Uniform Limit Theorem

Theorem
Let $\struct {M, d_M}$ and $\struct {N, d_N}$ be metric spaces.

Let $\sequence {f_n}$ be a sequence of mappings from $M$ to $N$ such that:
 * $\quad \forall n \in \N: f_n$ is continuous at every point of $M$

and:
 * $\quad \sequence {f_n}$ converges uniformly to $f$

Then $f$ is continuous at every point of $M$.

Proof
Let $a \in M$.

We are given that $d_N$ is a metric on $N$.

So by applying twice:
 * $(1): \quad \forall n \in \N: \forall x \in M: \map {d_N} {\map f x, \map f a} \le \map {d_N} {\map f x, \map {f_n} x} + \map {d_N} {\map {f_n} x, \map {f_n} a} + \map {d_N} {\map {f_n} a, \map f a}$

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

Since $\sequence {f_n}$ converges uniformly to $f$:

We are given that $\forall n \in \N: f_n$ is continuous.

So:
 * $(3): \quad \forall n \in \N: \exists \delta \in \R_{>0}: \forall x \in M: \map {d_M} {x, a} < \delta \implies \map {d_N} {\map {f_n} x, \map {f_n} a} < \dfrac \epsilon 3$

By combining $(1)$, $\text {(2 a)}$, $\text {(2 b)}$ and $(3)$:
 * $\exists \NN \in \R_{>0}: \forall n \in \N: n > \NN \implies \paren {\exists \delta \in \R_{>0}: \forall x \in M: \map {d_M} {x, a} < \delta \implies \map {d_N} {\map f x, \map f a} < \dfrac \epsilon 3 + \dfrac \epsilon 3 + \dfrac \epsilon 3 = \epsilon}$

As $a$ and $\epsilon$ are arbitrary, by Universal Instantiation of $n$ it follows that:
 * $\forall a \in M: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in M: \map {d_M} {x, a} < \delta \implies \map {d_N} {\map f x, \map f a} < \epsilon$

Hence, $f$ is continuous at every point of $M$.