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 $\mathscr{sup}\{|f(x)-f_n(x)|:x\in X\}<\epsilon$?

Could someone help me, please?


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 $n\geqslant m$,$$\sup_{x\in X}\bigl\lvert f(x)-f_n(x)\bigr\rvert\leqslant\frac\varepsilon2<\varepsilon.$$


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