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 function $g$ satisfying the hypotheses. 
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, and so is our desired function. 
== Proof ==
=== Existence ===
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.
Suppose now that $A$ is not closed.
==== 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 in a metric space, 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 function, 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.
From Uniformly Continuous Function is Continuous: Metric Space, $g$ is also continuous.
=== Uniqueness ===
Let $h$ be another continuous function satisfying the hypotheses. 
Then, for any $a \in A^-$ and sequence $\sequence {a_n}$ convergent sequence to $a$, we have: 
:$\map h {a_n} \to \map h a$
and:
:$\map f {a_n} \to \map g a$
by Sequential Continuity is Equivalent to Continuity in Metric Space.
But since $h$ coincides with $f$ on $A$, this gives:
:$\map f {a_n} \to \map h a$
from Convergent Sequence in Metric Space has Unique Limit, we obtain: 
:$\map h a = \map g a$
So we must have $g = h$. 
Category:Uniformly Continuous Functions
Category:Complete Metric Spaces
Category:Set Closures
Category:Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain