Let $A\subset$ $\mathbb{R}^n$ ($A\neq \emptyset$) and $f: \mathbb{A} \to \mathbb{R}$ uniformly continuous. Prove:
$a)$ If $\left\{x_n\right\}$ is a Cauchy sequence of points of $A$ $\implies$ $\left\{f(x_n)\right\}$ is a Cauchy sequence in $\mathbb{R}$.
$b)$ $\exists !$ continuous function $g:{\overline{A}}$ $\rightarrow$ $\mathbb{R}$ $|$ $g(a) = f(a)$ $\forall a \in A$
for the first
Let $\epsilon>0$ given.
$f$ is uniformly continuous at $A$ so $ \exists \eta>0\;:\; \forall x,y\in A $
$|x-y|<\eta\implies |f(x)-f(y)|<\epsilon$
$(x_n)$ is Cauchy $\implies$
$\exists N>0\;:\; \forall p>q>N |x_p-x_q|<\eta$
$\implies \forall p>q>N \;\;|f(x_p)-f(x_q)|<\epsilon$. qed.
for the second use the fact that
$x\in\overline{A}\implies x=\lim_{n\to+\infty}a_n$ with $a_n\in A$.