Let $X$ be be a set. Do there exist non-homeomorphic topologies $(X,\tau_1)$ and $(X, \tau_2)$ be on $X$ such that $$\prod_{i\in\mathbb{N}}(X, \tau_1)\cong \prod_{i\in\mathbb{N}}(X, \tau_2).$$ I ask because I recently learned that the infinite product of a discrete space with itself isn't necessarily discrete, and this led me to wondering what interesting things happen with infinite products of topological spaces. Further, I'm mostly interested in cases where $X$ is finite, but any examples work.
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6One trick is this: let $X$ be in bijection with $\Bbb R$, and therefore with $\Bbb R^2$ as well. You have a topologies $\tau_1,\tau_2$ such that $(X,\tau_1)$ is homeomorpic to the usual $\Bbb R$ and $(X,\tau_2)$ is homeomorphic to the usual $\Bbb R^2$. Those two spaces aren't homeomoprhic, but you have $\prod_{i\in\Bbb N}(X,\tau_1)\cong(\Bbb R^{\Bbb N},\text{product topology})\cong \prod_{i\in \Bbb N}(X,\tau_2)$. – Sassatelli Giulio Nov 01 '22 at 00:34
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@SassatelliGiulio This is great, thanks! Do you know if an example is possible for finite X? – FazeZizek Nov 01 '22 at 16:09
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1I don't. I was thinking this: take a set of 8 points $X$ and fix two partitions, $P_1$ of type $(1,1,1,1,2,2)$ and $P_2$ of type $(1,1,2,2,2)$; call $X_1$ the topological space that has $P_1$ as a basis and $X_2$ the same for $P_2$. I thought that maybe $X_1^{\Bbb N}\cong X_2^{\Bbb N}$, but I can't quite prove one way or the other. I even thought that it could be the same with 6 elements and partitions $(1,1,1,1,2)$ and $(1,1,2,2)$. If you have more that one $1$ in either partition, then you need more than one in both, because of the number of $G_\delta$ points in the product. – Sassatelli Giulio Nov 01 '22 at 16:55