# A function can be extended to a uniformly continuous function

Let $S$ be a subset of the metric space $E$ with the property that each point of $S^c$ is a cluster point of $S.$ Let $E'$ be a complete metric space and $f: S\to E'$ a uniformly continuous function. Prove that $f$ can be extended to a continuous function from $E$ into $E'$ in one and only one way, and that this extended function is also uniformly continuous.

Let $s \in S^c$ then since $s$ is a cluster point of $S$ then there is a sequence $s_n \in S$ that converges to $s$ thus the sequence $s_n$ is a Cauchy sequence. Since $f$ is uniformly continuous and $E'$ is complete we have that $f(x_n)$ is also a Cauchy sequence which converges in $E'$ to some $x.$

How can I now porve that $f$ can be extended, in one and only one way, to a continuous function from $E$?

Continue along the same lines : Define a function $g:E\to E'$ as follows :

a) If $s \in S$, write $g(s) = f(s)$

b) If $s\in S^c$, choose a sequence $s_n \to s$, with $s_n \in S$, and then you know that $f(s_n)$ converges in $E'$, so define $$g(s) = \lim f(s_n)$$

Now you need to show that this definition of $g(s)$ is independent of choice of $s_n$. This will be an extension of $f$, and it will be unique since $S$ is dense in $E$

• Why define $g(s) = f(s)$? I know $g(s)$ will be the unique limit for $f(s_n)$ but what is $f(s)$ and how $g(s)$ and $f(s)$ related? – Tom Oct 28 '13 at 5:05
• @Tom : $f$ is already defined on $S$, so for any $s\in S, f(s)$ makes sense. You want $g$ to be equal to $f$ on $S$ (this is what it means to extend a function on $S$) – Prahlad Vaidyanathan Oct 28 '13 at 5:18
• I didn't understand what it meant to extend a function but thank you very much! – Tom Oct 28 '13 at 5:20
• Prahlad, would this be true if it were not uniformly continuous instead just continuous? – Tom Oct 28 '13 at 6:03
• @Tom : No. the sequence $f(s_n)$ would not be Cauchy if $f$ were just continuous. For instance think of $f(x) = 1/x$ on $(0,1] \subset [0,1]$ - there is no continuous extension to $[0,1]$ – Prahlad Vaidyanathan Oct 28 '13 at 6:09