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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.

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

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

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  • $\begingroup$ 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

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