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Let ##(X,d)## be a metric space, ##D \subset X## a dense subset, and ##f: D→ℝ## a uniformly continuous function. Prove that f has a unique extension to all ##X##.

3. The Attempt at a Solution .

I have some ideas but not the complete proof. If ##x \in D##, then I define ##f'(x)=f(x)##, so let ##x \in X \setminus D##. ##D## is a dense subset, so there exists ##(x_n)_{n≥1}## ##\subset D## such that ##x_n→x##. I define ##f'(x)=\lim_{n\rightarrow +\infty} {x_n}##. Observe that this limit makes sense since ##(f(x_n)_n≥1## is a Cauchy sequence (to prove this one has to use the fact that f is uniformly continuous) in ##ℝ##. ##ℝ## is complete, so the sequence is convergent. Then I should check that ##f'## is well defined, in other words, if there are two different sequences ##(x_n)_{n≥1}## and ##(y_n)_{n≥1}##in ##D## converging to a point ##x##, then it must be ##\lim_{n\rightarrow +\infty} {x_n}##=##\lim_{n\rightarrow +\infty} {y_n}## (this is easy to prove). Now I want to prove that f' is uniformly continuous, this means that for every ##ε>0## ##\exists δ_ε>0##:

##d(x,y)<δ_ε## implies ##|f'(x)-f'(y)|<ε##. Here I got totally stuck. This is an exercise belonging to the topic of compact metric spaces so I suppose I have to use something about compact spaces, the problem is that ##ℝ## is not compact, so I don't know where should I use some information about compactness.