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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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1 Answer 1

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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}

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