10
$\begingroup$

Could you please provide an example for two metric spaces $X,Y$, a continuous function $f$ that maps $X$ to $Y$ and a Cauchy sequence in $X$, which is not mapped to a Cauchy sequence in $Y$ by $f$?

Does $f(x) = \frac{1}{x}$ work if $X$ is any metric space and $Y$ is the set of real numbers?

$\endgroup$
  • 1
    $\begingroup$ No, it doesn't work for any space, but you just need one example right? Take $X = (0,1), Y = \mathbb{R}$ and $f : x \mapsto 1/x$ $\endgroup$ – Prahlad Vaidyanathan Oct 22 '13 at 17:03
  • 1
    $\begingroup$ A function between metric spaces is called Cauchy-continuous if it takes Cauchy sequences to Cauchy sequences. Every Cauchy continuous function is continuous. The converse holds if $X$ is complete. $\endgroup$ – Stefan Hamcke Oct 22 '13 at 17:54
  • 1
    $\begingroup$ If $f$ is uniformly continuous between metric spaces, then Cauchy sequences are mapped to Cauchy sequences. $\endgroup$ – Ron Apr 30 '17 at 15:52
11
$\begingroup$

$X=(0,1),Y=\mathbb{R},f(x)={1\over x}, {1\over n}$ is cauchy in $X$ but $f({1\over n})=n$ which is not cauchy in $\mathbb{R}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.