Continuity of the closure of a set I have a homework question that I just could not figure out.
Suppose that $f:A \rightarrow \mathbb{R}$ is a uniformly continuous real-valued function on a subset $A$ of a metric space $X$. Show that there is a continuous real-valued function $g:\overline{A} \rightarrow \mathbb{R}$ defined on the closure $\overline{A}$ of $A$ such that, for each $p \in A,  g(p) = f(p)$. Show that $g$ is unique.
My idea: I proved in an earlier problem that if $\{p_n\}$ is a Cauchy sequence in $A$, then $\{f(p_n)\}$ is a Cauchy sequence in $\mathbb{R}$. I want to use that for the $p \in A' \setminus A $ but I don't know where to go from there.
Thanks for the help.

Choose $a \in {\overline A}$.  then there is a sequence $a_n$ in $A$ so that $a_n\to a$.  Let $\epsilon > 0$.  Since $f$ is uniformly continuous on $A$, there is a $\delta > 0$ so that $d(a,b) < \delta\implies d(f(a),f(b)) < \epsilon$.  Since the sequence $a_n$ converges, it is Cauchy. Choose $N\in\mathbb{N}$ so that $n\ge N \implies d(a_n, a_m) < \epsilon.$  by uniform continuity, $d(f(a_m), f(a_n)) <\epsilon$ whenever $m, n \ge N$.  The sequence $f(a_n)$ is Cauchy and therefore converges. 
The real numbers are complete, so the sequence $f(a_n)$ converges to some real number.  Now you can do a uniqueness argument that only one such number is possible.  The uniformity of the continuity will allow you to argue this extension is continuous.  Finally, it is easy to see that if functions are continuous and agree on a dense subset of a metric space, they must agree everywhere.
