**Nested Selection:** For every infinite set $G$ of pairwise disjoint infinite sets such that any two distinct elements $x,y$ of $G$ either "$y$ is a set of proper supersets of elements of $x$ and each element of $x$ has a proper superset of it in $y$" or "$x$ is a set of proper supersets of elements of $y$ and each element of $y$ has a proper superset of it in $x$"; there exists a set $cG$ that is a choice set on $G$ (i.e. $cG$ is a subset of $\bigcup G$ that has exactly one element from each element of $G$, among its elements) and such that for any two elements $a,b$ of $cG$ we have $a$ subset of $b$ or $b$ subset of $a$.



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> Is Nested selection equivalent to $AC$?
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The formal capture is a little bit messy. It is:


$\forall G: \operatorname {infinite} (G) \land \forall h \in G (\operatorname {infinite}(h)) \land \\ \forall k,l \in G (k \neq l \to k \cap l = \emptyset) \land \\ \forall x,y \in G (\forall z \in y [x] \exists u \in x [y] (z \supsetneq u) \land \\ \forall v \in x[y] \exists w \in y[x] (w \supsetneq v) ) \\ \implies \\ \exists cG: cG \subseteq \bigcup G \land \\\forall g \in G \exists! m (m \in g \land m \in cG) \land \\\forall a,b \in cG: a \subseteq b \lor b \subseteq a$



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> If the above nested selection is too particular theme to pace with Choice.Then is the following general form equivalent to choice?
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Define $\begin{align} Y \text { is } \Phi\text{-image of } X \iff &\forall a \in X \exists b \in Y: \Phi(a,b) \land \\ &\forall b \in Y \exists a \in X: \Phi(a,b)\end{align}$


**General Nested Selection:** If $\Phi$ is a transitive asymmetric binary relation, then if $G$ is an infinite set of pairwise disjoint infinite sets such that every two distinct sets $X,Y \in G$ either $ Y \text { is } \Phi\text{-image of } X $ or $ X \text { is } \Phi\text{-image of } Y$; then there exists a choice set $C$ on $G$ such that for any two distinct elements $a,b \in C$ either $ \Phi(a,b)$ or $\Phi(b,a)$.


By $C$ being a choice set on $G$ it means that $C \subseteq \bigcup G$ and $C$ shares exactly one element with each element of $G$.



Is Nested Selection equivalent to AC?
For a counterexample to Nested Selection, construct sets $A\_\alpha\subseteq\omega$ ($\alpha\lt\omega\_1$) so that, for $\alpha\lt\beta\lt\omega\_1$, we have $|A\_\alpha\setminus A\_\beta|\lt\aleph\_0=|A\_\beta\setminus A\_\alpha|$. Let $\mathcal S\_\alpha=\{X\subseteq\omega:|X\triangle A\_\alpha|\lt\aleph\_0\}$ and let $G=\{\mathcal S\_\alpha:\alpha\lt\omega\_1\}$. A choice set for $G$ which was totally ordered by inclusion would be a chain of length $\omega\_1$ in $\mathcal P(\omega)$, which doesn't exist.


**P.S.** To construct the sets $A\_\alpha$, first partition $\omega$ into infinitely many disjoint infinite sets $N\_i$ ($i\lt\omega$), and then define $A\_\alpha$ recursively for $\alpha\lt\omega\_1$ so that $|A\_\alpha\cap N\_i|\lt\aleph\_0$ for each $i\lt\omega$ and $|A\_\beta\setminus A\_\alpha|\lt\aleph\_0=|A\_\alpha\setminus A\_\beta|$ for each $\beta\lt\alpha$.


