I'm trying to understand an alternative proof of the idea that if $E$ is a dense subset of a metric space $X$, and $f\colon E\to\mathbb{R}$ is uniformly continuous, then $f$ has a uniform continuous extension to $X$.
I think I know how to do this using Cauchy sequences, but there is this suggested alternative. For each $p\in X$, let $V_n(p)$ be the set of $q\in E$ such that $d(p,q)<\frac{1}{n}$. Then prove that the intersections of the closures
$$
A=\bigcap_{n=1}^\infty\overline{f(V_n(p))}
$$
consists of a single point, $g(p)$, and so $g$ is the desired continuous extension of $f$. Why is this intersection a single point, and why is $g$ continuous?
This is what I did so far. Since $f$ is uniformly continuous, for given $\epsilon>0$, there is $\delta>0$ such that $\text{diam }f(V)<\epsilon$ whenever $\text{diam }V<\delta$. Since $V_n(p)$ has diameter at most $\frac{2}{n}$, taking $n>2/\delta$ would imply
$$
\text{diam }f(V_n(p))=\text{diam }\overline{f(V_n(p))}<\epsilon
$$
So I think $\lim_{n\to\infty}\text{diam }\overline{f(V_n(p))}=0$, which would imply $A$ consists of at most one point. I noticed that the closures form a descending sequence of closed sets, but I couldn't tell if they are bounded since $X$ is an arbitrary metric space, in order to conclude that the intersection is nonempty, and hence a single point.
Lastly, why is $g$ continuous at points $p\in X\setminus E$? I was trying to think of an argument with sequences converging to $p$ since $p$ is a limit point of $E$, but got stumping on how to show $g$ is actually continuous. Thanks.
Best Answer
I had a lot of help on this question in chat from users Srivatsan and t.b. the other day. I tried my best to write up what was said as an answer here.
Notice that the sets $\overline{f(V_n(p))}\supseteq\overline{f(V_{n+1}(p))}\supseteq\cdots$ form a nested sequence of closed sets. Moreover, let $\epsilon>0$ be given. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d(p,q) < \delta$ implies $d(f(p),f(q))<\epsilon$ for $p,q\in E$. Taking $n$ large enough so that $\frac{2}{n}<\delta$, then for $q,r\in V_n(p)$, $$ d(q,r)<d(q,p)+d(p,r)<\frac{2}{n}<\delta $$ so $d(f(q),f(r))<\epsilon$. Thus $f(V_n(p))$ is bounded in $\mathbb{R}$, so $\overline{f(V_n(p))}$ is bounded as well. Hence for large enough $n$ the sets form a compact nested sequence. Since $V_n(p)$ has diameter at most $\frac{2}{n}$, taking $n>2/\delta$ would imply $$ \operatorname{diam } f(V_n(p))=\operatorname{diam }\overline{f(V_n(p))}<\epsilon $$ So $\lim_{n\to\infty}\operatorname{diam }\overline{f(V_n(p))}=0$, and thus their intersection consists of a single point. Also, since $\operatorname{diam }f(V_n(p))\to 0$ as $n\to\infty$, and so by choosing points arbitrarily close to $p$, their images under $g$ are arbitrarily close to $g(p)$. (To be more explicit, letting $\delta$ be small enough such that for $x,y\in E$, then $d(x,y)<2\delta$ implies $d(f(x),f(y))<\epsilon/3$, choose $n$ large enough that $\frac{1}{n}<\delta$, and thus for any $x,y\in V_n(p)$, $d(x,y)<2/n<2\delta$, so $\operatorname{diam }f(V_n(p))<2\epsilon/3$, so $d(f(x),g(p))<2\epsilon/3$. Note also that this can be done for any $p$.)
I contend that $g$ is uniformly continuous. Let $\epsilon>0$ be given. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d(r,s)<\delta$ implies $d(f(r),f(s))<\epsilon/3$. Now let $p,q\in X$ be any points such that $d(p,q)<\delta/3$. By the above reasoning, choose $n$ large enough so that $n>\frac{3}{\delta}$, and both $d(g(r),g(p))<\epsilon/3$ and $d(g(s),g(q))<\epsilon/3$ for $r\in V_n(p)$ and $s\in V_n(q)$. Also, $$ d(r,s)<d(r,p)+d(p,q)+d(q,s)<\delta $$ so $d(f(r),f(s))=d(g(r),g(s))<\epsilon/3$. By the triangle inequality, $d(g(p),g(q))<\epsilon$, so $g$ is uniformly continuous, and thus continuous on $X$, and of course $g|_E=f$.