PatrickR
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 May 12 awarded Nice Question Feb 25 awarded Popular Question Jan 3 comment Infimum of two measures In the last sentence, I think you mean the only possible $A$ that works would be $B^c$. Jan 3 comment Infimum of two measures Nice general argument. Dec 31 accepted Infimum of two measures Dec 31 revised Infimum of two measures added 48 characters in body Dec 31 asked Infimum of two measures Dec 13 awarded Yearling Nov 6 awarded Revival Oct 2 awarded Self-Learner Jul 2 awarded Curious May 6 awarded Popular Question Dec 26 awarded Nice Question Dec 13 awarded Yearling Sep 24 comment Boolean algebras without atoms You are right. I had missed that part. Nice proof. Sep 24 comment Boolean algebras without atoms I am not sure this proof works. The $y_{i,j}$ may not be all distinct, for different values of $i$. And similarly for the $w_{i,j}$. But there is no a priori guarantee that the equalities between the $y_{i,j}$ match exactly the equalities between the $w_{i,j}$. So it's not clear that $\pi$ is well defined. Sep 6 comment nonisomorphic groups whose quotients are isomorphic Like you say, subgroups can be embedded in all sorts of ways into other groups. So if $C$ and $C'$ are isomorphic but not exactly the same, it's a simple matter of modifying $A$ and $B$ and identifying elements in certain ways to make $C$ and $C'$ actually identical. See Ittay Weiss's answer for example. Sep 6 answered nonisomorphic groups whose quotients are isomorphic Sep 2 answered Non-isomorphic countable Boolean algebras Jun 30 revised Show that $\langle G^+\rangle=G$ in a directed group Subgroup was meant here