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.