-1
$\begingroup$

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: $$C_R (A) = \{x \in A : ( \forall y \in A ) ( xRy ) \}.$$
$\endgroup$
3
  • $\begingroup$ I've re-tagged since this doesn't concern the axiom of choice, or proof theory (which, despite the name, is a specific subfield of logic). (And what is the connection with economics?) $\endgroup$ Sep 20, 2015 at 5:01
  • 1
    $\begingroup$ These concepts are under consumer choice in microeconomics theory. $\endgroup$ Sep 20, 2015 at 11:16
  • 1
    $\begingroup$ As it stands, this question has no background, motivation, or context. Such questions are often put on hold until that information is added. $\endgroup$ Sep 20, 2015 at 20:08

1 Answer 1

0
$\begingroup$

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 $C_R(\mathbb{Z})=\emptyset$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .