TITLE: Is the box topology good for anything?
QUESTION [64 upvotes]: 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.

REPLY [3 votes]: Consider the space $C\subseteq \mathbb{R}^\omega$ of real sequences $(a_n)$ such that $\sum a_n$ converges.  This space can inherit the product topology, the uniform topology, or the box topology from $\mathbb{R}^\omega$.  The function $\Sigma: C\rightarrow \mathbb{R}$ that maps a sequence to its sum is continuous if $C$ inherits the box topology, and not continuous if it inherits the product topology or the uniform topology.
In fact, if $U\subseteq C$ is a non-empty open set in the topology induced by the product topology or the uniform topology, then $\Sigma(U)=\mathbb{R}$.  I'll just show this in the uniform topology, since it is finer than the product topology so the similar result on the product topology follows.  Since $U$ is non-empty, there exists $s_1\in U$ with $\Sigma{s_1}=r_1$ for some $r_1\in\mathbb{R}$.  There exists a neighborhood $N\ni s_1$ of radius $\varepsilon < 1$ such that $N\subseteq U$.  Define $s_2$ as $s_1$ with $\delta$ added to the first $n$ elements, $\delta<\varepsilon$.  Then $s_2\in N\subseteq U$, and $\Sigma(s_2)=r_1+n\delta$, and any real number can be expressed as $r_1+n\delta$ for some $n\in\mathbb{N}$ and $\delta\in(0,\varepsilon)$.
To show that $\Sigma$ is not continuous, let $S\subseteq \mathbb{R}$ be a non-empty proper open subset.  Then $\Sigma^{-1}(S)$ is non-empty, since $\Sigma$ is surjective, but it cannot be open, because if it were open, $S=\Sigma(\Sigma^{-1}(S))=\mathbb{R}$.
However, if $C$ inherits the box topology, then for an open interval $I=(\Sigma(s)-\varepsilon, \Sigma(s)+\varepsilon)\subseteq \mathbb{R}$, the neighborhood $\prod_{i=1}^\infty (s_i-\frac{\varepsilon}{2^i}, s_i+\frac{\varepsilon}{2^i})\subseteq \Sigma^{-1}(I)$, so $\Sigma$ is continuous.