So, let (X, dx) and (Y, dy) be metric spaces, and let Y be complete. Let A be a subset of X, and f : A --> Y a uniformly continuous function. Prove that f can be uniquely extended to a uniformly continuous function g : Cl(A) --> Y.
The Attempt at a Solution
My first idea was to use Theorem 30.1. in Munkres.
If x is in Cl(A), since X is a metric space, there is a sequence of points xn of A converging to x. Further on, since f is continuous, the sequence f(xn) converges to f(x). But is x is not in A, i.e. if it is in Cl(A)\A, f(x) may not be defined. This is a problem I don't know how to remedy.
Uniform continuity of f implies Cauchy-continuity, i.e. for any Cauchy sequence xn in A, the sequence f(xn) is Cauchy in Y. Since Y is complete, f(xn) must converge.
I feel I'm close, but I can't add the pieces together. :uhh: