User:Fake Proof/Sandbox/Uniform Limit Theorem

Theorem
Let $\struct {S, \tau}$ be a topological space.

Let $\struct {M, d}$ be a metric space.

Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of mappings from $S$ to $M$ such that:
 * $(1): \quad \forall n \in \N: f_n$ is continuous at every point of $S$
 * $(2): \quad \sequence {f_n}$ converges uniformly to $f$

Then $f$ is continuous at every point of $S$.

Proof
Let $a \in S$.

Let $V$ be a neighborhood of $\map f a$ in $\struct {M, d}$.

Then there exists $\epsilon \in \R_{>0}$ such that:
 * $(3): \quad \map {B_\epsilon} {\map f a} \subseteq V$

where $\map {B_\epsilon} {\map f a}$ denotes the open $\epsilon$-ball of $\map f a$ in $\struct {M, d}$.

From $(1)$, there exists a sequence $\sequence {U_n}_{n \mathop \in \N}$ of neighborhoods of $a$ in $\struct {S, \tau}$ such that:
 * $(4): \quad \forall n \in \N: \forall x \in U_n: \map d {\map {f_n} x, \map {f_n} a} < \dfrac \epsilon 3$

From $(2)$, there exists $N \in \R_{>0}$ such that:

Then:

Hence $f$ is continuous at every point of $S$.