Is a Cauchy sequence - preserving (continuous) function is (uniformly) continuous? 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?
 A: 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. 
A: 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.
A: Consider $f:(0,1)\to \mathbb R$, defined by $f(x)=1/x$. $f$ is continuous. Sequence $1/n$ is Cauchy in $(0,1)$, while $f(1/n)=n$ is not a Cauchy sequence.
