Uniformly Convergent Sequence Evaluated on Convergent Sequence

Theorem
Let $X = \struct {A, d_X}$ and $Y = \struct {B, d_Y}$ be metric spaces.

Let $K$ be a subspace of $X$.

Let $f: X \to Y$ be a mapping.

Let $\FF = \sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to Y$ that converges to $f$ uniformly on $K$.

Let $\sequence {a_n}$ be a convergent sequence in $K$ with limit $a \in K$.

Then the sequence $\sequence {\map {f_n} {a_n} }$ is convergent such that:
 * $\ds \lim_{n \mathop \to \infty} \map {f_n} {a_n} = \map f a$

Proof
We want to show that:
 * $\map {d_Y} {\map {f_n} {a_n}, \map f a} \to 0$ as $n \to \infty$

Let $\epsilon \in \R_{>0}$ be fixed.

From the Triangle Equality, it follows that:
 * $\map {d_Y} {\map {f_n} {a_n}, \map f a} \le \map {d_Y} {\map {f_n} {a_n} , \map f {a_n} } + \map {d_Y} {\map f {a_n} , \map f a}$

From the Uniform Limit Theorem, it follows that $f$ is continuous.

Since $a_n \to a$, Sequential Continuity is Equivalent to Continuity in Metric Space tells us that:
 * $\exists M \in \N : \forall n \ge M : \map {d_Y} {\map f {a_n}, \map f a} < \dfrac \epsilon 2$

Since $\sequence {f_n}$ is uniformly convergent on $K$, it follows that:
 * $\exists N \in \N : \forall n \ge N : \map {d_Y} {\map {f_n} {a_n}, \map f {a_n} } < \dfrac \epsilon 2$

So if $n \ge \max \set {M, N}$, it follows that:

which shows that $\map {d_Y} {\map {f_n} {a_n}, \map f a} \to 0$ as $n \to \infty$.