Although the box topology is a topology worth studying and is similar to the strong topology in differential topology, the box topology is in many regards very badly behaved since the box product of even nice spaces has many undesirable properties. For instance, unlike ordinary products, non-trivial box products are never connected, metrizable, nor locally compact. I want to know what properties of topological spaces are preserved under taking box products. Let me list the more obvious properties.

$\textbf{Preserved under box products:}$ Separation axioms such as $T_{0},T_{1}$, Hausdorffness, regularity, complete regularity; zero-dimensionality, total disconnectedness(otherwise known as hereditary disconnectedness), total separation (two distinct points can always be separated by a clopen set), other disconnectedness properties, negative properties in general, $P$-spaces, discreteness, complete uniformizability.

$\textbf{Not preserved under box products:}$ Normality, paracompactness, ultraparacompactness, compactness, other compactness properties, connectedness, path connectedness, other connectedness properties, . . .

See chapter 4 in the Handbook of set-theoretic topology for more information on box products and a few other properties preserved under box products that I neglected to mention in this question.

For which other topological properties $P$ does it hold that if $X_{i}$ has property $P$ for all $i\in I$, then $\prod_{i\in I}^{\textrm{Box}}X_{i}$ also has property $P$? I am looking for specific examples of such properties since I do not believe that there is a nice general characterization of all such properties. I am looking for positive properties rather than the negation of certain well known properties.

  • 3
    $\begingroup$ The notation $\models$ is not widely known among people who haven't studied mathematical logic. (I had to look at the code of your question, as if I were to edit it, to know how to typeset it.) Would you consider rewriting the last paragraph so it is understandable to those without familiarity with that notation by writing the question in plain English? $\endgroup$ – KConrad Jun 19 '15 at 1:50
  • $\begingroup$ @KConrad: I took a shot at writing it out in prose. $\endgroup$ – Nate Eldredge Jun 19 '15 at 3:17
  • 1
    $\begingroup$ That's a better version. $\endgroup$ – KConrad Jun 19 '15 at 4:24
  • $\begingroup$ (I was wondering whether meta-compactness might be an exception to your 'negative' list of other compactness properties.) $\endgroup$ – Dominic van der Zypen Jun 19 '15 at 7:22
  • $\begingroup$ Related question: mathoverflow.net/questions/319920/homotopy-type-of-box-topology $\endgroup$ – Gro-Tsen Jan 2 at 18:10

A space $X$ is discretely generated (DG) if for every non-closed set $A \subset X$ and for every point $x \in \overline{A} \setminus A$ there is a discrete set $D \subset A$ such that $x \in \overline{D}$.

For being a convergence-type property (note that Fréchet-Urysohn and even radial spaces are DG) discrete generability is pretty weak: every scattered space is DG, every compact space of countable tightness is DG in ZFC and every monotonically normal space is DG (see Dow, A.; Tkachenko, M.G.; Tkachuk, V.V.; Wilson, R.G., Topologies generated by discrete subspaces, Glas. Mat., III. Ser. 37, No.1, 187-210 (2002). ZBL1009.54005.).

There are examples of DG spaces with a non-DG square (see Murtinová, Eva, On products of discretely generated spaces, Topology Appl. 153, No. 18, 3402-3408 (2006). ZBL1107.54006.), but in a few notable special case discrete generability is indeed preserved in box products:

THEOREM (Tkachuk, 2012): Every box product of monotonically normal spaces is discretely generated.

THEOREM (Barriga-Acosta and Hernández-Hernández, 2016): Every box product of regular first-countable spaces is discretely generated.


A space is said to be hyperconnected if every two non-empty open sets have non-empty intersection.

The axiom of choice implies that the box product of hyperconnected spaces is hyperconnected.


Some local network or base properties are preserved by countable box-products of topological spaces. In particular:

1) the existence of a countable $cs$-network at each point;

2) the existence of a countable $cs^*$-network at each point;

3) the existence of a countable $s^*$-network at each point;

4) the existence of an $\omega^\omega$-base at each point.

Now I recall the corresponding definitions.

Let $X$ be a topological space and $x$ be a point of $X$. A family $\mathcal F$ of subsets of $X$ is called a

$\bullet$ a $cs$-network at $x$ if for any sequence $\{x_n\}_{n\in\omega}\subset X$ convergent to $x$ and any neighborhood $O_x\subset X$ of $x$ there exists a set $F\in\mathcal F$ such that $F\subset O_x$ and $F$ contains all but finitely many elements of the sequence $(x_n)$;

$\bullet$ a $cs^*$-network at $x$ if for any sequence $\{x_n\}_{n\in\omega}\subset X$ convergent to $x$ and any neighborhood $O_x\subset X$ of $x$ there exists a set $F\in\mathcal F$ such that $F\subset O_x$ and $F$ contains infinitely many elements of the sequence $(x_n)$;

