Uniform Limit Theorem

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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 $(\text M 2)$: Triangle Inequality 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 $M$.

$\blacksquare$


Sources