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Suppose $f$ is a uniformly continuous mapping of a metric space $X$ into a metric space $Y$ and prove that $\{f(x_n)\}$ is a Cauchy sequence in $Y$ for every Cauchy sequence $\{x_n\}$ in $X$.

Proof: Let $\varepsilon>0$ be given then $\exists \delta>0$ such that $\forall x,y\in X$ and $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\varepsilon.$

Since $\{x_n\}$ is a Cauchy sequence then for above $\delta>0$ $\exists N$ such that $n,m\geqslant N$ implies $d_X(x_n,x_m)<\delta.$

Combine these statements we get: for given $\varepsilon>0$ $\exists N$ such that $n,m\geqslant N$ implies $d_Y(f(x_n),f(x_m))<\varepsilon.$

Is my proof true?

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    $\begingroup$ Yes. you are right RFZ. $\endgroup$ – Groups Sep 25 '15 at 11:33
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Your proof is correct and very clear.

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