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

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Theorem

Let $S \subseteq \R$.

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

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


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


Proof

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

Since $f_n \to f$ uniformly, there exists some $N \in \N$ such that:

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

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

Then if $x \in S$ and $n, m > N$, we have:

\(\ds \size {\map {f_n} x - \map {f_m} x}\) \(=\) \(\ds \size {\map {f_n} x - \map f x - \paren {\map {f_m} x - \map f x} }\)
\(\ds \) \(=\) \(\ds \size {\map {f_n} x - \map f x} + \size {\map {f_m} x - \map f x}\) Triangle Inequality for Real Numbers
\(\ds \) \(<\) \(\ds \epsilon\)

Since $\epsilon$ was arbitrary, $\sequence {f_n}$ is uniformly Cauchy on $S$.

$\blacksquare$