# Is the box topology good for anything?

In point-set topology, one always learns about the box topology: the topology on an infinite product $X = \prod_{i \in I} X_i$ generated by sets of the form $U = \prod_{i \in I} U_i$, where $U_i \subset X_i$ is open. This seems naively like a “good” topology to use for $X$. However, one quickly learns that this is not so; that the product topology is the natural one.

The box topology has many strange properties that make it a good source for counterexamples, but I am not aware of it having any other applications. So I would like to know:

Are there examples of using the box topology to prove interesting “positive” statements?

Edit: And to pursue a comment of Jim Conant:

Are there “non-artificial” problems where the box topology arises naturally?

Edit: The title is perhaps too flippant. I don’t mean to minimize the obvious significance of the box topology as a counterexample. However, for the purposes of this question I am interested in positive results. I’m not looking to be convinced that counterexamples are useful; I know that they are.

A year after, I have somehow ended up here and Nate’s comment made me want to post an answer (this answer is to elaborate dfeuer’s comment). Box topology is indeed useful for other purposes as well other than generating counter-examples. One may show that on a function space where the codomain is a bounded metric space (or, e.g. in $C(X)$ with compact $X$ since continuous functions are bounded on compact sets), we have that the sup-metric topology is coarser than box-topology. So indeed, if anything can be shown in Box topology then it applies as well in the uniform:
Suppose we have a bounded metric space $(Y,d)$ and any set $J$. Denote the product set as $X=Y^{J}$, which is the collection of all functions $x:J\to Y$. Since $Y$ is bounded we may consider the sup-metric in $X$
which generates a topology $\tau_{\sup}$ with the following basis elements for $x\in X$ and $\varepsilon>0$
Every such is Box-open: take $y\in U(x,\varepsilon)$, whence $\delta:=\sup_{j\in J}d(x(j),y(j))<\varepsilon$. Choose $r=\frac{\delta+\varepsilon}{2}$, whence $\delta. Now since $d(x(j),y(j))\leq \delta < r$ for all $j$, then $y\in \Pi_{j\in J}B(x(j),r)\subset U(x,\varepsilon)$ and $\Pi_{j\in J}B(x(j),r)$ is certainly box-open, which shows that $U(x,\varepsilon)$ is open in the Box topology. So $\tau_{\sup}$ indeed is coarser than box topology, since each basis of $\tau_{\sup}$ is box-open. Moreover we have by definition of $d_{\sup}$ that $x_{n}\to x$ uniformly (as functions) if and only if $x_{n}\to x$ in $\tau_{\sup}$.