# 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 Metric Space Axiom $\text M 2$ twice:

 $\text {(3)}: \quad$ $\ds \forall n \in \N, \forall x \in M: \,$ $\ds \map {d_N} {\map f x, \map f a}$ $\le$ $\ds \map {d_N} {\map f x, \map {f_n} x} + \map {d_N} {\map {f_n} x, \map {f_n} a} + \map {d_N} {\map {f_n} a, \map f a}$

Let $\epsilon \in \R_{>0}$.

Since $\sequence {f_n}$ converges uniformly to $f$:

 $\text {(4 a)}: \quad$ $\ds \exists \NN \in \R_{>0}: \forall n \in \N_{>\NN}: \forall x \in M: \,$ $\ds \map {d_N} {\map f x, \map {f_n} x}$ $<$ $\ds \frac \epsilon 3$ $\text {(4 b)}: \quad$ $\ds \leadsto \ \$ $\ds \exists \NN \in \R_{>0}: \forall n \in \N_{>\NN}: \,$ $\ds \map {d_N} {\map f a, \map {f_n} a}$ $<$ $\ds \frac \epsilon 3$ Universal Instantiation of $x$

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

Hence:

 $\text {(5)}: \quad$ $\ds \forall n \in \N: \exists \delta \in \R_{>0}: \forall x \in M: \,$ $\ds \map {d_M} {x, a}$ $<$ $\ds \delta$ $\ds \implies \ \$ $\ds \map {d_N} {\map {f_n} x, \map {f_n} a}$ $<$ $\ds \frac \epsilon 3$

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

 $\ds \exists \NN \in \R_{>0}: \forall n \in \N_{>\NN}: \exists \delta \in \R_{>0}: \forall x \in M: \,$ $\ds \map {d_M} {x, a}$ $<$ $\ds \delta$ $\ds \implies \ \$ $\ds \map {d_N} {\map f x, \map f a}$ $<$ $\ds \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3$ $\ds$ $=$ $\ds \epsilon$

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

 $\ds \forall a \in M: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in M: \,$ $\ds \map {d_M} {x, a}$ $<$ $\ds \delta$ $\ds \implies \ \$ $\ds \map {d_N} {\map f x, \map f a}$ $<$ $\ds \epsilon$

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

$\blacksquare$