Show for $f:A \to Y$ uniformly continuous exists a unique extension to $\overline{A}$, which is uniformly continuous

Working on the following problem from Munkres:

Let $(X, d_{X})$ and $(Y, d_Y)$ be metric spaces; let $Y$ be complete. Let $A \subset X$. Show that if $f:A \to Y$ is uniformly continuous, then $f$ can be uniquely extended t a continuous function $g: \overline{A} \to Y$, and $g$ is uniformly continuous.

If there exists such an extension, it must be unique, and $f(x) = \lim_{n \to \infty}f(x_n)$ for $x_n$ a sequence in $A$ converging to $x \in \overline{A}$. To show this limit exists, let $x_n \to x$, then $f(x_n)$ is Cauchy, since uniformly continuous functions take Cauchy sequences to Cauchy sequences. Then $f(x_n)$ converges by completeness, and I claim it converges to $f(x)$.

Now, to show $g$ is uniformly continuous, I kind of brute-forced it, and I was wondering if anyone had a more elegant method. (I consider the proof up to now pretty elegant.) What I did was: Suppose for contradiction that $\exists \epsilon >0$ s.t. $\forall \delta >0\,\, \exists x_\delta, y_\delta \in \overline{A}$ s.t. $d(x_\delta, y_\delta)< \delta$ and $d(g(x_\delta), g(y_\delta))> \epsilon$. Let $\delta := \delta_{\epsilon}/2$, where $\delta_\epsilon$ is taken from the uniform continuity of $f$. Now $d(g(\cdot), g(\cdot))$ is continuous...in particular, it is continuous in each variable separately. Therefore we can vary $x_{\delta}$ and $y_{\delta}$ a little bit, to get $x^*,\, y^* \in A$ such that $d(x^*, y^*)<\delta_{\epsilon}$ and $d(f(x^*), f(y^*))>\epsilon$, a contradiction.

Does anyone know a more elegant method to show the uniform continuity?

• You can actually give a direct $\varepsilon/3$-argument. You can try again, or give me a while and I'll post something. =) – Pedro Tamaroff May 30 '14 at 0:52
• For a different take you can check this question – Guillaume F. Jul 19 '18 at 5:42

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

• Why do we need to show that $g$ is independent of the sequence used to define it? – Al Jebr Oct 31 '17 at 3:21
• @AlJebr: So the limiting value is independent of the particular sequence. – copper.hat Oct 31 '17 at 3:24
• If two Cauchy sequences converge to the same value, then this means they are equivalent and so $g$ would be independent of the sequence converging to some point $x$. Is this correct? – Al Jebr Oct 31 '17 at 4:40
• The point is that if $x_n \to x$ and $y_n \to x$ we must be sure that $f(x_n)$ and $f(y_n)$ both converge and that they both converge to the same value. – copper.hat Oct 31 '17 at 4:47
• @copper.hat Why when we take $\lim d(f(x_n),f(y_n))<\frac{1}{2} \epsilon$ we have $d(g(x),g(y)) \le \frac{1}{2}\epsilon$ and not $d(g(x),g(y))< \frac{1}{2} \epsilon$? – idk Nov 19 '17 at 11:50