The uniform topology on $\Bbb{R}^J$ is finer than the product topology and coarser than the box topology; these three topologies are all different if $J$ is infinite.
My attempt: let $B\in \mathcal{B}_p$ and $x=(x_\alpha )\in B$. $B=\prod_{\alpha \in J}U_\alpha ,U_\alpha \in \mathcal{T}_\lt$ and $U_\alpha =\Bbb{R}, \forall \alpha \in J\setminus F$. So $x_\alpha \in U_\alpha, \forall \alpha \in F$. By definition of standard/order topology on $\Bbb{R}$, $\exists B_\alpha =(a_\alpha ,b_\alpha)\in \mathcal{B}_s$(basis for standard topology) such that $x_\alpha \in (a_\alpha ,b_\alpha)\subseteq U_\alpha, \forall \alpha \in F$. $\forall \alpha ,\exists \epsilon_\alpha \in \Bbb{R}^{+}$ such that $x_\alpha \in (x_\alpha -\epsilon_\alpha , x_\alpha +\epsilon_\alpha)\subseteq (a_\alpha ,b_\alpha)$. Choose/take $\epsilon =\min (\{ \epsilon_\alpha \} \cup \{ 1\})$. Claim: $B_{\overline{\rho}}(x,\epsilon )\subseteq B$. Proof: let $y\in B_{\overline{\rho}}(x,\epsilon)$. Then $\overline{d} (x_\alpha, y_\alpha) \leq \overline{\rho}(x,y) \lt \epsilon \leq 1$. So $\overline{d}(x_\alpha ,y_\alpha)\lt 1\Rightarrow \overline{d}(x_\alpha ,y_\alpha) = d(x_\alpha ,y_\alpha), \forall \alpha \in F$. Which implies $d(x_\alpha ,y_\alpha) \lt \epsilon$, $y_\alpha \in (x_\alpha -\epsilon, x_\alpha +\epsilon)\subseteq (x_\alpha -\epsilon_\alpha , x_\alpha +\epsilon_\alpha)\subseteq (a_\alpha ,b_\alpha), \forall \alpha \in F$ Since $\epsilon \leq \epsilon_\alpha ,\forall \alpha \in F$. Thus $y=(y_\alpha)\in B=\prod U_\alpha$. Is this proof correct? This proof is slight variation of proof given in Munkres and it is also little bit clever. Munkres proof: $x=(x_\alpha)\in \prod U_\alpha$. Since $U_\alpha \in \mathcal{T}_\lt$ and $\mathcal{T}_\lt =\mathcal{T}_d = \mathcal{T}_{\overline{d}}$(first equality follows from theorem 20.3 and second equality follows from theorem 20.1) ,$\forall \alpha \in F$, we have $\exists B_{\overline{d}}(p_\alpha, r_\alpha)$ such that $x_\alpha \in B_{\overline{d}}(p_\alpha, r_\alpha)\subseteq U_\alpha$. Since $B_{\overline{d}}(p_\alpha, r_\alpha)$ is open in metric space notion, $\exists \delta_\alpha$ such that $B_{\overline{d}}(x_\alpha, \delta_\alpha)\subseteq B_{\overline{d}}(p_\alpha, r_\alpha) \subseteq U_\alpha, \forall \alpha \in F$. Now choose $\epsilon =\min_{\alpha \in F}\{ \delta_\alpha \}$. So that $x\in B_{\overline{\rho}} (x,\epsilon)\subseteq \prod U_\alpha$, our desired result.
I don’t understand the proof of coarser part. Munkres didn’t “properly” use the lemma 13.3. Let take general basis element of uniform topology, $B_{\overline{\rho}}(x,\epsilon)$. Let $y\in B_{\overline{\rho}}(x,\epsilon)$. Then what to take for basis element of box topology, $B^\prime$, such that $y\in B^\prime \subseteq B_{\overline{\rho}}(x,\epsilon)$?
Claim: If $J$ is infinite, then uniform topology is strictly finer than product topology i.e. both topology are different. Proof: Assume toward the contradiction, i.e. $\mathcal{T}_{\overline{\rho}}\subseteq \mathcal{T}_p$. Let $B_{\overline{\rho}}(0,1)$ and $y\in B_{\overline{\rho}}(0,1)$. Then $\overline{\rho}(0,y)=\sup \{ \min \{ |y_\alpha|,1 \} |\alpha \in J\}\lt 1$. So $|y_\alpha|\lt 1$. Assume $\exists B^\prime =\prod U_\alpha \in \mathcal{T}_p$ such that $y\in B^\prime \subseteq B_{\overline{\rho}}(0,1)$. Take $z=(z_\alpha)$ so that $y_\alpha =z_\alpha, \forall \alpha \in F$ and $z_\alpha =1, \forall \alpha \in J\setminus F$. Then clearly $z\in B^\prime =\prod U_\alpha$ but $z\notin B_{\overline{\rho}}(0,1)$. Thus we reached contradiction. $\nexists B^\prime$ such that …… Hence they are different. Is this proof correct?
I haven’t proved(cause I don’t know) the same thing for uniform topology and box topology. I hope proof is easy as previous one.
If you have time, then please also help me in Theorem 20.3 of Munkres’ Topology post.