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 $S$.
$\blacksquare$