# Uniformly continuous Extension functions and limit points.

A book is asking the following question, and there seems to me to be a contradiction in the question. However, author has a lot more credibility than I do and I'm confident that I'm mistaken. I just don't see how.

Here is the question:

Consider a uniformly continuous function $$f: S_1 \rightarrow M,$$ such that $$S_1$$ is a subset of the complete metric space $$M.$$ A function $$g:S_2 \rightarrow M$$ extends $$f$$ if $$S_1 \subset S_2$$ and $$g(S_1) = f(S_1).$$

Prove that $$f$$ extends to a uniformly continuous function $$\bar{f}: \bar{S} \rightarrow M$$. Show that $$\bar{f}$$ and is the unique function that extends $$f$$ and maintains continuity (that is, all other functions break the continuity of $$f$$).

The contradiction?: As $$f$$ is continuous, it maintains the sequential convergence property, that is, if a sequence $$\{p_n\} \subset S_1$$ in $$S_1$$ converges to $$p,$$ then the sequence $$\{f(p_n)\} \subset M$$ converges to $$f(p)$$ (that is, $$f(p)$$ must exist, that is, the function $$f$$ must be defined at $$p$$). Is it not the case that $$S_1$$ is closed and therefore $$S_1 = \bar{S_1}$$ because $$f$$ is uniformly continuous. It follows from that that there is no extension of $$f$$ to $$\bar{f}?$$

if a sequence $$\{p_n\} \subset S_1$$ in $$S_1$$ converges to $$p,$$ then the sequence $$\{f(p_n)\} \subset M$$ converges to $$f(p)$$

What makes you think so? Take $$f:(0,1)\to\mathbb{R}$$, $$f(x)=x$$. Clearly $$1/n$$ converges to $$0$$. But why do you think that implies that $$f$$ is defined at $$0$$? It can be extended to $$0$$, and indeed $$\overline{f}:[0,1]\to\mathbb{R}$$ given by $$\overline{f}(x)=x$$ is the unique extension. And this is what the extension argument is all about.

So here's how you do it in general. For any $$p\in \overline{S}\backslash S$$ there is a sequence $$(p_n)\subseteq S$$ convergent to $$p$$. You define

$$\overline{f}:\overline{S}\to M$$ $$\overline{f}(x)=f(x)\text{ for }x\in S$$ $$\overline{f}(p)=\lim f(p_n)$$

Now you have to prove that:

(a) $$f(p_n)$$ is convergent. This is not trivial, for example take $$f(x)=1/x$$ and note that $$f(1/n)$$ is not convergent even though $$1/n$$ is. And indeed $$f(x)$$ cannot be (continuously) extended to $$0$$. But $$f(x)$$ is not uniformly continuous. And indeed the convergence follows from the fact that $$f$$ is uniformly continuous (and so $$f$$ maps Cauchy sequences to Cauchy sequences) and $$M$$ is complete.

(b) $$\overline{f}(p)$$ is well defined, i.e. it doesn't depend on the choice of $$(p_n)$$.

(c) $$\overline{f}$$ is (uniformly) continuous.

• @RafaelVergnaud Yes, but your $f$ is not defined in $p$. Again, have a look at $f:(0,\infty)\to\mathbb{R}$ given by $f(x)=1/x$. Consider sequence $p_n=1/n$. Obviously $p_n\to 0$ but is $f(p_n)$ convergent? Can $f$ be (continuously) extended to $0$, i.e. is there a way to define $f(0)$ while preserving continuity? Oct 11, 2018 at 20:57
• Thanks for your response Freakish! Oct 11, 2018 at 20:57
• Hey freakish: last question. WOuldn't the proof for any given $(a_n)$ generalize to any convergent sequence $(a_n) \subset S$? Oct 11, 2018 at 21:02
• @RafaelVergnaud you mean to define $f(p)=\lim f(p_n)$ for any $p\in\oveline{S}$? Yeah, I guess so. Simplifies a bit. Oct 11, 2018 at 21:04
• Ya, and that it is well-defined follows immediately, for it does not depend on the choice of $(p_n).$ Oct 11, 2018 at 21:04