Prove that if $\displaystyle f:A \to R $ is uniformly continuous on A and $\displaystyle x_n $ is a cauchy sequence in A then $\displaystyle f(x_n) $ is a cauchy sequence in R .
R=real numbers
Fix $\displaystyle \epsilon>0$. Then by uniform continuity, $\displaystyle \exists\delta>0$ s.t. $\displaystyle |x_n-x_m|<\delta \implies |f(x_n)-f(x_m)|<\epsilon$.
Since $\displaystyle \{x_n\}$ is Cauchy, $\displaystyle \forall\delta>0$, $\displaystyle \exists N$ s.t. $\displaystyle n,m>N \implies |x_n-x_m|<\delta$
So...