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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.

Thanks for your help.

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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 $\Bbb R$.

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  • $\begingroup$ I was just writing the exact same example. You need to shift the sequence as the first term is not in $A$. $\endgroup$ – Michael Albanese Oct 5 '12 at 3:17
  • $\begingroup$ @Michael: Thanks for catching it. (I expanded $A$ instead.) $\endgroup$ – Brian M. Scott Oct 5 '12 at 3:21
  • $\begingroup$ Why do you note the sequence with angle brackets? $\endgroup$ – Pedro Tamaroff Oct 5 '12 at 3:26
  • $\begingroup$ @Peter: Because I like the convention that indicates sequences and tuples with angle brackets: curly braces are only for sets, and parentheses are already overloaded. I probably picked it up from the set theorists when I was in grad school. $\endgroup$ – Brian M. Scott Oct 5 '12 at 3:53
  • $\begingroup$ @BrianM.Scott OK. Just curious. $\endgroup$ – Pedro Tamaroff Oct 5 '12 at 3:57

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