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Doubt about eqivalence .

Would you please give me an example to show that $(P(X), C_R)$ may be a choice structure even if $R$ is not rational (*i.e.*, complete and transitive).

Clarification:

- For any nonempty set $X$, let $P(X)$ denote the set of all nonempty subsets of X.
- For any nonempty subset $B$ of $P(X)$, a function $c: B \rightarrow P(X)$ is called a
*choice function*iff $c(A) \subset A$ for all $A \in B$. The pair $(B, c)$ is called a*choice structure*. - For any binary relation $R$ on $X$, define the function $C_R : P(X) \rightarrow P(X) \cup \{\emptyset\}$ as follows: $$C_R (A) = \{x \in A : ( \forall y \in A ) ( xRy ) \}.$$

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No non-complete relation $R$ can yield a choice structure in this way. Suppose neither $aRb$ nor $bRa$ holds. Then the map $C_R$ is not a choice function: $C_R(\{a, b\})=\emptyset$.

On the other hand, note that $R$ being rational isn’t even enough: if $R$ is the usual ordering on $\mathbb{R}$, then $C_R(\mathbb{Z})=\emptyset$.

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