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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 $g:{\overline{A}}$ $\rightarrow$ $\mathbb{R}$ $|$ $g(a) = f(a)$ $\forall a \in A$

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1 Answers 1

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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 $a_n\in A$.