Why is the box topology finer than the product topology?

Question

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.

Answer 1

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 $\alpha$.

Answer 2

In box topology, the open sets are defined through a basis which involves product of open sets. Now product of the open sets we can do with any size of index set.

BUT~!

In product topology, we define the topology's sub basis as pre-image of the projections. Now, the thing is that when we generate the topology back from sub basis, we can only take finite intersections!

Now, it is clear that we can't hit many of the thing in product of some infinite sequence of open sets by just finite intersections, hence it is intuitive that box is bigger than product. QED.