If $A$ is compact and $f : A \to \mathbb{R}$ is continuous and $(x_n) $ is Cauchy in $A$ show that $(f(x_n))$ is Cauchy and moreover show that if $A = \mathbb{R}$, $(f(x_n))$ is Cauchy
I can see this is leading to uniform continuity but we haven't covered that yet. Just making sure I have the correct idea. thanks.
Since $(x_n)$ is Cauchy in $A$, $\forall \delta > 0, \exists N \in \mathbb{N} \,\,s.t. n,m \ge N \implies |x_n-x_m| < \delta$ and if $A$ is compact then $A$ is closed and bounded, which gives that $(x_n)$ converges in $A$ (b/c Cauchy implies convergent)
since $f$ is continuous and $A$ is compact then $f(A)$ is compact. So it follows the $f(x_n), f(x_m) \in f(A)$, $(x_n \in A)(\forall \epsilon > 0) (\exists \delta > 0 \wedge x_m \in A) \implies|f(x_n)-f(x_m)| < \epsilon$ So $f(x_n)$ is Cauchy
And if $A = \mathbb{R}$ then this seems to follow from the above but since $\mathbb{R}$ is not compact, just use that $(x_n)$ is Cauchy implies it is convergent then use a $\delta - \epsilon$ proof and any convergent sequence in the reals is Cauchy.