Uniform Limit Theorem

Theorem
Let $(M,d_M)$ and $(N,d_N)$ be metric spaces and $M^N$ be the set of all mappings from $M$ to $N$.

Let $\langle f_n\rangle\subset M^N:\forall n, f_n$ is continuous at every point of $M$ and $f_n\to f$ uniformly.

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

Proof
Let $a\in M$

$d_N$ is a metric on $M$, so $\forall x,y,z\in M$, we have $d_N(x,z)\leq d_N(x,y)+d_N(y,z)$.

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, $\exists\mathcal{N}\in\R:\forall n\geq \mathcal{N},\forall x\in M$,

$\forall n\in\N,f_n$ 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 $M$.