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 can apply this property twice to assert that $\forall n\in\N,\forall x\in M$, we have:
 * $d_N(f(x),f(a))\leq d_N(f(x),f_n(x))+d_N(f_n(x),f_n(a))+d_N(f_n(a),f(a)). \hspace{10mm} (1)$

Let $\epsilon >0$.

By the definition of uniform convergence, we know $\exists\mathcal{N}\in\R$ such that $\forall n\geq \mathcal{N},\forall x\in M$, we have:
 * $\displaystyle d_N(f(a),f_n(a))<\frac{\epsilon}{3}$, and $\displaystyle d_N(f(x),f_n(x))<\frac{\epsilon}{3}. \hspace{10mm}(2)$

$\forall n\in\N$, by the definition of continuity on metric spaces, we know $\exists \delta >0$ such that $\forall x \in M$, we have:
 * $\displaystyle d_M(x,a)<\delta \implies d_N(f_n(x),f_n(a))<\frac{\epsilon}{3}. \hspace{10mm}(3)$

By combining $(1),(2)$ and $(3)$, we see that $\exists \delta >0$ and $\exists n$ sufficiently large such that $\forall x\in M$,
 * $\displaystyle d_M(x,a)<\delta \implies d_N(f(x),f(a))\leq \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon.$

As $a$ and $\epsilon$ are arbitrary, it follows that, $\forall a\in M,\forall \epsilon >0,\exists \delta>0,\forall x\in M$,
 * $\displaystyle d_M(x,a)<\delta \implies d_N(f(x),f(a))\leq \epsilon.$

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