Uniformly Continuous Function to Complete Metric Space has Unique Continuous Extension to Closure of Domain/Lemma 2

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.

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

Then there exists a function $L : A^- \to Y$ such that:


 * $\ds \lim_{n \to \infty} \map f {a_n} = \map L a$

That is, the limit of $\sequence {\map f {a_n} }$ is dependent only on the limit of $\sequence {a_n}$.

Proof
Let $\sequence {a_n}$ and $\sequence {b_n}$ be sequences in $A$ such that $a_n \to a$ and $b_n \to a$, with:


 * $\map f {a_n} \to L_1$

and:


 * $\map f {b_n} \to L_2$

We have, by the Triangle Inequality:


 * $\map {d'} {\map f {a_n}, L_2} \le \map {d'} {\map f {a_n}, \map f {b_n} } + \map {d'} {\map f {b_n}, L_2}$

We want to show that we can find $K$ such that:


 * $\map {d'} {\map f {a_n}, L_2} < \epsilon$

for $n > K$.

Note that since $f$ is uniformly continuous, we can find $\delta_1$ such that for:


 * $\map d {a_n, b_n} < \delta_1$

we have:


 * $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/2$

By the Triangle Inequality, we can write:


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

Since $a_n \to a$ and $b_n \to a$, we can find $K_1, K_2$ such that for $n > K_1$ we have:


 * $\map d {a_n, a} < \delta_1/2$

and for $n > K_2$ we have:


 * $\map d {b_n, a} < \delta_1/2$

So, for $n > \max \{K_1, K_2\}$, we have:


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

and so:


 * $\map {d'} {\map f {a_n}, \map f {b_n} } < \epsilon/2$

for $n > \max \{K_1, K_2\}$.

Since $\map f {b_n} \to L_2$, we can pick $K_3$ such that:


 * $\map {d'} {\map f {b_n}, L_2} < \epsilon/2$

for $n > K_3$.

So, setting $K = \max \{K_1, K_2, K_3\}$, we have:


 * $\map {d'} {\map f {b_n}, L_2} < \epsilon/2$

for $n > K$.

So:


 * $\map {d'} {\map f {a_n}, L_2} < \epsilon$

for $n > K$.

So $\map f {a_n} \to L_2$.

From Convergent Sequence in Metric Space has Unique Limit, we have:


 * $L_1 = L_2$

So, if $a_n \to a$, the sequence $\sequence {\map f {a_n} }$ converges to some limit $\map L a$ dependent only on $a$.