Uniformly Convergent Sequence Evaluated on Convergent Sequence

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Theorem

Let $X = \left({A, d}\right)$ and $Y = \left({B, \rho}\right)$ be metric spaces.

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

Let $\mathcal F = \left \langle{f_n}\right \rangle$ be a sequence of continuous mappings $f_n: X \to Y$ uniformly convergent on $K$.

Let $\left \langle{a_n}\right \rangle$ be a Convergent Sequence in $K$ with limit $a \in K$.


Then $\left \langle{f_n\left({a_n}\right)}\right \rangle$ is convergent and

$\displaystyle \lim_{n \to \infty} f_n \left({a_n}\right) = f(a)$.


Proof

We want to show that:

$\left|{ f_n\left({a_n}\right) - f(a) }\right| \to 0$ as $n \to \infty$

From the Triangle Inequality:

$\left|{ f_n\left({a_n}\right) - f(a) }\right| \leq \left|{ f_n\left({a_n}\right) - f\left({a_n}\right) }\right| + \left|{ f\left({a_n}\right) - f(a) }\right|$

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:

$\exists M \in \N: n \geq M \implies \left|{ f_n\left({a_n}\right) - f\left({a_n}\right) }\right| < \dfrac{\epsilon}{2}$

Since $\left \langle{f_n}\right \rangle$ is uniformly convergent:

$\exists N \in \N: n \geq N \implies \left|{ f_n\left({a_n}\right) \to f\left({a_n}\right) }\right| < \dfrac{\epsilon}{2}$

So if $n \geq \max(M, N)$:

$\left|{ f_n\left({a_n}\right) - f(a) }\right| \leq \left|{ f_n\left({a_n}\right) - f\left({a_n}\right) }\right| + \left|{ f\left({a_n}\right) - f(a) }\right| < \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon$

This completes the proof.

$\blacksquare$