Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

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}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.