I must answer this question: Assume that for the complete metric space $N$, $f: S \rightarrow N$ is a uniformly continuous function defined on a subset $S$ of a metric space $M$. Prove that $f$ extends to a uniformly continuous function $\bar{f}: \bar{S} \rightarrow N$. Then show that $\bar{f}$ is the unique continuous function defined on $\bar{S}$ such that $\bar{f}(x) = f(x)$ for all $x \in S$.
I start my solution by noting that if $x \in S$, then set $\bar{f} = f$. I'm not quite sure how to extend for $\bar{S} - S$, though. My intuition says that I should use the fact that $\bar{S} = \lim S$, but I'm not quite sure how to apply it. That still leaves the matter of uniqueness, which I don't have the foggiest clue of how to prove either.
For $x\in\bar S\setminus S$, pick a sequence $x_i\in S$ that converges to $x$. Because $f$ is uniformly continuous, $f(x_n)$ converges to a finite value $L(x)$. Set $f(x)=L(x)$. Similarly, because $f$ is uniformly continuous, $f(x_i)\to L(x)$ for ANY sequence $x_i\to x$ - hence $\bar f$ is unique and by construction continuous. [Edit] It is uniformly continuous since its modulus of continuity is the same as that of $f$. [\Edit].