# About the definition of uniform convergence

Following a reference from "General Topology" by Ryszard Engelking.

Uniform convergence

Let be $$X$$ a topological space and $$\{f_n\}_{n\in\mathbb{N}}$$ a sequence of functions from $$X$$ to $$I\subseteq\mathbb{R}$$. So we say that the sequence $$\{f_n\}_{n\in\mathbb{N}}$$ is uniformly convergent to a real-valued function $$f$$ if for every $$\epsilon>0$$ there exist a $$m$$ such that we have $$|f(x)-f_n(x)|<\epsilon$$ for every $$x\in X$$ and $$n\ge m$$.

Anyway in other text it is used an another definition of uniform convergence. Infact if $$\{f_n\}_{n\in\mathbb{N}}$$ is a sequence of continuous function from $$X$$ to $$I\subseteq\mathbb{R}$$ we can regard it as an element of the space of the continuous function from $$X$$ to $$I$$, or rather $$\{f_n\}_{n\in\mathbb{N}}\in\mathbf{C}(I^X)$$, that is a metric space where the distance function is defined as $$d(f,g)_\infty=\mathscr{sup}\{|f(x)-g(x)|:x\in X\}$$ for any $$f,g\in\mathbf{C}(I^X)$$ and so using the topology that is induced by the metric $$d_\infty$$ we can say that the sequence $$\{f_n\}_{n\in\mathbb{N}}$$ converge to some $$f\in\mathbf{C}(I^X)$$ if for any open basic $$B(f,\epsilon)=\{g\in\mathbf{C}(I^X):d_\infty(f,g)<\epsilon\}$$ there exist $$m\in\mathbb{N}$$ such that $$f_n\in B(f,\epsilon)$$ for every $$n\ge m$$.

Well obviously this definition implies the uniform convergence since by the definition of supremum it is $$|f(x)-f_n(x)|<\mathscr{sup}\{|f(x)-f_n(x)|:x\in X\}$$. Howewer unfortunately I don't be able to prove that the two definition are equivalent: infact if for any $$\epsilon>0$$ there exist $$m\in\mathbb{N}$$ such that $$|f(x)-f_n(x)|<\epsilon$$ for any $$x\in X$$ and $$n\ge m$$, then $$\epsilon$$ is an upper bound of $$\{|f(x)-f_n(x)|:x\in X\}$$ and so by definition of supremum it is that $$\mathscr{sup}\{|f(x)-f_n(x)|:x\in X\}\le\epsilon$$. So how prove that it is $$\mathscr{sup}\{|f(x)-f_n(x)|:x\in X\}<\epsilon$$?

Take $$m\in\mathbb N$$ such that$$(\forall x\in X)(\forall n\in\mathbb N):n\geqslant m\implies\bigl\lvert f(x)-f_n(x)\bigr\rvert<\frac\varepsilon2.$$Then, if $$n\geqslant m$$,$$\sup_{x\in X}\bigl\lvert f(x)-f_n(x)\bigr\rvert\leqslant\frac\varepsilon2<\varepsilon.$$