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Dec 14, 2018 at 20:33 history closed YCor
Monroe Eskew
Andreas Blass
Ramiro de la Vega
Mark Wildon
Not suitable for this site
Dec 10, 2018 at 21:34 comment added YCor The corrected version seems indeed as interpreted by Nik. It simply restates as: does the poset $\mathcal{P}(\omega)/\mathrm{fin}$ have a decreasing chain of type $\omega_1$? this is actually very standard and even in this version might be more suitable for MathSE.
Dec 10, 2018 at 21:28 comment added Andreas Blass Since @YCor has answered (in a comment) the question as asked and has suggested the probably intended modification to make it nontrivial, and since Nik Weaver has answered the nontrivial version, I'm voting to close. (I'd favor reopening if (and only if) a different nontrivial version were proposed.)
Dec 10, 2018 at 18:05 comment added YCor Well, I have to insist that it's still trivial (in ZFC): for every non-successor cardinal $\alpha<2^{\aleph_0}$, choose an infinite subset $B_\alpha$, so that the chosen subsets $B_\alpha$ have pairwise infinite symmetric difference. Define $B_{\alpha+n}$ as $B_{\alpha+n-1}$ minus a point. Finally define $A_\alpha=B_\alpha\times\omega\subseteq\omega^2$.
Dec 10, 2018 at 17:59 history edited Forever Mozart CC BY-SA 4.0
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Dec 10, 2018 at 17:51 history edited Forever Mozart CC BY-SA 4.0
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Dec 10, 2018 at 17:48 comment added Asaf Karagila Obligatory keyword reference: $\frak t$, by the way.
Dec 10, 2018 at 17:46 history edited Forever Mozart CC BY-SA 4.0
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Dec 10, 2018 at 17:45 comment added Forever Mozart @YCor I would like the elements of the sequence to be distinct if possible
Dec 10, 2018 at 17:45 review Close votes
Dec 14, 2018 at 20:33
Dec 10, 2018 at 17:43 comment added YCor As regards the current question, the constant sequence $A_\alpha=A$ works... Second, your axioms on the sequence make no relation between $A_\alpha$ and $A_\beta$ except when $\alpha$ and $\beta$ differ by addition by an integer. Please write your question more carefully. (Your last edit doesn't address my second sentence. Currently it's obvious to arrange a sequence of length $2^{\aleph_0}$.)
Dec 10, 2018 at 17:42 answer added Nik Weaver timeline score: 2
Dec 10, 2018 at 17:37 comment added Forever Mozart @AsafKaragila I would like for $A_{\alpha+1}$ to be contained in $A_\alpha$, up to some finite error.
Dec 10, 2018 at 17:36 comment added Asaf Karagila Do you want $\subseteq^*$ or just finite differences? Because the answer is yes in either case, but for different reasons. Please settle on a version before I put any more energy into writing an answer...
Dec 10, 2018 at 17:31 history edited Forever Mozart CC BY-SA 4.0
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Dec 10, 2018 at 17:17 history asked Forever Mozart CC BY-SA 4.0