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## Proof: (related to "Uniform Convergence Criterion of Cauchy")

### $”\Leftarrow”$

• Assume, for every $\epsilon > 0$ there is an index $N$ such that the supremum norm $||f_n-f_m||_\infty < \epsilon$ for all $n,m\ge N.$
• This means that $|f_n(x)-f_m(x)|<\epsilon$ for all $x\in D$ and all $n,m\ge N.$
• By definition of Cauchy sequences, this means that $(f_n(x))_{n\in\mathbb N}$ is a Cauchy sequence for all $x\in D.$
• By the completeness principle, this means that there is a limit $y:=\lim_{n\to\infty}f_n(x)$ for all $x\in D.$
• In other words, there is a function $f:D\to\mathbb F$ to which the sequence of functions $f_n:D\to\mathbb F$ is uniformly convergent.
q.e.d

| | | | created: 2020-02-25 18:45:07 | modified: 2020-02-25 18:47:39 | by: | references: [586]