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:
 * $(1): \quad \forall n \in \N: f_n$ is continuous at every point of $M$
 * $(2): \quad \sequence {f_n}$ 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 $N$.

By applying twice:

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.

Hence:

Combining $(3)$, $\text {(4 a)}$, $\text {(4 b)}$ and $(5)$:

As $a$ and $\epsilon$ are arbitrary, it follows by Universal Instantiation of $n$ that:

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