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**TaylorM0192** Let $\displaystyle E$ $ be a subset of $\displaystyle X $, and $\displaystyle f : E -> Y $ be uniformly continuous, where $\displaystyle Y$ is complete. Show that there exists a continuous function which maps the closure of $\displaystyle E $ to $\displaystyle Y $ such that $\displaystyle F(x) = f(x). $

So basically, we are trying to prove that a function can be extended to its domains closure, and still retain the same values on the original set.

It's late, so maybe I need to spend more time pondering this question, but I'm frankly a little stumped - of the continuity problems I have tackled so far from chapter 4 of Rudin, this is the first involving the hypothesis of a mapping into a *complete* metric space, so it suggests (at first to me anyway) a different method of proof.

I guess I would start off with first establishing that the closure of E is (obviously) a closed set, with a limit point in X (in fact, in E closure by definition). I can than approach this limit point with a convergent Cauchy sequence, and then the image of this sequence will again be Cauchy in Y and thus convergent in Y. These facts follow from the hypothesis that the function f is uniformly continuous, and Y is complete. Then I suppose I could somehow use the sequential characterization of continuity to conclude the theorem...but rigorizing this argument, and then actually proving that this "extension" big F is equal to little f on E, is something I'm not sure about.

Anyway, any other ideas, or modifications to mine, for this problem?