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Suppose that $(\mathcal X,d)$ and $(\mathcal Y, \rho)$ are two metric spaces.

I am trying to show that if $f: (\mathcal X,d) \mapsto (\mathcal Y, \rho) $ is uniformly continuous and $(x_n)_n$ is a Cauchy Sequence in $\mathcal X$, show that $(f(x_n))_n$ is a Cauchy sequence in $\mathcal Y$.

A sequence $X = (x_n)$ of real numbers is said to be a Cauchy Sequence if: $\forall \epsilon > 0, \exists$ $H(\epsilon)\in \Bbb N$ such that $\forall$ $ m,n \ge H(\epsilon), |x_n - x_m| < \epsilon $

I'm just not sure how to begin, any pointers are appreciated!

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  • $\begingroup$ What is the definition of uniform continuity? $\endgroup$ – Yunus Syed Dec 2 '15 at 3:19
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Begin by letting $\epsilon >0$. Get a $\delta >0$ by uniform continuity of $f$. Use this $\delta$ to get an $H(\epsilon)$ for the Cauchy sequence $(x_n)$. See if the same $H(\epsilon)$ will work to conclude that $(f(x_n))$ is Cauchy.

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Let $\epsilon>0$. Since $f$ is uniformly continuous, there is a $\delta>0$ such that $d(x,y)<\delta\implies\rho (f(x),f(y))<\epsilon$ for any $x,y\in X$. Since, $\{x_n\}$ is Cauchy, there is an $N\in\Bbb N$ for which $n,m\ge N\implies d(x_n,x_m)<\delta$. Can you get it from here?

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