2

Let $X$ be a completely regular topological space and $X^n$ its $n$-ary Cartesian product for positive integer $n$.

Let $V$ be a subset of $X^n$ which is homeomorphic to $X^m$.

Does it imply $m\leq n$?

I expect the answer to be simple, but at the moment I am stumped to find it, as well as I don't seem to find an existing solution for the question in this generality.

Anton
  • 21
  • 2
    Hi, welcome to MSE. I think this fact answers your question. – Izaak van Dongen Feb 19 '24 at 15:25
  • 2
    @Izaak van Dongen's counterexample is much more complicated than it needs to be. One can directly see that $\mathbb N\cong\mathbb N^2$ since both are countable, discrete spaces. – tth2507 Feb 19 '24 at 15:33
  • Thank you! Do you know other examples? – Anton Feb 19 '24 at 17:55
  • 1
    I thought a bit about your question, and I must say that I expect this to basically never work. If you take a “random” topological space (or even completely regular Hausdorff), the product topology is frankly a mess. Your claim holds obviously for finite spaces, and it is a hard theorem that it holds for $\mathbb R$ (or any manifold). Other than that I could not find examples of your claim holding. – tth2507 Feb 19 '24 at 18:09
  • The two solutions are exactly what I was looking for. But out of curiosity: is there an if and only if theorem for this phenomenon? Meaning: "the implication stands if and only if X is ..." . – Anton Feb 19 '24 at 18:13
  • 1
    A related question is discussed here. This also gives a list of counterexamples to your claim. I do not know of an equivalent condition, but this is certainly an interesting question — likely a very hard one. – tth2507 Feb 19 '24 at 18:13
  • 1
    Yes, finite sets are obvious :) Discrete topology also. Thank you for the list! Rationals though are aleady ideally suited for my purpose. – Anton Feb 19 '24 at 18:18

0 Answers0