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Let $(X,d)$ and $(Y,\rho)$ be metric spaces and $f:X\to Y$ be a function and suppose for any Cauchy sequence $(a_n)$ in $X$, $(f(a_n))$ is a Cauchy sequence in $Y$.

Is $f$ continuous?

Let $f$ be continuous, is it uniformly continuous?

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Yes if $f$ sends Cauchy sequences to Cauchy sequences then it is continuous:

Let $x\in X$. Assume for the sake of contradiction that $f$ is not continuous at $x$. Then exists an $\epsilon>0$ and a sequence $(a_n)_{n\in\mathbb N}$ in $X$ such that $a_n\rightarrow x$ but $\rho(f(a_n),f(x))>\epsilon$ for all $n\in\mathbb N$.

To finish the proof consider the sequence $$ b_n= \begin{cases} a_n, \ n\text{ even},\\ \\ x, \ n\text{ odd}. \end{cases} $$ The sequence $(b_n)_{n\in\mathbb N}$ is Cauchy but $(f(b_n))_{n\in\mathbb N}$ it isn't.

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  • $\begingroup$ If a sequence diverges, then is it necessarily non-Cauchy? $\endgroup$ – Error 404 May 21 '15 at 9:48
  • $\begingroup$ @VikrantDesai This is true only when the metric space $X$ is complete. $\endgroup$ – P.. May 22 '15 at 8:13
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    $\begingroup$ @VikrantDesai The sequence $(f(b_n))_{n\in\mathbb N}$ is not Cauchy because $\rho(f(b_{2n}),f(b_{2n-1}))=\rho(f(a_{2n}),f(x))>\epsilon$ for all $n\in\mathbb N$. $\endgroup$ – P.. May 22 '15 at 8:19
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    $\begingroup$ No the contrapositive is "If $f$ is not continuous then it is not true that 'sends Cauchy sequences to Cauchy sequences'" or equivalently "If $f$ is not continuous then there is at least one Cauchy sequence $(a_n)_{n\in\mathbb N}$ such that $(f(a_n))_{n\in\mathbb N}$ is not a Cauchy sequence" $\endgroup$ – P.. May 22 '15 at 8:38
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    $\begingroup$ Now all of my doubts have been cleared. :) $\endgroup$ – Error 404 May 22 '15 at 8:39
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No it is not necessary to uniformly continuous. Take function $ f(x) =x^2 $ on real line. note that if you take any cauchy sequence then it is contained in some closed bounded interval and there function is uniformly continuous so image sequence must cauchy. But function is not uniformly continuous on whole real line.

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