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?

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    $\begingroup$ This seems unlikely. If $x\in X$ is a non-isolated point of $X$ and $\kappa$ is the minimum possible cardinality for a neighborhood base at $x$, then the point $(x,x,\ldots)$ has a neighborhood base of cardinality $\kappa\cdot |I|$ under the product topology, but the cardinality of the smallest neighborhood base for this point in the box topology should be something like $\kappa^{|I|}$. $\endgroup$
    – Jim Belk
    Apr 24, 2022 at 13:15
  • $\begingroup$ Well, for $X$ antidiscrete the topologies coincide, no? $\endgroup$
    – freakish
    May 6, 2022 at 11:43
  • $\begingroup$ @freakish I said nontrivial in the first sentence :) $\endgroup$ May 6, 2022 at 11:54
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  • $\begingroup$ @MartinSleziak Thank you all for your work! $\endgroup$ Oct 9, 2022 at 10:24


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