Are these two metric spaces complete? Let $J$ be an arbitrary non-empty set of indices, and let $\mathbb{R}^J$ denote the set of all the $J$-tuples of real numbers (i.e. the set of all the functions $x \colon J \to \mathbb{R}$, where we denote $x(\alpha)$ by $x_\alpha$, for each $\alpha \in J$). Now let the function $\tilde{\rho} \colon \mathbb{R}^J \times \mathbb{R}^J \to \mathbb{R}$ be defined as follows: 
$$ \tilde{\rho}( x, y) \colon= \sup \left\{ \ \min \left\{ \ \vert x_\alpha - y_\alpha \vert, \ 1 \ \right\} \ \colon \ \alpha \in J \ \right\} \ \mbox{ for all } \ x \colon= \left(x_\alpha \right)_{\alpha \in J}, \ y \colon= \left(y_\alpha \right)_{\alpha \in J} \in \mathbb{R}^J.$$
This function $\tilde{\rho}$ is of course a metric on $\mathbb{R}^J$, called the uniform metric. Is the metric space $\left( \ \mathbb{R}^J, \  \tilde{\rho} \ \right)$ complete?
My effort: 
Let $\left( x_n \right)_{n \in \mathbb{N}}$ be a Cauchy sequence in $\left( \ \mathbb{R}^J, \  \tilde{\rho} \ \right)$, and let $x_n \colon= \left( x_{\alpha n} \right)_{\alpha \in J}$, for each $n = 1, 2, 3, \ldots$. 
Then, given any real number $\epsilon > 0$, we can find a natural number $N = N(\epsilon)$ such that 
$$ \tilde{\rho}(x_m, x_n ) < \epsilon \ \mbox{ for all } \ m, n \in \mathbb{N} \ \mbox{ such that } \ m> N \ \mbox{ and } \ n> N. $$
Let's assume that $0< \epsilon < 1$. 
Let $\beta \in J$. Then for all $m, n \in \mathbb{N}$ such that $m > N$ and $n > N$, we have 
$$\epsilon > \tilde{\rho}(x_m, x_n) \geq  \min \left\{ \  \left\vert x_{\beta m} - x_{\beta n} \right\vert, \ 1 \ \right\} = \left\vert x_{\beta m} - x_{\beta n} \right\vert,$$
which shows that the sequence $\left(x_{\beta n} \right)_{n \in \mathbb{N}}$ is a Cauchy sequence in $\mathbb{R}$, and so this sequence converges to some real number $x_\beta$. 
Now since our $\beta \in J$ was arbitrary, let us define
$x \colon= \left( x_\alpha \right)_{\alpha \in J}$.
Is this logic correct so far? If so, then how to proceed from here, especially if the index set  $J$ is infinite?
Now let's take $J$ to be the set $\mathbb{N}$ of natural numbers, and let's denote $\mathbb{R}^J$ by $\mathbb{R}^\omega$ and define the function $D \colon \mathbb{R}^\omega \times \mathbb{R}^\omega \to \mathbb{R}$ by the formula
$$ D(x, y) \colon = \sup \left\{ \frac{ \min \left\{ \ \left\vert x_\alpha - y_\alpha  \right\vert, \ 1 \ \right\}}{\alpha } \ \colon \ \alpha  \in \mathbb{N} \ \right\} \ \mbox{ for all } \ x \colon= \left( x_\alpha  \right)_{\alpha \in \mathbb{N}}, \ y \colon= \left( y_\alpha \right)_{ \alpha \in \mathbb{N}} \in \mathbb{R}^\omega.$$
Then $D$ is also a metric on $\mathbb{R}^\omega$, and the topology induced by $D$ is the same as the product topology on $\mathbb{R}^\omega$. Is the metric space $\left( \ \mathbb{R}^\omega, \ D \ \right)$ complete? 
Let $\left(x_n \right)_{n \in \mathbb{N}}$, where $x_n \colon= \left( x_{\alpha n} \right)_{\alpha \in \mathbb{N}}$ for each $n \in \mathbb{N}$, be a Cauchy sequence in $\left( \ \mathbb{R}^\omega, \ D \ \right)$. Let $\beta$ be a fixed natural number. 
Then, given any real number $\epsilon > 0$, we can find a natural number $N = N(\epsilon)$ such that
$$ D(x_m, x_n) < \frac{ \epsilon}{ \beta} \ \mbox{ for all } \ m, n \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \  n > N. $$
Let's assume that $0 < \epsilon < 1$. Then for all natural numbers $m$ and $n$ such that $m, n > N$, we have 
$$\left\vert x_{\beta m} - x_{\beta n} \right\vert < \epsilon,$$
from which it follows that the sequence $\left( x_{\beta n} \right)_{ n \in \mathbb{N}}$ is a Cauchy sequence of real numbers and therefore this sequence converges to some real number $x_\beta$. 
Since our $\beta \in \mathbb{N}$ was arbitrary, we can define an element $x \in \mathbb{R}^\omega$ as $$x \colon= \left( x_\alpha \right)_{\alpha \in \mathbb{N}}.$$
Is my logic correct so far? If so, then how to proceed from here? 
What if, in either of the above two cases, we replace $\mathbb{R}$ by an arbitrary complete metric space (and replace $\vert \cdot \vert$ by the corresponding distance of course)? Do we still have a metric in either case? And, if so, is (are) this (these) metric(s) complete? 
 A: Yes, your logic is correct so far, for both metrics. Actually, your questions are answered (positively) here. There (in theorem 7.11) you can also find the argument to show the Cauchy sequence $(x_n)_n\subset \mathbb{R}^J$ converges to $x$, and thus, conclude $(\mathbb{R}^J,\tilde{\rho})$ is complete. 
To conclude that $(\mathbb{R}^\omega,D)$ is a complete metric space, let $x=(x_\alpha)_{\alpha\in\mathbb{N}}$ be the pointwise limit of the Cauchy sequence $(x_n)_n\subset \mathbb{R}^\omega$. Let $1>\epsilon>0$ and $\alpha_0$ s.t. $\frac{1}{\alpha_0}<\epsilon$. Since 
$$
D(x_n,x)=\sup_\alpha \left\{\frac{\min\{1,|x_{n\alpha}-x_\alpha|\}}{\alpha}\right\}
$$
we only need to worry about $|x_{n\alpha}-x_\alpha|$ for $\alpha\in\{1,...\alpha_0\}$. Since $(x_{n\alpha})_n$ converges to $x_\alpha$ for all those indices, then $\exists N,\forall n>N, \forall \alpha \in \{1,...,\alpha_0\},|x_{n\alpha}-x_\alpha|<\epsilon$. Hence, we have shown that
$$
\forall \epsilon>0, \exists N, \forall n>N, D(x_n,x)<\epsilon 
$$
ie $x_n\to x$ in $(\mathbb{R}^\omega,D)$.
