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Oct 5, 2018 at 8:05 answer added KP Hart timeline score: 3
Sep 24, 2018 at 6:37 answer added Yifan timeline score: 1
Sep 23, 2018 at 14:39 answer added Andreas Blass timeline score: 4
Sep 22, 2018 at 19:27 vote accept Tomasz Kania
Sep 22, 2018 at 19:10 answer added Don Monk timeline score: 4
Sep 22, 2018 at 17:45 answer added Jing Zhang timeline score: 2
Sep 22, 2018 at 16:54 comment added YCor By the way this is purely ring-theoretic: in a ring $R$ (say associative unital commutative), you considers ideals $I$ and subsets $\mathcal{A}\subset R\smallsetminus I$ such that for all $a\neq b\in\mathcal{A}$ we have $ab\in I$, and require that all such subsets are countable. In a Boolean algebra $a,b\notin I,a-b\in I$ implies $ab\notin I$, so this forces $\mathcal{A}$ to embed into $A/I$. Beyond the Boolean case, this applies when $A/I$ is reduced.
Sep 22, 2018 at 16:51 comment added YCor You have all ideals $I$ such that $P(N)/I$ is countable (but this forces $P(N)/I$ finite).
Sep 22, 2018 at 16:51 history edited Tomasz Kania CC BY-SA 4.0
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Sep 22, 2018 at 16:50 comment added YCor "only the empty family": you mean "only the empty family or singletons". By the way, what you call "family" is usually called a "subset". Of course, every set can be thought of as a family $(a)_{a\in A}$...
Sep 22, 2018 at 16:38 history asked Tomasz Kania CC BY-SA 4.0