# Extension of a Uniformly Continuous Function between Metric Spaces

Let $(X,d)$ and $(Y,d')$ be metric spaces with $(Y,d')$ complete. Let $A\subseteq X$. I need to show that if $f:A\to Y$ is uniformly continuous, then $f$ can be uniquely extended to $\bar{A}$ maintaining the uniform continuity.

My attempt at this has involved taking each point $a\in \bar{A}-A$ and forming a Cauchy sequence to it by considering open balls $B_{\frac{1}{n}}(a)-B_{\frac{1}{n+1}}(a)$ beginning with $n$ large enough so there is such a sequence, and defining $g(a)$ to be the limit in $Y$. The uniqueness seems to be obvious just by thinking about the uniqueness of limits (referring to the sequence in $Y$), but I have to admit I don't know how to rigorously show it. The uniform continuity seems natural, but I don't know how to show it, either.

This seems to be correct, but I'm not entirely sure... Any help would be very appreciated!

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If $a \in \overline{A}$ then $a = \lim_n a_n$ where $a _n \in A$. Then, $a_n$ is Cauchy and as $f$ is uniformily continuous n $A$, $f(a_n)$ is Cauchy and as $(Y,d´)$ is complete you can define $f(a) : = \lim_n f(a_n)$. For example, if $b \in A$ you have that \begin{eqnarray} d(f(a),f(b)) &\le& d(f(a_n),f(b)) + d(f(a),f(a_n)) \end{eqnarray} and by definition of uniformily continuous you can see that $c$ continue being uniformily continuous. Analogously, if $b \in \overline{A}, b =\lim_n b_n$ where $b_n \in \overline{A}$ and \begin{eqnarray} d(f(a),f(b)) &\le& d(f(a_n),f(a)) + d(f(b_n),f(a_n)) + d(f(b_n), f(b)) \end{eqnarray}