# Continuous Functions and Cauchy Sequences

We know that if a function $f: A \mapsto \mathbb{R}$, $A \subseteq \mathbb{R}$, is uniformly continuous on $A$ then, if $(x_n)$ is a Cauchy sequence in $A$, then $(f(x_n))$ is also a Cauchy sequence.

I would like an example of continuous function $g: A \mapsto \mathbb{R}$ such that for a Cauchy sequence $(x_n)$ in $A$, it is not true that $f(x_n)$ is a Cauchy sequence.

Take $A=(0,1]$ with the usual metric and $f:(0,1]\to\Bbb R:x\mapsto \frac1x$; the sequence $$\left\langle \frac1n:n\in\Bbb Z^+\right\rangle$$ is Cauchy in $(0,1]$, but its image under $f$ is $\langle n:n\in\Bbb Z^+\rangle$, which is very far from being Cauchy in $\Bbb R$.
I was just writing the exact same example. You need to shift the sequence as the first term is not in $A$. – Michael Albanese Oct 5 '12 at 3:17
@Michael: Thanks for catching it. (I expanded $A$ instead.) – Brian M. Scott Oct 5 '12 at 3:21