box topology vs product topology Given a countably infinite family of topological spaces $(X_1,\tau_1),..,(X_n,\tau_n),...$, and their product $X$, I read that the box topology has as its basis the family:
$$
B' = \prod_{i=1}^{\infty}O_i : O_i \in \tau_i
$$
... and that the box topology is not compact, but the product topology is compact. The example where box topology is not compact is when each $(X_i,\tau_i)$ is a compact finite discrete space - which leads to a non-compact infinite discrete topology over $X$.
But under the  product topology (where $O_i\neq X_i$ for only finitely many $i$ and all $X_i$ are compact), suppose that we use the basis $(O_1,...,O_{i-1},U_i,O_{i+1},...)$, where $U_i \neq X_i$, but $O_j=X_j$ for all $j \neq i$. A countable intersection of these bases is $(U_1,U_2,U_3,...,U_n,...)$ which can possibly lead to an infinite set of open covers with no finite subcovers.
But am I right that  $(U_1,U_2,U_3,...,U_n,...)$ is no longer open in the product topology? --i.e. since the infinite intersection of open sets is not necessarily open. This implies that we won't get the 'non-compactness' of a box topology.
 A: $(1).$ Unless some $U_i$ is empty, the set $U=\prod_{i\in \Bbb N}U_i$ is not open in the  product topology:
Suppose $U$ is open and $p\in U.$ Then $p\in C\subset U$ for some basic open set $C=\prod_{i\in \Bbb N}C_i$ where each $C_i$ is non-empty & open in $X_i$ and the set $D=\{i:C_i\ne X_i\}$ is finite.
The key is that D is not empty. Consider, for some (any) $j\in D,$ the projection $p_j:\prod_{i\in \Bbb N} X_i\to X_j$  where $p_j(x_1, x_2, x_3,...)=x_j.$ The image $p_j(C)$ of $C$ under $p_j$ is a subset of the image $p_j(U)$ because $C\subset U.$ But then $X_j=p_j(C)\subset p_j(U)=U_j,$ which is false.
In other words: If $U$ is not empty then for any $j,$ the set of the $j$-th co-ordinates of the members of $U$ is $U_j,$ but if $U$ is open and non-empty then for any $j\in D,$ the set of the $j$-th co-ordinates of the members of $U$ is $X_j.$
$(2).$ It is possible that the box topology is compact. For example if $X_i$ is the only non-empty open subset of each $X_i$ then the only non-empty open subset of $X=\prod_{i\in \Bbb N}X_i$ in the box topology is $X$ itself.
On the other hand, suppose each $X_i$ is a two-point discrete space (which is obviously compact). Then the box topology on $\prod_{i\in \Bbb N}X_i$ is an infinite discrete space (which is obviously not compact). 
A: Assume $U=(U_1,U_2,U_3,...,U_n,...)$ is open , then there must be an topology basis $O \subset U$ . This leads to a contradiction since only finte coordinate of $O$ can be subset of $U_i$ . 
