Exercise 4.13

Following the hint, for $p\in X$ let $V_n(p)$ be the set of $q\in E$ such that $d(p,q)<1∕n$. By Exercise 9, there is an integer $N$ such that $\mathit{diamf}\left(V_N(p)\right)<1$ so the closure $F_n$ of $f\left(V_n(p)\right)$ for $n\ge N$ is a compact subset of $R$. Since $F_N\supset F_{N+1}\supset \cdots$, the intersection ${\text{∩}}_N^{\infty }{F}_n$ is nonempty by Theorem 2.36. This intersection consists of a single point. For suppose $x,y\in \cap F_n$ and $|x-y|=𝜀>0$. Let $n\ge N$ be large enough so that if $q_1,q_2\in V_n(p)$ then $|f(q_1)-f(q_2)|<𝜀∕3$. Since both $x$ and $y$ are in $F_n$, they have open neighborhoods of radius $𝜀∕3$ which intersect $f\left(V_n(p)\right)$, let $q_1,q_2\in V_n(p)$ such that $f(q_1)$ and $f(q_2)$ are in these neighborhoods of $x$ and $y$, respectively. Then

which is a contradiction.

Let $g(p)$ be this single point in $\cap F_n$. If $p\in E$, then $f(p)\in F_n$ for all $n$, so $g(p)=f(p)$, hence $g$ is an extension of $f$ to all of $X$. To show that $g$ is continuous, let $𝜀>0$. Since $f$ is uniformly continuous on $E$, there is $\delta >0$ so that $|f(q_1)-f(q_2)|<𝜀∕3$ if $q_1,q_2\in E$ and $d(q_1,q_2)<\delta$. Let $p_1,p_2\in X$ such that $d(p_1,p_2)<\delta ∕3$, and let $n$ be a large enough integer so that $1∕n<\delta ∕3$ and large enough so that, if $q_1\in V_n(p_1)$ and $q_2\in V_n(p_2)$, then $|g(p_1)-f(q_1)|<𝜀∕3$ and $|g(p_2)-f(q_2)|<𝜀∕3$. Then

so that

Hence $g$ is uniformly continuous on $X$.

The only topological property of $R$ that was used was that closed and bounded sets are compact, so the assertion can probably also be extended to $R^k$. Also, the $F_n$ will also be compact if the target space is compact, so the assertion is probably also true for that case. For a complete metric space, the assertion was proved in the answer to Exercise 11. The assertion is probably not true for a general metric space, but I can’t come up with an example.