Let $f$ be a uniformly continuous function on A of $\Bbb{R}$. How do I show that if $a_n$ is Cauchy, then $f(a_n)$ is Cauchy.

This is what I have worked on, but it does not quite make sense since I feel like I didn't really use the given condition that $f$ is uniformly continuous.

Let $\epsilon>0$, $f$ is uniformly continuous, so there exists$\delta>0$ st $|f(x)-f(y)|<\epsilon$ for $|x-y|<\delta$.

Since $a_n$ is Cauchy, there exists $N>0$ such that $|a_n-a_m|<\delta$ for $m,n>N$

Hence$|f(a_n)-f(a_m)|<\epsilon$ for $m,n>N$. So $f(a_n)$ is Cauchy

  • $\begingroup$ Just to clarify, I'm assuming when you say "$A$ of $\Bbb{R}$" you mean "a subset $A$ of $\Bbb{R}$". $\endgroup$ – icurays1 Dec 7 '12 at 2:03

Your proof looks good. You use uniform continuity to claim that for every $n,m>N$, $\vert f(a_n)-f(a_m)\vert<\epsilon$. If you only had continuity, $\delta$ would depend on $x$ and hence you couldn't claim that $\vert a_n-a_m\vert<\delta\Longrightarrow\vert f(a_n)-f(a_m)\vert<\epsilon$ for all $n,m>N$.

Additional comment

If $A$ is assumed to be closed, then uniform continuity is not necessary (only "pointwise" continuity). This is since if $A$ is closed, $(a_n)$ is guaranteed to converge to some $a\in A$. Then, $f$ will be continuous at $a$ and the proof can be modified accordingly without using uniform continuity.

  • $\begingroup$ Couldn't you use the fact that $\Bbb R$ is complete to say that $a_n$ Cauchy $\implies$ it converges, and then use the $\delta$ that works for its limit? $\endgroup$ – Ben Millwood Dec 6 '12 at 21:58
  • $\begingroup$ If I'm interpreting OP's language correctly, we only have $f$ defined on an arbitrary subset $A\subset\Bbb{R}$. For instance, $A$ could be $(0,1)$ which is not complete (hence $f$ might not be continuous at the limit point of $(a_n)$). $\endgroup$ – icurays1 Dec 7 '12 at 1:58
  • $\begingroup$ @icurays1: I do not agree with you idea. The uniformly continuous mapping necessarily preservs the Cauchy sequence from one metric space to another regardless of closed or open while pointwise continuous can't! The key point is that if A is closed with pointwise continuous, then {an} will still converge but not be a Cauchy sequence! $\endgroup$ – Bear and bunny Nov 24 '13 at 23:36
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    $\begingroup$ @icurays1 Yes, since a continuous function on a closed subset is always automatically uniformly continuous. You aren't really making the conditions less stringent, just rewording them. $\endgroup$ – Michael Angelo Apr 16 '16 at 8:31
  • $\begingroup$ @MichaelAngelo continuous function on closed set is uniformly continuous??? R is closed in R, no? $\endgroup$ – Hrit Roy Jun 12 at 14:23

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