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 $\bar{f}?$


1 Answer 1


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.

  • $\begingroup$ @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? $\endgroup$
    – freakish
    Oct 11, 2018 at 20:57
  • $\begingroup$ Thanks for your response Freakish! $\endgroup$ Oct 11, 2018 at 20:57
  • $\begingroup$ Hey freakish: last question. WOuldn't the proof for any given $(a_n)$ generalize to any convergent sequence $(a_n) \subset S$? $\endgroup$ Oct 11, 2018 at 21:02
  • $\begingroup$ @RafaelVergnaud you mean to define $f(p)=\lim f(p_n)$ for any $p\in\oveline{S}$? Yeah, I guess so. Simplifies a bit. $\endgroup$
    – freakish
    Oct 11, 2018 at 21:04
  • $\begingroup$ Ya, and that it is well-defined follows immediately, for it does not depend on the choice of $(p_n).$ $\endgroup$ Oct 11, 2018 at 21:04

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