Let $(X,\rho)$ and $(Y,\sigma)$ be two metric spaces. Assume ${x_n}$ is Cauchy in $X$, and that $f:X \rightarrow Y$ is uniformly continuous. Prove that $f(x_n)$ is Cauchy in $Y$.
Take $\frac{\epsilon}{3}>0$ there exists $\delta>0$ such that if $\rho(x,y)<\delta$ then $\sigma(f(x),f(y))<\frac{\epsilon}{3}$ since the function is uniformly continuous.
Since ${x_n}$ is Cauchy for $\frac{\epsilon}{3}>0$ there exists $N$ such that $\forall n \geq N, \rho(x_n,x)<\frac{\epsilon}{3}$
At this point I want to use the triangle inequality to say something like $\sigma (f((x_n),f(x))) \leq \sigma(f(x),f(y)) + \rho(x_n,x)...\leq \epsilon $
I know I'm missing something here, but I can't think of anything to tie it together. Any advice would be appreciated.
Let $\epsilon > 0$ be given. There is a $\delta > 0$ such that $\sigma(f(x),f(y)) < \epsilon$ if $x \in X$, $y \in X$ and $\rho(x,y)<\delta$. If $(x_n)_{n=1}^\infty$ is a Cauchy sequence in $X$, there exists an integer $N$ such that $\rho(x_m,x_n)<\delta$ whenever $m$ and $n$ are $\geqq N$. Hence, $m \geqq N$ and $n \geqq N$ imply $\sigma(f(x_m),f(x_n))<\epsilon$. Thus $(f(x_n))_{n=1}^\infty$ is Cauchy sequence in $Y$.