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Sep 15, 2013 at 19:21 vote accept Asaf Karagila
Aug 30, 2013 at 21:41 answer added Monroe Eskew timeline score: 6
Aug 30, 2013 at 3:54 answer added Garrett Ervin timeline score: 4
Aug 29, 2013 at 18:42 comment added Asaf Karagila Thanks Paul, that's quite helpful. This seems to be easily generalized to large cardinalities. Moreover if we take $A'_n=A_1\cup\ldots\cup A_n$ then we get the same result for an increasing sequence, so both my questions have a negative answer. (Nice seeing you here, by the way!)
Aug 29, 2013 at 18:03 comment added Paul McKenney You don't need a gap to show that suprema don't exist in $\mathcal{P}(\omega)/\mathrm{Fin}$; take any sequence $A_n$ ($n < \omega$) of pairwise-disjoint, infinite subsets of $\omega$. If $A$ is a $\subseteq_*$-upper bound for this sequence, then one can find $B\subseteq A$ such that $A\setminus B$ is infinite and $B$ is still an upper bound, by removing one element from each intersection $A\cap A_n$ . Hence the sequence $A_n$ ($n < \omega$) has no least upper bound.
Aug 29, 2013 at 8:46 history asked Asaf Karagila CC BY-SA 3.0