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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
S Jan 10, 2020 at 23:01 history bounty ended CommunityBot
S Jan 10, 2020 at 23:01 history notice removed CommunityBot
Jan 8, 2020 at 18:46 comment added Bjørn Kjos-Hanssen $n$ is $1+\log_2 n$-unsimplifiable for finite $n$. Fix any order $B$, $|A|>|B|$. Ch picks $a_1\in A$ near the middle of $A$. Then Du picks some $b_1\in B$. Either (i) $|A_{<a_1}|>|B_{<b_1}|$ or (ii) $|A_{>a_1}|>|B_{>b_1}|$. Ch picks $a_2<a_1$, near the middle of $A_{<a_1}$, in case (i), and $a_2>a_1$, near the middle of $A_{>a_2}$, in case (ii). Continuing, we maintain the property that $|A_{<a_k}|>|B_{<b_k}|$ or $|A_{>a_k}|>|B_{>b_k}|$. So Ch always goes where Du has little room left. Eventually Du runs out of room. Ch picks the greatest element of $X$ not yet picked.
Jan 8, 2020 at 8:26 comment added Bjørn Kjos-Hanssen Or need a weird sense in which $2^{\omega^2}=\omega_1^{CK}$
Jan 8, 2020 at 8:09 comment added Bjørn Kjos-Hanssen I think $n$ is $1+\log_2 n$-unsimplifiable. (Thinking of $k$-unsimplifiable as "revealed in time $k$" basically.) If the analogy holds then I guess $X\ge\omega$ is enough...
S Jan 2, 2020 at 21:49 history bounty started Noah Schweber
S Jan 2, 2020 at 21:49 history notice added Noah Schweber Draw attention
Dec 26, 2019 at 23:59 history asked Noah Schweber CC BY-SA 4.0