Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain

Theorem
Let $\tuple {X, d}$ be a metric space.

Let $\tuple {Y, d'}$ be a complete metric space.

Let $A \subseteq X$.

Let $f : A \to Y$ be a uniformly continuous function.

Then there exists a unique continuous function $g : A^- \to Y$ such that:


 * $\map g a = \map f a$

for all $a \in A$, where $A^-$ denotes the topological closure of $A$.

Furthermore, $g$ is uniformly continuous.

Proof Outline
Suppose we had a uniformly continuous $g$. We know that from Sequential Continuity is Equivalent to Continuity in Metric Space, that if $\sequence {a_n}$ is a sequence in $A$ converging to $a \in A^-$, we must have:


 * $\map g {a_n} \to \map g a$

Since $\sequence {a_n}$ is a sequence in $A$, this is equivalent to:


 * $\map f {a_n} \to \map g a$

So our first goal is to "fill in" the missing values by considering limits of sequences of the form $\sequence {\map f {a_n} }$ where $a_n \to a$. To ensure uniqueness, we show that the limit is the same regardless of the choice of $\sequence {a_n}$.

We show that the thus constructed function is uniformly continuous, at which point we are done.

Proof
Note that if $A$ is closed, then from Set is Closed iff Equals Topological Closure, we have:


 * $A^- = A$

So taking $g = f$ suffices in this case.

Lemma 1
We now show that the limit of $\sequence {\map f {a_n} }$ is independent of the particular $\sequence {a_n}$ chosen.

Lemma 2
Now, define the function $g : A^- \to Y$ by:


 * $\map g a = \begin{cases}\map L a & a \in A^- \setminus A \\ \map f a & \text{otherwise}\end{cases}$

We now show this function is uniformly continuous.

Let $\epsilon > 0$.

We want to show that there exists some $\delta' > 0$ such that whenever $a, b \in A^-$ and:


 * $\map d {a, b} < \delta'$

we have:


 * $\map {d'} {\map g a, \map g b} < \epsilon$

From Point in Closure of Subset of Metric Space iff Limit of Sequence:


 * there exists a sequence $\sequence {a_n}$ in $A$ such that $a_n \to a$

and:


 * there exists a sequence $\sequence {b_n}$ in $A$ such that $b_n \to b$.

We then have, by the Triangle Inequality:


 * $\map {d'} {\map g a, \map g b} \le \map {d'} {\map g a, \map f {a_n} } + \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, \map g b}$

Recall by construction of $g$:


 * $\map f {a_n} \to \map g a$

and:


 * $\map f {b_n} \to \map g b$

From the definition of convergence, we can find $N_1 \in \N$ such that:


 * $\map {d'} {\map g a, \map f {a_n} } < \epsilon/3$

for $n > N_1$.

We can also find $N_2 \in \N$ such that:


 * $\map {d'} {\map g b, \map f {b_n} } < \epsilon/3$

for $n > N_2$.

We now wish to find $N_3 \in \N$ such that:


 * $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$

for $n > N_3$.

Recall that since $f$ is uniformly continuous, we can find $\delta > 0$ such that whenever:


 * $\map d {a_n, b_n} < \delta$

we have:


 * $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$

Then by the Triangle Inequality we have:


 * $\map d {a_n, b_n} \le \map d {a_n, a} + \map d {a, b} + \map d {b, b_n}$

Since $a_n \to a$ we can pick $M_1 \in \N$ such that:


 * $\map d {a_n, a} < \delta/3$

for $n > M_1$.

Since $b_n \to b$ we can pick $M_2 \in \N$ such that:


 * $\map d{b, b_n} < \delta/3$

for $n > M_2$.

So, if:


 * $\map d {a, b} < \delta/3$

and:


 * $n > \max \set {M_1, M_2} = N_3$

we have:


 * $\map d {a_n, b_n} < \delta$

and so:


 * $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/3$

Now, let:


 * $N = \max \set {N_1, N_2, N_3}$

Then for $n > N$ and whenever $\map d {a, b} < \delta/3$, we have:


 * $\map {d'} {\map g a, \map g b} \le \map {d'} {\map g a, \map f {a_n} } + \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, \map g b} < \epsilon$

So:


 * $\map {d'} {\map g a, \map g b} < \epsilon$

whenever:


 * $\map d {a, b} < \delta/3$

So $g$ is uniformly continuous.