Product space, product and uniform topology -- open sets?

Question

Consider a countable product of (compact) spaces $X= \Pi_{j\in J}X_j$. Let it have the uniform topology. An open ball in $X$ is then given by

$$B^u_\epsilon(x):=\{x'\in X: \sup_{j\in J}d(x_j,x'_j)<\epsilon \}$$

We know that the uniform topology is strictly finer than the product topology on $X$.

My question is how to represent any given open ball $B^u_\epsilon$ in terms of open balls that are open also in the product topology?

Answer 1

For $j\in J$ and $\delta>0$ let $U_j(\delta)=\{x'\in X:d_j(x_j,x_j')<\delta\}$; these sets are open in the product topology. For $\delta>0$ let $B(\delta)=\bigcap_{j\in J}U_j(\delta)$; these sets are open in the box topology. Then

$$B_\epsilon^u(x)=\bigcup_{0<\delta<\epsilon}B(\delta)=\bigcup_{0<\delta<\epsilon}\bigcap_{j\in J}U_j(\delta)\;.$$

Answer 2

I believe, it not because take $(0,1)^J$ where is $J$ any set. Note it is open in unform metric which is not open that product