# 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 good tame example of a counterexample to the second claim is $f(x)=x^2$ on $\mathbb{R}$. Commented Mar 28, 2013 at 5:39
• Answers are here. And more can be found here. Commented Mar 28, 2013 at 5:49
• Commented May 27, 2015 at 8:34

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.

• If a sequence diverges, then is it necessarily non-Cauchy? Commented May 21, 2015 at 9:48
• @VikrantDesai This is true only when the metric space $X$ is complete.
– P..
Commented May 22, 2015 at 8:13
• @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 $n\in\mathbb N$.
– P..
Commented May 22, 2015 at 8:19
• 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"
– P..
Commented May 22, 2015 at 8:38
• Now all of my doubts have been cleared. :) Commented May 22, 2015 at 8:39

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.

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.

• You take the question wrong! Commented Mar 4, 2021 at 9:26