# functions defined on metric spaces

How do we prove that if a function $f:A\rightarrow \mathbb{R}$ is uniformly continuous ($A\subset M$ is a dense subset of $M$ where $M$ is a metric space), then there exists a uniformly continuous function: $g:M\rightarrow \mathbb{R}$ such that : $\forall x\in A:g(x)=f(x)$

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Some books: Searcoid: Metric spaces, p.176, Giaquinat, Modica: Mathematical Analysis, vol.1 p.206, Bridges: Foundations of Real and Abstract Analysis, p.143 Ammann, Escher: Analysis II, p.10. Usually you can find one of two approaches: one is using Cauchy sequences, another one intersection of images of balls with decreasing diameter. – Martin Sleziak Apr 27 '12 at 8:30

Hint: try define $F(x) = \displaystyle \lim_{n\to \infty} f(\xi_n)$ where $\{\xi_n\}^{\infty}_{n=1}$ is a sequence in that dense subset $A$ which has limit $x\in M$, the existence of this sequence depends on the density condition, which serves as an interpretation of "closure of $A$ is equal to $M$". And somehow playing $\epsilon-\delta$ argument with $|f(\xi_n) - f(\zeta_n)|$ would suffice, in which $\displaystyle\lim_{n\to\infty} \xi_n = x$, $\displaystyle\lim_{n\to\infty} \zeta_n = y$, $\xi_n$ and $\zeta_n$ are both sequences in $A$, also by this definition $F(a) = f(a)$ for $a\in A$.

EDIT: First step for any fixed $\epsilon>0$, consider if $\epsilon/3 >|F(x)-F(\xi_n)|=| \displaystyle \lim_{n\to \infty}f(\xi_n) - f(\xi_n)|$, then by the definition of the limit, there exists $N_1>0$, for any $n>N_1$, such that $| \displaystyle \lim_{n\to \infty}f(\xi_n) - f(\xi_n)| < \epsilon/3$, then apply the same argument on $|F(y)-F(\zeta_n)|$ and get $N_2$.

Secondly for $|f(\xi_n) - f(\zeta_n)|< \epsilon/3$, apply the uniform continuity of $f$ here, ie once $\|\xi_n-\zeta_n\|< \delta/3$ we have $|f(\xi_n) - f(\zeta_n)|< \epsilon/3$.

Now consider the exploitation of $$\|x-y\|\leq \|x-\xi_n\| + \|y-\zeta_n\| + \|\xi_n-\zeta_n\|$$ Use $\displaystyle\lim_{n\to\infty} \xi_n = x$, and $\displaystyle\lim_{n\to\infty} \zeta_n = y$, hence there exists $N_3>0$ such that for any $n>N_3$, $\|x-\xi_n\| < \delta/3$ and $\|y-\zeta_n\|< \delta/3$.

Lastly, for the $\epsilon$ in the beginning, use triangle inequality again: $$|F(x)-F(y)|\leq |F(x)-F(\xi_n)| + |F(y)-F(\zeta_n)| + |f(\xi_n) - f(\zeta_n)|<\epsilon$$ for $n>\max\{N_1,N_2,N_3\}$ and $\|x-y\|<\delta$.

In summary, for any $\epsilon$, we have constructed $\delta$: for $\|x-y\|\leq \|x-\xi_n\| + \|y-\zeta_n\| + \|\xi_n-\zeta_n\|<\delta$, $|F(x)-F(y)|\leq |F(x)-F(\xi_n)| + |F(y)-F(\zeta_n)| + |f(\xi_n) - f(\zeta_n)|<\epsilon$, since both left side doesn't depend on $n$, we get the uniform continuity.

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@mchris619 I edited more into my answer. – Shuhao Cao Apr 27 '12 at 0:09