If $f$ is a real-valued uniformly continuous function on $A$, then for every Cauchy sequence $(x_n)$ in $A$, $(f(x_n))$ is a real Cauchy sequence.

But why do we need the uniform continuity of $f$ for this? If $f$ is continuous on $A$ then for any point $c$ in $A$, and any Cauchy sequence $(x_n)$ converging to $c$, $(f(x_n))$ converges to $f(c)$, thus making it a Cauchy sequence.


marked as duplicate by Pedro, tomasz, The Phenotype, Shailesh, Leucippus Jan 28 '18 at 0:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ A continuous function maps convergent sequences to Cauchy sequences. But - for general domain $A$ - not all Cauchy sequences are convergent. $\endgroup$ – Daniel Fischer Jan 27 '18 at 15:52
  • $\begingroup$ @Pedro thank you for the link $\endgroup$ – Hrit Roy Jan 27 '18 at 16:04

Take $f(x)=\dfrac{1}{x}$ on $(0,\infty)$ which is continuous but not uniformly and $a_n=\dfrac{1}{n}$

  • $\begingroup$ Oh. So if the point of convergence of $(a_n)$ lies outside the domain we may face this issue. $\endgroup$ – Hrit Roy Jan 27 '18 at 15:55
  • $\begingroup$ In fact we need compactness in this case and what you argued is true... $\endgroup$ – Mostafa Ayaz Jan 27 '18 at 15:57

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