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Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$. If {$x_n$} is a Cauchy sequence in $X$, then {$f(x_n)$} is a Cauchy sequence in $Y$.

Give an example to show that this statement is not true if $f$ is continuous, but not uniformly continuous.

I was thinking maybe something like $\frac{x}{n}$? Not really sure if I understand this problem correctly. Any help would be great. Thank you.

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    $\begingroup$ For the first part, use what it means for a sequence to be Cauchy and a function to be uniformly continuous. For the second, you need to find metric spaces $X,Y$, a continuous function $f:X\to Y$, and a Cauchy sequence $(x_n)$ in $X$ such that $f(x_n)$ is not Cauchy in $Y$. $\endgroup$ – John Griffin Oct 19 '17 at 21:16
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Consider $a_n = \frac{1}{n}$ This is clearly cauchy, as it converges. Now consider $f: \mathbb{R/\{0\}} \to \mathbb{R/\{0\}}$, $$f(x) = \frac{1}{x}$$

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