7
$\begingroup$

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.

  • 0
    See also: http://mathoverflow.net/questions/27901/does-cauchy-continuity-imply-uniform-continuity-no2015-05-27

1 Answers 1

8

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 $\Bbb R$.

  • 0
    I was just writing the exact same example. You need to shift the sequence as the first term is not in $A$.2012-10-05
  • 0
    @Michael: Thanks for catching it. (I expanded $A$ instead.)2012-10-05
  • 0
    Why do you note the sequence with angle brackets?2012-10-05
  • 0
    @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.2012-10-05
  • 0
    @BrianM.Scott OK. Just curious.2012-10-05