Is a Cauchy sequence - preserving (continuous) function is (uniformly) continuous?

Question

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?

Answer 1

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.

Answer 2

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.