Sequence of Functions is Uniformly Cauchy iff Uniformly Convergent/Sufficient Condition

Theorem
Let $S \subseteq \R$.

Let $\sequence {f_n}$ be a sequence of real functions $S \to \R$.

Let $\sequence {f_n}$ be uniformly Cauchy on $S$.

Then $\sequence {f_n}$ is uniformly convergent on $S$.

Proof
Let $\epsilon \in \R_{> 0}$ be arbitrary.

As $\sequence {f_n}$ is uniformly Cauchy on $S$, there exists $N \in \N$ such that:


 * $\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$

for all $n, m > N$ and $x \in S$.

Let $x \in S$ be fixed.

Then:


 * $\size {\map {f_n} x - \map {f_m} x} < \epsilon$

for all $n, m > N$.

As $\epsilon$ was arbitrary, the sequence $\sequence {\map {f_n} x}$ is therefore Cauchy.

By Cauchy's Convergence Criterion on Real Numbers, it follows that $\sequence {\map {f_n} x}$ is convergent.

Define a function $f : S \to \R$ by:


 * $\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$

for all $x \in S$.

We aim to show that $f_n \to f$ uniformly.

For all $x \in S$, we have:


 * $\ds \lim_{m \mathop \to \infty} \size {\map {f_n} x - \map {f_m} x} = \size {\map {f_n} x - \map f x}$

We established that for all $\epsilon \in \R_{> 0}$ we can find $N \in \N$ such that:


 * $\size {\map {f_n} x - \map {f_m} x} < \dfrac \epsilon 2$

for $x \in S$ and $n, m > N$.

We therefore have:


 * $\size {\map {f_n} x - \map f x} \le \dfrac \epsilon 2 < \epsilon$

for all $x \in S$ and $n > N$.

So $\sequence {f_n}$ converges uniformly to $f$ on $S$.