# Uniformly Convergent Sequence Evaluated on Convergent Sequence

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## Theorem

Let $X = \struct {A, d}$ and $Y = \struct {B, \rho}$ be metric spaces.

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

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

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

Then $\sequence {\map {f_n} {a_n} }$ is convergent such that:

- $\displaystyle \lim_{n \mathop \to \infty} \map {f_n} {a_n} = \map f a$

## Proof

We want to show that:

- $\size {\map {f_n} {a_n} - \map f a} \to 0$ as $n \to \infty$

From the Triangle Inequality:

- $\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a}$

Now fix $\epsilon \in \R^{>0}$.

Since $f$ is continuous and $a_n \to a$, Sequential Continuity is Equivalent to Continuity in Metric Space tells us that:

- $\exists M \in \N: n \ge M \implies \size {\map {f_n} {a_n} - \map f {a_n} } < \dfrac \epsilon 2$

Since $\sequence {f_n}$ is uniformly convergent on $K$:

- $\exists N \in \N: n \ge N \implies \size {\map {f_n} {a_n} \to \map f {a_n} } < \dfrac \epsilon 2$

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

- $\size {\map {f_n} {a_n} - \map f a} \le \size {\map {f_n} {a_n} - \map f {a_n} } + \size {\map f {a_n} - \map f a} < \dfrac \epsilon 2 + \dfrac \epsilon 2 = \epsilon$

This completes the proof.

$\blacksquare$