Is the box topology good for anything?
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 $X$ \begin{align*} d_{\sup}(x,y)=\sup_{j\in J}d(x(j),y(j)), \end{align*} which generates a topology $\tau_{\sup}$ with the following basis elements for $x\in X$ and $\varepsilon>0$ \begin{align*} U(x,\varepsilon)=\{y\in X:\sup_{j\in J}d(x(j),y(j))<\varepsilon\}. \end{align*} 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 $\delta<r<\varepsilon$. 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 $\tau_{\sup}$.
There is a positive answer to Jim Conant's query. A closure system is a set $X$ together with a collection $\mathcal{C}$ of subsets of $X$ that satisfies
- We have $X \in \mathcal{C}$.
- For all $\mathcal{A} \subseteq \mathcal{C}$ we have $\cap \mathcal{A} \in \mathcal{C}$.
For all $A \subseteq X$ define $c(A) = \cap \{ C \in \mathcal{C} \colon A \subseteq C \} $.
The elements of $\mathcal{C}$ will be called closed sets. In practice if you are interested in this sort of thing it is better to change the definitions to convex structure and convex set because the word "closed" has too many meanings. The definitions above are the standard definitions. Sometimes people want $\varnothing$ to be a closed set.
Suppose that $E \subseteq A \subseteq X$. We will say that $E$ is an extreme subset of $A$ if and only if for all $D \subseteq A$ we have $E \cap c(D) = E \cap c(E \cap D)$. If you wish you can check that in a real vector space this is equivalent to the usual notion of an extreme subset.
Suppose that $A, S \subseteq X$. we will say that $S$ selvages $A$ if and only if the following conditions are satisfied:
- $A \cap S = \varnothing$.
- The set $S$ is an extreme subset of $A$.
One more definition. Suppose that $O$ is a closed set. We will say that $O$ is selvagable if and only if for every closed $D$ that satisfies neither $D \cap O$ nor $D \setminus O$ is empty there exists a nonempty closed $S \subseteq D$ that selvages $O$. The selvagable sets form a basis for a topology on $X$. If $X$ is a finite dimensional real vector space and $\mathcal{C}$ is the collection of convex sets then the resulting topology is the usual topology. If $\{ \left( X_{i}, {\mathcal{C}}_{i} \right) \colon i \in I \} $. is a collection of closure systems and we provide $\Pi X_{i}$ with the smallest closure system so that the inverse image of a closed set in a factor is closed then the selvagable sets in this product is the box topology. Unfortunately this is not a nice topology because the projection maps are not, in general, continuous.