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It is well known that if $X$ is a first countable topological space and $Y$ is a topological space, then $f : X \rightarrow Y$ is continuous iff $$\forall x \in {\rm map}(\mathbb{N},X),\forall p \in X \quad x_{n} \rightarrow p \Rightarrow f(x_{n}) \rightarrow f(p)$$

It is also well known that if $X$ and $Y$ are metric spaces and $f : X \rightarrow Y$ is uniformly continuous, then $f$ maps Cauchy sequences to Cauchy sequences.

By analogy it seems plausible that if a function between metric spaces maps Cauchy sequences to Cauchy sequences then it must be uniformly continuous. However mimicking the proof of the analogous result for continuous maps doesn't work, which makes me think the result if false. Does anyone know any counterexamples?

Also on the uniform continuity wikipedia page, it says that the result is true if $X$ and $Y$ are subsets of $\mathbb{R}^{n}$. EDIT: It actually doesn't say this, I misread the page.

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    $\begingroup$ I wrote the portion of the wikipedia page en.wikipedia.org/wiki/Uniform_continuity dealing with the extension problem and the condition of Cauchy continuity. Formerly the page only implied (in the usual mathematical sense, but also in the conventional sense of suggested, rather than said explicitly) that Cauchy continuity does not imply uniform continuity, since any continuous, non-uniformly continuous function on a complete space gives a counterexample, and such a function, $x \mapsto x^2$, was provided. I just edited the article to be more explicit on this point. $\endgroup$
    – Pete L. Clark
    Commented Jun 12, 2010 at 17:08
  • $\begingroup$ @Daniel Barter: you write "it says that the result is true if $X$ and $Y$ are subsets of $\mathbb{R}^n$". Please be more specific, so we can tell if there is an error and fix it if necessary: which page says that what result is true? $\endgroup$
    – Pete L. Clark
    Commented Jun 12, 2010 at 17:09
  • $\begingroup$ Summary of the question: "A implies B", by analogy it's plausible that "B implies A". Is it true? A="function is uniformly continuous", B="function is Cauchy continuous". $\endgroup$
    – Victor Protsak
    Commented Jun 12, 2010 at 21:51
  • $\begingroup$ @Pete: I was under the impression that the uniform continuity wikipedia page said that if a function between two subsets of euclidean space mapped Cauchy sequences to Cauchy sequences then it was uniformly continuous. However i misread. $\endgroup$
    – Daniel Barter
    Commented Jun 12, 2010 at 23:50

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No it's not true.

f(x) = x^2 on whole real line.

It maps Cauchy sequences to Cauchy sequences but it's not uniformly continuous on the whole real line.

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Take for $X$ the disjoint union of a sequence $A_n$ of two point sets. Set $d(p,q)=1/n$ if $p$ and $q$ are different points in $A_n$ and $d(p,q)=1$ if $p\in A_n$ and $q\in A_m$ with $m\not=n$. Every Cauchy sequence in $X$ is eventually constant. Consider the identity mapping from $X$ with this metric to $X$ with the discrete metric.

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    $\begingroup$ X in this metric is also a discrete space, as the ball around x with radius 1/n, when x is in A_n, is {x}. Any function from a discrete space is continuous. $\endgroup$
    – Henno Brandsma
    Commented Jun 12, 2010 at 6:51
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    $\begingroup$ It is a counterexample to the statement "a map that preserves Cauchy sequences needs to be uniformly continuous", as required by the OP. The above counterexample shows that the statement fails to be true even with the identity map on the same topological space, with respect to 2 different metrics. $\endgroup$
    – Pietro Majer
    Commented Jun 12, 2010 at 7:11
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Several people have already given examples to the effect that preservation of Cauchyness is not enough to prove that a map is uniformly continuous. It is still possible however, to characterize uniform continuity in terms of sequences (for metric spaces only, for uniform spaces you would need nets (or filters)). In case you are interested here goes the result.

Theorem: Let $f:X\to Y$ be a map between metric spaces (both metrics denoted by $d$). Then $f$ is uniformly continuous iff, for every pair of sequences $(x_n)$ and $(z_n)$ in $X$ such that $d(x_n, z_n)$ converges to $0$, then $d(f(x_n), f(z_n))$ converges to zero.

Proof: exercise left to the reader.

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Well, here is [most probably another] Wikipedia page: http://en.wikipedia.org/wiki/Cauchy-continuous_function. HTH.

Alternatively, you may use the function $f:\mathbb{R\rightarrow\mathbb{R}}$ , expressed by $f\left(x\right)=x^{2}$ to show that it is possible for a continuous function to send Cauchy sequences to Cauchy sequences without being uniformly continuous.

As for the continuity question: the function must be continuous. For, embed $Y$ into its completion, say $Y^{\sim}$, let $p\in X$ , and let $\left(x_{n}\right)$ be a sequence in $X$ converging to $p$ . Then the sequence $\left(x_{1}, p, x_{2}, p,...\right)$ is Cauchy in $X$ , isn't it ? And, therefore, its image by $f$ should be a Cauchy sequence in $Y$ , hence convergent in $Y^{\sim}$ . Yet, that image contains a constant subsequence... isn't it ?

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  • $\begingroup$ I am a bit sceptical about this page, for two reasons. No link to a proof is given and secondly, they have hyper linked the term "if and only if"... $\endgroup$
    – Daniel Barter
    Commented Jun 12, 2010 at 5:49
  • $\begingroup$ @Daniel: It is certainly a good idea to be somewhat skeptical about wikipedia articles. But this one seems basically sound. They do give a reference, to a book even. Regarding the hyperlink, well, the style and quality of the writing on wikipedia varies, to put it mildly. $\endgroup$
    – Harald Hanche-Olsen
    Commented Jun 12, 2010 at 11:54
  • $\begingroup$ An answer which, in fact, is not a truly answer, is over voted, due to the name of the author. Some annoying comments about this there were discreetly erased. This is due to the "young wolves" mentality: "Let’s catch a big name for our site". Another answer, which is correct, but partial, and following my answer and being included in this, is also fully voted. Still, my answer is not yet voted. In conclusion, I'll give myself a "-1", just as a protest, not because I'm frustated, but since I'm sickened of hypocrisy. $\endgroup$
    – Ady
    Commented Jun 12, 2010 at 12:09
  • $\begingroup$ OK, since I can't, I'll ask someone downvoting me. $\endgroup$
    – Ady
    Commented Jun 12, 2010 at 12:11
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    $\begingroup$ You are right to be critical of my answer, Ady, as $x^2$ is quicker and more elementary. I was thinking that the implication is true for totally bounded spaces and wanted a more or less minimal and simple counterexample for bounded (and complete) spaces. $\endgroup$
    – Bill Johnson
    Commented Jun 12, 2010 at 15:49

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