Box topology is finer than the uniform topology on $\mathbb{R}^\mathbb{N}$ This time, I wish to show that the box topology is finer than the uniform topology on countable Cartesian products on $\mathbb{R}$ denoted $\mathbb{R}^\mathbb{N}$
However, the problem here is that the box topology is not metrizable, so we cannot compare metric balls...
All is not lost, because we know what those basic open sets are


*

*The base on the box topology on $\mathbb{R}^\mathbb{N}$ is:


$$B_b = \Bigg\{\prod_{n \in \mathbb{N}} U_n: U_n \subset \mathbb{R} \text{ is open }\Bigg\}$$


*

*The base on the uniform topology on $\mathbb{R}^\mathbb{N}$ :


$$B_u =  \Bigg\{B_\epsilon^u(x):x \in \mathbb{R}^\mathbb{N},  \epsilon > 0  \Bigg\}$$
where $$B_\epsilon^u(x) = \{y \in \mathbb{R}^\mathbb{N}|d_u(x,y)<\epsilon\} = \{y \in \mathbb{R}^\mathbb{N}|  \sup_{n\in \mathbb{N}}(\min\{1, |x_n-y_n|\})< \epsilon\}$$
$$d_u(x,y) = \sup_{n\in \mathbb{N}}(\min\{1, |x_n-y_n|\})$$

We wish to show that $\mathcal{T}_u \subseteq \mathcal{T}_b
 \Leftrightarrow B_b \subseteq B_u$

Proof attempt: 

  
*
  
*We want to take $U \in B_b$ and show that $U \in B_u$. 
  
*Let $U \in B_b$, then $U = \prod_{n \in \mathbb{N}} U_n$, for some    countable collection of $\{U_n\}_{n \in \mathbb{N}}$
     (err....we have  a basic open set defined in term of components, but    the basic open sets in uniform topology is not)
  
*Then for each $x \in U$, $\exists \epsilon_x > 0$, such that    $B_{\epsilon_x}^u(x) \subset U$. 
  
*Claim: $\bigcup_{x \in U} B_{\epsilon_x}^u(x) \in B_u$ 
  
*Then $U \subset \bigcup_{x \in U} B_{\epsilon_x}^u(x) \in B_u$

I am not confident that this proof works, because it seems we can also pick any ball in the uniform topology and find a collection of $\{U_n\}$ such that their union contains the ball...How do I fix this proof. Thanks!
 A: You appear to be trying to prove the wrong thing. To show that the box topology is at least as fine as the uniform topology, you need to show that $\mathscr{T}_u\subseteq\mathscr{T}_b$, and if you want to show that it’s strictly finer, you have to show further that $\mathscr{T}_b\setminus\mathscr{T}_u\ne\varnothing$.
Let $U\in\mathscr{T}_u$; if $U=\varnothing$, then certainly $U\in\mathscr{T}_b$, so assume that $U\ne\varnothing$, and let $x\in U$; there is an $\epsilon(x)\in(0,1)$ such that $B_{2\epsilon(x)}^u(x)\subseteq U$. Let 
$$V_x=\big(x-\epsilon(x),x+\epsilon(x)\big)^{\Bbb N}\;;$$
clearly $V_x\in\mathscr{T}_b$, and $d_u(x,y)\le\epsilon<2\epsilon$ for each $y\in V_x$, so $x\in V_x\subseteq U$. Thus, 
$$U=\bigcup_{x\in U}V_x\in\mathscr{T}_b\;,$$ 
and $\mathscr{T}_u\subseteq\mathscr{T}_b$.
To show that $\mathscr{T}_b\setminus\mathscr{T}_u\ne\varnothing$, let
$$\begin{align*}
U&=\prod_{n\in\Bbb N}\left(-\frac1{n+1},\frac1{n+1}\right)\\
&=(-1,1)\times\left(-\frac12,\frac12\right)\times\left(-\frac13,\frac13\right)\times\ldots\;;
\end{align*}$$
I’ll leave it to you to verify that $U\notin\mathscr{T}_u$.
A: My verification for the last comment from Brian M. Scott.
Will this work?
There does not exist an $\epsilon$ small enough such that the $B_{\bar{\rho}}(x_n,\epsilon)$ as $n \to \infty$ is a superset of $(\frac{-1}{n+1}, \frac{1}{n+1})$ as $n \to \infty.$
