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Sep 20, 2018 at 15:04 comment added KP Hart Or follow Sierpinski and define the sequence explicitly for irrational $r>1$: let $r_n=\frac1n\lfloor nr\rfloor$
Mar 31, 2015 at 13:09 comment added Asaf Karagila @Andreas: Ha! It didn't even cross my mind. Thanks!
Mar 31, 2015 at 13:06 comment added Andreas Blass It might be worth pointing out that this example works whether you use the set-theorist's definition of antichain (pairwise incompatible) or everybody else's definition (pairwise incomparable). If you wanted only the latter, weaker sort of antichain, you could take the sets $\{k:r<q_k<r+1\}$ for all reals $r$.
Mar 23, 2015 at 14:41 history edited Asaf Karagila CC BY-SA 3.0
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Mar 23, 2015 at 14:39 history edited Asaf Karagila CC BY-SA 3.0
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Mar 23, 2015 at 9:08 vote accept Dominic van der Zypen
Mar 23, 2015 at 8:42 history answered Asaf Karagila CC BY-SA 3.0