# A uniformly continuous function can be extended on the boundary

Suppose $$X$$ a metric space, $$Y$$ a complete metric space and $$f: S \rightarrow Y$$ a uniformly continuous function from $$S \subseteq X$$ to $$Y$$. Prove that $$f$$ can be extended to a uniformly continuous function on $$\overline{S}$$.

I've no problem in showing that $$f$$ can be uniquely extended to a $$\overline{f}$$ continuous on $$\overline{S}$$, but i can't prove that $$\overline{f}$$ is uniformly continuous.

I know that this question is alredy be answered many times but in every argument there is some step that I don't understand.

EDIT

In Show for 𝑓:𝐴→𝑌 uniformly continuous exists a unique extension to 𝐴, which is uniformly continuous there is the following answer of copper.hat:

(Your proof above should explicitly show that $$g$$ is independent of the sequence used to define it. This is the key point of the proof.)

Let $$\epsilon>0$$, then you have some $$\delta>0$$ such that if $$d(x,y) < \delta$$, then $$d(f(x),f(y)) < {1 \over 2}\epsilon$$.

Pick $$x,y \in \overline{A}$$ such that $$d(x,y) < \delta$$, and let $$x_n,y_n$$ be sequences in $$A$$ such that $$x_n \to x,y_n \to y$$. By construction above, $$g(x) = \lim_n f(x_n)$$ and similarly for $$g(y)$$.

For sufficiently large $$n$$, we have $$d(x_n,y_n) < \delta$$, and so $$d(f(x_n),f(y_n)) < {1 \over 2}\epsilon$$.

Taking limits we have $$d(g(x),g(y)) \le {1 \over 2}\epsilon < \epsilon$$.

I can't get the last step, how can we be sure that the logic implication $$d(x_n,y_n) < \delta \implies d(f(x_n),f(y_n)) < {1 \over 2}\epsilon$$ is still true under the limit process?

Assume that $$x,y \in cl(S)$$

$$d(x,y) < \delta/3$$

therefore you can find two sequence $$x_n ,y_n$$ such that they converge to x ,y respectively .

therefore for a good n $$d(x_n,y_n ) \le d(x_n,x) + d(x,y_n) < \delta/3 + d(x,y) + d(y,y_n) < \delta /3 + \delta /3+ \delta /3=\delta$$

and therefore by the hypothesis of uniform continuity :

$$d(f(x_n),f(y_n) ) <1/2 \epsilon$$

and by go to the limit :

$$d(g(x),g(y) ) \le 1/2 \epsilon <\epsilon$$

• Assume that $x,y \in cl(S)$ $d(x,y) < \delta/3$ therefore you can find two sequence $x_n ,y_n$ such that they converge to x ,y respectively . therefore for a good n $d(x_n,y_n ) \le d(x_n,x) + d(x,y_n) < \delta/3 + d(x,y) + d(y,y_n) < \delta /3 + \delta /3+ \delta /3=\delta$ and therefore by the hypothesis of uniform continuity $d(f(x_n),f(y_n) ) <1/2 \epsilon$ and by go to the limit $d(g(x),g(y) ) \le 1/2 \epsilon <\epsilon$ – Anonyme Apr 25 at 10:12
• @Giovanni Barbarani – Anonyme Apr 25 at 10:18
• Now i can fully understand thanks. – Giovanni Barbarani Apr 25 at 10:25