Continuous extension of a uniformly continuous function from a dense subset.

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.

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Essentially by construction $g$ is uniformly continuous... (You've done the hard work, you should now just piece the information together). – t.b. Jan 5 '12 at 14:01
To answer why the sequence consists of bounded sets: any set with finite diameter is bounded, and the closure of a bounded set is again bounded. – Srivatsan Jan 5 '12 at 14:16
@Srivatsan Thanks, but isn't it true that the image of a bounded subset of an arbitrary metric space under a uniformly continuous function need not be bounded? I just looked it up on wiki, so I think my approach may be wrong there. – yunone Jan 5 '12 at 14:27
@yuone: The usual counterexample is a space from which you map endowed with the discrete metric. The way you contrive your intersection, the set E is obviously totally bounded (take any ball and the triangular inequality). The image of a uniformly continuous function from a totally bounded set is bounded. For overall continuity, use triangular inequality again. – gnometorule Jan 5 '12 at 15:08
Congrats: Two badges + 10k'er. – Asaf Karagila Aug 16 '12 at 19:31

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 $g|_E=f$.

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I follow what you are saying, but what is this function $g$? – user38268 Jan 8 '12 at 8:03
@BenjaminLim The value of the function $g$ on a point $p\in X$ is determined by the explanation in the body of question. The goal is that $g$ is a continuous extension of $f$ to all of $X$, where $f$ happens to only be defined on a dense subset of $X$. – yunone Jan 8 '12 at 8:21
@BenjaminLim Since $E$ is dense, and $p\in X$ is either in $E$ or a limit point of $E$, so the $V_n(p)$ are nonempty for every $n$, and an application of essentially the nested intervals theorem shows that $\bigcap_{n=1}^\infty\overline{f(V_n(p))}$ is a singleton. So defining $g(p)$ to be this point is a well defined function on $X$. – yunone Jan 8 '12 at 8:35

As Srivatsan notes, any set with a finite diameter is bounded. You have shown that for a fixed $\varepsilon$, ${\rm diam} \overline {f(V_n(p))} < \varepsilon$, starting from some $n$. So starting from some $n$, $\overline {f(V_n(p))}$ are all bounded and hence compact. Furthermore the image of a real uniformly continuous function on the bounded set is bounded. This is Exercise 4.8 in Rudin.

For continuity on $p \in X$ you do the following. For $\varepsilon > 0$ you essentially want a $\delta = {1 \over n}$ so small that ${\rm diam} \overline {f(V_n(p))} < \varepsilon$. Now if $q \in X$ is such that $d(p, q) < {1 \over n}$, then you can pick $m$ so large that $V_m(q) \subseteq V_n(p)$. Then $\overline {f(V_m(q))} \subseteq \overline {f(V_n(p))}$ and so $g(p), g(q) \in \overline {f(V_n(p))}$. This implies that $d(g(p), g(q)) < \varepsilon$. By uniform continuity of $f$ you can pick $\delta$ independent of $p$ and this gives you uniform continuity.

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Thanks for your answer, Levon. – yunone Jan 7 '12 at 13:56

Here is a perhaps an alternative way of looking at what happens before you take the infinite intersection. We know that $\textrm{diam} V_n(p) < \frac{2}{n}$. So by uniform continuity of $f$, for all $\epsilon > 0$ there exists $\delta > 0$ (and hence an $n\in \Bbb{N}$ such that $0 < \frac{1}{2n} < \delta$ by the Archimedean property) such that

$$\textrm{diam} f(V_n(p)) < \epsilon.$$

In other words we have that $\textrm{diam} f(V_n(p)) \rightarrow 0$. Now Theorem 3.10(a) of Rudin tells you that for any subset $E$ of a metric space $X$, $\textrm{diam} E = \textrm{diam} \overline{E}$. Applying it here we have $\textrm{diam} \overline{f(V_n(p))} \rightarrow 0$. Now we have a nested sequence of closed sets

$$\overline{f(V_1(p))} \supset \overline{f(V_2(p))} \supset \ldots$$

Instead of using compactness, you can pick for each $n$ an $x_n \in \overline{f(V_n(p))}$. It is easy to see that $x_n$ is a cauchy sequence so by completeness of $\Bbb{R}$ must convergence to some $x$. This $x$ is easily seen to be in the infinite intersection and the diameter going to zero means that this is the only element left in it.

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From where is $\textrm{diam} \, f(V_n(p)) \rightarrow 0$ drawn? I understand $\textrm{diam} \, f(V_n(p)) < \epsilon$, but not the jump to it approaching zero as $n\rightarrow \infty$. – AnonSubmitter85 Apr 23 '13 at 7:32