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There exist families of sets whose elements are comparable in terms of $\in$, like for example the set of finite von Neumann ordinals, there exist such that their elements are incomparable in terms of $\in$ (in the sense that for no two elements $x$ and $y$: $x\in y$ or $y\in x$), like for example $\{\{1\},\{2\},\{3\},\ldots\}$.

My question is: can we construct for an arbitrary cardinality $\kappa$, a family $\mathcal{A}$ of sets of cardinality $\kappa$, whose elements are pairwise incomparable in terms of $\in$: \[ (\forall {x,y\in\mathcal{A}})\,(x\notin y\wedge y\notin x)\ ? \]

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    $\begingroup$ By the way, your example, $\{\{0\},\{1\},\dots\}$ does not work, because $\{0\}=1\in\{1\}$. $\endgroup$ – Andrés E. Caicedo Aug 7 '14 at 14:26
  • $\begingroup$ Right, corrected. $\endgroup$ – Mad Hatter Aug 7 '14 at 14:28
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If you take any set of size $\kappa$, $X$, that none of its elements are singletons, then $\{\{x\}\mid x\in X\}$ is such set. For example, take $X=\kappa\setminus\{1\}$.

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  • $\begingroup$ More than one whole minute this time. :-) $\endgroup$ – Andrés E. Caicedo Aug 7 '14 at 14:22
  • $\begingroup$ Well, I just woke up twenty minutes ago from a very long sleep. I'm trying to size up my environment and get a hold of my mind... so it's expect that I type slower than usually. ;-) $\endgroup$ – Asaf Karagila Aug 7 '14 at 14:23
  • $\begingroup$ But it does not work :-) $\{0\}\in\{1\}$ and nonsense like that. $\endgroup$ – Andrés E. Caicedo Aug 7 '14 at 14:24
  • $\begingroup$ Yeah, that's another side effect... :-P $\endgroup$ – Asaf Karagila Aug 7 '14 at 14:25

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