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Let $M_1,M_2$ be metric spaces such that $M_2$ is complete. Let $f$ be a uniformly continuous function from a subset $X$ of $M_1$ into $M_2$. Suppose that $\overline{X}=M_1$. Prove that $f$ has a unique uniformly continuous extension from $M_1$ into $M_2$ (that is, prove that there exists a unique uniformly continuous function $g$ from $M_1$ into $M_2$ such that $g|X=f$.)

I'm not sure where to start on this one... how can I extend a uniformly continuous function from $X$ to $M_1$?

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Because $X$ is dense in $M_1$, any $x \in M_1$ is the limit of a sequence $(x_n)$ in $X$.

  • Show that $(f(x_n))$ is a Cauchy sequence. In particulier, $(f(x_n))$ converges.
  • Show that if $(y_n)$ is another sequence converging to $x$, then $(f(x_n))$ and $(f(y_n))$ have the same limit.

Thus, you can extend $f$ by $f(x)=\lim\limits_{n \to + \infty} f(x_n)$.

  • To conclude, show that if $g$ is another extension of $f$, then $f=g$ (the key property is that $X$ is dense in $M_1$).
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  • $\begingroup$ Ok, I think I got it. Quite nice! $\endgroup$
    – PJ Miller
    Jun 20 '13 at 16:25

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