Let $I$ be an interval and let $f: I\to \mathbb{ R}$ be uniformly continuous on I. Suppose that $\{a_n\}$ is a Cauchy sequence in $I$. Prove that $\{f(a_n)\} $is a Cauchy sequence.
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1Your title doesn't make sense. The question does, however. Write down the definitions of a Cauchy sequence and a uniformly continuous function right next to each other. Look at them for a bit. – Daniel Fischer Nov 03 '13 at 13:51
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How do I prove a uniformly continuous function preserves Cauchy sequences? – Martin Sleziak Nov 03 '13 at 15:32
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Fix $\epsilon>0$, Since $f$ is uniformly continuous,$\exists \delta>0 \ni |f(x)-f(y)|<\epsilon$ Whenever $|x-y|<\delta$.
$a_n$ is Cauchy, so $\exists N\in\mathbb{N}\ni |a_n-a_m|<\delta \forall m,n>N$
So $|f(a_n)-f(a_m)|<\epsilon \forall m,n>N$. So $\{f(a_n)\}$ is Cauchy
Myshkin
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