Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain/Lemma 1

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

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

Let $A \subseteq X$ be a set that is not closed.

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

Let $\sequence {a_n}$ be a sequence in $A$ convergent to $a \in A^-$.

Then $\sequence {\map f {a_n} }$ converges.

Proof
From Set is Closed iff Equals Topological Closure, $A^- \setminus A$ is non-empty.

Let $a \in A^- \setminus A$.

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


 * there exists a sequence $\sequence {a_n}$ in $A$ converging to $a$.

Consider now the sequence $\sequence {\map f {a_n} }$ in $Y$.

Note that since $Y$ is complete, if we can show that $\sequence {\map f {a_n} }$ is a Cauchy sequence, then we know that it converges.

We therefore want to show that there exists an $N$ such that for $n, m > N$ we have:


 * $\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$

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


 * $\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$

whenever:


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

Since $\sequence {a_n}$ is a Cauchy sequence, we can find $N$ such that for $n, m > N$ we have:


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

giving:


 * $\map {d'} {\map f {a_n}, \map f {a_m} } < \epsilon$

for $n, m > N$.

So $\sequence {\map f {a_n} }$ is a Cauchy sequence, and so is convergent.