, m\ge n,$ and hence $(f(x_n))_{n=1}^{\infty}$ is a Cauchy sequence as well.