# Can ($X^I$, product topology) and ($X^I$, box topology) be homeomorphic for some nontrivial $X$ and infinite $I$?

This is a verbatim repost of this question by Jianing Song. A few months ago I placed a bounty on the question but there were no answers, so I am reposting it here.

Let $$X$$ be a nontrivial topological space, $$I$$ be a infinite set, we can endow $$X^I$$ (the set of all functions $$I\to X$$) with either the product topology or the box topology. We know that the box topology is strictly finer than product topology, but since a topology can be homeomorphic to a strictly finer topology, can the product topology be homeomorphic to the box topology anyway?

EDIT: Here I interpret the term "nontrivial" by that the topology endowed on $$X$$ is not the trivial topology. I do not ask that $$X$$ is Hausdorff, but I think restricting the problem to Hausdorff spaces is also a question worth asking.

• In case it is useful, this is equivalent to proving that for non trivial spaces $X,Y$ we cannot have that $X^I\cong(Y^J)_{box}$ for some infinite cardinals $I,J$: if so, then $Z:=X^I$ would satisfy $Z^\mathbb{N}\cong(Z^\mathbb{N})_{box}$ Oct 2, 2022 at 18:02
• Just shooting from the hip. In the Hausdorff case, you can assume that you have a two point discrete space--then consider compactness. Otherwise, look at the two point indiscrete case. Oct 3, 2022 at 5:30
• I would guess this is what OP meant by "nontrival topological space," but I guess it's not clear. Oct 6, 2022 at 16:23
• @TarasBanakh I believe the antidiscrete topology is also commonly referred to as the trivial topology, so yes we do not include that in our example. I will write that in the post just to be clear. Oct 6, 2022 at 23:11
• In case anyone needs a reminder, or to restrict the search for an example, it is shown in theorem 1.3 of Scott Williams's chapter “Box Product” in the Handbook of Set-Theoretic Topology (Kunen & Vaughan eds., 1984), that the box product of an infinite family of infinite non-discrete Hausdorff completely regular topological spaces is never any of the following: (i) locally compact, (ii) separable, (iii) connected or locally connected, (iv) first-countable or (v) perfect (as in: “every closed set is $G_\delta$”). Oct 7, 2022 at 7:46

This answer was supposed to contain a counterexample, but there was a mistake in it. In any case I think the following class of topological spaces may be useful to find a counterexample:

Suppose that we have a set $$X$$ and a collection $$\mathcal{A}$$ of subsets of $$X$$ closed under finite unions. Then we can define a topology on $$2^X$$ by the following basis of clopens: let $$A\in\mathcal{A}$$ and let $$g:A\to\{0,1\}$$ be a function. Then we define $$C_{A,g}=\{f:X\to\{0,1\};f\text{ is an extension of }g\}$$. Then $$\{C_{A,g};A\in\mathcal{A},g:A\to\{0,1\}\}$$ is a basis for a topology on the set $$2^X$$.

For example, $$2^\mathbb{N}$$ is obtained by setting $$X=\mathbb{N}$$ and $$\mathcal{A}=\{$$ finite subsets of $$\mathbb{N}\}$$. Similarly, $$(2^\mathbb{N})^\mathbb{N}_\text{box}$$ can be obtained by setting $$X=\mathbb{N}^2$$, and $$\mathcal{A}=\{A\subseteq\mathbb{N}^2;A\cap(\mathbb{N}\times\{n\})\text{ is finite for every }n\in\mathbb{N}\}$$, and $$(((2^\mathbb{N}_\text{box})^\mathbb{N})^\mathbb{N}_\text{box})^\mathbb{N}$$ can be expressed by setting $$X=\mathbb{N}^4$$ and saying that a set $$A$$ is in $$\mathcal{A}$$ if $$\{a_1\in\mathbb{N};(a_1,a_2,a_3,a_4)\in A\}$$ is bounded and for every pair $$(x_1,x_2)$$, the set $$\{a_3\in\mathbb{N};(x_1,x_2,a_3,a_4)\in A\}$$ is bounded.

In general, if we let $$2^X_\mathcal{A}$$ be the space we defined, then for any infinite set $$I$$:

• $$(2^X_\mathcal{A})^I_\text{box}\cong 2^Y_\mathcal{B}$$, where $$Y=X\times I$$ and $$\mathcal{B}=\{B\subseteq X\times I;B\cap (X\times\{i\})\in\mathcal{A}\;\forall i\in I\}$$. (Here we identified $$X\times\{i\}$$ with $$X$$)

• $$(2^X_\mathcal{A})^I\cong 2^Y_\mathcal{B}$$, where $$Y=X\times I$$ and $$\mathcal{B}=\{B\subseteq X\times I;B\cap (X\times\{i\})\in\mathcal{A}\;\forall i, B\cap (X\times\{i\})=\varnothing\text{ for all i except finite}\}$$.

Moreover, suppose we have two spaces $$2^X_\mathcal{A}$$ and $$2^Y_\mathcal{B}$$ and we have a bijection $$\phi:X\to Y$$ which induces a bijection $$\mathcal{A}\to\mathcal{B}$$. Then the function $$F:2^X_\mathcal{A}\to2^Y_\mathcal{B};F(f)=f\circ\phi^{-1}$$ is a homeomorphism, just by how we have defined the topologies.

• Thanks for the answer. Do you mind elaborating on why the function $F$ at the end works? Oct 8, 2022 at 11:35
• I was planning on writing the details of that today and I have noticed a mistake in my reasoning, $F$ seems to take $\mathcal{D}$ to $\mathcal{C}$ but not the other way around, so it does not work :( sorry. I will leave the first part of the answer in case it is useful to anyone to operate with products and box products Oct 8, 2022 at 12:41