Below is the definition of box topology:

Given an indexed family of topological spaces $X_\alpha $, the collection of all sets of the form $$\prod_{\alpha\in J} U_\alpha,$$ where $U_\alpha$ is open in $X_\alpha$ forms a basis for the box topology on $$\prod_{\alpha\in J} X_\alpha.$$

Does this definition require axiom of choice to be well-defined?

I am thinking so since each $X_\alpha$ contains possibly uncountably many open sets, and we choose one open set from each $X_\alpha$, but am I correct?

  • 1
    $\begingroup$ The definition is about declaring all products $\prod U_{\alpha}$ as open subsets, regardless of such products existing or not. As in we can talk about the set of choice functions without assuming AC. $\endgroup$
    – Cihan
    May 26, 2015 at 4:07

1 Answer 1


This is an issue if you don't allow the empty set to be a topological space (and if you do allow it, then there might be issues elsewhere).

But generally if the product is empty, then it is empty. Note that this has nothing to do with the particular topology (box, or otherwise). If the entire product is empty, then every subset of it is also empty, because even if somehow there is a sequence of open sets which does admit a choice function, what really happens is this: the basis is defined as $\{f\in\prod X_\alpha\mid \forall\alpha:f(\alpha)\in U_\alpha\}$.

In either case, this is a well-defined construction. If you allow the empty set to be a space, then it's also a compact space; if you disallow it to be a space, then you might say that it is undefined because it's not a space. But it has nothing to do with the topology.

Do note, however, that as with all things, being well-defined and having the same properties as the objects have in $\sf ZFC$ universes are two different things. For example, it might be that $\prod X_\alpha$ is non-empty, but for some sequence of open sets, $U_\alpha$, you have that $\prod U_\alpha$ is in fact empty. Topologically it doesn't matter since the empty set is open.

But look at this strange example. For each $n\in\Bbb N$ consider $P_n$ to be a pair, such that $\prod P_n=\varnothing$ over any infinite subset of $\Bbb N$ (this is essentially Russell's socks). Now let $X_n=P_n\cup\{n\}$ and give it the topology $\{\varnothing,X_n,P_n\}$.

What is the product of these spaces? It's not empty, since we can always choose the natural number from $X_n$. But an open set is non-empty if and only if it only chooses from $P_n$ on finitely many coordinates. Namely, the box product and the Tychonoff product coincide. Strange!

  • $\begingroup$ Sorry I am not fully following. So, if I do not allow empty set to be a topological space, then the axiom of choice is used? Is that what you are saying? $\endgroup$
    – user74261
    May 26, 2015 at 4:03
  • $\begingroup$ Yes, because you just need the space not to be empty. The rest follows on its own. It just might not satisfy the same properties as you might expect. $\endgroup$
    – Asaf Karagila
    May 26, 2015 at 4:04
  • $\begingroup$ I've added some strange example that might happen. $\endgroup$
    – Asaf Karagila
    May 26, 2015 at 4:10
  • $\begingroup$ Hehe, that's lovely. $\endgroup$ May 26, 2015 at 4:25
  • $\begingroup$ @Noah: Thanks. It came to me when I was writing the first comment on this answer. (And by the way, congrats on finally putting your real name here.) $\endgroup$
    – Asaf Karagila
    May 26, 2015 at 4:36

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