# How do I prove a uniformly continuous function preserves Cauchy sequences?

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

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Just to clarify, I'm assuming when you say "$A$ of $\Bbb{R}$" you mean "a subset $A$ of $\Bbb{R}$". – 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 $n,m>N$.
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.
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? – Ben Millwood Dec 6 '12 at 21:58
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 $(a_n)$). – icurays1 Dec 7 '12 at 1:58