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

Solution 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.

Solution 2:

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.