TITLE: How to prove that the uniform topology is different from both the product and the box topology?
QUESTION [8 upvotes]: Let $J$ be an arbitrary index set. Then how to prove that the uniform topology on the Cartesian product $\mathbf{R}^J$ of the set $\mathbf{R}$ of real numbers with itself is different from both the product and the box topologies when $J$ is infinite? 
It's already known to me how the uniform topology is finer than the product topology and coarser than the box topology. 
The uniform metric $d\colon \mathbf{R}^J \times \mathbf{R}^J \to \mathbf{R}$ is given by $$ d(x,y) := \sup_{\alpha\in J} \min(1, |x_{\alpha}-y_{\alpha}|) $$ for any two points $x := (x_{\alpha})_{\alpha\in J}$, $\,$ $y := (y_{\alpha})_{\alpha\in J}$ $\,$ in $\mathbf{R}^J$.

REPLY [2 votes]: Here is an example to demonstrate that the uniform topology is different from the box topology when $J = \omega$. It is from the exercise 6 of section 20 titled "The Metric Topology" of Munkres's Topology.

Given $\mathbf{x} = (x_1, x_2, \ldots) \in \mathbb{R}^{\omega}$ and given $0 < \epsilon < 1$, let $$U(\mathbf{x,\epsilon}) = (x_1 - \epsilon, x_1 + \epsilon) \times \cdots \times (x_n - \epsilon, x_n + \epsilon) \times \cdots.$$

$U(\mathbf{x}, \epsilon)$ is open in box product, but not in the uniform topology.
For the reason why $U(\mathbf{x}, \epsilon)$ is not open in the uniform topology, consider the particular point $(x_1 + \epsilon/2, x_2 + 2\epsilon/3, x_3 + 3\epsilon/4, \cdots)$ which is obviously in $U(\mathbf{x}, \epsilon)$. However, there is no ball (in the uniform topology) centered around it that is in $U(\mathbf{x}, \epsilon)$.