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I was always thinking, in an informal way, that the powerset lattices ${\cal P}(X)$ (where $X$ is an infinite set) are the "widest" bounded distributive lattices with respect to their height. (In ${\cal P}(\omega)$ every chain is at most countably infinite, but there are uncountable collections of infinite subsets of $\omega$ such that the pairwise intersection of the subsets is finite.)

Since I am not sure whether there are always chains of maximum cardinality in bounded distributive lattices, I want to put the thought I started this post with, in a loose, but (hopefully) precise manner.

Question. If $(L,\leq)$ is a bounded, distributive lattice, is it true that for every antichain $A\subseteq L$ there is a chain $C\subseteq L$ such that $2^{|C|} \geq |A|$?

Note. An antichain of a lattice $L$ is meant to be a set $A\subseteq L$ such that for $a\neq b\in A$ we have both $a\not\leq b$ and $b\not\leq a$.

Note 2. My remark above about countable chains is false, mind-bogglingly so as I think, see Andreas Blass' comment.

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    $\begingroup$ Let $L=P_{\omega}(A)\cup\{A\}$. Then $\{\{a\}|a\in A\}$ is an antichain of size $|A|$ but there is no chain of size greater than $|A|=|L|,$ $\endgroup$ – Joseph Van Name Oct 7 '17 at 14:32
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    $\begingroup$ If you're using "chain" with its usual meaning, "linearly ordered subset," then it is not true that "in $\mathcal P(\omega)$ every chain is at most countably infinite." $\mathcal P(\omega)$ includes chains of the cardinality of the continuum. For example, let $(r_n)_{n\in\omega}$ be an enumeration of the rationals and, for each real number $z$, let $A_z=\{n:r_n<z\}$. These sets $A_z$ constitute a chain, because $A_z\subsetneq A_x$ whenever $z<x$. $\endgroup$ – Andreas Blass Oct 7 '17 at 14:51
  • $\begingroup$ @JosephVanName In your lattice there is in fact no uncountable chain, since below every element (except $A$ itself) there are only finitely many elements. So why not write this as an anwer? $\endgroup$ – Goldstern Oct 7 '17 at 17:00
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    $\begingroup$ @AndreasBlass That's mind-boggling, this fact about uncountable chains -- thanks, will correct my post! $\endgroup$ – Dominic van der Zypen Oct 7 '17 at 19:17

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