# Uniform Limit Theorem

## Theorem

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

Let $\left \langle{f_n}\right \rangle$ 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 $M$
$(2): \quad \langle f_n\rangle$ 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:

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

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

 $(1):\quad$ $\displaystyle d_N \left({f \left({x}\right), f \left({a}\right)}\right)$ $\le$ $\displaystyle d_N \left({f \left({x}\right), f_n \left({x}\right)}\right)$ $\displaystyle$ $+$ $\displaystyle d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right)$ $\displaystyle$ $+$ $\displaystyle d_N \left({f_n \left({a}\right), f \left({a}\right)}\right)$

Let $\epsilon >0$.

Since $f_n \to f$ uniformly:

$\exists \mathcal N \in \R: \forall n \ge \mathcal N: \forall x \in M$:
 $(2 a):\quad$ $\displaystyle d_N \left({f \left({a}\right), f_n \left({a}\right)}\right)$ $<$ $\displaystyle \frac \epsilon 3$ $(2 b):\quad$ $\displaystyle d_N \left({f \left({x}\right), f_n \left({x}\right)}\right)$ $<$ $\displaystyle \frac \epsilon 3$

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

So, for any $n\in\N$, $\exists \delta > 0: \forall x \in M$:

 $(3):\quad$ $\displaystyle d_M \left({x, a}\right)$ $<$ $\displaystyle \delta$ $\displaystyle \implies \ \$ $\displaystyle d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right)$ $<$ $\displaystyle \frac \epsilon 3$

By combining $(1)$, $(2 a)$, $(2 b)$ and $(3)$:

$\exists \delta > 0$ and $\exists n$ sufficiently large such that $\forall x \in M$:
 $\displaystyle d_M \left({x, a}\right)$ $<$ $\displaystyle \delta$ $\displaystyle \implies \ \$ $\displaystyle d_N \left({f \left({x}\right), f \left({a}\right)}\right)$ $\le$ $\displaystyle \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3$ $\displaystyle$ $=$ $\displaystyle \epsilon$

As $a$ and $\epsilon$ are arbitrary, it follows that:

$\forall a \in M: \forall \epsilon > 0: \exists \delta > 0: \forall x \in M$:
 $\displaystyle d_M \left({x, a}\right)$ $<$ $\displaystyle \delta$ $\displaystyle \implies \ \$ $\displaystyle d_N \left({f \left({x}\right), f \left({a}\right)}\right)$ $\le$ $\displaystyle \epsilon$

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

$\blacksquare$