Once the notion of the uniform metric on $\mathbb{R}^{X}$ is clarified, the rest is an easy implication. By definition, $\bar{\rho}(f,g)=\sup_{x\in X}\{\min\{|f(x)-g(x)|,1\}\}$ so that for every $\epsilon\le1$ , $g\in B_{\overline{\rho}}(f,\epsilon)$ iff there is some $0<\delta<\epsilon$ such that for every $x\in X$ , $|f(x)-g(x)|<\delta$ . Now, it is clear that if for every $0<\epsilon\le1$ starting from some $N$ , $f_{n}\in B_{\bar{\rho}}(f,\epsilon)$ , then for $n\ge N$ , $|f_{n}(x)-f(x)|<\epsilon$ for every $x\in X$ , which is just the definition of uniform convergence. And vice versa, if $f_{n}$ converges uniformly to $f$ then for every $\epsilon>0$ starting from some $N$ , $|f_{n}(x)-f(x)|<\tfrac{\epsilon}{2}$ for every $x\in X$ , then for $n\ge N$ , $\bar{\rho}(f_{n},f)=\sup_{x\in X}\{\min\{|f_{n}(x)-f(x)|,1\}\}\le\tfrac{\epsilon}{2}<\epsilon$ and $f_{n}\rightarrow f$ . Generally speaking, the definition of uniform convergence is (almost) the same as the definition of convergence in the uniform metric.