# Cauchy Criterion for Uniform Convergence of Functions

I have proved the following and I would like to know if I have made any mistakes:

A sequence of functions $$(f_n)$$ defined on a set $$A\subseteq\mathbb{R}$$ converges uniformly on $$A$$ if and only if for every $$\varepsilon>0$$ there exists an $$N\in\mathbb{N}$$ such that $$|f_n(x)-f_m(x)|<\varepsilon$$ whenever $$m, n\geq N$$ and $$x\in A$$.

My proof (NOTE: edited leftward implication according to answer below):

$$\Rightarrow$$ Let $$\varepsilon>0$$: then by hypothesis there exists a limit function $$f$$ and $$N\in\mathbb{N}$$ such that $$|f_n(x)-f(x)|<\frac{\varepsilon}{2}$$ for all $$n\geq N$$ and $$x\in A$$ so if we pick $$m\geq N$$ we have also that $$|f_m(x)-f(x)|<\frac{\varepsilon}{2}$$ for all $$m\geq N$$ and $$x\in A$$ so $$|f_n(x)-f_m(x)|=|f_n(x)-f(x)+f(x)-f_m(x)|\leq |f_n(x)-f(x)|+|f(x)-f_m(x)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$ for all $$m,n\geq N$$ and $$x\in A$$, as desired.

$$\Leftarrow$$ Fix $$x\in A$$: then $$(f_n(x))_{n\in\mathbb{N}}$$ is a Cauchy sequence of real numbers which is convergent by the Cauchy Criterion for sequences so the function $$f:A\to\mathbb{R}, f(x):= \lim_{n\to\infty} f_n(x)$$ is well defined for every $$x\in A$$. Now, let $$\varepsilon>0$$: by hypothesis we have that there is $$N\in\mathbb{N}$$ such that $$|f_n(x)-f_m(x)|<\frac{\varepsilon}{2}$$ for all $$m,n\geq N$$ and $$x\in A$$ so $$|f(x)-f_n(x)|=\lim_{m\to\infty}|f_m(x)-f_n(x)|\leq\frac{\varepsilon}{2}<\varepsilon$$ for $$n\geq N$$ and $$x\in A$$, as desired.

Thank you.

In the $$\Leftarrow$$ direction, note that your $$N$$ may depend on $$x$$. You write (red additions by me):
So, if $$x\in A$$ and $$\varepsilon>0$$, there exists $$N\color{red}{(x,\varepsilon)}\in\mathbb{N}$$ such that $$|f_n(x)-f(x)|<\frac{\varepsilon}{2}$$ for all $$n\geq N$$ (since $$\lim_{n\to\infty} f_n(x)=f(x)$$ for $$x\in A$$).
You have to show that there is an $$N \in \mathbb N$$ that does the trick for any $$x \in A$$. Let $$\varepsilon > 0$$, by the uniform Cauchy condition there is $$N \in \mathbb N$$ such that $$\left|f_n(x)- f_m(x) \right| < \varepsilon, \qquad n, m \ge N, x \in A.$$ (Note that $$N$$ is independent of $$x$$). Now let $$x \in A$$. As $$f_m(x) \to f(x)$$ for $$m \to \infty$$, we have $$\left|f_n(x)- f(x) \right| = \lim_{m \to \infty} \left|f_n(x)- f_m(x) \right|\le \varepsilon$$ That is $$\left|f_n(x)- f(x) \right| \le \varepsilon, \qquad n \ge N, x \in A.$$