$\bullet$ an $s^*$-network at $x$ if for any sequence $\{x_n\}_{n\in\omega}\subset X$ accumulating at $x$ and any neighborhood $O_x\subset X$ of $x$ there exists a set $F\in\mathcal F$ such that $F\subset O_x$ and $F$ contains infinitely many elements of the sequence $(x_n)$;

$\bullet$ an $\omega^\omega$-base at $x$ if each set $F\in\mathcal F$ is a neighborhood of $x$ and $\mathcal F$ can be written as $\mathcal F=\{F_\alpha\}_{\alpha\in\omega^\omega}$ so that $F_\alpha\subset F_\beta$ for any elements $\beta\le\alpha$ of $\omega^\omega$ (endowed with the ccordinatewise partial order).

Those notions are studied in this paper and often appear in Topological Algebra and Functional Analysis.

For a topological space $X$ and a point $x\in X$ we have the implications:

($X$ has an $\omega^\omega$-base at $x$) $\Rightarrow$ ($X$ has a countable $s^*$-network at $x$) $\Rightarrow$

$\Rightarrow$ ($X$ has a countable $cs^*$-network at $x$) $\Leftrightarrow$ ($X$ has a countable $cs$-network at $x$).


A space is called Alexandrov if every open set is clopen. If $X_i$ are Alexandrov spaces, then so is $\prod_{i\in I}^{\textrm{Box}}X_{i}$.

  • 1
    $\begingroup$ An Alexandrov space usually refers to a space such that the intersection of arbitrarily many open sets is open. This is a weaker property than saying that every open set is clopen. In fact, the spaces where every open set is clopen are precisely the spaces whose $T_{0}$-reflection is discrete. $\endgroup$ – Joseph Van Name Jun 22 '15 at 15:33

Caution: Answer contains a serious mistake (see comments below by Joseph van Name), but is not deleted for the record.

A space $(X,\tau)$ is called a $D$-space if if whenever one is given a neighborhood $N(x)$ of $x$ for each $x\in X$, then there is a closed discrete subset $D\subseteq X$ such that $\{N(x): x\in D\}$ covers $X$. This is a survey of $D$-spaces and their properties.

A routine verification shows that whenever $X_i$ are $D$-spaces for all $i$, then so is $\prod_{i\in I}^{\textrm{Box}}X_{i}$.

EDIT: Verification of the claim above.

Let $X_i, i\in I$ be $D$-spaces and assume $\{N\big((x_i)\big):(x_i) \in\prod_{i\in I} X_i\}$ is an assignment of a neighborhood to each point of $\prod_{i\in I} X_i$ with the box product topology. We can assume that $N\big((x_i)\big) = \prod_{i\in I} N_i(x_i)$ for each $i$ where $N_i(x_i)$ is an open neighborhood of $x_i\in X_i$. Since each $X_i$ is a $D$-space let $D_i$ be a discrete set in $X_i$ such that $\bigcup \{N_i(d): d\in D_i\} = X_i$, for each $i$. So we set $$D = \prod_{i\in I} D_i.$$

Let $(y_i)\in \prod_{i\in I} X_i$. For each $i\in I$ we have $y_i \in N_i(d_i)$ for some $d_i\in X_i$ because by assumption $\bigcup \{N_i(d): d\in D_i\} = X_i$. So let $d:= (d_i)$, so $d\in D$. Then $(y_i) \in N\big((d_i)\big)$. This implies that $$\bigcup \{N\big((x_i)\big): (x_i)\in D\} = \prod_{i\in I} X_i.$$ Therefore $\prod_{i\in I} X_i$ with the box product topology is a $D$-space.

  • $\begingroup$ Dominic van der Zypen. That same paper that you referenced claims that not even finite products of $D$-spaces are $D$-spaces. In general, the notion of a $D$-space is not very well behaved. $\endgroup$ – Joseph Van Name Jun 19 '15 at 21:04
  • $\begingroup$ OK interesting... so I assume there is a mistake above in the argument? $\endgroup$ – Dominic van der Zypen Jun 20 '15 at 6:52
  • $\begingroup$ Dominic van der Zypen. The mistake began when you said "we can assume that $N((x_{i})_{i\in I})=\prod_{i\in I}N_{i}(x_{i})$." Here the $N_{i}(x_{i})$ depends on the entire function $(x_{i})_{i\in I}$ instead of the individual point $x_{i}$. Even though the notion of a $D$-space is a covering property like paracompactness, the notion of a $D$-space is hardly preserved under any constructions and is difficult to work with. $\endgroup$ – Joseph Van Name Jun 20 '15 at 21:22
  • $\begingroup$ Oh - that's right. I leave this (wrong) answer for "instructive purposes" and mark it as such. $\endgroup$ – Dominic van der Zypen Jun 22 '15 at 6:57

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.