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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?

Martin Sleziak
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3 Answers3

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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.

P..
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  • If a sequence diverges, then is it necessarily non-Cauchy? – Error 404 May 21 '15 at 09:48
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    @VikrantDesai This is true only when the metric space $X$ is complete. – P.. May 22 '15 at 08:13
  • I didn't understand the latter part of your proof. I mean why $ (f(b_n))_{n \in \Bbb N} $ is not Cauchy? – Error 404 May 22 '15 at 08:16
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    @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.. May 22 '15 at 08:19
  • Got you now. Nice proof. – Error 404 May 22 '15 at 08:23
  • Just for the curiosity I want to ask that why we took a specific Cauchy sequence? It should be true for all Cauchy sequences right? – Error 404 May 22 '15 at 08:26
  • For which one do you mean specific. The $(a_n){n\in\mathbb N}$ or the $(b_n){n\in\mathbb N}$? If you mean the $(b_n){n\in\mathbb N}$ then the answer is No is not true for all the Cauchy sequences. For example if you take a constant sequence then is not true. The $(a_n){n\in\mathbb N}$ is not specific is general. – P.. May 22 '15 at 08:29
  • I meant $ (b_n)_{n \in \Bbb N} $. I said so because we want to prove that "If $f$ sends Cauchy sequences to Cauchy sequences then $f$ is continuous." so we proved it's contrapositive which is "If $f$ is not continuous then it doesn't send any Cauchy sequence to a Cauchy sequence". Is my thinking right? – Error 404 May 22 '15 at 08:34
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    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.. May 22 '15 at 08:38
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    Now all of my doubts have been cleared. :) – Error 404 May 22 '15 at 08:39
  • Under what conditions can one tell

    : Cauchy Continuous $\implies$ Uniformly Continuous?

     – Saikat Dec 11 '20 at 08:39
  • I thing you are wrong because you should use the axiom of choice to construct $a_{n}$ by constructing a new sequence in the codomain of $f$. – maths and chess Jun 22 '23 at 14:33
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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.

CHOUDHARY bhim sen
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  • Note that if you take any Cauchy sequence, it is not necessarily contained in a closed and bounded subset of the space.

    For example, let $X = (0, \infty)$ and $Y = \mathbb{R}$, both equipped with the usual metric. Define $f : X \to Y$ by $f(x) = x^2$ for all $x \in X$. Now consider the sequence ${ \frac{1}{n} }_{n \in \mathbb{N}}$ in $X$.

    This sequence is Cauchy, but it is not contained in any closed and bounded subset of $X$.

     – Hamza Ayub Jun 13 '25 at 18:35
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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.

Infinite
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