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!
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 $U\notin\mathscr{T}_u$.