# Existence of families of sets whose elements are incomparable in terms of $\in$

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 $\in$: $(\forall {x,y\in\mathcal{A}})\,(x\notin y\wedge y\notin x)\ ?$

• By the way, your example, $\{\{0\},\{1\},\dots\}$ does not work, because $\{0\}=1\in\{1\}$. – Andrés E. Caicedo Aug 7 '14 at 14:26
• Right, corrected. – Mad Hatter Aug 7 '14 at 14:28

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 $X=\kappa\setminus\{1\}$.
• But it does not work :-) $\{0\}\in\{1\}$ and nonsense like that. – Andrés E. Caicedo Aug 7 '14 at 14:24