Let $(S,d_S)$ and $(T,d_T)$ be two metric spaces with $T$ complete and suppose that a mapping $f:A\subseteq S\to T$ is uniformly continuous on its domain. I want to show that $f$ can be uniquely extended to the closure $\overline{A}$ with the extension uniformly continuous on $\overline{A}$.
Actually, this is a homework exercise from my introductory-analysis class, and before this exercise, several subsidiary questions had been deployed so I kind of cracked this question. For example, with every $x\in\overline{A}$, there comes a sequence $\{x_n\}_{n=1}^\infty$ in $A$ s.t. $x_n\to x$. This sequence is mapped to a Cauchy sequence $\{f(x_n)\}_{n=1}^\infty$ since $f$ is uniformly continuous on $A$. Then, thanks to completeness of $T$, we are assured of convergence of this Cauchy sequence, and hence able to designate the limit to be the value of a newborn $g:\overline{A}\to T$ at $x$. This $g$ will be the desired extension once we justify the following claims.
First, we have to show that $g$ is well-defined. By this, I mean another sequence $\{x_n\}_{n=1}^\infty\subseteq A$ converging to $x$ does not yield a different limit of $\{f(x_n)\}_{n=1}^\infty$. After establishing this claim, we need to show that $g$ is uniformly continuous on its domain, which requires employing uniform continuity of $f$. Last but not least, we have to show that $g$ is the restriction of $f$ to $A$. This can be done by recalling that $f$ is a continuous mapping.
So far, I've been able to resolve all the doubts about the claims mentioned above, but as indicated in the first paragraph, I fail to explain why a uniformly continuous extension of $f$ to $\overline{A}$ is unique. Does anyone have an idea? Thank you.
Best Answer
Uniqueness (Alternative)
Suppose $f_1, f_2$ be two uniform extension of $f$ on $\overline{A}$
Then $f_1=f=f_2$ on $A$
Let $D=\{x\in \overline{A} : f_1(x) =f_2(x) \}$
Then clearly $D$ is a closed set i.e $\overline{D}=D$ (the set of all points where two continuous map agrees is a closed set, provided the target space is Hausdorff.)
$ A\subset D\subset \overline{A}$
Then $\overline{A}\subset\overline{ D}\subset \overline{A}$
Implies $\overline{A}\subset{ D}\subset \overline{A}$
Hence $D=\overline{A}$ , implies $f_1=f_2$ on $\overline{A}$.