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.

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    $\begingroup$ Being a good source for counterexamples is a very good thing indeed. It helps us in our attempt to understand where, how and why certain properties break in the general case. $\endgroup$ – Asaf Karagila May 2 '11 at 13:00
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    $\begingroup$ Kunen-Vaughan, Handbook of set-theoretic topology, Chapter 4 (written by S.P. Williams) contains a lengthy survey on box products. The results there seem to be more of the nature that Asaf points out. $\endgroup$ – t.b. May 2 '11 at 13:03
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    $\begingroup$ I like this question. For example, are there any cases where a topological space is given from some other context, and it turns out that it is homeomorphic to a box product? $\endgroup$ – Cheerful Parsnip May 2 '11 at 13:51
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    $\begingroup$ There is something called the "Box product problem", which asks whether the box product of countably many copies of the real line is normal; it spurred some interest in the 70s, at least, with Mary Ellen Rudin proving that the Continuum Hypothesis implies this is the case, and E.K. van Douwen proving that in general the box product of normal spaces need not be normal. From a quick look through "Counterexamples in Topology" and a quick google search, problems involving the box topology (in the infinite index set) seem to be closely connected to problems of set-theory (CH, AC, etc). $\endgroup$ – Arturo Magidin May 2 '11 at 15:39
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    $\begingroup$ @Arturo: I believe that was one of the things that sparked general topology as it is today. $\endgroup$ – Asaf Karagila May 2 '11 at 16:51

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 $\tau_{\sup}$.

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    $\begingroup$ I don't think the identification of topologies is correct. Take $J = \mathbb{N}$ and $Y=[-1,1]$. Then $U = \prod_{n=1}^\infty \left(-\frac{1}{n},\frac{1}{n}\right)$ is open in $X = Y^{\mathbb{N}}$ with the box topology while it doesn't contain any uniform ball around zero. $\endgroup$ – t.b. May 31 '12 at 12:04
  • $\begingroup$ @t.b. Consider taking two balls each of radius one and center the first ball at (1-1/n) and the second ball at (1/n-1). Then the intersection of these two balls is your open set. $\endgroup$ – JSchlather May 31 '12 at 12:13
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    $\begingroup$ That's true $\tau_\sup$ is certainly contained in the box topology. @Jacob: that doesn't work since $0$ is not contained in either of the two open balls of radius $1$ (the $\sup_{n}|(1-1/n)-0|$ is equal to one, not strictly less than one). We have a metric topology, so $(0,0,\ldots)$ is an interior point of every open set, so if $U = \prod (-1/n,1/n)$ were open, it would have to contain a $2\varepsilon$-ball, but such a $2\varepsilon$-ball contains the constant sequence $(\varepsilon,\varepsilon,\ldots)$, which is not contained in $U$. $\endgroup$ – t.b. May 31 '12 at 12:34
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    $\begingroup$ It is a common misconception that $$U(x,\epsilon) = \prod_{j\in J} B(x(j), \epsilon)$$ but this is not true (even worse: the set on the right hand side is not even open in the sup-topology). Take $J = \mathbb N$ and $x = (0,0,0, \dots) \in {[0,1]}^{\mathbb N}$. Then the product of 1-balls around $0$ contains the element $y = (1-1/2, 1-1/3, 1-1/4, \dots)$, but $\sup_{j\in J} |x(j) - y(j)| = 1$. So $y$ is not in the 1-ball under the sup-metric (and no $\delta$-ball around $y$ is contained in $\prod_{j\in J} B(0,1)$). I think this remark is (at least implicitely) already in t.b.'s comment. $\endgroup$ – Sam May 31 '12 at 13:00
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    $\begingroup$ Sure; and thanks for doing it. Such things which require correction are important to point out:) $\endgroup$ – T. Eskin May 31 '12 at 13:47

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

  1. We have $X \in \mathcal{C}$.
  2. 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:

  1. $A \cap S = \varnothing$.
  2. 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.


Here is a situation where the box topology can be useful.

Free topological groups are important objects in general topology and topological algebra that date back to A.A. Markov. While it is convenient that the underlying groups are just the usual free groups, the topological structure of the free (Markov) topological group $F_M(X)$ on a general Tychonoff space $X$ is fairly complicated to describe. For instance, one might hope $F_M(X)$ is the quotient of the free topological monoid $\bigoplus_{n\geq 0}(X\oplus \{e\}\oplus X^{-1})^n$ under the usual reduction of words but this is not the case, e.g. $X=\mathbb{Q}$! Still, it would be nice to have a characterization of these beasts as quotient of some space with simple operations. One way to do this is to use certain $\sigma$-products with the box topology.

In the paper The topology of free topological groups (an english version can be found in the Journal of Mathematical Sciences), O.V. Sipacheva gives a very general characterization of the topology of free topological groups (on Tychonoff spaces) as the quotient of a certain $\sigma$-product with the box topology. The characterization is quite helpful in understanding the topology of free topological groups; an instructive paragraph on why this certain construction is appropriate is on pp. 5800 in the english version.


May I (rather belatedly) suggest the following situation where box topologies make a natural appearance, namely in the context of functional analysis. Suppose we have a collection of Banach spaces---for simplicty I will assume that it is a sequence. On the Cartesian product we can consider three topologies. The product topology corrresponds to the usual locally convex product of the underlying spaces. Another natural topology is that generated by the product of the unit balls as unit ball---this corresponds to the sup-norm. This is not a linear topology on the Cartesian product, but only on the subspace which is absorbed by this ball. This is precisely the $\ell^\infty$-sum (more prosaically, the family of bounded sequences in the Cartesian product) which is the natural product of the spaces in the category of Banach spaces. A similar phenomenon occurs with the box topology---it is is not a linear topology on the Cartesian product. Once again, the natural thing to do is to restrict it to the largest subspace where it is a linear topology. This is precisely the direct sum of the spaces, i.e., the sequences which are zero almost eveywhere. Thus the three natural topologies on the product make it natural to introduce these three corresponding product constructions in the theory of locally convex spaces. Once one has direct sums, then it is a short step to the inductive limits which played and play such an important role in the Schwartzian theory of distributiohs.


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.

  • $\begingroup$ Interesting fact, but why is it useful? If you want summation to be continuous, then the "right" topology is the $\ell^1$ topology, it seems to me. The box topology doesn't give you a topological vector space, which takes out all the fun for functional analysts. Can one use this topology on this space to prove some results of independent interest? $\endgroup$ – Nate Eldredge Oct 21 '16 at 19:00
  • $\begingroup$ Doesn't my argument work on the $l^1$ topology as well? Also, if it would be more interesting, my argument works for the space of convergent sequences. $\endgroup$ – Joshua Meyers Oct 27 '16 at 6:57

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