PatrickR
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 Feb25 awarded Popular Question Jan3 comment Infimum of two measures In the last sentence, I think you mean the only possible $A$ that works would be $B^c$. Jan3 comment Infimum of two measures Nice general argument. Dec31 accepted Infimum of two measures Dec31 revised Infimum of two measures added 48 characters in body Dec31 asked Infimum of two measures Dec13 awarded Yearling Nov6 awarded Revival Oct2 awarded Self-Learner Jul2 awarded Curious May6 awarded Popular Question Dec26 awarded Nice Question Dec13 awarded Yearling Sep24 comment Boolean algebras without atoms You are right. I had missed that part. Nice proof. Sep24 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. Sep6 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. Sep6 answered nonisomorphic groups whose quotients are isomorphic Sep2 answered Non-isomorphic countable Boolean algebras Jun30 revised Show that $\langle G^+\rangle=G$ in a directed group Subgroup was meant here Jun30 suggested approved edit on Show that $\langle G^+\rangle=G$ in a directed group