Let $X=\{x=(x_i)_{i \in \mathbb{N}} \in \mathbb{R}^{\mathbb{N}} \ \vert \ \exists N \in \mathbb{N} : x_i \geq 0 \ \ \forall i \geq N\}$ and let $\bar{\rho}$ be the uniform metric on $\mathbb{R}^{\mathbb{N}}$. Is $(X, \bar{\rho})$ complete?
My effort: The uniform metric is $\bar{\rho} \colon \mathbb{R}^{\mathbb{N}}\times \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$ be defined as follows: $$ \tilde{\rho}( x, y) \colon= \sup \left\{ \ \min \left\{ \ \vert x_\alpha - y_\alpha \vert, \ 1 \ \right\} \ \colon \ \alpha \in \mathbb{N}\ \right\} \ \mbox{ for all } \ x \colon= \left(x_\alpha \right)_{\alpha \in \mathbb{N}}, \ y \colon= \left(y_\alpha \right)_{\alpha \in \mathbb{N}} \in \mathbb{R}^\mathbb{N}.$$ I have tried to prove that it is complete. Let $(x_n) \subseteq X$ be a Cauchy Sequence, and let $\varepsilon>0$, then $\exists K \in \mathbb{N}$ such that if $n,m \geq K$ then $\bar{\rho}(x_n,x_m)< \varepsilon$. Since $x_n \in X$, for all $n \in \mathbb{N}$ we have that $x_n=(x_i^{(n)})_{i \in \mathbb{N}}$ then $\exists N_n \in \mathbb{N}$ such that $x_i^{(n)} \geq 0$ for all $i \geq N_n$ and by the definition of supreme we have that $$\min\{ \vert x_i^{(n)}- x_i^{(m)} \vert, 1 \}< \varepsilon, \quad \forall i \in \mathbb{N}.$$ And this is what I have at the moment. Am I on the right way or do you think this space is not complete?