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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.

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    $\begingroup$ 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 $Z^\mathbb{N}\cong(Z^\mathbb{N})_{box}$ $\endgroup$
    – Saúl RM
    Oct 2, 2022 at 18:02
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    $\begingroup$ 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. $\endgroup$
    – terceira
    Oct 3, 2022 at 5:30
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    $\begingroup$ I would guess this is what OP meant by "nontrival topological space," but I guess it's not clear. $\endgroup$ Oct 6, 2022 at 16:23
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    $\begingroup$ @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. $\endgroup$ Oct 6, 2022 at 23:11
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    $\begingroup$ 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 $G_\delta$”). $\endgroup$
    – Gro-Tsen
    Oct 7, 2022 at 7:46

1 Answer 1

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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.

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  • $\begingroup$ Thanks for the answer. Do you mind elaborating on why the function $F$ at the end works? $\endgroup$ Oct 8, 2022 at 11:35
  • $\begingroup$ 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 $\endgroup$
    – Saúl RM
    Oct 8, 2022 at 12:41

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