Is the continuous extension theorem true when the range space of $f$ is not complete? So the problem is Exercise $13$, Chap. $4$ of Principles of Mathematical Analysis by Rudin:
Problem
Let $E$ be a dense subset of metric space $X$, and let $f$ be a uniformly continuous real function defined on $E$. Prove that $f$ has a continuous extension from $E$ to $X$. 
Following the hint given by the author, I am able to prove this result and understand that the result still holds when the range space of $f$ is replaced by any complete metric space. My question is, does the result also hold when the range space is any metric space? If yes, how to prove, if no, are there any illuminating counterexamples?
 A: Suppose $f:\Bbb Q\to \Bbb Q,\, f(x)=x$. Were there some continuous extension $g:\Bbb R\to \Bbb Q$, would it be possible to define $g(\sqrt 2)$?
PS: I just did exactly this same problem in Baby Rudin yesterday :)
A: The uniformity is necessary. If a sequence in E satisfies the Cauchy criterion,with respect to the metric on X, the uniform continuity of f implies that the image of this sequence also satisfies the Cauchy condition so the image sequence converges to a limit point y.Now if the sequence in E as no limit point inX,there is nothing to do,but if it has a limit point p in X\E, we define f(p)=y.This value is independent of which which sequence in E,converging to p,was chosen.To show that f is continuous on all of X, consider that if a sequence in X converges to a point q, then each term in the sequence is arbitrarily close to a member of E, and use the uniformity of f on E to show that the image sequence converges to f(p) whether p is in E or in X/E.(I think you can finish this). The completeness (or not) of X is irrelevant..... On the other hand,if you have already proved the theorem for complete metric spaces, and X is not complete, then let cX be the metric completion of X. Now E is dense in X and X is dense in cX so E is dense in cX. So extend f from E to cX and then restrict it to X..... The theorem is not true when the range is not a complete metric space,because a sequence in E may converge to a point in X/E when the image of the sequence may be a Cauchy sequence with no point to converge to.For example, if X =R (the reals, with the usual metric)),E=R/{0} and E is also the range of f,with f(x)=x for x in E.
