Suppose $X$ a metric space, $Y$ a complete metric space and $f: S \rightarrow Y$ a uniformly continuous function from $S \subseteq X$ to
$Y$. Prove that $f$ can be extended to a uniformly continuous function
on $\overline{S}$.
I've no problem in showing that $f$ can be uniquely extended to a $\overline{f}$ continuous on $\overline{S}$, but i can't prove that $\overline{f}$ is uniformly continuous.
I know that this question is alredy be answered many times but in every argument there is some step that I don't understand.
EDIT
In Show for 𝑓:𝐴→𝑌 uniformly continuous exists a unique extension to 𝐴, which is uniformly continuous there is the following answer of copper.hat:
(Your proof above should explicitly show that $g$ is independent of
the sequence used to define it. This is the key point of the proof.)Let $\epsilon>0$, then you have some $\delta>0$ such that if $d(x,y) < \delta$, then $d(f(x),f(y)) < {1 \over 2}\epsilon$.
Pick $x,y \in \overline{A}$ such that $d(x,y) < \delta$, and let
$x_n,y_n$ be sequences in $A$ such that $x_n \to x,y_n \to y$. By
construction above, $g(x) = \lim_n f(x_n)$ and similarly for $g(y)$.For sufficiently large $n$, we have $d(x_n,y_n) < \delta$, and so
$d(f(x_n),f(y_n)) < {1 \over 2}\epsilon$.Taking limits we have $d(g(x),g(y)) \le {1 \over 2}\epsilon < \epsilon$.
I can't get the last step, how can we be sure that the logic implication $d(x_n,y_n) < \delta \implies d(f(x_n),f(y_n)) < {1 \over 2}\epsilon$ is still true under the limit process?
Best Answer
Assume that $ x,y \in cl(S) $
$ d(x,y) < \delta/3 $
therefore you can find two sequence $x_n ,y_n$ such that they converge to x ,y respectively .
therefore for a good n $d(x_n,y_n ) \le d(x_n,x) + d(x,y_n) < \delta/3 + d(x,y) + d(y,y_n) < \delta /3 + \delta /3+ \delta /3=\delta$
and therefore by the hypothesis of uniform continuity :
$d(f(x_n),f(y_n) ) <1/2 \epsilon $
and by go to the limit :
$d(g(x),g(y) ) \le 1/2 \epsilon <\epsilon$