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.

I’ll get you started. If $x\in\operatorname{cl}A$, there is a sequence $\langle a_n:n\in\Bbb N\rangle$ in $A$ converging to $x$. This sequence is Cauchy, so its image under $f$ is Cauchy; and since $\Bbb R$ is complete in the usual metric, $\langle f(a_n):n\in\Bbb N\rangle$ converges to some real number, $r$. Clearly one would like to set $g(x)=r$. In order to do this, however, one must show that $r$ is well-defined. Is it possible that if we’d chosen a different sequence in $A$ converging to $x$, this approach would have given us a different $r$? You need to show that it would not. Once you’ve done that, you need to verify that $g$ is continuous, and that $g\upharpoonright A=f$.
Finally, once all that’s done, you still have to show that $g$ is unique: if $h:\operatorname{cl}A\to\Bbb R$ is a continuous function such that $h\upharpoonright A=f$, then $h=g$. For this you’ll want to use the fact that $A$ is dense in $\operatorname{cl}A$.
