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Nov 27, 2023 at 7:19 comment added Dominic van der Zypen Right -- apologies for the duplicate
Nov 27, 2023 at 7:18 vote accept Dominic van der Zypen
Nov 26, 2023 at 19:53 comment added YCor @JoelDavidHamkins yes that's exactly what's used in the original answer to the question, implying that there's a subset $X$ in this group whose centralizer is not equal to the centralizer of any countable subset of $X$. This implies that this group doesn't embed into any group with a metrizable separable topology.
Nov 26, 2023 at 19:00 history closed Emil Jeřábek
Jeremy Rickard
Joseph Van Name
Sam Hopkins
Yemon Choi
Duplicate of Is $S_\omega/F_\omega$ embeddable to $S_\omega$?
Nov 26, 2023 at 18:11 comment added YCor Exact duplicate of your own question!
Nov 26, 2023 at 17:42 review Close votes
Nov 26, 2023 at 19:02
Nov 26, 2023 at 16:58 answer added Account will self-destruct timeline score: 5
Nov 26, 2023 at 16:25 comment added Joel David Hamkins Perhaps we can somehow use the fact that every collection of disjoint sets in $\omega$ is countable, but there are uncountable collections of sets that are disjoint mod finite.
Nov 26, 2023 at 15:09 history asked Dominic van der Zypen CC BY-SA 4.0