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Let $X,Y$ be metric spaces and let $f: X \to Y$ be uniformly continuous. If $(x_n)$ is a Cauchy sequence in $X$, show that $(f(x_n))$ is a Cauchy sequence in $Y$. On the other hand, if the assumption of uniform continuity is dropped, the result is false: give an example of a continuous function $f : (0, 1) \to \mathbb{R}$ that does not map Cauchy sequences to Cauchy sequences.

I don't know where to start with this. Any hints?

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  • $\begingroup$ This has been asked many times on this site; even a quick look at the related questions will provide all the answers to your questions. $\endgroup$ – Alex Provost Mar 11 '16 at 17:05
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Let $\epsilon > 0$ and choose $\delta > 0$ so $d_X(x,y) < \delta \implies d_Y(f(x), f(y)) < \epsilon$. Since the sequence $\{x_n\}$ is cauchy, you can choose $N\in \mathbb{N}$ so that m,$n\ge N\implies d_X(x_m, x_n)< \delta$. We now see that $d_Y(f(x_m), f(x_n) < \epsilon$ for $m, n \ge N$.

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Hint: You want to prove that $(f(x_n))_{n \geq 1}$ is a Cauchy sequence. Then use the definition: take $\epsilon > 0$, pick the suitable $\delta$ (that depends only on $\epsilon$) using uniform continuity, and for that $\delta$, pick the suitable $n_0$ using the definition of $(x_n)_{n \geq 1}$ being a Cauchy sequence. Everything will fit.

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