Why is the box topology finer than the product topology? I am having a hard time understanding why the box topology is finer the the product topology. Of course I know that with finite product, the two are the same. With product topology, the basis elements are defined so that the $U_{\alpha}=X_\alpha$ except for finitely many $\alpha$. So how can this topology be contained in the box topology which has only the product of the  $U_{\alpha}$ as basis. I think it should have been the other way round. Am I missing something here? I need help with this because I have been confused over it for a long time.

Fix a set $X=\prod_{\alpha\in I} X_\alpha$ where $I$ is infinite, and consider the two topologies on it.
The box topology is finer than the product topology because if $U_\alpha\subset X_\alpha$ is a proper subset, then $U=\prod_\alpha U_\alpha$ will be open in the box topology (and is basic), but it is not open in the product topology.  Such a set cannot be open in the product topology, because every open set in the product topology is the union basic open sets, each of which has $U_\alpha=X_\alpha$ for all but finitely many terms.  In particular, this means any open set of $X$ in the product topology must contain a subset of the form $\prod_\alpha U_\alpha$ where $U_\alpha=X_\alpha$ for almost all $\alpha$.
