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We know that under uniformly continuous function a Cauchy sequence goes to a Cauchy sequence. Let $f:A\rightarrow \mathbb{R}^m$ be uniformly continuous function where $A\subseteq\mathbb{R}^n$, I need to prove $\lim_{n\rightarrow\infty}f(x_n)$ exists where $x_n$ is a Cauchy sequence in $A$, well If I write coordinate wise of the element of $f(x_n)$ then coordinatewise each sequence is Cauchy and hence they will converge to some element in $\mathbb{R}^m$

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  • $\begingroup$ $\mathbb{R}^n$ is complete, so every Cauchy sequence is convergent $\endgroup$ – Norbert May 4 '12 at 6:26
  • $\begingroup$ yes that I should have written, :P $\endgroup$ – Marso May 4 '12 at 6:28
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Since $\{x_n\}$ is Cauchy in a complete metric space, it converges to some $\hat x$. Since $f$ is continuous, you have $\lim_{n \rightarrow \infty}f(x_n) = f(\hat x)$.

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