Uniform Limit Theorem

Theorem
Let $\left({M, d_M}\right)$ and $\left({N, d_N}\right)$ be metric spaces.

Let $M^N$ be the set of all mappings from $M$ to $N$.

Let $\left \langle{f_n}\right \rangle \subset M^N$ be a sequence of mappings from $M$ to $N$ such that:
 * $(1): \quad \forall n \in \N: f_n$ is continuous at every point of $M$
 * $(2): \quad \forall n \in \N: f_n$ converges uniformly $f$.

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

Proof
Let $a \in M$.

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

So:
 * $\forall x, y, z \in M: d_N \left({x, z}\right) \le d_N \left({x, y}\right) + d_N \left({y, z}\right)$

We apply this property twice to assert that $\forall n \in \N, \forall x \in M$, we have:

Let $\epsilon >0$.

Since $f_n \to f$ uniformly:
 * $\exists \mathcal N \in \R: \forall n \ge \mathcal N: \forall x \in M$:

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

So $\exists \delta > 0: \forall x \in M$:

By combining $(1)$, $(2 a)$, $(2 b)$ and $(3)$:
 * $\exists \delta > 0$ and $\exists n$ sufficiently large such that $\forall x \in M$:

As $a$ and $\epsilon$ are arbitrary, it follows that:
 * $\forall a \in M: \forall \epsilon > 0: \exists \delta > 0: \forall x \in M$:

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